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Copy constructors and copy assignment operators (C++)

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Starting in C++11, two kinds of assignment are supported in the language: copy assignment and move assignment . In this article "assignment" means copy assignment unless explicitly stated otherwise. For information about move assignment, see Move Constructors and Move Assignment Operators (C++) .

Both the assignment operation and the initialization operation cause objects to be copied.

Assignment : When one object's value is assigned to another object, the first object is copied to the second object. So, this code copies the value of b into a :

Initialization : Initialization occurs when you declare a new object, when you pass function arguments by value, or when you return by value from a function.

You can define the semantics of "copy" for objects of class type. For example, consider this code:

The preceding code could mean "copy the contents of FILE1.DAT to FILE2.DAT" or it could mean "ignore FILE2.DAT and make b a second handle to FILE1.DAT." You must attach appropriate copying semantics to each class, as follows:

Use an assignment operator operator= that returns a reference to the class type and takes one parameter that's passed by const reference—for example ClassName& operator=(const ClassName& x); .

Use the copy constructor.

If you don't declare a copy constructor, the compiler generates a member-wise copy constructor for you. Similarly, if you don't declare a copy assignment operator, the compiler generates a member-wise copy assignment operator for you. Declaring a copy constructor doesn't suppress the compiler-generated copy assignment operator, and vice-versa. If you implement either one, we recommend that you implement the other one, too. When you implement both, the meaning of the code is clear.

The copy constructor takes an argument of type ClassName& , where ClassName is the name of the class. For example:

Make the type of the copy constructor's argument const ClassName& whenever possible. This prevents the copy constructor from accidentally changing the copied object. It also lets you copy from const objects.

Compiler generated copy constructors

Compiler-generated copy constructors, like user-defined copy constructors, have a single argument of type "reference to class-name ." An exception is when all base classes and member classes have copy constructors declared as taking a single argument of type const class-name & . In such a case, the compiler-generated copy constructor's argument is also const .

When the argument type to the copy constructor isn't const , initialization by copying a const object generates an error. The reverse isn't true: If the argument is const , you can initialize by copying an object that's not const .

Compiler-generated assignment operators follow the same pattern for const . They take a single argument of type ClassName& unless the assignment operators in all base and member classes take arguments of type const ClassName& . In this case, the generated assignment operator for the class takes a const argument.

When virtual base classes are initialized by copy constructors, whether compiler-generated or user-defined, they're initialized only once: at the point when they are constructed.

The implications are similar to the copy constructor. When the argument type isn't const , assignment from a const object generates an error. The reverse isn't true: If a const value is assigned to a value that's not const , the assignment succeeds.

For more information about overloaded assignment operators, see Assignment .

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21.12 — Overloading the assignment operator

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Copy assignment operator.

A copy assignment operator of class T is a non-template non-static member function with the name operator = that takes exactly one parameter of type T , T & , const T & , volatile T & , or const volatile T & . For a type to be CopyAssignable , it must have a public copy assignment operator.

[ edit ] Syntax

[ edit ] Explanation

• Typical declaration of a copy assignment operator when copy-and-swap idiom can be used.
• Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used (non-swappable type or degraded performance).
• Forcing a copy assignment operator to be generated by the compiler.
• Avoiding implicit copy assignment.

The copy assignment operator is called whenever selected by overload resolution , e.g. when an object appears on the left side of an assignment expression.

[ edit ] Implicitly-declared copy assignment operator

If no user-defined copy assignment operators are provided for a class type ( struct , class , or union ), the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T & T :: operator = ( const T & ) if all of the following is true:

• each direct base B of T has a copy assignment operator whose parameters are B or const B & or const volatile B & ;
• each non-static data member M of T of class type or array of class type has a copy assignment operator whose parameters are M or const M & or const volatile M & .

Otherwise the implicitly-declared copy assignment operator is declared as T & T :: operator = ( T & ) . (Note that due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument.)

A class can have multiple copy assignment operators, e.g. both T & T :: operator = ( const T & ) and T & T :: operator = ( T ) . If some user-defined copy assignment operators are present, the user may still force the generation of the implicitly declared copy assignment operator with the keyword default . (since C++11)

The implicitly-declared (or defaulted on its first declaration) copy assignment operator has an exception specification as described in dynamic exception specification (until C++17) exception specification (since C++17)

Because the copy assignment operator is always declared for any class, the base class assignment operator is always hidden. If a using-declaration is used to bring in the assignment operator from the base class, and its argument type could be the same as the argument type of the implicit assignment operator of the derived class, the using-declaration is also hidden by the implicit declaration.

[ edit ] Deleted implicitly-declared copy assignment operator

A implicitly-declared copy assignment operator for class T is defined as deleted if any of the following is true:

• T has a user-declared move constructor;
• T has a user-declared move assignment operator.

Otherwise, it is defined as defaulted.

A defaulted copy assignment operator for class T is defined as deleted if any of the following is true:

• T has a non-static data member of non-class type (or array thereof) that is const ;
• T has a non-static data member of a reference type;
• T has a non-static data member or a direct or virtual base class that cannot be copy-assigned (overload resolution for the copy assignment fails, or selects a deleted or inaccessible function);
• T is a union-like class , and has a variant member whose corresponding assignment operator is non-trivial.

[ edit ] Trivial copy assignment operator

The copy assignment operator for class T is trivial if all of the following is true:

• it is not user-provided (meaning, it is implicitly-defined or defaulted) , , and if it is defaulted, its signature is the same as implicitly-defined (until C++14) ;
• T has no virtual member functions;
• T has no virtual base classes;
• the copy assignment operator selected for every direct base of T is trivial;
• the copy assignment operator selected for every non-static class type (or array of class type) member of T is trivial;

A trivial copy assignment operator makes a copy of the object representation as if by std::memmove . All data types compatible with the C language (POD types) are trivially copy-assignable.

[ edit ] Implicitly-defined copy assignment operator

If the implicitly-declared copy assignment operator is neither deleted nor trivial, it is defined (that is, a function body is generated and compiled) by the compiler if odr-used . For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove ). For non-union class types ( class and struct ), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using built-in assignment for the scalars and copy assignment operator for class types.

The generation of the implicitly-defined copy assignment operator is deprecated (since C++11) if T has a user-declared destructor or user-declared copy constructor.

[ edit ] Notes

If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either a prvalue such as a nameless temporary or an xvalue such as the result of std::move ), and selects the copy assignment if the argument is an lvalue (named object or a function/operator returning lvalue reference). If only the copy assignment is provided, all argument categories select it (as long as it takes its argument by value or as reference to const, since rvalues can bind to const references), which makes copy assignment the fallback for move assignment, when move is unavailable.

It is unspecified whether virtual base class subobjects that are accessible through more than one path in the inheritance lattice, are assigned more than once by the implicitly-defined copy assignment operator (same applies to move assignment ).

See assignment operator overloading for additional detail on the expected behavior of a user-defined copy-assignment operator.

[ edit ] Example

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Copy assignment operator

A copy assignment operator of class T is a non-template non-static member function with the name operator = that takes exactly one parameter of type T , T & , const T & , volatile T & , or const volatile T & . A type with a public copy assignment operator is CopyAssignable .

[ edit ] Syntax

[ edit ] Explanation

• Typical declaration of a copy assignment operator when copy-and-swap idiom can be used
• Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used
• Forcing a copy assignment operator to be generated by the compiler
• Avoiding implicit copy assignment

The copy assignment operator is called whenever selected by overload resolution , e.g. when an object appears on the left side of an assignment expression.

[ edit ] Implicitly-declared copy assignment operator

If no user-defined copy assignment operators are provided for a class type ( struct , class , or union ), the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T & T :: operator = ( const T & ) if all of the following is true:

• each direct base B of T has a copy assignment operator whose parameters are B or const B& or const volatile B &
• each non-static data member M of T of class type or array of class type has a copy assignment operator whose parameters are M or const M& or const volatile M &

Otherwise the implicitly-declared copy assignment operator is declared as T & T :: operator = ( T & ) . (Note that due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument)

A class can have multiple copy assignment operators, e.g. both T & T :: operator = ( const T & ) and T & T :: operator = ( T ) . If some user-defined copy assignment operators are present, the user may still force the generation of the implicitly declared copy assignment operator with the keyword default .

Because the copy assignment operator is always declared for any class, the base class assignment operator is always hidden. If a using-declaration is used to bring in the assignment operator from the base class, and its argument type could be the same as the argument type of the implicit assignment operator of the derived class, the using-declaration is also hidden by the implicit declaration.

[ edit ] Deleted implicitly-declared copy assignment operator

The implicitly-declared or defaulted copy assignment operator for class T is defined as deleted in any of the following is true:

• T has a non-static data member that is const
• T has a non-static data member of a reference type.
• T has a non-static data member that cannot be copy-assigned (has deleted, inaccessible, or ambiguous copy assignment operator)
• T has direct or virtual base class that cannot be copy-assigned (has deleted, inaccessible, or ambiguous move assignment operator)
• T has a user-declared move constructor
• T has a user-declared move assignment operator

[ edit ] Trivial copy assignment operator

The implicitly-declared copy assignment operator for class T is trivial if all of the following is true:

• T has no virtual member functions
• T has no virtual base classes
• The copy assignment operator selected for every direct base of T is trivial
• The copy assignment operator selected for every non-static class type (or array of class type) memeber of T is trivial

A trivial copy assignment operator makes a copy of the object representation as if by std:: memmove . All data types compatible with the C language (POD types) are trivially copy-assignable.

[ edit ] Implicitly-defined copy assignment operator

If the implicitly-declared copy assignment operator is not deleted or trivial, it is defined (that is, a function body is generated and compiled) by the compiler. For union types, the implicitly-defined copy assignment copies the object representation (as by std:: memmove ). For non-union class types ( class and struct ), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using, using built-in assignment for the scalars and copy assignment operator for class types.

The generation of the implicitly-defined copy assignment operator is deprecated (since C++11) if T has a user-declared destructor or user-declared copy constructor.

[ edit ] Notes

If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either prvalue such as a nameless temporary or xvalue such as the result of std:: move ), and selects the copy assignment if the argument is lvalue (named object or a function/operator returning lvalue reference). If only the copy assignment is provided, all argument categories select it (as long as it takes its argument by value or as reference to const, since rvalues can bind to const references), which makes copy assignment the fallback for move assignment, when move is unavailable.

[ edit ] Copy and swap

Copy assignment operator can be expressed in terms of copy constructor, destructor, and the swap() member function, if one is provided:

T & T :: operator = ( T arg ) { // copy/move constructor is called to construct arg     swap ( arg ) ;     // resources exchanged between *this and arg     return * this ; }   // destructor is called to release the resources formerly held by *this

For non-throwing swap(), this form provides strong exception guarantee . For rvalue arguments, this form automatically invokes the move constructor, and is sometimes referred to as "unifying assignment operator" (as in, both copy and move).

[ edit ] Example

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Copy Constructor vs Assignment Operator in C++

• How to Create Custom Assignment Operator in C++?
• Assignment Operators In C++
• Why copy constructor argument should be const in C++?
• Advanced C++ | Virtual Copy Constructor
• Move Assignment Operator in C++ 11
• Self assignment check in assignment operator
• Is assignment operator inherited?
• Copy Constructor in C++
• How to Implement Move Assignment Operator in C++?
• Default Assignment Operator and References in C++
• How to Define a Move Constructor in C++?
• Passing a vector to constructor in C++
• Can a constructor be private in C++ ?
• When is a Copy Constructor Called in C++?
• C++ Assignment Operator Overloading
• std::move in Utility in C++ | Move Semantics, Move Constructors and Move Assignment Operators
• Assignment Operators in C
• Copy Constructor in Python
• Copy Constructor in Java

Copy constructor and Assignment operator are similar as they are both used to initialize one object using another object. But, there are some basic differences between them:

Consider the following C++ program. 

Explanation: Here, t2 = t1;  calls the assignment operator , same as t2.operator=(t1); and   Test t3 = t1;  calls the copy constructor , same as Test t3(t1);

Must Read: When is a Copy Constructor Called in C++?

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SettingWithCopyWarning in pandas: Views vs Copies

Table of Contents

Prerequisites

Example of a settingwithcopywarning, understanding views and copies in numpy, understanding views and copies in pandas, indexing in numpy: copies and views, indexing in pandas: copies and views, chained indexing and settingwithcopywarning, impact of data types on views, copies, and the settingwithcopywarning, hierarchical indexing and settingwithcopywarning, change the default settingwithcopywarning behavior.

NumPy and pandas are very comprehensive, efficient, and flexible Python tools for data manipulation. An important concept for proficient users of these two libraries to understand is how data are referenced as shallow copies ( views ) and deep copies (or just copies ). pandas sometimes issues a SettingWithCopyWarning to warn the user of a potentially inappropriate use of views and copies.

In this article, you’ll learn:

• What views and copies are in NumPy and pandas
• How to properly work with views and copies in NumPy and pandas
• Why the SettingWithCopyWarning happens in pandas
• How to avoid getting a SettingWithCopyWarning in pandas

You’ll first see a short explanation of what the SettingWithCopyWarning is and how to avoid it. You might find this enough for your needs, but you can also dig a bit deeper into the details of NumPy and pandas to learn more about copies and views.

Free Bonus: Click here to get access to a free NumPy Resources Guide that points you to the best tutorials, videos, and books for improving your NumPy skills.

To follow the examples in this article, you’ll need Python 3.7 or 3.8 , as well as the libraries NumPy and pandas . This article is written for NumPy version 1.18.1 and pandas version 1.0.3. You can install them with pip :

If you prefer Anaconda or Miniconda distributions, you can use the conda package management system. To learn more about this approach, check out Setting Up Python for Machine Learning on Windows . For now, it’ll be enough to install NumPy and pandas in your environment :

Now that you have NumPy and pandas installed, you can import them and check their versions:

That’s it. You have all the prerequisites for this article. Your versions might vary slightly, but the information below will still apply.

Note: This article requires you to have some prior pandas knowledge. You’ll also need some knowledge of NumPy for the later sections.

To refresh your NumPy skills, you can check out the following resources:

• NumPy Quickstart Tutorial
• Look Ma, No for Loops: Array Programming With NumPy
• Python Plotting With Matplotlib

To remind yourself about pandas, you can read the following:

• 10 minutes to pandas
• pandas DataFrames 101
• The pandas DataFrame: Make Working With Data Delightful
• Using pandas and Python to Explore Your Dataset
• Python pandas: Tricks & Features You May Not Know

Now you’re ready to start learning about views, copies, and the SettingWithCopyWarning !

If you work with pandas, chances are that you’ve already seen a SettingWithCopyWarning in action. It can be annoying and sometimes hard to understand. However, it’s issued for a reason.

The first thing you should know about the SettingWithCopyWarning is that it’s not an error . It’s a warning . It warns you that you’ve probably done something that’s going to result in unwanted behavior in your code.

Let’s see an example. You’ll start by creating a pandas DataFrame :

This example creates a dictionary referenced by the variable data that contains:

• The keys "x" , "y" , and "z" , which will be the column labels of the DataFrame
• Three NumPy arrays that hold the data of the DataFrame

You create the first two arrays with the routine numpy.arange() and the last one with numpy.array() . To learn more about arange() , check out NumPy arange(): How to Use np.arange() .

The list attached to the variable index contains the strings "a" , "b" , "c" , "d" , and "e" , which will be the row labels for the DataFrame.

Finally, you initialize the DataFrame df that contains the information from data and index . You can visualize it like this:

Here’s a breakdown of the main information contained in the DataFrame:

• Purple box: Data
• Blue box: Column labels
• Red box: Row labels

The DataFrame stores additional information, or metadata, including its shape, data types, and so on.

Now that you have a DataFrame to work with, let’s try to get a SettingWithCopyWarning . You’ll take all values from column z that are less than fifty and replace them with zeros. You can start by creating a mask, or a filter with pandas Boolean operators :

mask is an instance of a pandas Series with Boolean data and the indices from df :

• True indicates the rows in df in which the value of z is less than 50 .
• False indicates the rows in df in which the value of z is not less than 50 .

df[mask] returns a DataFrame with the rows from df for which mask is True . In this case, you get rows a , c , and d .

If you try to change df by extracting rows a , c , and d using mask , you’ll get a SettingWithCopyWarning , and df will remain the same:

As you can see, the assignment of zeros to the column z fails. This image illustrates the entire process:

Here’s what happens in the code sample above:

• df[mask] returns a completely new DataFrame (outlined in purple). This DataFrame holds a copy of the data from df that correspond to True values from mask (highlighted in green).
• df[mask]["z"] = 0 modifies the column z of the new DataFrame to zeros, leaving df untouched.

Usually, you don’t want this! You want to modify df and not some intermediate data structure that isn’t referenced by any variable. That’s why pandas issues a SettingWithCopyWarning and warns you about this possible mistake.

In this case, the proper way to modify df is to apply one of the accessors .loc[] , .iloc[] , .at[] , or .iat[] :

This approach enables you to provide two arguments, mask and "z" , to the single method that assigns the values to the DataFrame.

An alternative way to fix this issue is to change the evaluation order:

This works! You’ve modified df . Here’s what this process looks like:

Here’s a breakdown of the image::

• df["z"] returns a Series object (outlined in purple) that points to the same data as the column z in df , not its copy.
• df["z"][mask] = 0 modifies this Series object by using chained assignment to set the masked values (highlighted in green) to zero.
• df is modified as well since the Series object df["z"] holds the same data as df .

You’ve seen that df[mask] contains a copy of the data, whereas df["z"] points to the same data as df . The rules used by pandas to determine whether or not you make a copy are very complex. Fortunately, there are some straightforward ways to assign values to DataFrames and avoid a SettingWithCopyWarning .

Invoking accessors is usually considered better practice than chained assignment for these reasons:

• The intention to modify df is clearer to pandas when you use a single method.
• The code is cleaner for readers.
• The accessors tend to have better performance, even though you won’t notice this in most cases.

However, using accessors sometimes isn’t enough. They might also return copies, in which case you can get a SettingWithCopyWarning :

In this example, as in the previous one, you use the accessor .loc[] . The assignment fails because df.loc[mask] returns a new DataFrame with a copy of the data from df . Then df.loc[mask]["z"] = 0 modifies the new DataFrame, not df .

Generally, to avoid a SettingWithCopyWarning in pandas, you should do the following:

• Avoid chained assignments that combine two or more indexing operations like df["z"][mask] = 0 and df.loc[mask]["z"] = 0 .
• Apply single assignments with just one indexing operation like df.loc[mask, "z"] = 0 . This might (or might not) involve the use of accessors, but they are certainly very useful and are often preferable.

With this knowledge, you can successfully avoid the SettingWithCopyWarning and any unwanted behavior in most cases. However, if you want to dive deeper into NumPy, pandas, views, copies, and the issues related to the SettingWithCopyWarning , then continue on with the rest of the article.

Views and Copies in NumPy and pandas

Understanding views and copies is an important part of getting to know how NumPy and pandas manipulate data. It can also help you avoid errors and performance bottlenecks. Sometimes data is copied from one part of memory to another, but in other cases two or more objects can share the same data, saving both time and memory.

Let’s start by creating a NumPy array :

Now that you have arr , you can use it to create other arrays. Let’s first extract the second and fourth elements of arr ( 2 and 8 ) as a new array. There are several ways to do this:

Don’t worry if you’re not familiar with array indexing. You’ll learn more about these and other statements later . For now, it’s important to notice that both statements return array([2, 8]) . However, they have different behavior under the surface:

This might seem odd at the first sight. The difference is in the fact that arr[1:4:2] returns a shallow copy , while arr[[1, 3]] returns a deep copy . Understanding this difference is essential not only for dealing with the SettingWithCopyWarning but also for manipulating big data with NumPy and pandas.

In the sections below, you’ll learn more about shallow and deep copies in NumPy and pandas.

Views in NumPy

A shallow copy or view is a NumPy array that doesn’t have its own data. It looks at, or “views,” the data contained in the original array. You can create a view of an array with .view() :

You’ve obtained the array view_of_arr , which is a view, or shallow copy, of the original array arr . The attribute .base of view_of_arr is arr itself. In other words, view_of_arr doesn’t own any data—it uses the data that belongs to arr . You can also verify this with the attribute .flags :

As you can see, view_of_arr.flags.owndata is False . This means that view_of_arr doesn’t own data and uses its .base to get the data:

The image above shows that arr and view_of_arr point to the same data values.

Copies in NumPy

A deep copy of a NumPy array, sometimes called just a copy , is a separate NumPy array that has its own data. The data of a deep copy is obtained by copying the elements of the original array into the new array. The original and the copy are two separate instances. You can create a copy of an array with .copy() :

As you can see, copy_of_arr doesn’t have .base . To be more precise, the value of copy_of_arr.base is None . The attribute .flags.owndata is True . This means that copy_of_arr owns data:

The image above shows that arr and copy_of_arr contain different instances of data values.

Differences Between Views and Copies

There are two very important differences between views and copies:

• Views don’t need additional storage for data, but copies do.
• Modifying the original array affects its views, and vice versa. However, modifying the original array will not affect its copy.

To illustrate the first difference between views and copies, let’s compare the sizes of arr , view_of_arr , and copy_of_arr . The attribute .nbytes returns the memory consumed by the elements of the array:

The amount of memory is the same for all arrays: 48 bytes. Each array looks at six integer elements of 8 bytes (64 bits) each. That’s 48 bytes in total.

However, if you use sys.getsizeof() to get the memory amount directly attributed to each array, then you’ll see the difference:

arr and copy_of_arr hold 144 bytes each. As you’ve seen previously, 48 bytes out of the 144 total are for the data elements. The remaining 96 bytes are for other attributes. view_of_arr holds only those 96 bytes because it doesn’t have its own data elements.

To illustrate the second difference between views and copies, you can modify any element of the original array:

As you can see, the view is also changed, but the copy stays the same. The code is illustrated in the image below:

The view is modified because it looks at the elements of arr , and its .base is the original array. The copy is unchanged because it doesn’t share data with the original, so a change to the original doesn’t affect it at all.

pandas also makes a distinction between views and copies. You can create a view or copy of a DataFrame with .copy() . The parameter deep determines if you want a view ( deep=False ) or copy ( deep=True ). deep is True by default, so you can omit it to get a copy:

At first, the view and copy of df look the same. If you compare their NumPy representations, though, then you may notice this subtle difference:

Here, .to_numpy() returns the NumPy array that holds the data of the DataFrames. You can see that df and view_of_df have the same .base and share the same data. On the other hand, copy_of_df contains different data.

You can verify this by modifying df :

You’ve assigned zeros to all elements of the column z in df . That causes a change in view_of_df , but copy_of_df remains unmodified.

Rows and column labels also exhibit the same behavior:

df and view_of_df share the same row and column labels, while copy_of_df has separate index instances. Keep in mind that you can’t modify particular elements of .index and .columns . They are immutable objects.

Indices and Slices in NumPy and pandas

Basic indexing and slicing in NumPy is similar to the indexing and slicing of lists and tuples . However, both NumPy and pandas provide additional options to reference and assign values to the objects and their parts.

NumPy arrays and pandas objects ( DataFrame and Series ) implement special methods that enable referencing, assigning, and deleting values in a style similar to that of containers :

• .__getitem__() references values.
• .__setitem__() assigns values.
• .__delitem__() deletes values.

When you’re referencing, assigning, or deleting data in Python container-like objects, you often call these methods:

• var = obj[key] is equivalent to var = obj.__getitem__(key) .
• obj[key] = value is equivalent to obj.__setitem__(key, value) .
• del obj[key] is equivalent to obj.__delitem__(key) .

The argument key represents the index, which can be an integer , slice , tuple, list, NumPy array, and so on.

NumPy has a strict set of rules related to copies and views when indexing arrays. Whether you get views or copies of the original data depends on the approach you use to index your arrays : slicing, integer indexing, or Boolean indexing.

One-Dimensional Arrays

Slicing is a well-known operation in Python for getting particular data from arrays, lists, or tuples. When you slice a NumPy array, you get a view of the array:

You’ve created the original array arr and sliced it to get two smaller arrays, a and b . Both a and b use arr as their bases and neither has its own data. Instead, they look at the data of arr :

The green indices in the image above are taken by slicing. Both a and b look at the corresponding elements of arr in the green rectangles.

Note: When you have a large original array and need only a small part of it, you can call .copy() after slicing and delete the variable that points to the original with a del statement . This way, you keep the copy and remove the original array from memory.

Though slicing returns a view, there are other cases where creating one array from another actually makes a copy.

Indexing an array with a list of integers returns a copy of the original array. The copy contains the elements from the original array whose indices are present in the list:

The resulting array c contains the elements from arr with the indices 1 and 3 . These elements have the values 2 and 8 . In this case, c is a copy of arr , its .base is None , and it has its own data:

The elements of arr with the chosen indices 1 and 3 are copied into the new array c . After the copying is done, arr and c are independent.

You can also index NumPy arrays with mask arrays or lists. Masks are Boolean arrays or lists of the same shape as the original. You’ll get a copy of the original array that contains only the elements that correspond to the True values of the mask:

The list mask has True values at the second and fourth positions. This is why the array d contains only the elements from the second and fourth positions of arr . As in the case of c , d is a copy, its .base is None , and it has its own data:

The elements of arr in the green rectangles correspond to True values from mask . These elements are copied into the new array d . After copying, arr and d are independent.

Note: Instead of a list, you can use another NumPy array of integers, but not a tuple .

To recap, here are the variables you’ve created so far that reference arr :

Keep in mind that these examples show how you can reference data in an array. Referencing data returns views when slicing arrays and copies when using index and mask arrays. Assignments , on the other hand, always modify the original data of the array.

Now that you have all these arrays, let’s see what happens when you alter the original:

You’ve changed the second value of arr from 2 to 64 . The value 2 was also present in the derived arrays a , b , c , and d . However, only the views a and b are modified:

The views a and b look at the data of arr , including its second element. That’s why you see the change. The copies c and d remain unchanged because they don’t have common data with arr . They are independent of arr .

Chained Indexing in NumPy

Does this behavior with a and b look at all similar to the earlier pandas examples? It might, because the concept of chained indexing applies in NumPy, too:

This example illustrates the difference between copies and views when using chained indexing in NumPy.

In the first case, arr[1:4:2] returns a view that references the data of arr and contains the elements 2 and 8 . The statement arr[1:4:2][0] = 64 modifies the first of these elements to 64 . The change is visible in both arr and the view returned by arr[1:4:2] .

In the second case, arr[[1, 3]] returns a copy that also contains the elements 2 and 8 . But these aren’t the same elements as in arr . They’re new ones. arr[[1, 3]][0] = 64 modifies the copy returned by arr[[1, 3]] and leaves arr unchanged.

This is essentially the same behavior that produces a SettingWithCopyWarning in pandas, but that warning doesn’t exist in NumPy.

Multidimensional Arrays

Referencing multidimensional arrays follows the same principles:

• Slicing arrays returns views.
• Using index and mask arrays returns copies.

Combining index and mask arrays with slicing is also possible. In such cases, you get copies.

Here are a few examples:

In this example, you start from the two-dimensional array arr . You apply slices for rows. Using the colon syntax ( : ), which is equivalent to slice(None) , means that you want to take all rows.

When you work with the slices 1:3 and 1:4:2 for columns, the views a and b are returned. However, when you apply the list [1, 3] and mask [False, True, False, True] , you get the copies c and d .

The .base of both a and b is arr itself. Both c and d have their own bases unrelated to arr .

As with one-dimensional arrays, when you modify the original, the views change because they see the same data, but the copies remain the same:

You changed the value 2 in arr to 100 and altered the corresponding elements from the views a and b . The copies c and d can’t be modified this way.

To learn more about indexing NumPy arrays, you can check out the official quickstart tutorial and indexing tutorial .

You’ve learned how you can use different indexing options in NumPy to refer to either actual data (a view, or shallow copy) or newly copied data (deep copy, or just copy). NumPy has a set of strict rules about this.

pandas heavily relies on NumPy arrays but offers additional functionality and flexibility. Because of that, the rules for returning views and copies are more complex and less straightforward. They depend on the layout of data, data types, and other details. In fact, pandas often doesn’t guarantee whether a view or copy will be referenced.

Note: Indexing in pandas is a very wide topic. It’s essential for using pandas data structures properly. You can use a variety of techniques:

• Dictionary-like notation
• Attribute-like (dot) notation
• The accessors .loc[] , .iloc[] , .at[] , and .iat

For more information, check out the official documentation and The pandas DataFrame: Make Working With Data Delightful .

In this section, you’ll see two examples of how pandas behaves similarly to NumPy. First, you can see that accessing the first three rows of df with a slice returns a view:

This view looks at the same data as df .

On the other hand, accessing the first two columns of df with a list of labels returns a copy:

The copy has a different .base than df .

In the next section, you’ll find more details related to indexing DataFrames and returning views and copies. You’ll see some cases where the behavior of pandas becomes more complex and differs from NumPy.

Use of Views and Copies in pandas

As you’ve already learned, pandas can issue a SettingWithCopyWarning when you try to modify the copy of data instead of the original. This often follows chained indexing.

In this section, you’ll see some specific cases that produce a SettingWithCopyWarning . You’ll identify the causes and learn how to avoid them by properly using views, copies, and accessors.

You’ve already seen how the SettingWithCopyWarning works with chained indexing in the first example . Let’s elaborate on that a bit.

You’ve created the DataFrame and the mask Series object that corresponds to df["z"] < 50 :

You already know that the assignment df[mask]["z"] = 0 fails. In this case, you get a SettingWithCopyWarning :

The assignment fails because df[mask] returns a copy. To be more precise, the assignment is made on the copy, and df isn’t affected.

You’ve also seen that in pandas, evaluation order matters . In some cases, you can switch the order of operations to make the code work:

df["z"][mask] = 0 succeeds and you get the modified df without a SettingWithCopyWarning .

Using the accessors is recommended, but you can run into trouble with them as well:

In this case, df.loc[mask] returns a copy, the assignment fails, and pandas correctly issues the warning.

In some cases, pandas fails to detect the problem and the assignment on the copy passes without a SettingWithCopyWarning :

Here, you don’t receive a SettingWithCopyWarning and df isn’t changed because df.loc[["a", "c", "e"]] uses a list of indices and returns a copy, not a view.

There are some cases in which the code works, but pandas issues the warning anyway:

In these two cases, you select the first three rows with slices and get views. The assignments succeed both on the views and on df . But you still receive a SettingWithCopyWarning .

The recommended way of performing such operations is to avoid chained indexing. Accessors can be of great help with that:

This approach uses one method call, without chained indexing, and both the code and your intentions are clearer. As a bonus, this is a slightly more efficient way to assign data.

In pandas, the difference between creating views and creating copies also depends on the data types used. When deciding if it’s going to return a view or copy, pandas handles DataFrames that have a single data type differently from ones with multiple types.

Let’s focus on the data types in this example:

You’ve created the DataFrame with all integer columns. The fact that all three columns have the same data types is important here! In this case, you can select rows with a slice and get a view:

This mirrors the behavior that you’ve seen in the article so far. df["b":"d"] returns a view and allows you to modify the original data. That’s why the assignment df["b":"d"]["z"] = 0 succeeds. Notice that in this case you get a SettingWithCopyWarning regardless of the successful change to df .

If your DataFrame contains columns of different types, then you might get a copy instead of a view, in which case the same assignment will fail:

In this case, you used .astype() to create a DataFrame that has two integer columns and one floating-point column. Contrary to the previous example, df["b":"d"] now returns a copy, so the assignment df["b":"d"]["z"] = 0 fails and df remains unchanged.

When in doubt, avoid the confusion and use the .loc[] , .iloc[] , .at[] , and .iat[] access methods throughout your code!

Hierarchical indexing , or MultiIndex , is a pandas feature that enables you to organize your row or column indices on multiple levels according to a hierarchy. It’s a powerful feature that increases the flexibility of pandas and enables working with data in more than two dimensions.

Hierarchical indices are created using tuples as row or column labels:

Now you have the DataFrame df with two-level column indices:

• The first level contains the labels powers and random .
• The second level has the labels x and y , which belong to powers , and z , which belongs to random .

The expression df["powers"] will return a DataFrame containing all columns below powers , which are the columns x and y . If you wanted to get just the column x , then you could pass both powers and x . The proper way to do this is with the expression df["powers", "x"] :

That’s one way to get and set columns in the case of multilevel column indices. You can also use accessors with multi-indexed DataFrames to get or modify the data:

The example above uses .loc[] to return a DataFrame with the rows a and b and the columns x and y , which are below powers . You can get a particular column (or row) similarly:

In this example, you specify that you want the intersection of the rows a and b with the column x , which is below powers . To get a single column, you pass the tuple of indices ("powers", "x") and get a Series object as the result.

You can use this approach to modify the elements of DataFrames with hierarchical indices:

In the examples above, you avoid chained indexing both with accessors ( df.loc[["a", "b"], ("powers", "x")] ) and without them ( df["powers", "x"] ).

As you saw earlier, chained indexing can lead to a SettingWithCopyWarning :

Here, df["powers"] returns a DataFrame with the columns x and y . This is just a view that points to the data from df , so the assignment is successful and df is modified. But pandas still issues a SettingWithCopyWarning .

If you repeat the same code, but with different data types in the columns of df , then you’ll get a different behavior:

This time, df has more than one data type, so df["powers"] returns a copy, df["powers"]["x"] = 0 makes a change on this copy, and df remains unchanged, giving you a SettingWithCopyWarning .

The recommended way to modify df is to avoid chained assignment. You’ve learned that accessors can be very convenient, but they aren’t always needed:

In both cases, you get the modified DataFrame df without a SettingWithCopyWarning .

The SettingWithCopyWarning is a warning, not an error. Your code will still execute when it’s issued, even though it may not work as intended.

To change this behavior, you can modify the pandas mode.chained_assignment option with pandas.set_option() . You can use the following settings:

• pd.set_option("mode.chained_assignment", "raise") raises a SettingWithCopyException .
• pd.set_option("mode.chained_assignment", "warn") issues a SettingWithCopyWarning . This is the default behavior.
• pd.set_option("mode.chained_assignment", None) suppresses both the warning and the error.

For example, this code will raise a SettingWithCopyException instead of issuing a SettingWithCopyWarning :

In addition to modifying the default behavior, you can use get_option() to retrieve the current setting related to mode.chained_assignment :

You get "raise" in this case because you changed the behavior with set_option() . Normally, pd.get_option("mode.chained_assignment") returns "warn" .

Although you can suppress it, keep in mind that the SettingWithCopyWarning can be very useful in notifying you about improper code.

In this article, you learned what views and copies are in NumPy and pandas and what the differences are in their behavior. You also saw what a SettingWithCopyWarning is and how to avoid the subtle errors it points to.

In particular, you’ve learned the following:

• Indexing-based assignments in NumPy and pandas can return either views or copies .
• Both views and copies can be useful, but they have different behaviors .
• Special care must be taken to avoid setting unwanted values on copies.
• Accessors in pandas are very useful objects for properly assigning and referencing data.

Understanding views and copies is an important requirement for using NumPy and pandas properly, especially when you’re working with big data. Now that you have a solid grasp on these concepts, you’re ready to dive deeper into the exciting world of data science!

If you have questions or comments, then please put them in the comment section below.

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14.7.4 Implicit Copy-Assignment for Virtual Bases ¶

When a base class is virtual, only one subobject of the base class belongs to each full object. Also, the constructors and destructors are invoked only once, and called from the most-derived class. However, such objects behave unspecified when being assigned. For example:

The C++ standard specifies that ‘ Base::Base ’ is only called once when constructing or copy-constructing a Derived object. It is unspecified whether ‘ Base::operator= ’ is called more than once when the implicit copy-assignment for Derived objects is invoked (as it is inside ‘ func ’ in the example).

G++ implements the “intuitive” algorithm for copy-assignment: assign all direct bases, then assign all members. In that algorithm, the virtual base subobject can be encountered more than once. In the example, copying proceeds in the following order: ‘ name ’ (via strdup ), ‘ val ’, ‘ name ’ again, and ‘ bval ’.

If application code relies on copy-assignment, a user-defined copy-assignment operator removes any uncertainties. With such an operator, the application can define whether and how the virtual base subobject is assigned.

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What is the concept of Assignment of Copyright?

In this blog post, Sayan Mukherjee, a student of University of Calcutta, who is currently pursuing a  Diploma in Entrepreneurship Administration and Business Laws from NUJS , Kolkata, discusses the concept of assignment in copyright along with the most probable disputes related to it.

Introduction

Copyright, a unique intellectual property meant for the creative brothers and sisters around the world is res incorporalis. In that sense, it has no tangible existence but is a proprietary right and can be disposed of.

In modern life, every individual is aware of the concept of Copyright because of the expansion of media and communication throughout the world. Today’s world has no shortage of ideas, thoughts, modes of expression, and its distribution, which the world media has upheld through the gift of technology coupled with a wider scope of communication and share. This very thing has directed out attention towards the creative world, their rights and obligations, along with their grievances in the form of disputes faced by the creators.

The Copyright Act, 1957 as amended in 2012 is the current vehicle to settle and guide the creators towards betterment and give them some pecuniary opportunities so that they are further encouraged to bless the world with their creativity.

Assignment of Copyright                                  

Nobody is entitled to copy, reproduce, publish or sell an original writing, painting, dramatic production, sculpture, etc. without the permission of the creator. Thus, law provides a right to the owner of the copyright (i.e. the creator) to transfer the ownership of the copyright to a third party. For instance, in the case of making a complete movie – all the creative persons with their idea turned into relevant works come to a producer, assign their rights that subsist in their work in return for a royalty. These works are then summed up to form a complete movie. Yes, the process isn’t that easy and involves many questions that arise both at the time of assignment and especially after it.

Facets of Copyright Assignment      

• It is a pecuniary opportunity for the first owner of copyright. The assignment must specify the amount of copyright [vide Section 19(3) of the Copyright Act]. The creator shall not assign or waive the right to receive royalties to be shared on an equal basis with the assignee of copyright, subject to certain conditions. [vide Section 18(1) proviso of The Copyright Act, 1957 [1] ]
• In the case of an assignment of copyright in any future work, it shall take effect only when the work comes into existence. In this regard, “assignee” includes the legal representatives of the assignee, if he dies before the work comes into existence. [vide Section 18(1) proviso of the Copyright Act]
• The ownership may be assigned either wholly or only for a part of the work in question. [vide Section 18(1) of the Act]
• The Copyright Assignment must be in writing and signed by the assignor or by his duly authorized agent. [vide Section 19(1) of the Act]
• The duration of assignment must also be specified. The Delhi High Court recognized Section 19(5) and stated that if the assignment deed is silent about the duration, it shall be deemed to be 5 years from the date of assignment [2] .
• The agreement deed may specify the territorial extent of such assignment. If silent, it shall be presumed to extend within India. [vide Section 19(6) of the Act]
• The assignment shall be subject to revision, extension, or termination on terms mutually agreed upon by the parties. [vide Section 19(3) of the Act]
• Where the assignee fails to exercise his rights within one year from the date of assignment, the assignment in respect of such right shall be deemed to have lapsed, unless otherwise specified in the assignment deed. [vide Section 19(4) of the Act]
• If the assignment is in contrary to the terms and conditions of the rights already assigned to a copyright society to which the creator is a member, it shall be deemed void. [vide Section 19(8) of the Copyright Act [3] ]
• The creator is entitled to subsequent royalties in the course of future exploitation of a cinematographic film, which includes his work, other than by way of exhibitions in a cinema hall. For example, the creator will be entitled to subsequent royalties for satellite right, home video, internet rights, the etc. Similar clause has been added for the case of sound recording. [vide Section 19(9) and 19(10) of the Copyright Act [4] ]
• In the case of a manuscript, the copyright being a personal property of the owner can be transmitted by testamentary disposition. [vide Section 20 of the Act ]
• The equitable assignment is just the agreement to assign.
• The assignee has the rights of- translation, abridgment, adaptation, dramatic and filmmaking in the work.
• For relinquishment of work, the author has to give notice in prescribed form to the Registrar of Copyrights or by way of public notice. On its receipt, Registrar shall publish it in the Official Gazette. With 14 days of the publication, the Registrar shall post the notice on the official website of Copyright Office, so that such notice remains in the public domain for not less than three years. Such right shall cease to exist from the date of the notice. [vide section 21 of the Copyright Act]

It may be noted in this context, that the author has an alternative for the shortcomings or confusions of assignment of copyright. They can register their work with a copyright society and thereafter license it to whomsoever they desire.

Moral Rights involved in Copyright Assignment

Moral rights are independent of the author’s copyright and shall remain with the author even if he has assigned his copyright.

• The creator of work has the right to claim ownership thereof;
• In case of any distortion, modification or mutilation of the original work, he shall have the right to claim damages;
• If harm is being caused to the goodwill of the creator by commission or omission of any act by the assignee, he shall have the right to damages provided such an act is done before the expiration of the term of assignment.

Foreseeable disputes

• The first dispute which may arise is that as regards the period of copyright assignment. The statute is very particular that an assignment has to be for a specified period even if there is an agreement in contrary [5] . [vide Section 19(2) of the Copyright Act]
• Again, in a situation where assignee fails to exercise his rights assigned to him, and the assignor’s actions do not influence such failure, then, the statute empowers the Copyright Board, on receipt of a complaint from the assignor, to take cognizance of the case and make necessary inquiries as it may deem fit. It further gives a discretionary power to the Board where it can revoke such assignment. [vide Section 19A(1) of the Copyright Act]
• In the case of a monetary dispute over a copyright assignment, the Copyright Board has the power on of a complaint from the aggrieved party, to hold an inquiry and pass necessary order including an order for the recovery of any royalty payable [vide Section 19A(2) of the Copyright Act]. Any such final order must be passed within a period of six months from the date of receipt of the complaint. Delay in compliance shall oblige the Board to record the reasons thereof. [vide Section 19A(3) of the Copyright Act [6] ]

The feasibility of Copyright Assignment is highly questioned because of the rising counts of Copyright Infringement cases. The sole objective of assignment process is to provide both pecuniary as well as distribution benefits to the original work of the creator. It cannot be used to deprive the original owner permanently from his creation.

Copyright Assignment is an inevitable necessity in this dynamic world. People can’t be self-sufficient in every respect. For the better frame of the Art, the ownership right of the creation needs to change hands and bring out the full potential of the original work by exploring various tiers of creativity.

[1] Inserted by Copyright (Amendment) Act, 2012.

[2] Pine Labs Private Limited vs. Gemalto Terminals India Private Limited and others (FAO 635 of 2009 and FAO 636 of 2009)

[3] Inserted by Copyright (Amendment) Act, 2012.

[4] Inserted by Copyright (Amendment) Act, 2012.

[5] Saregama India Ltd. V. Suresh Jindal AIR 2006 Cal. 340.

[6] Inserted by 2012 Amendment.

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Numpy array assignment with copy

For example, if we have a numpy array A , and we want a numpy array B with the same elements.

What is the difference between the following (see below) methods? When is additional memory allocated, and when is it not?

B[:] = A (same as B[:]=A[:] ?)

numpy.copy(B, A)

3 Answers 3

All three versions do different things:

This binds a new name B to the existing object already named A . Afterwards they refer to the same object, so if you modify one in place, you'll see the change through the other one too.

This copies the values from A into an existing array B . The two arrays must have the same shape for this to work. B[:] = A[:] does the same thing (but B = A[:] would do something more like 1).

This is not legal syntax. You probably meant B = numpy.copy(A) . This is almost the same as 2, but it creates a new array, rather than reusing the B array. If there were no other references to the previous B value, the end result would be the same as 2, but it will use more memory temporarily during the copy.

Or maybe you meant numpy.copyto(B, A) , which is legal, and is equivalent to 2?

• 27 @Mr_and_Mrs_D: Numpy arrays work differently than lists do. Slicing an array does not make a copy, it just creates a new view on the existing array's data. –  Blckknght Commented Sep 29, 2017 at 10:35
• 1 What is meant by but B = A[:] would do something more like 1 ? According to this stackoverflow.com/a/2612815 new_list = old_list[:] is also a copy. –  mrgloom Commented May 23, 2018 at 12:13
• 5 @mrgloom: Numpy arrays work differently than lists when it comes to slicing and copying their contents. An array is a "view" of an underlying block of memory where the numeric values are stored. Doing a slice like some_array[:] will create a new array object, but that new object will be a view of the same memory as the original array, which won't have been copied. That's why I said it's more like B = A . It takes only O(1) space and time, rather than the O(n) of each a real copy would need. –  Blckknght Commented May 23, 2018 at 23:19
• The preferred 'copy' method, according to the documentation, is B = A.copy() . However this form doesn't preserve order by default, you need B = A.copy(order='k') . –  Howard Lovatt Commented Oct 15, 2020 at 6:37
• B=A creates a reference
• B[:]=A makes a copy
• numpy.copy(B,A) makes a copy

the last two need additional memory.

To make a deep copy you need to use B = copy.deepcopy(A)

• 2 Refering to your second example: B[:] = A does not make a deep copy of arrays of object-type, e.g. A = np.array([[1,2,3],[4,5]]); B = np.array([None,None], dtype='O') . Now try B[:] = A; B[0][0]=99 , this will change the first element in both A and B ! To my knowledge, there is no other way to guarantee a deep copy, even of a numpy-array, than copy.deepcopy –  Rolf Bartstra Commented May 1, 2018 at 12:38

This is the only working answer for me:

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Lesson Plan

June 17, 2024, 6:05 a.m.

Lesson plan: History of Juneteenth and why it became a national holiday

A Juneteenth celebration in Richmond, Virginia, 1905. Library of Congress

This lesson was originally published on June 16, 2021, and was updated on June 16, 2024.

For a Google version of this lesson plan, click here . (Note: you will need to make a copy of the document to edit it).

In this lesson, students will explore and discuss the history and context around the Juneteenth holiday in the United States. Topics explored will include the history of racial injustice in the U.S., the Civil War and the limitations of the Emancipation Proclamation. Additionally, students will be encouraged to explore the modern significance of Juneteenth and its long-term impact.

Estimated time

One 50-60 minute class period

Grade Level

Grades 6-12

On June 15, 2021, the Senate unanimously approved a bill approving June 19 as a federal holiday for “Juneteenth National Independence Day.” The House passed the bill one day later. Still, many Americans are still unaware of the history and significance of June 19.

On Jan. 1, 1863, President Abraham Lincoln issued the Emancipation Proclamation declaring “that all persons held as slaves” in the Confederacy “shall be free.” While this may have freed some enslaved people on paper, the reality was much more complicated.

Source: PBS NewsHour via Associated Press

For instance, the Emancipation Proclamation only freed those slaves held under the Confederacy, not in border states loyal to the Union, including Kentucky, West Virginia and Delaware, where slavery was still legal after the Emancipation Proclamation. In fact, slavery was still legal in Kentucky until Dec. 1865, when the 13th Amendment was passed, though Kentucky voted against ratifying the amendment.

Confederate states and slaveholders also resisted emancipation, and many people remained enslaved in Confederate states after the proclamation, even as many enslaved people fought for their freedom or escaped behind Union lines. On June 19, 1865, Major General Gordon Granger of the Union issued an order in Galveston, Texas, alerting all enslaved persons that they were legally free.

At this point in 1865, Texas was the westernmost state in America and one of the last Confederate states to be occupied by the Union. Many slaveholders had fled Union advances in other parts of the South to Texas, along with the people they had enslaved.

While it took time for the logistics of “freeing” enslaved people to come into effect, the importance of June 19, or “Juneteenth” lived on. Considering how complicated emancipation was, many dates were considered for holding celebrations of emancipation, but over 150 years later, June 19 remains.

What originally was a holiday mainly observed by Texans has grown to be recognized all over the country. Each year on “Juneteenth,” (or more formally Juneteenth National Freedom Day), communities all around the United States gather and celebrate and reflect on the history of slavery and struggle for civil rights and equality, including the work that still remains after conditional advances such as the Emancipation Proclamation.

Warm up activity

As a class, watch the BrainPop video (8 minutes) below found here introducing Juneteenth. While watching the video, answer the following discussion questions.

Source: BrainPop

Discussion questions:

• What is “Juneteenth”? What does it celebrate?
• Why did it take so long for enslaved peoples in Texas to finally be free? What obstacles existed?
• What were some of the forms of discrimination against newly freed people mentioned in the video?
• What is the Great Migration?
• How did Juneteenth become a national, not just regional, celebration?

After watching the video, separate into groups of 3-4 to discuss the focus questions (5 minutes).

Main activities:

• Why was June 19th chosen as the date to celebrate the freedom of all Americans? What were some of the drawbacks to other dates? Can you make an argument for why you think a different date might have been better and/or worse?
• Gates describes several reasons why Juneteenth struggled to be remembered at times, and why it was able to endure. Compare and contrast what the BrainPop video included as reasons why Juneteenth struggled and endured with what Gate’s emphasizes. What do you think were the most important factors in Juneteenth’s momentum and remembrance continuing?
• “When did they start recognizing Juneteenth, if at all?”
• “What was the process of Juneteenth becoming a holiday in my state?”
• If Juneteenth isn’t recognized in your state, see if you can answer, “Why is Juneteenth not recognized?”
• This search engine for state and local government websites
• The Library of Congress

Additional activities

• Brainstorm or plan a Juneteenth celebration activity. This can be decorating a common area, bringing in a relevant local speaker or planning a refreshment break for your school. Juneteenth celebrations can be in the home, at school or in community locations. For more inspiration see these resources:
• “ How to Celebrate ” from Juneteenth.com
• See how others are celebrating Juneteenth on Twitter .
• Some activists feel ambivalent about Juneteenth becoming a national holiday, or reject the idea. To learn more about the nuances surrounding making Juneteenth a federal holiday, watch this NewsHour interview with Dr. Mark Anthony Neal recorded in 2020 amid the George Floyd protests.

• The day now known as Juneteenth was formally recognized as a national holiday in 2021, due in large part to the activism of retired teacher Opal Lee. Learn more about Lee's activism and the message of Juneteenth in this NewsHour interview with Opal Lee.

If classrooms finish and plan a celebratory activity, please share your ideas with us on social media @NewsHourEXTRA on Twitter.

This lesson was written by Cecilia Curran, NewsHour Classroom intern, while she was a rising sophomore at Amherst College. This lesson was edited by NewsHour Classroom's education producer and former history teacher Vic Pasquantonio.

Fill out this form to share your thoughts on Classroom’s resources. Sign up for NewsHour Classroom’s ready-to-go Daily News Lessons delivered to your inbox each morning.

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