# Python Set intersection_update() Method Example

Python set intersection_update() is an inbuilt method that is similar to the python intersection() method. Still, the difference is that in the intersection() method, the caller set remains unchanged, but in the intersection_update() method, the caller set is updated with the result value. Means, the caller set is updated with the intersection value of two sets.

**Python Set intersection_update() Method**

Before going through an intersection_update() method, we should know what an intersection is. It refers to the common elements of two given sets. Means if there are two sets and they have some common elements then those common elements are known as the intersection of both the element.

Python **intersection_update()** updates the **set** calling **intersection_update() method** with the intersection of **sets**. The intersection of two or more **sets** is the **set** of items that are common to all **sets**. To learn more, visit the **Python set** **Intersection**.

**How intersection_update() works in Python**

Let’s say we have two sets A and B, then A.intersection_update(B) operation updates set A with common elements in set A and B. For example, A=set([1,2,3]) and B=set([4,2,3]) now after taking **A.intersection_update(B)**, value of set A will be [2,3].

**Syntax**

First_Set.intersection_update(Second_Set)

Here, **First_Set** is called a caller set, which is updated with the result value, and **Second_Set** is the set with which we find an intersection.

**Return Value**

The intersection_update() method returns None. But it updates the value of the caller set with the result value.

See the following code example.

# app.py # Declaring two sets # Even nums between 2 and 10 set1 = {2, 4, 6, 8, 10} # Multiple of 3 between 1 to 10 set2 = {3, 6, 9} # priting both the sets print("Set1 is: ", set1) print("Set2 is : ", set2) # Now we will find intersection of these two sets result = set1.intersection_update(set2) print("Result value is: ", result) print("Value of set1 is: ", set1) print("Value of set2 is : ", set2)

**Output**

Set1 is: {2, 4, 6, 8, 10} Set2 is : {9, 3, 6} Result value is: None Value of set1 is: {6} Value of set2 is : {9, 3, 6}

From the output, we can see that, before calling the intersection_update() method, the value of set1 was different, but after calling the intersection_update() method, the value is changed. This is because, as we learned, this method updates the value of the caller set ( which is set1 ) with the result value.

Also, we can see that the value of the result is None as this method returns None and set2 value is unchanged.

**Python intersection_update() with Two Parameters**

See the following code.

# app.py A = {1, 2, 3, 4} B = {2, 3, 4, 5, 6} C = {4, 5, 6, 7, 8} result = C.intersection_update(B, A) print('result =', result) print('C =', C) print('B =', B) print('A =', A)

**Output**

python3 app.py result = None C = {4} B = {2, 3, 4, 5, 6} A = {1, 2, 3, 4}

In the above example, we have three Sets A, B, and C. Here, we are passing more than one parameter to the intersection_update() method.

To find the intersection between more than two sets, we can pass the additional sets in the intersection_update() method, and we got the output.

**Equivalent Operator(&=) in Python**

You can achieve the same result of the intersection_update() method by using the **&=** augmented assignment operator.

See the following code.

# app.py A = {1, 2, 3, 4} B = {2, 3, 4, 5, 6} A &= B print('A =', A)

**Output**

python3 app.py A = {2, 3, 4}

Finally, Python Set intersection_update() Method Example is over.