>>> s = {1, 2, 3, 4} >>> t = {3, 4, 5} >>> s.union(t) {1, 2, 3, 4, 5}

## Syntax

set.union(*sets)

## Return Value of Set union()

## Advanced Examples Set Union

>>> {1, 2, 3}.union({1, 2}) {1, 2, 3}

>>> {1, 2}.union({1, 2, 3}) {1, 2, 3}

>>> {1, 2, 3}.union(set()) {1, 2, 3}

## Set Union Multiple Set Arguments

>>> {1, 2, 3, 4, 5, 6}.union({0, 2}, {42, 3, 4}, {33, 3, 5}) {0, 1, 2, 3, 4, 5, 6, 33, 42}

## Python Set Union | Operator

>>> {1, 2}.union({3, 4}) {1, 2, 3, 4} >>> {1, 2} | {3, 4} {1, 2, 3, 4}

>>> {1, 2} | {3, 4} | {5, 6} {1, 2, 3, 4, 5, 6}

## Python Set Update vs Union

Both `set.update()`

and `set.union()`

perform the union operation. However, `set.update()`

adds all missing elements to the set on which it is called whereas `set.union()`

creates a new set. Consequently, the return value of `set.update()`

is `None`

(with side effects) and the return value of `set.union()`

is a set (without side effects).

Here’s an example of the `set.update()`

method:

>>> s = {1, 2, 3} >>> s.update({4, 5}) >>> s {1, 2, 3, 4, 5}

The original set `s`

is modified and now contains five elements after the update. There is no return value, so you need to separately print out the set.

Here’s an example of the `set.union()`

method:

>>> s = {1, 2, 3} >>> s.union({4, 5}) {1, 2, 3, 4, 5}

Both sets remain unchanged. However, a new set has been created—and this set is the return value of the operation!

## What is the Time Complexity of Set Union in Python?

import matplotlib.pyplot as plt import time sizes = [i * 10**5 for i in range(50)] runtimes = [] for size in sizes: s = set(range(size)) t = set(range(0, size, 2)) # Start track time ... t1 = time.time() s.union(t) t2 = time.time() # ... end track time runtimes.append(t2-t1) plt.plot(sizes, runtimes) plt.ylabel('Runtime (s)') plt.xlabel('Set Size') plt.show()

## Other Python Set Methods

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