Python’s set.isdisjoint(set) returns True if no element from this set is a member of the specified set. Sets are disjoint if and only if their intersection is the empty set.

Here’s a minimal example that checks whether sets s and t are disjoint:

>>> s = {1, 2, 3, 4}
>>> t = {'Alice', 'Bob'}
>>> s.isdisjoint(t)
True

Syntax

Let’s dive into the formal syntax of the set.isdisjoint() method.

set.isdisjoint(set)

Argument

Data Type

Explanation

set

A set or iterable

The set against which the elements of this set should be checked

Return Value of set.isdisjoint()

The return value of set.isdisjoint() is a Boolean whether the set is disjoint with the set defined as an argument.

Advanced Examples set.isdisjoint()

There are some subtleties you need to understand regarding the set disjoint method. Let’s dive into them by example!

We start with a simple and trivial example of two disjoint sets:

>>> {1, 2, 3}.isdisjoint({'Alice', 'Bob'})
True

? Can you also pass a list as an argument to the set.isdisjoint() method? The answer is yes—the method takes any iterable.

>>> {1, 2, 3}.isdisjoint(['Alice', 'Bob'])
True

Let’s have an example of a non-disjoint set combination that intersect in one element—can you figure out which?

What is the Time Complexity of set.isdisjoint() in Python?

The worst-case runtime complexity of the set.disjoint() method is the same as the set.intersection() method because it first computes the intersection of both sets and then checks if the intersection is empty to determine whether the sets are disjoint. For a set with n elements and a set argument with m elements, the runtime complexity is O(min(n, m)) because you need to check for the smaller set whether each of its elements is a member of the larger set. However, in practice, the runtime can be much faster because the first common element found already determines the answer and the whole computation can be aborted.

You can see this in the following simple experiment where we run the set method multiple times for increasing set sizes:

I ran this experiment on my Acer Aspire 5 notebook (I know) with Intel Core i7 (8th Gen) processor and 16GB of memory. Here’s the code of the experiment:

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.isdisjoint(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()

The runtime complexity feels constant. Only if we slightly modify the experiment to ensure that the sets are not disjoint will we see an increase of the runtime complexity with increasing set sizes:

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(size, 2* size, 1))
# Start track time ...
t1 = time.time()
s.isdisjoint(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

All set methods are called on a given set. For example, if you created a set s = {1, 2, 3}, you’d call s.clear() to remove all elements of the set. We use the term “this set” to refer to the set on which the method is executed.

Create and return a new set containing all elements of this set except the ones in the given set arguments. The resulting set has at most as many elements as any other.

Replace this set with the symmetric difference, i.e., elements in either this set or the specified set argument, but not elements that are members of both.

Update this set with all elements that are in this set, or in any of the specified set arguments. The resulting set has at least as many elements as any other.