How can I compare two lists in python and return matches

python

list-comparison

data-structures

performance-testing

byNikita Barsukov·Dec 29, 2024

Retrieving common elements in two lists using set intersection:

matches = list(set(list1) & set(list2)) # Set it and forget it

Or find matches through a powerful list comprehension:

matches = [x for x in list1 if x in list2] # List, meet comprehension

Both snippets yield the matching elements right away.

Python list comparison in-depth

Keep the pace: comparing ordered lists

Maintain element order while finding matches, only when lists are of the same size:

matches_in_order = [a for a, b in zip(list1, list2) if a == b] # Zip zap zapped

This respects the original order of your lists.

Playing heavy: efficiency with large lists

Handling big data efficiently by harnessing set powers when lists are super large:

list1_set, list2_set = set(list1), set(list2)
matches = list(list1_set & list2_set) # Big Brother is efficient

A significantly faster treatment for long lists with lots of items.

Dealing the hand: duplicates and empty lists

Defend against duplicates and empty lists like a poker pro:

matches = list(set(list1) & set(list2)) # Sets don't like clones

Sets will strip off duplicates, leaving only the essential matches. And for empty lists? You'll get an empty list in return - no surprises there.

Tackling oddballs: types and edge cases

Comparing apples and oranges: heterogeneous lists

Python isn't easily confused - compare lists with mixed data types, no sweat:

string_numbers = ["1", "2", "3"]
integers = [1, 2, 3, 4, 5]

matches = [int(x) for x in string_numbers if int(x) in integers] # From words to numbers in a snap

Staying ahead of the count: match frequency

Rarely thought of, but sometimes you need to know the number of match occurrences:

from collections import Counter
matches_counter = Counter(list1) & Counter(list2)
matches_with_frequency = list(matches_counter.elements()) # Count on me

Here you can see how often a match happens.

Putting your solution to the test: performance and robustness

Measuring speed: performance testing of solutions

Because no one likes to wait, test the performance of your stuck-in-a-list loops:

import timeit
set_time = timeit.timeit('list(set(list1) & set(list2))', setup='list1=[...]; list2=[...]', number=10000)
list_time = timeit.timeit('[x for x in list1 if x in list2]', setup='list1=[...]; list2=[...]', number=10000)

print(f"Set intersection time: {set_time} \nList comprehension time: {list_time}") // Fast and furious, Python style