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Suppose I have two lists, L and M. Now I want to know if they share an element. Which would be the fastest way of asking (in python) if they share an element? I don't care which elements they share, or how many, just if they share or not.
For example, in this case
L = [1,2,3,4,5,6] M = [8,9,10] I should get False, and here:
L = [1,2,3,4,5,6] M = [5,6,7] I should get True.
I hope the question's clear. Thanks!
Manuel
547
Using set's intersection inbuilt function. a_set. intersection(b_set) returns a positive integer if there is at least one element in common, else it returns 0.
Another approach to find, if two lists have common elements is to use sets. The sets have unordered collection of unique elements. So we convert the lists into sets and then create a new set by combining the given sets. If they have some common elements then the new set will not be empty.
Python count intersection of sets To count the intersection of sets in Python, we will use “len(set(set1) & set(set2))”. Here, ” & “ is an intersection element common to both. It will return the count as “3” because “10, 8, and 6” are common to both the sets.
Or more concisely
if set(L) & set(M): # there is an intersection else: # no intersection If you really need True or False
bool(set(L) & set(M)) After running some timings, this seems to be a good option to try too
m_set=set(M) any(x in m_set for x in L) If the items in M or L are not hashable you have to use a less efficient approach like this
any(x in M for x in L) Here are some timings for 100 item lists. Using sets is considerably faster when there is no intersection, and a bit slower when there is a considerable intersection.
M=range(100) L=range(100,200) timeit set(L) & set(M) 10000 loops, best of 3: 32.3 µs per loop timeit any(x in M for x in L) 1000 loops, best of 3: 374 µs per loop timeit m_set=frozenset(M);any(x in m_set for x in L) 10000 loops, best of 3: 31 µs per loop L=range(50,150) timeit set(L) & set(M) 10000 loops, best of 3: 18 µs per loop timeit any(x in M for x in L) 100000 loops, best of 3: 4.88 µs per loop timeit m_set=frozenset(M);any(x in m_set for x in L) 100000 loops, best of 3: 9.39 µs per loop # Now for some random lists import random L=[random.randrange(200000) for x in xrange(1000)] M=[random.randrange(200000) for x in xrange(1000)] timeit set(L) & set(M) 1000 loops, best of 3: 420 µs per loop timeit any(x in M for x in L) 10 loops, best of 3: 21.2 ms per loop timeit m_set=set(M);any(x in m_set for x in L) 1000 loops, best of 3: 168 µs per loop timeit m_set=frozenset(M);any(x in m_set for x in L) 1000 loops, best of 3: 371 µs per loop To avoid the work of constructing the intersection, and produce an answer as soon as we know that they intersect:
m_set = frozenset(M) return any(x in m_set for x in L) Update: gnibbler tried this out and found it to run faster with set() in place of frozenset(). Whaddayaknow.
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