CODEWITHSUNDEEP
python
doglas
doglas

Reputation: 115

Python: Better way to compare two lists?

Please dont give me minus. I want to ask a better method to solve this problem, because what I am doing now is becoming a huge burden for me.

Hier is my question: I have two lists. I want to make sure that none item from one list is in the other list.

In Python I have been working with the following lines... (Assuming List_s has 3 items.)

if List_s[0] not in List_big and List_s[1] not in List_big and List_s[2] not in List_big:   #none item from List_s should be in List_big
    do something
else:
    pass

These lines were actually OK for me, till I suddenly realize that I have to work with lists with >200 length. And I have a lot of lists to compare.

So how could I change the code? Thank you very much for your help!

Upvotes: 1

Views: 105

Answers (3)

Keerthana Prabhakaran
Keerthana Prabhakaran

Reputation: 3787

For very large list, 

import timeit
import random

L=[random.randrange(2000000) for x in xrange(1000000)]
M=[random.randrange(2000000) for x in xrange(1000000)]

start_time = timeit.default_timer()
print any(x in M for x in L)
#True
print timeit.default_timer() - start_time
#0.00981207940825

start_time = timeit.default_timer()
print not set(L).isdisjoint(M)
#True
print timeit.default_timer() - start_time
#0.164795298542

start_time = timeit.default_timer()
print True if set(L) & set(M) else False
#True
print timeit.default_timer() - start_time
#0.436377859225

start_time = timeit.default_timer()
print True if set(L).intersection(M) else False
#True
print timeit.default_timer() - start_time
#0.368563831022

Clearly,

print any(x in M for x in L)

is much more efficient

Upvotes: 1

Tom Lynch
Tom Lynch

Reputation: 913

You will get the same results whether you use:

set(List_s).isdisjoint(List_big)

or:

not set(List_s).intersection(List_big)

But set.isdisjoint is much faster. On my computer, isdisjoint takes about 150 nanoseconds to run against test data, while intersection takes 95 microseconds to run against the same data--about 630 times slower!

Upvotes: 2

krock
krock

Reputation: 29629

You can convert one of the lists to a set and use set.intersection:

if not set(List_s).intersection(List_big):
    print('no common items between lists')

Note that the elements in both lists must be hashable.

Upvotes: 2

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