How can I compare two lists in python and return matches.

Question

I want to take two lists and find the values that appear in both.

a = [1, 2, 3, 4, 5]
b = [9, 8, 7, 6, 5]

returnMatches(a, b)

would return [5], for instance.

Answer 1

Not the most efficient one, but by far the most obvious way to do it is:

>>> a = [1, 2, 3, 4, 5]
>>> b = [9, 8, 7, 6, 5]
>>> set(a) & set(b)
{5}

if order is significant you can do it with list comprehensions like this:

>>> [i for i, j in zip(a, b) if i == j]
[5]

(only works for equal-sized lists, which order-significance implies).

Answer 2

The easiest way to do that is to use sets:

>>> a = [1, 2, 3, 4, 5]
>>> b = [9, 8, 7, 6, 5]
>>> set(a) & set(b)
set([5])

Answer 3

Use set.intersection(..).

aa = set(a)
print aa.intersection(b)
# => set([5])

Answer 4

Quick way:

list(set(a).intersection(set(b)))

Answer 5

Do you want duplicates? If not maybe you should use sets instead:

>>> set([1, 2, 3, 4, 5]).intersection(set([9, 8, 7, 6, 5]))
set([5])

Answer 6

You can use

def returnMatches(a,b):
       return list(set(a) & set(b))

Answer 7

how to get the unmatched entries?

Answer 8

s = ['a','b','c']
f = ['a','b','d','c']
ss= set(s)
fs =set(f)
print ss.intersection(fs)
set(['a', 'c', 'b'])
print ss.union(fs)
set(['a', 'c', 'b', 'd'])
print ss.union(fs) - ss.intersection(fs)
set(['d'])

Answer 9

Setz' answer seems complicated. Perhaps you would rather do

set(f) - set(s)

:)

Answer 10

I prefer the set based answers, but here's one that works anyway

[x for x in a if x in b]