How can I compare two lists in python and return not matches

Question

I would like to return values from both lists that not in the other one:

bar = [ 1,2,3,4,5 ]
foo = [ 1,2,3,6 ]

returnNotMatches( a,b )

would r

Answer 1

You could use a list comprehension and zip

''.join([i[0] for i in zip(a, a.lower()) if i[0] == i[1]])

Answer 2

This should do

def returnNotMatches(a, b):
    a = set(a)
    b = set(b)
    return [list(b - a), list(a - b)]

And if you don't care that the result should be a list you could just skip the final casting.

Answer 3

One of the simplest and quickest is:

new_list = list(set(list1).difference(list2))

Answer 4

Just use a list comprehension:

def returnNotMatches(a, b):
    return [[x for x in a if x not in b], [x for x in b if x not in a]]

Answer 5

I might rely on the stdlib here...

from itertools import tee, izip
def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

import difflib

def returnNotMatches(a, b):
    blocks = difflib.SequenceMatcher(a=a, b=b).get_matching_blocks()
    differences = []
    for b1, b2 in pairwise(blocks):
        d1 = a[b1.a + b1.size: b2.a]
        d2 = b[b1.b + b1.size: b2.b]
        differences.append((d1, d2))
    return differences

print returnNotMatches([ 1,2,3,4,5 ], [ 1,2,3,6 ])

which prints: [([4, 5], [6])]

This compares the sequences as streams and finds the differences in the streams. It takes order into account, etc. If order and duplicates don't matter, then sets are by far the way to go (so long as the elements can be hashed).