如何比较python中的两个列表并返回不匹配

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 return

[[ 4,5 ],[ 6 ]]

Answer 1

One of the simplest and quickest is:

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

BONUS! If you want an intersection (return the matches):

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

Answer 2

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 3

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 4

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).

Answer 5

您可以使用列表理解和 zip

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