在python列表中查找相同索引的不匹配项

Question

I have following lists

l1 = ['a','b','c','a']
l2 = ['a','d','c','c']

I want to find out elements from l2 which do not match to elements of l1 at the same index. eg: output for above list would be ['d','c']

as l2 should have 'b' at 2nd position.

I can do this by iterating over the list and finding the mismatch.

l3 = []
for i in range(len(l1)):
    if l1[i] != l2[i]: l3.append(l2[i])
print l3

Is there any better way to do this. Thanks.

Answer 1

missing = [b for a,b in itertools.izip_longest(l1,l2,fillvalue=object()) if a != b]

比pythonic多一点...但是基本上是一样的

Answer 2

Python list comprehension, without needing to import any modules like itertools.

l3 = [b for a,b in zip(l1,l2) if b != a]

Whenever you find yourself wanting to do a for loop wherein at each iteration you may be appending to something that started as an empty list, think of using a list comprehension.

Answer 3

If you have the same size lists you can use enumerate :

l1 = ['a','b','c','a']
l2 = ['a','d','c','c']
print [ele for ind,ele in enumerate(l2) if ele != l1[ind]]
['d', 'c']