python remove duplicates from 2 lists

问题

I am trying to remove duplicates from 2 lists. so I wrote this function:

a = ["abc", "def", "ijk", "lmn", "opq", "rst", "xyz"]

b = ["ijk", "lmn", "opq", "rst", "123", "456", ]

for i in b:
    if i in a:
        print "found " + i
        b.remove(i)

print b

But I find that the matching items following a matched item does not get remove.

I get result like this:

found ijk
found opq
['lmn', 'rst', '123', '456']

but i expect result like this:

['123', '456']

How can I fix my function to do what I want?

Thank you.

回答1:

Here is what's going on. Suppose you have this list:

['a', 'b', 'c', 'd']

and you are looping over every element in the list. Suppose you are currently at index position 1:

['a', 'b', 'c', 'd']
       ^
       |
   index = 1

...and you remove the element at index position 1, giving you this:

['a',      'c', 'd']
       ^
       |
    index 1

After removing the item, the other items slide to the left, giving you this:

['a', 'c', 'd']
       ^
       |
    index 1

Then when the loop runs again, the loop increments the index to 2, giving you this:

['a', 'c', 'd']
            ^ 
            |
         index = 2

See how you skipped over 'c'? The lesson is: never delete an element from a list that you are looping over.

回答2:

Your problem seems to be that you're changing the list you're iterating over. Iterate over a copy of the list instead.

for i in b[:]:
    if i in a:
        b.remove(i)


>>> b
['123', '456']

But, How about using a list comprehension instead?

>>> a = ["abc", "def", "ijk", "lmn", "opq", "rst", "xyz"]
>>> b = ["ijk", "lmn", "opq", "rst", "123", "456", ]
>>> [elem for elem in b if elem not in a ]
['123', '456']

回答3:

What about

b= set(b) - set(a)

If you need possible repetitions in b to also appear repeated in the result and/or order to be preserved, then

b= [ x for x in b if not x in a ] 

would do.

回答4:

You asked to remove both the lists duplicates, here's my solution:

from collections import OrderedDict
a = ["abc", "def", "ijk", "lmn", "opq", "rst", "xyz"]
b = ["ijk", "lmn", "opq", "rst", "123", "456", ]

x = OrderedDict.fromkeys(a)
y = OrderedDict.fromkeys(b)

for k in x:
    if k in y:
        x.pop(k)
        y.pop(k)


print x.keys()
print y.keys()

Result:

['abc', 'def', 'xyz']
['123', '456']

The nice thing here is that you keep the order of both lists items

回答5:

or a set

set(b).difference(a)

be forewarned sets will not preserve order if that is important

回答6:

One way of avoiding the problem of editing a list while you iterate over it, is to use comprehensions:

a = ["abc", "def", "ijk", "lmn", "opq", "rst", "xyz"]
b = ["ijk", "lmn", "opq", "rst", "123", "456", ]
b = [x for x in b if not x in a]

回答7:

There are already many answers on "how can you fix it?", so this is a "how can you improve it and be more pythonic?": since what you want to achieve is to get the difference between list b and list a, you should use difference operation on sets (operations on sets):

>>> a = ["abc", "def", "ijk", "lmn", "opq", "rst", "xyz"]
>>> b = ["ijk", "lmn", "opq", "rst", "123", "456", ]
>>> s1 = set(a)
>>> s2 = set(b)
>>> s2 - s1
set(['123', '456'])

来源：https://stackoverflow.com/questions/18194968/python-remove-duplicates-from-2-lists