python从2个列表中删除重复项
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python remove duplicates from 2 lists
提问by michael
I am trying to remove duplicates from 2 lists. so I wrote this function:
我正在尝试从 2 个列表中删除重复项。所以我写了这个函数：
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.
但是我发现匹配项之后的匹配项不会被删除。
I get result like this:
我得到这样的结果：
found ijk
found opq
['lmn', 'rst', '123', '456']
but i expect result like this:
但我希望结果是这样的：
['123', '456']
['123', '456']
How can I fix my function to do what I want?
我怎样才能修复我的功能来做我想做的事？
Thank you.
谢谢你。
回答by Sukrit Kalra
Your problem seems to be that you're changing the list you're iterating over. Iterate over a copy of the list instead.
您的问题似乎是您正在更改正在迭代的列表。而是迭代列表的副本。
for i in b[:]:
if i in a:
b.remove(i)
>>> b
['123', '456']
But, How about using a list comprehension instead?
但是，如何使用列表理解来代替？
>>> 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by Joran Beasley
or a set
或一组
set(b).difference(a)
be forewarned sets will not preserve order if that is important
预先警告如果这很重要，集合将不会保留顺序
回答by Mario Rossi
What about
关于什么
b= set(b) - set(a)
If you need possible repetitions in bto also appear repeated in the result and/or order to be preserved, then
如果您需要可能的重复b出现在结果中重复出现和/或要保留的顺序，那么
b= [ x for x in b if not x in a ]
would do.
会做。
回答by Anthony Perot
You asked to remove both the lists duplicates, here's my solution:
您要求删除两个列表重复项，这是我的解决方案：
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:
结果：
['abc', 'def', 'xyz']
['123', '456']
The nice thing here is that you keep the order of both lists items
这里的好处是您保持两个列表项的顺序
回答by 7stud
Here is what's going on. Suppose you have this list:
这是发生了什么。假设你有这个列表：
['a', 'b', 'c', 'd']
and you are looping over every element in the list. Suppose you are currently at index position 1:
并且您正在遍历列表中的每个元素。假设您当前位于索引位置 1：
['a', 'b', 'c', 'd']
^
|
index = 1
...and you remove the element at index position 1, giving you this:
...然后您删除索引位置 1 处的元素，为您提供：
['a', 'c', 'd']
^
|
index 1
After removing the item, the other items slide to the left, giving you this:
删除项目后，其他项目向左滑动，为您提供：
['a', 'c', 'd']
^
|
index 1
Then when the loop runs again, the loop increments the index to 2, giving you this:
然后当循环再次运行时，循环将索引增加到 2，给你这个：
['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.
看看你是如何跳过“c”的？教训是：永远不要从您正在循环的列表中删除元素。
回答by Mayur Patel
One way of avoiding the problem of editing a list while you iterate over it, is to use comprehensions:
避免在迭代列表时编辑列表问题的一种方法是使用推导式：
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by Vincenzo Pii
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 band list a, you should use difference operation on sets (operations on sets):
已经有很多关于“你如何修复它？”的答案，所以这是一个“你如何改进它并变得更加pythonic？”：因为你想要实现的是获得 listb和 list之间的区别a，你应该对集合使用差分操作（对集合的操作）：
>>> 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'])
