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
但是我发现匹配项之后的匹配项不会被删除。
我得到这样的结果：
found ijk
found opq
['lmn', 'rst', '123', '456']
但我希望结果是这样的：
['123', '456']
我怎样才能修复我的功能来做我想做的事？
谢谢你。
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.
您的问题似乎是您正在更改您正在迭代的列表。而是迭代列表的副本。
for i in b[:]:
if i in a:
b.remove(i)
>>> b
['123', '456']
但是，如何使用列表推导来代替？
>>> 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']
关于什么
b= set(b) - set(a)
如果您需要可能的重复在b结果和/或要保留的顺序中也出现重复，那么
b= [ x for x in b if not x in a ]
会做。
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()
结果：
['abc', 'def', 'xyz']
['123', '456']
这里的好处是您保持两个列表项的顺序
或一组
set(b).difference(a)
如果这很重要，则预先警告集合将不会保持秩序
您可以使用 lambda 函数。
f = lambda list1, list2: list(filter(lambda element: element not in list2, list1))
list2 中的重复元素将从 list1 中删除。
>>> a = ["abc", "def", "ijk", "lmn", "opq", "rst", "xyz"]
>>> b = ["ijk", "lmn", "opq", "rst", "123", "456"]
>>> f(a, b)
['abc', 'def', 'xyz']
>>> f(b, a)
['123', '456']
避免在迭代列表时编辑列表问题的一种方法是使用推导：
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]
关于“如何解决它？”已经有很多答案了，所以这是一个“如何改进它并变得更加 Pythonic？”：既然你想要实现的是得到 listb和 list之间的区别a，你应该对集合使用差异操作（集合上的操作）：
>>> 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'])
您可以使用综合列表
a = ["abc", "def", "ijk", "lmn", "opq", "rst", "xyz"]
b = ["ijk", "lmn", "opq", "rst", "123", "456", ]
从 a 中删除的重复值
c=[value for value in a if value not in b]
从 b 中删除重复值
c=[value for value in b if value not in a]
按照 7stud 的思路，如果您以相反的顺序浏览列表，则不会遇到您遇到的问题：
a = ["abc", "def", "ijk", "lmn", "opq", "rst", "xyz"]
b = ["ijk", "lmn", "opq", "rst", "123", "456", ]
for i in reversed(b):
if i in a:
print "found " + i
b.remove(i)
print b
Output:
found rst
found opq
found lmn
found ijk
['123', '456']
一个简单的解决方法是迭代一个范围，查看索引处的元素，删除该元素，然后将计数器减 1。
模拟未经测试的代码
for i in range(0, len(b)):
if b[i] in a:
del b[i]
i -= 1