[英]Python - alternative to list.remove(x)?
I wish to compare two lists. 我想比较两个清单。 Generally this is not a problem as I usually use a nested for loop and append the intersection to a new list.
通常这不是问题,因为我通常使用嵌套的for循环并将交集附加到新列表中。 In this case, I need to delete the intersection of A and B from A.
在这种情况下,我需要从A中删除A和B的交集。
A = [['ab', 'cd', 'ef', '0', '567'], ['ghy5'], ['pop', 'eye']]
B = [['ab'], ['hi'], ['op'], ['ej']]
My objective is to compare A and B and delete A intersection B from A, ie, delete A[0][0] in this case. 我的目标是比较A和B并从A删除A交集B,即在这种情况下删除A [0] [0]。
I tried: 我试过了:
def match():
for i in A:
for j in i:
for k in B:
for v in k:
if j == v:
A.remove(j)
list.remove(x) throws a ValueError. list.remove(x)引发ValueError。
If possible (meaning if the order and the fact that you have "sublists" does not matter), I would first flatten the lists , create sets and then you can easily remove the elements from A
that are in B
: 如果可能(意味着顺序和您拥有“子列表”的事实无关紧要),我首先将列表展平 ,创建集合 ,然后可以轻松地从
A
中删除B
中的元素:
>>> from itertools import chain
>>> A = [['ab', 'cd', 'ef', '0', '567'], ['ghy5'], ['pop', 'eye']]
>>> B = [['ab'], ['hi'], ['op'], ['ej']]
>>> A = set(chain(*A))
>>> B = set(chain(*B))
>>> A-B
set(['ghy5', 'eye', 'ef', 'pop', 'cd', '0', '567'])
Or if the order and structure of A
matters, you can do (thanks and credits to THC4k ): 或者,如果
A
的顺序和结构A
重要,则可以这样做(感谢并感谢THC4k ):
>>> remove = set(chain(*B))
>>> A = [[x for x in S if x not in remove] for S in A].
But note: This only works under the assumption that A
and B
will be always lists of lists. 但是请注意:这仅在
A
和B
始终为列表列表的前提下起作用。
A naive approach using sets and itertools. 使用集合和迭代工具的幼稚方法。 You can tweak this further according to your requirements:
您可以根据自己的要求进一步进行调整:
#!/usr/bin/env python
a = [['ab', 'cd', 'ef', '0', '567'], ['ghy5'], ['pop', 'eye']]
b = [['ab'], ['hi'], ['op'], ['ej']]
from itertools import chain
# this results in the intersection, here => 'ab'
intersection = set(chain.from_iterable(a)).intersection(
set(chain.from_iterable(b)))
def nested_loop(iterable):
"""
Loop over arbitrary nested lists.
"""
for e in iterable:
if isinstance(e, list):
nested_loop(e)
else:
if e in intersection:
iterable.remove(e)
return iterable
print nested_loop(a)
# =>
# [['cd', 'ef', '0', '567'], ['ghy5'], ['pop', 'eye']]
Edit: To use remove in a situation like this, you can't directly remove j ('ab') in this case from the list a, since it is a nested list. 编辑:在这种情况下要使用删除,在这种情况下,您不能直接从列表a中删除j('ab'),因为它是一个嵌套列表。 You would have to use A.remove(['ab']) or A.remove([j]) to accomplish this.
您必须使用A.remove(['ab'])或A.remove([j])来完成此操作。
Another possibility is the pop(int) method. 另一种可能性是pop(int)方法。 So A.pop(index) should actually work as well.
因此A.pop(index)实际上也应该工作。
Source: http://docs.python.org/tutorial/datastructures.html 资料来源: http : //docs.python.org/tutorial/datastructures.html
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