删除python中的阵列元素

发布于 2024-12-05 07:06:06 字数 451 浏览 3 评论 0原文

所有，我想从另一个数组中删除一个特定的数组元素。这是一个例子。阵列是一长串单词。

A = ['at','in','the']
B = ['verification','at','done','on','theresa']

我想删除出现在B中的单词。

B = ['verification','done','theresa']

这是我到目前为止尝试的错误

for word in A:
    for word in B:
        B = B.replace(word,"")

：我遇到了一个错误：

attributeError：'list'对象没有属性'替换'

我应该用什么来获得它？

原文

All, I want to delete specific array elements of one array from the other. Here is an example. The arrays are a long list of words though.

A = ['at','in','the']
B = ['verification','at','done','on','theresa']

I would like to delete words that appear in A from B.

B = ['verification','done','theresa']

Here is what I tried so far

for word in A:
    for word in B:
        B = B.replace(word,"")

I am getting an error:

AttributeError: 'list' object has no attribute 'replace'

What should I use to get it?

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计㈡愣 2024-12-12 07:06:06

使用列表理解来获得完整的答案：

[x for x in B if x not in A]

但是，您可能想进一步了解替换，因此...

Python列表没有替换方法。如果您只想从列表中删除元素，请将相关切片设置为空列表。例如：

>>> print B
['verification', 'at', 'done', 'on', 'theresa']
>>> x=B.index('at')
>>> B[x:x+1] = []
>>> print B
['verification', 'done', 'on', 'theresa']

请注意，尝试使用值b [x]进行相同的操作不会从列表中删除元素。

Using a list comprehension to get the full answer:

[x for x in B if x not in A]

However, you may want to know more about replace, so ...

python list has no replace method. If you just want to remove an element from a list, set the relevant slice to be an empty list. For example:

>>> print B
['verification', 'at', 'done', 'on', 'theresa']
>>> x=B.index('at')
>>> B[x:x+1] = []
>>> print B
['verification', 'done', 'on', 'theresa']

Note that trying to do the same thing with the value B[x] will not delete the element from the list.

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撩人痒 2024-12-12 07:06:06

如果您可以删除 B 中的重复项并且不关心顺序，则可以坚持使用集合：

>>> A = ['at','in','the']
>>> B = ['verification','at','done','on','theresa']
>>> list(set(B).difference(A))
['on', 'done', 'theresa', 'verification']

在这种情况下，您将获得显着的加速，因为集合中的查找比集合中的查找快得多。列表。实际上，在这种情况下最好设置 A 和 B

If you are ok to delete duplicates in B and don't care about order, you can stick up with sets:

>>> A = ['at','in','the']
>>> B = ['verification','at','done','on','theresa']
>>> list(set(B).difference(A))
['on', 'done', 'theresa', 'verification']

In this case you'll get a significant speedup as lookup in set is much faster than in list. Actually, in this case it's better A and B to be sets

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就是爱搞怪 2024-12-12 07:06:06

您还可以尝试从 B 中删除元素，例如：

A = ['at','in','the']
B = ['verification','at','done','on','theresa']
print B
for w in A:
    #since it can throw an exception if the word isn't in B
    try: B.remove(w)
    except: pass
print B

You could also try removing the elements from B, ex :

A = ['at','in','the']
B = ['verification','at','done','on','theresa']
print B
for w in A:
    #since it can throw an exception if the word isn't in B
    try: B.remove(w)
    except: pass
print B

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