如何从 Python 中的一组字符串中删除特定的子字符串？
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How to remove specific substrings from a set of strings in Python?
提问by controlfreak
I have a set of strings set1, and all the strings in set1have a two specific substrings which I don't need and want to remove.
Sample Input:
set1={'Apple.good','Orange.good','Pear.bad','Pear.good','Banana.bad','Potato.bad'}
So basically I want the .goodand .badsubstrings removed from all the strings.
What I tried:
我有一组 strings set1，并且其中的所有字符串set1都有两个特定的子字符串，我不需要并且想要删除它们。
示例输入：
set1={'Apple.good','Orange.good','Pear.bad','Pear.good','Banana.bad','Potato.bad'}
所以基本上我希望从所有字符串中删除.good和.bad子字符串。
我试过的：
for x in set1:
x.replace('.good','')
x.replace('.bad','')
But this doesn't seem to work at all. There is absolutely no change in the output and it is the same as the input. I tried using for x in list(set1)instead of the original one but that doesn't change anything.
但这似乎根本不起作用。输出绝对没有变化，它与输入相同。我尝试使用for x in list(set1)而不是原始的，但这并没有改变任何东西。
回答by Reut Sharabani
Strings are immutable. string.replace(python 2.x) or str.replace(python 3.x) creates a newstring. This is stated in the documentation:
字符串是不可变的。string.replace(python 2.x) 或str.replace(python 3.x) 创建一个新字符串。这在文档中说明：
Return a copyof string s with all occurrences of substring old replaced by new. ...
返回字符串 s的副本，其中所有出现的子字符串 old 都被 new 替换。...
This means you have to re-allocate the set or re-populate it (re-allocating is easier with set comprehension):
这意味着您必须重新分配集合或重新填充它（使用集合理解重新分配更容易）：
new_set = {x.replace('.good', '').replace('.bad', '') for x in set1}
回答by Alex Hall
>>> x = 'Pear.good'
>>> y = x.replace('.good','')
>>> y
'Pear'
>>> x
'Pear.good'
.replacedoesn't changethe string, it returns a copy of the string with the replacement. You can't change the string directly because strings are immutable.
.replace不更改字符串，它返回带有替换字符串的副本。您不能直接更改字符串，因为字符串是不可变的。
You need to take the return values from x.replaceand put them in a new set.
您需要从中获取返回值x.replace并将它们放入新集合中。
回答by gueeest
All you need is a bit of black magic!
你所需要的只是一点黑魔法！
>>> a = ["cherry.bad","pear.good", "apple.good"]
>>> a = list(map(lambda x: x.replace('.good','').replace('.bad',''),a))
>>> a
['cherry', 'pear', 'apple']
回答by Vivek
You could do this:
你可以这样做：
import re
import string
set1={'Apple.good','Orange.good','Pear.bad','Pear.good','Banana.bad','Potato.bad'}
for x in set1:
x.replace('.good',' ')
x.replace('.bad',' ')
x = re.sub('\.good$', '', x)
x = re.sub('\.bad$', '', x)
print(x)
回答by user140259
I did the test (but it is not your example) and the data does not return them orderly or complete
我做了测试（但它不是你的例子）并且数据没有有序或完整地返回它们
>>> ind = ['p5','p1','p8','p4','p2','p8']
>>> newind = {x.replace('p','') for x in ind}
>>> newind
{'1', '2', '8', '5', '4'}
I proved that this works:
我证明这是有效的：
>>> ind = ['p5','p1','p8','p4','p2','p8']
>>> newind = [x.replace('p','') for x in ind]
>>> newind
['5', '1', '8', '4', '2', '8']
or
或者
>>> newind = []
>>> ind = ['p5','p1','p8','p4','p2','p8']
>>> for x in ind:
... newind.append(x.replace('p',''))
>>> newind
['5', '1', '8', '4', '2', '8']
回答by cs95
When there are multiple substrings to remove, one simple and effective option is to use re.subwith a compiled pattern that involves joining all the substrings-to-remove using the regex OR (|) pipe.
当要删除多个子字符串时，一个简单而有效的选择是使用re.sub编译模式，该模式涉及使用正则表达式 OR ( |) 管道连接所有要删除的子字符串。
import re
to_remove = ['.good', '.bad']
strings = ['Apple.good','Orange.good','Pear.bad']
p = re.compile('|'.join(map(re.escape, to_remove))) # escape to handle metachars
[p.sub('', s) for s in strings]
# ['Apple', 'Orange', 'Pear']
回答by rsc05
If list
如果列出
I was doing something for a list which is a set of strings and you want to remove all lines that have a certain substring you can do this
我正在为一个列表做一些事情，它是一组字符串，你想删除所有具有某个子字符串的行，你可以这样做
import re
def RemoveInList(sub,LinSplitUnOr):
indices = [i for i, x in enumerate(LinSplitUnOr) if re.search(sub, x)]
A = [i for j, i in enumerate(LinSplitUnOr) if j not in indices]
return A
where subis a patter that you do not wish to have in a list of lines LinSplitUnOr
sub您不希望在行列表中出现的模式在哪里LinSplitUnOr
for example
例如
A=['Apple.good','Orange.good','Pear.bad','Pear.good','Banana.bad','Potato.bad']
sub = 'good'
A=RemoveInList(sub,A)
Then Awill be
然后A将是
