如何从Python中的一组字符串中删除特定的子字符串?

问题:如何从Python中的一组字符串中删除特定的子字符串?

我有一组字符串set1,并且其中的所有字符串set1都有两个不需要并且想要删除的特定子字符串。
输入示例: 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.bad子字符串。
我试过的

for x in set1:
    x.replace('.good','')
    x.replace('.bad','')

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)而不是原始版本,但没有任何改变。

I have a set of strings set1, and all the strings in set1 have 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 .good and .bad substrings removed from all the strings.
What I tried:

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.


回答 0

字符串是不可变的。string.replace(python 2.x)或str.replace(python 3.x)创建一个字符串。在文档中对此进行了说明:

返回字符串s 的副本,其中所有出现的子字符串old都替换为new。…

这意味着您必须重新分配集合或重新填充集合(使用集合推导更容易进行重新分配

new_set = {x.replace('.good', '').replace('.bad', '') for x in set1}

Strings are immutable. string.replace (python 2.x) or str.replace (python 3.x) creates a new string. This is stated in the documentation:

Return a copy of string s with all occurrences of substring old replaced by new. …

This means you have to re-allocate the set or re-populate it (re-allocating is easier with set comprehension):

new_set = {x.replace('.good', '').replace('.bad', '') for x in set1}

回答 1

>>> x = 'Pear.good'
>>> y = x.replace('.good','')
>>> y
'Pear'
>>> x
'Pear.good'

.replace不会更改字符串,而是返回字符串的副本并替换。您不能直接更改字符串,因为字符串是不可变的。

您需要从中获取返回值x.replace并将其放入新集合中。

>>> x = 'Pear.good'
>>> y = x.replace('.good','')
>>> y
'Pear'
>>> x
'Pear.good'

.replace doesn’t change the string, it returns a copy of the string with the replacement. You can’t change the string directly because strings are immutable.

You need to take the return values from x.replace and put them in a new set.


回答 2

您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']

All you need is a bit of black magic!

>>> a = ["cherry.bad","pear.good", "apple.good"]
>>> a = list(map(lambda x: x.replace('.good','').replace('.bad',''),a))
>>> a
['cherry', 'pear', 'apple']

回答 3

您可以这样做:

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)

You could do this:

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 4

我进行了测试(但这不是您的示例),并且数据未按顺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'}

我证明这可行:

>>> ind = ['p5','p1','p8','p4','p2','p8']
>>> newind = [x.replace('p','') for x in ind]
>>> newind
['5', '1', '8', '4', '2', '8']

要么

>>> newind = []
>>> ind = ['p5','p1','p8','p4','p2','p8']
>>> for x in ind:
...     newind.append(x.replace('p',''))
>>> newind
['5', '1', '8', '4', '2', '8']

I did the test (but it is not your example) and the data does not return them orderly or complete

>>> 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:

>>> ind = ['p5','p1','p8','p4','p2','p8']
>>> newind = [x.replace('p','') for x in ind]
>>> newind
['5', '1', '8', '4', '2', '8']

or

>>> 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 5

当有多个要删除的子字符串时,一种简单有效的选择是re.sub与已编译模式一起使用,该模式涉及使用regex 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']

When there are multiple substrings to remove, one simple and effective option is to use re.sub with a compiled pattern that involves joining all the substrings-to-remove using the regex OR (|) pipe.

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 6

如果清单

我正在为包含一组字符串的列表做某事,并且您想要删除具有特定子字符串的所有行,可以执行此操作

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

这里sub是一个图案,你不希望在行的列表LinSplitUnOr

例如

A=['Apple.good','Orange.good','Pear.bad','Pear.good','Banana.bad','Potato.bad']
sub = 'good'
A=RemoveInList(sub,A)

然后A

If list

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

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 sub is a patter that you do not wish to have in a list of lines LinSplitUnOr

for example

A=['Apple.good','Orange.good','Pear.bad','Pear.good','Banana.bad','Potato.bad']
sub = 'good'
A=RemoveInList(sub,A)

Then A will be


回答 7

如果您从列表中删除某些内容,则可以使用以下方式:(方法子区分大小写)

new_list = []
old_list= ["ABCDEFG","HKLMNOP","QRSTUV"]

for data in old_list:
     new_list.append(re.sub("AB|M|TV", " ", data))

print(new_list) // output : [' CDEFG', 'HKL NOP', 'QRSTUV']

if you delete something from list , u can use this way : (method sub is case sensitive)

new_list = []
old_list= ["ABCDEFG","HKLMNOP","QRSTUV"]

for data in old_list:
     new_list.append(re.sub("AB|M|TV", " ", data))

print(new_list) // output : [' CDEFG', 'HKL NOP', 'QRSTUV']