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
和.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)创建一个新字符串。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
将
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']
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