在字母前后删除字符串中的空格

Question

I have a quite a few sums of strings that look like this: "a name / another name / something else".

I want to get to this: "a name/another name/something else".

Basically removing the spaces before and after the forward slashes only (not between the words themselves).

I know nothing about programming but I looked and found that this can be done with Python and Regex. I was a bit overwhelmed though with the amount of information I found.

Answer 1

You can use the pattern:

(?:(?<=\/) | (?=\/))

• (?: Non capturing group.
• (?<=\\/) Lookbehind for / .
• | OR
• (?=\\/) Positive lookahead for / .
• ) Close non capturing group.

You can try it live here .

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Python snippet:

import re
str = 'a name / another name / something else'
print(re.sub(r'(?:(?<=\/) | (?=\/))','',str))

Prints:

a name/another name/something else

Answer 2

Here is an answer without using regex that I feel is easier to understand

string = "a name / another name / something else"
edited = "/".join([a.strip() for a in string.split("/")])
print(edited)

output:

a name/another name/something else

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.join() joins elements of a sequence by a given seperator, docs

.strip() removes beginning and trailing whitespace, docs

.split() splits the string into tokens by character, docs

Answer 3

There's no need for regex here, since you're simply replacing a string of literals.

str = "a name / another name / something else"
print(str.replace(" / ", "/"))

Answer 4

This pattern will match for any amount of whitespace surrounding / and remove it. I think the regex is relatively easy to understand

\s*([\/])\s*

Has a capturing group that matches the backslash (that's what the middle part is). The s* parts match whitespace (at least one up to any amount of times).

You can then replace these matched strings with just a / to get rid of all the whitespace.

Answer 5

str1是您的字符串：

re.sub(" / ", "/" ,str1)

Answer 6

Use the following code to remove all spaces before and after the / character:

import re
str = 'a name /  another name / something else'
str = re.sub(r'(?:(?<=\/)\s*|\s*(?=\/))','', str)

Check this document for more information.