英文:
Regex: keep part of pattern multiple times
问题
我想使用正则表达式将","替换为", "(逗号后跟空格)，但不要在字符串末尾：
"a,b,c,d," 替换为 "a, b, c, d,"
同时也包括
"Berlin,London,Offenbach,Gera," 替换为 "Berlin, London, Offenbach, Gera,"
或者
"123,2345,653,12," 替换为 "123, 2345, 653, 12,"
我在Python中尝试了以下代码：
import re
re.sub(r'([1-9a-zA-Z]),([1-9a-zA-Z])', r', ', "a,b,c,d,")
但是得到了结果：
"a, b,c, d,"（"b"和"c"之间的逗号后没有空格）
有什么问题？
英文:
I want to use regex for replacing "," with ", "(comma followed by space) but not at the end of the string:
"a,b,c,d," to "a, b, c, d,"
but also
"Berlin,London,Offenbach,Gera," to "Berlin, London, Offenbach, Gera,"
or
"123,2345,653,12," to "123, 2345, 653, 12,"
I tried in Python
import re
re.sub(r'([1-9a-zA-Z]),([1-9a-zA-Z])', r', ', "a,b,c,d,")
but get the result:
'a, b,c, d,' (space after comma between "b" and "c" is missing)
What's wrong?
答案1
得分: 1
你可以简单地替换 ,(?=.)，这只匹配不出现在输入末尾的逗号。
inp = "a,b,c,d,"
output = re.sub(r',(?=.)', ', ', inp)
print(output) # a, b, c, d,
英文:
You may simply replace on ,(?=.), which will only match commas not occurring at the end of the input.
<!-- language: python -->
inp = "a,b,c,d,"
output = re.sub(r',(?=.)', ', ', inp)
print(output) # a, b, c, d,
答案2
得分: 1
你可以使用 ,(?!$) 来匹配除了行尾的逗号之外的任何逗号。
import re
print(re.sub(r',(?!\s*$)', ', ', "a,b,c,d,")+'|')
a, b, c, d,|
（| 用来显示缺少尾随空格）
解释：
(?! ... )是负向前瞻。它检查前一个匹配不是紧跟着此组的内容。$- 表示字符串结束的元字符。
英文:
You can use ,(?!$) to match any comma except in the end of line.
import re
print(re.sub(r',(?!$)', r', ', "a,b,c,d,")+'|')
a, b, c, d,|
(| to see lack of trailing space)
Explanation:
(?! ... )is a negative lookahead. It checks that previous match is not followed by content of this group.$- meta symbol meaning end of string.
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