Python正则表达式：查找并替换引号之间的逗号

Question

我有一个字符串，

line = '12/08/2013,3,"9,25",42:51,"3,08","12,9","13,9",159,170,"3,19",437,'

并且我想在引号之间找到逗号并将其替换为“：”。寻找结果

line = '12/08/2013,3,9:25,42:51,3:08,12:9,13:9,159,170,3:19,437,'

到目前为止，我已经能够匹配此模式，

import re
re.findall('(\"\d),(.+?\")', line)

但是，我想我应该使用

re.compile(...something..., line)
re.sub(':', line)

有人知道该怎么做吗？谢谢，labjunky

Answer 1

>>> import re
>>> line = '12/08/2013,3,"9,25",42:51,"3,08","12,9","13,9",159,170,"3,19",437,'
>>> re.sub(r'"(\d+),(\d+)"', r'\1:\2', line)
'12/08/2013,3,9:25,42:51,3:08,12:9,13:9,159,170,3:19,437,'

[\1，\2是指匹配的组。

---

非正则表达式解决方案：

>>> ''.join(x if i % 2 == 0 else x.replace(',', ':')
            for i, x in enumerate(line.split('"')))
'12/08/2013,3,9:25,42:51,3:08,12:9,13:9,159,170,3:19,437,'

Answer 2

import re
line = '12/08/2013,3,"9,25",42:51,"3,08","12,9","13,9",159,170,"3,19",437,'
r = ""
for t in re.split(r'("[^"]*")', line):
    if t[0] == '"': 
        t = t.replace(",", ":")[1:-1]
    r += t
print r

打印：

12/08/2013,3,9:25,42:51,3:08,12:9,13:9,159,170,3:19,437,

Answer 3

还有一种通用的正则表达式解决方案，用于替换双引号（或单引号）之间的任何类型的固定（和非固定的）模式：用相应的模式匹配双引号或单引号的子字符串，并使用可调用的作为re.sub的替代参数，您可以在其中操纵比赛：

1. 用逗号替换双引号之间的逗号，并删除双引号（当前OP方案）：

   re.sub(r'"([^"]*)"', lambda x: x.group(1).replace(',', ':'), line)（demo）# => 12/08/2013,3,9:25,42:51,3:08,12:9,13:9,159,170,3:19,437,

2. 用逗号将双引号之间的逗号替换，并将keep用双引号引起来：

   re.sub(r'"[^"]*"', lambda x: x.group(0).replace(',', ':'), line)（demo）# => 12/08/2013,3,"9:25",42:51,"3:08","12:9","13:9",159,170,"3:19",437,

3. 用逗号将双引号（keep]替换为逗号：re.sub(r''''[^']*'|"[^"]*"''', lambda x: x.group(0).replace(',', ':'), '''0,1,"2,3",'4,5',''')（demo）# => 0,1,"2:3",'4:5',

   [此外，如果需要处理转义的单引号和双引号，请考虑使用r"'[^\\']*(?:\\.[^\\']*)*'"（对于单引号的子字符串），r'"[^\\"]*(?:\\.[^\\"]*)*"'（对于双引号的子字符串）或同时使用-r''''[^\\']*(?:\\.[^\\']*)*'|"[^\\"]*(?:\\.[^\\"]*)*"'''，而不是上面的模式。] >