使用正则表达式拆分单个冒号而不是双冒号

I have a string like this

我有这样的字符串

"yJdz:jkj8h:jkhd::hjkjh"

I want to split it using colon as a separator, but not a double colon. Desired result:

我想用冒号作为分隔符拆分它，但不是双冒号。期望的结果：

("yJdz", "jkj8h", "jkhd::hjkjh")

I'm trying with:

我正在尝试：

re.split(":{1}", "yJdz:jkj8h:jkhd::hjkjh")

but I got a wrong result.

但是我得到了错误的结果。

In the meanwhile I'm escaping "::", with string.replace("::", "$$")

同时我用string.replace（“::”，“$$”）转义“::”

2 个解决方案

#1

You could split on (?<!:):(?!:). This uses two negative lookarounds (a lookbehind and a lookahead) which assert that a valid match only has one colon, without a colon before or after it.

你可以拆分（？<！:) :(？！:)。这使用了两个负面的外观（一个lookbehind和一个前瞻），它断言有效匹配只有一个冒号，在它之前或之后没有冒号。

To explain the pattern:

解释模式：

(?<!:)  # assert that the previous character is not a colon
:       # match a literal : character
(?!:)   # assert that the next character is not a colon

Both lookarounds are needed, because if there was only the lookbehind, then the regular expression engine would match the first colon in :: (because the previous character isn't a colon), and if there was only the lookahead, the second colon would match (because the next character isn't a colon).

两个外观是必需的，因为如果只有lookbehind，那么正则表达式引擎将匹配::中的第一个冒号（因为前一个字符不是冒号），如果只有前瞻，第二个冒号将匹配（因为下一个字符不是冒号）。

#2

You can do this with lookahead and lookbehind, if you want:

如果你想要的话，你可以通过前瞻和后瞻来做到这一点：

>>> s = "yJdz:jkj8h:jkhd::hjkjh"
>>> l = re.split("(?<!:):(?!:)", s)
>>> print l
['yJdz', 'jkj8h', 'jkhd::hjkjh']

This regex essentially says "match a : that is not followed by a : or preceded by a :"

这个正则表达式基本上是说“匹配a：后面没有a：或者前面跟着：”

#1