Python:分割字符串，尊重和保留引号[重复]

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这个问题已经有了答案:

Split a string by spaces — preserving quoted substrings — in Python 14 answers
在Python 14中，用空格分隔字符串——保留引用的子字符串

Using python, I want to split the following string:

使用python，我想拆分如下字符串:

a=foo, b=bar, c="foo, bar", d=false, e="false"

This should result in the following list:

这应导致下列清单:

['a=foo', 'b=bar', 'c="foo, bar"', 'd=false', 'e="false'"']

When using shlex in posix-mode and splitting with ", ", the argument for cgets treated correctly. However, it removes the quotes. I need them because false is not the same as "false", for instance.

当在posix模式下使用shlex并使用“，”拆分时，clex的参数得到了正确的处理。但是，它删除了引号。我需要它们，因为false和false不一样。

My code so far:

到目前为止我的代码:

import shlex

mystring = 'a=foo, b=bar, c="foo, bar", d=false, e="false"'

splitter = shlex.shlex(mystring, posix=True)
splitter.whitespace += ','
splitter.whitespace_split = True
print list(splitter) # ['a=foo', 'b=bar', 'c=foo, bar', 'd=false', 'e=false']

2 个解决方案

#1

>>> s = r'a=foo, b=bar, c="foo, bar", d=false, e="false", f="foo\", bar"'
>>> re.findall(r'(?:[^\s,"]|"(?:\\.|[^"])*")+', s)
['a=foo', 'b=bar', 'c="foo, bar"', 'd=false', 'e="false"', 'f="foo\\", bar"']

The regex pattern "[^"]*" matches a simple quoted string.
正则表达式模式”[^]*”一个简单的引用字符串匹配。
"(?:\\.|[^"])*" matches a quoted string and skips over escaped quotes because \\. consumes two characters: a backslash and any character.
“(?:\ \ |。[^])*”匹配引用字符串和跳过逃脱了引号,因为\ \。使用两个字符:反斜杠和任何字符。
[^\s,"] matches a non-delimiter.
[]匹配non-delimiter ^ \ s。”
Combining patterns 2 and 3 inside (?: | )+ matches a sequence of non-delimiters and quoted strings, which is the desired result.
将模式2和3组合在一起(?: |)+匹配一个非分隔符和引号字符串序列，这是期望的结果。

#2

Regex can solve this easily enough:

Regex可以很容易地解决这个问题:

import re

mystring = 'a=foo, b=bar, c="foo, bar", d=false, e="false"'

splitString = re.split(',?\s(?=\w+=)',mystring)

The regex pattern here looks for a whitespace followed by a word character and then an equals sign which splits your string as you desire and maintains any quotes.

这里的regex模式寻找空格，后面跟着一个单词字符，然后是一个等号，这个等号可以按照您的意愿分割字符串并保留任何引号。

#1

>>> s = r'a=foo, b=bar, c="foo, bar", d=false, e="false", f="foo\", bar"'
>>> re.findall(r'(?:[^\s,"]|"(?:\\.|[^"])*")+', s)
['a=foo', 'b=bar', 'c="foo, bar"', 'd=false', 'e="false"', 'f="foo\\", bar"']

The regex pattern "[^"]*" matches a simple quoted string.
正则表达式模式”[^]*”一个简单的引用字符串匹配。
"(?:\\.|[^"])*" matches a quoted string and skips over escaped quotes because \\. consumes two characters: a backslash and any character.
“(?:\ \ |。[^])*”匹配引用字符串和跳过逃脱了引号,因为\ \。使用两个字符:反斜杠和任何字符。
[^\s,"] matches a non-delimiter.
[]匹配non-delimiter ^ \ s。”
Combining patterns 2 and 3 inside (?: | )+ matches a sequence of non-delimiters and quoted strings, which is the desired result.
将模式2和3组合在一起(?: |)+匹配一个非分隔符和引号字符串序列，这是期望的结果。

#2