Python: Split a string, respect and preserve quotes [duplicate]

Question

Using python, I want to split the following string:

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

This should result in the following list:

['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.

My code so far:

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']

Answer 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"']

1. The regex pattern "[^"]*" matches a simple quoted string.
2. "(?:\\.|[^"])*" matches a quoted string and skips over escaped quotes because \\. consumes two characters: a backslash and any character.
3. [^\s,"] matches a non-delimiter.
4. Combining patterns 2 and 3 inside (?: | )+ matches a sequence of non-delimiters and quoted strings, which is the desired result.