python - 按分隔符分割时保持引用 block 完整

Question

给定一个示例字符串 s = '嗨，我的名字是 Humpty-Dumpty，来自“爱丽丝，爱丽丝镜中奇遇记”'，我想将其分成以下 block :

# To Do: something like {l = s.split(',')}
l = ['Hi', 'my name is Humpty-Dumpty', '"Alice, Through the Looking Glass"']

我不知道在哪里可以找到多少分隔符。

这是我最初的想法，它很长，而且不准确，因为它删除了所有分隔符，而我希望引号内的分隔符保留下来:

s = 'Hi, my name is Humpty-Dumpty, from "Alice, Through the Looking Glass"'
ss = []
inner_string = ""
delimiter = ','

for item in s.split(delimiter):
    if not inner_string: 
        if '\"' not in item: # regullar string. not intersting
            ss.append(item)
        else:
            inner_string += item # start inner string

    elif inner_string:
        inner_string += item

        if '\"' in item:  # end inner string
            ss.append(inner_string)
            inner_string = ""
        else:            # middle of inner string
            pass

print(ss)
# prints ['Hi', ' my name is Humpty-Dumpty', ' from "Alice Through the Looking Glass"'] which is OK-ish

Answer 1

您可以使用 re.split 按正则表达式进行拆分:

>>> import re
>>> [x for x in re.split(r'([^",]*(?:"[^"]*"[^",]*)*)', s) if x not in (',','')]

当s等于:

'Hi, my name is Humpty-Dumpty, from "Alice, Through the Looking Glass"'

它输出:

['Hi', ' my name is Humpty-Dumpty', ' from "Alice, Through the Looking Glass"']

正则表达式解释:

(
    [^",]*          zero or more chars other than " or ,
    (?:             non-capturing group
        "[^"]*"     quoted block
        [^",]*      followed by zero or more chars other than " or ,
    )*              zero or more times
)