按分隔符拆分时保持引用块完整答案

【问题标题】：keep quoted blocks intact when splitting by delimiter按分隔符拆分时保持引用块完整
【发布时间】：2018-11-20 11:13:31
【问题描述】：

给定一个示例字符串s = 'Hi, my name is Humpty-Dumpty, from "Alice, Through the Looking Glass"'，我想将它分成以下几个块：

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

我不知道我会在哪里找到多少个分隔符。

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能够保留：

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

【问题讨论】：

标签： python python-3.x split

【解决方案1】：

您可以通过正则表达式使用re.split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
)

【讨论】：

【解决方案2】：

我通过完全避免split 解决了这个问题：

s = 'Hi, my name is Humpty-Dumpty, from "Alice, Through the Looking Glass"'
l = []
substr = ""
quotes_open = False

for c in s:
    if c == ',' and not quotes_open: # check for comma only if no quotes open
        l.append(substr)
        substr = ""
    elif c == '\"':
        quotes_open = not quotes_open
    else:
        substr += c

l.append(substr)

print(l)

输出：

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

更通用的函数可能类似于：

def custom_split(input_str, delimiter=' ', avoid_between_char='\"'):
    l = []
    substr = ""
    between_avoid_chars = False
    for c in s:
        if c == delimiter and not between_avoid_chars:
            l.append(substr)
            substr = ""
        elif c == avoid_between_char:
            between_avoid_chars = not between_avoid_chars
        else:
            substr += c
    l.append(substr)
    return l

【讨论】：

【解决方案3】：

这将适用于这种特定情况，并且可以提供一个起点。

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

cut = re.search('(".*")', s)

r = re.sub('(".*")', '$VAR$', s).split(',')
res = []
for i in r:
    res.append(re.sub('\$VAR\$', cut.group(1), i))

输出

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

【讨论】：