我正在解析一个包含如下行的文件
类型（“书”）标题（“金苹果”）页数（10-35 70 200-234）cmets（“好读”）我想将其拆分为单独的字段。
在我的示例中，有四个字段：类型、标题、页面和 cmets。
拆分后想要的结果是
['type("book")', 'title("golden apples")', 'pages(10-35 70 200-234)', 'cmets("good read")]很明显，简单的字符串拆分是行不通的，因为它只会在每个空格处拆分。 我想分割空格，但保留括号和引号之间的任何内容。
如何拆分？
【问题讨论】：
标签： python string-parsing
让我添加一个非正则表达式的解决方案：
line = 'type("book") title("golden apples") pages(10-35 70 200-234) comments("good read")'
count = 0 # Bracket counter
last_break = 0 # Index of the last break
parts = []
for j,char in enumerate(line):
if char is '(': count += 1
elif char is ')': count -= 1
elif char is ' ' and count is 0:
parts.append(line[last_break:(j)])
last_break = j+1
parts.append(line[last_break:]) # Add last element
parts = tuple(p for p in parts if p) # Convert to tuple and remove empty
for p in parts:
print(p)
一般情况下，您 cannot do with regular expressions 会处理某些事情，并且可能会导致严重的性能损失（尤其是前瞻和后视），这可能导致它们不是特定问题的最佳解决方案。
还有；我想我会提到可用于创建自定义文本解析器的 pyparsing 模块。
【讨论】：
拆分") " 并将) 添加回除最后一个元素之外的每个元素。
【讨论】：
我会尝试使用积极的后视断言。
r'(?<=\))\s+'
例子：
>>> import re
>>> result = re.split(r'(?<=\))\s+', 'type("book") title("golden apples") pages(10-35 70 200-234) comments("good read")')
>>> result
['type("book")', 'title("golden apples")', 'pages(10-35 70 200-234)', 'comments(
"good read")']
【讨论】：
test test test等输入文本中没有括号将不起作用。
这个正则表达式应该适合你\s+(?=[^()]*(?:\(|$))
result = re.split(r"\s+(?=[^()]*(?:\(|$))", subject)
解释
r"""
\s # Match a single character that is a “whitespace character” (spaces, tabs, and line breaks)
+ # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
(?= # Assert that the regex below can be matched, starting at this position (positive lookahead)
[^()] # Match a single character NOT present in the list “()”
* # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
(?: # Match the regular expression below
# Match either the regular expression below (attempting the next alternative only if this one fails)
\( # Match the character “(” literally
| # Or match regular expression number 2 below (the entire group fails if this one fails to match)
$ # Assert position at the end of a line (at the end of the string or before a line break character)
)
)
"""
【讨论】：
re.split(r"\s+(?=[^()]*(?:\(|$))", subject)