Python 将空格替换为逗号忽略引号中的空格
给定一个由空格分隔的字符串。需要用逗号替换空白,忽略引号中的空白Python 将空格替换为逗号忽略引号中的空格,python,algorithm,Python,Algorithm,给定一个由空格分隔的字符串。需要用逗号替换空白,忽略引号中的空白 >>> some_string = 'one two "three four" five "six seven"' >>> replace_func(some_string) 'one,two,"three four",five,"six seven"' 以下是一个简单的决定: def replace_func(some_str): lines = [] i = 1 f
>>> some_string = 'one two "three four" five "six seven"'
>>> replace_func(some_string)
'one,two,"three four",five,"six seven"'
以下是一个简单的决定:
def replace_func(some_str):
lines = []
i = 1
for l in struct.split('"'):
if i % 2:
lines.append(l.replace(' ', ',')
else:
lines.append(l)
i += 1
parsed_struct = '"'.join(lines)
有什么建议吗?这可以通过以下帮助轻松完成: 我需要保留引号,您可以这样做:
>>> ','.join(['"{0}"'.format(fragment) if ' ' in fragment else fragment
... for fragment in shlex.split(some_string)])
'one,two,"three four",five,"six seven"'
或者,您可以尝试使用正则表达式实现更简单的解决方案:
>>> import re
>>> ','.join(re.findall('\"[^\"]*\"|\S+', some_string))
'one,two,"three four",five,"six seven"'
result = re.sub(' (?=(?:[^"]*"[^"]*")*[^"]*$)', ",", subject)
使用正则表达式的替代方法:
>>> import re
>>> ','.join(re.findall('\"[^\"]*\"|\S+', some_string))
'one,two,"three four",five,"six seven"'
result = re.sub(' (?=(?:[^"]*"[^"]*")*[^"]*$)', ",", subject)
它匹配一个空格,并且仅当其后有偶数个引号时才用逗号替换它。因此,它将只在字符串之外匹配。Pyparsing通常比正则表达式更容易阅读和理解:
>>> some_string = 'one two "three four" five "six seven"'
>>> from pyparsing import OneOrMore, quotedString, Word, printables
>>> ','.join(OneOrMore(quotedString | Word(printables)).parseString(some_string))
'one,two,"three four",five,"six seven"'
这也将删除双引号,这不是所谓的wants@MohamedKhamis谢谢你的评论。我更新了我的答案以考虑到这一点。太好了!幸好我在发布自己的乱七八糟之前看到了这一点!非常感谢thanx,我需要深入研究标准python库,就像正则表达式一样,thanx是您的解决方案