Question

像这样

text = "  \t  hello there\n  \t  how are you?\n  \t HHHH"
      hello there
      how are you?
     HHHH

我可以通过正则表达式获取公共前缀子字符串吗？

我试着

In [36]: re.findall(r"(?m)(?:(^[ \t]+).+[\n\r]+\1)", "  \t  hello there\n  \t  how are you?\n  \t HHHH")
Out[36]: ['  \t  ']

但显然公共前缀子字符串是'\ t' 我想用于dedent函数，如python textwrap模块。

Answer 1

这是一个在文本中找到公共前缀的表达式：

r'^(.+).*(\n\1.*)*$'

示例：

import re

text = (
    "No Red Leicester\n"
    "No Tilsit\n"
    "No Red Windsor"
)

m = re.match(r'^(.+).*(\n\1.*)*$', text)
if m:
    print 'common prefix is', m.group(1)
else:
    print 'no common prefix'

请注意，此表达式涉及大量回溯，因此请明智地使用它，尤其是在大型输入上。

找出最长的共同空间＆＃34;前缀，只需找到它们并应用len：

def dedent(text):
    prefix_len = min(map(len, re.findall('(?m)^\s+', text)))
    return re.sub(r'(?m)^.{%d}' % prefix_len, '', text)

text = (
    "     No Red Leicester\n"
    "    No Tilsit\n"
    "\t\t   No Red Windsor"
)

print dedent(text)

Answer 2

我建议

match = re.search(r'(?m)\A(.*).*(?:\n?^\1.*$)*\n?\Z', text)

请参阅 this demo 。

Answer 3

我对Python不太好，所以，也许这段代码对于语言看起来并不惯用，但从算法来看，它应该是好的：

>>> import StringIO
...
>>> def strip_common_prefix(text):
...     position = text.find('\n')
...     offset = position
...     match = text[: position + 1]
...     lines = [match]
...     while match and position != len(text):
...         next_line = text.find('\n', position + 1)
...         if next_line == -1: next_line = len(text)
...         line = text[position + 1 : next_line + 1]
...         position = next_line
...         lines.append(line)
...         i = 0
...         for a, b in zip(line, match):
...             if i > offset or a != b: break
...             i += 1
...         offset = i
...         match = line[: offset]
...     buf = StringIO.StringIO()
...     for line in lines:
...         if not match: buf.write(line)
...         else: buf.write(line[offset :])
...     text = buf.getvalue()
...     buf.close()
...     return text
... 
>>> strip_common_prefix("  \t  hello there\n  \t  how are you?\n  \t HHHH")
' hello there\n how are you?\nHHHH'
>>>

正则表达式会有很多开销。

Answer 4

import os
#not just  for paths...

text = "  \t  hello there\n  \t  how are you?\n  \t HHHH"
li = text.split("\n")

common = os.path.commonprefix(li)

li = [i[len(common):] for i in li]
for i in li:
    print i

=＆GT;

 hello there
 how are you?
HHHH

通过Regex获取公共前缀子字符串

4 个答案: