CODEWITHSUNDEEP
pythonregex
Dreampuf
Dreampuf

Reputation: 1181

Get the common prefix substring through Regex

like this 

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

Could I get the common prefix substring through regex?

I try to 

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

But apparently that common prefix substring is ' \t '
I want use for dedent function like python textwrap module.

Upvotes: 0

Views: 385

Answers (4)

JL Peyret
JL Peyret

Reputation: 12164

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

=>

 hello there
 how are you?
HHHH

Upvotes: 0

georg
georg

Reputation: 214959

Here's an expression that finds a common prefix in a text:

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

Example:

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'

Note that this expression involves a lot of backtracking, so use it wisely, especially on large inputs.

To find out the longest common "space" prefix, just find them all and apply 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)

Upvotes: 1

user797257
user797257

Reputation:

I'm not that good with Python, so, maybe this code doesn't look idiomatic for the language, but algorithmically, it should be good:

>>> 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'
>>>

Regular expression will have a lot of overhead on top of this.

Upvotes: 0

Ωmega
Ωmega

Reputation: 43673

I suggest

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

See this demo.

Upvotes: 1

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