Tom Plunket wrote:
> Frederic Rentsch wrote:
>
>
>>> Well, there is that small problem that there are leading tabs that I
>>> want stripped. I guess I could manually replace all tabs with eight
>>> spaces (as opposed to 'correct' tab stops), and then replace them when
>>> done, but it's probably just as easy to write a non-destructive dedent.
>>>
>> This should do the trick:
>>
>> >>> Dedent = re.compile ('^\s+')
>> >>> for line in lines: print Dedent.sub ('', line)
>>
>
> The fact that this doesn't do what dedent() does makes it not useful.
> Stripping all leading spaces from text is as easy as calling lstrip() on
> each line:
>
My goodness! How right your are.
> text = '\n'.join([line.lstrip() for line in text.split('\n')])
>
> alas, that isn't what I am looking for, nor is that what
> textwrap.dedent() is intended to do.
>
> -tom!
>
>
Following a call to dedent () it shouldn't be hard to translate leading
groups of so many spaces back to tabs. But this is probably not what you
want. If I understand your problem, you want to restore the dedented
line to its original composition if spaces and tabs are mixed and this
doesn't work because the information doesn't survive dedent (). Could
the information perhaps be passed around dedent ()? Like this: make a
copy of your lines and translate the copy's tabs to so many (8?) marker
bytes (e.g. ascii 0). Dedent the originals. Left-strip each of the
marked line copies to the length of its dedented original and translate
the marked groups back to tabs.
Frederic
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