Partial directory search question
rami.chowdhury at gmail.com
Wed Sep 30 06:34:10 CEST 2009
"Never attributed to malice that which can be attributed to stupidity." --
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On Tuesday 29 September 2009 19:54:17 chad wrote:
> On Sep 29, 7:52 pm, chad <cdal... at gmail.com> wrote:
> > On Sep 29, 7:20 pm, Tim Chase <python.l... at tim.thechases.com> wrote:
> > > > What's the sanest way to print out all the files in the directory
> > > > that start with the underscore? Ie, I just want to list _1, _2, _3,
> > > > _4.
> > >
> > > I'd use a string's join() method to combine the results of a
> > > list-comprehension or generator that filtered the output of
> > > os.listdir() based on the startswith() method of the strings.
> > >
> > > Left intentionally oblique and code-free because this sounds a
> > > bit like a home-work problem. If you're a python coder, that
> > > should make pretty decent sense and be a one-liner to implement.
> > >
> > > -tkc
> > Okay, sorry for the delay to the response. I got side tracked trying
> > to stalk, I mean talk to the 59 year old neighbor girl. Anyways, I
> > couldn't get it to one in one line. Here is what I did...
> > % more rec.py
> > #!/usr/local/bin/python
> > import os
> > import time
> > for filename in os.listdir("/usr/bbs/confs/september"):
> > #stat = os.stat(filename)
> > if filename.startswith("_"):
> > print filename
> > ./rec.py
> > _1
> > _2
> > _3
> > _4
> > _5
> > _6
> > _7
> > _8
> > It correctly prints out all the files in the directory that start with
> > an underscore.
> er *couldn't get it into a one liner*.
To get it into one line, I suggest:
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