Larry Bates kirjoitti:
> Wolfgang Draxinger wrote:
>> Jussi Salmela wrote:
>>
>>> I'm not claiming the following to be more elegant, but I would
>>> do it like this (not tested!):
>>>
>>> src_file_paths = dict()
>>> prefix = sourcedir + os.sep
>>> for fname in os.listdir(sourcedir):
>>> if mat
Wolfgang Draxinger wrote:
> I got, hmm not really a problem, more a question of elegance:
>
> In a current project I have to read in some files in a given
> directory in chronological order, so that I can concatenate the
> contents in those files into a new one (it's XML and I have to
> concatena
Tim Williams wrote:
> Are you running on windows?
No. Of course I could use some installed tool, like ls, but I
want a portable solution.
> HTH :)
Not really. Windows is only a niche for desktop systems. In my
case the python script is running on a heterogenous grid
monitoring system accessing
On 20/02/07, Wolfgang Draxinger <[EMAIL PROTECTED]> wrote:
> H folks,
>
> I got, hmm not really a problem, more a question of elegance:
>
> In a current project I have to read in some files in a given
> directory in chronological order, so that I can concatenate the
> contents in those files into a
Larry Bates wrote:
> 3) You didn't handle the possibility that there is s
> subdirectory
>in the current directory. You need to check to make sure it
>is a file you are processing as os.listdir() returns files
>AND directories.
Well, the directory the files are in is not supposed to
Wolfgang Draxinger wrote:
> However this code works (tested) and behaves just like listdir,
> only that it sorts files chronologically, then alphabetically.
>
> def listdir_chrono(dirpath):
> import os
> files_dict = dict()
> for fname in os.listdir(dirpath):
>
Wolfgang Draxinger kirjoitti:
> Jussi Salmela wrote:
>
>> I'm not claiming the following to be more elegant, but I would
>> do it like this (not tested!):
>>
>> src_file_paths = dict()
>> prefix = sourcedir + os.sep
>> for fname in os.listdir(sourcedir):
>> if match_fname_pattern(fname):
>>
Wolfgang Draxinger <[EMAIL PROTECTED]> writes:
> src_file_paths = dict()
> for fname in os.listdir(sourcedir):
> fpath = sourcedir+os.sep+fname
> if not match_fname_pattern(fname): continue
> src_file_paths[os.stat(fpath).st_mtime] = fpath
> for ftime in src_file_paths.keys(
Wolfgang Draxinger wrote:
> Jussi Salmela wrote:
>
>> I'm not claiming the following to be more elegant, but I would
>> do it like this (not tested!):
>>
>> src_file_paths = dict()
>> prefix = sourcedir + os.sep
>> for fname in os.listdir(sourcedir):
>> if match_fname_pattern(fname):
>>
Jussi Salmela wrote:
> I'm not claiming the following to be more elegant, but I would
> do it like this (not tested!):
>
> src_file_paths = dict()
> prefix = sourcedir + os.sep
> for fname in os.listdir(sourcedir):
> if match_fname_pattern(fname):
> fpath = prefix + fname
>
Wolfgang Draxinger kirjoitti:
> H folks,
>
> I got, hmm not really a problem, more a question of elegance:
>
> In a current project I have to read in some files in a given
> directory in chronological order, so that I can concatenate the
> contents in those files into a new one (it's XML and I ha
H folks,
I got, hmm not really a problem, more a question of elegance:
In a current project I have to read in some files in a given
directory in chronological order, so that I can concatenate the
contents in those files into a new one (it's XML and I have to
concatenate some subelements, about 4
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