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list files in a directory

Wed, 28 May 2003 08:16:55 -0700

hi, i just started with python. i want to get a list of files in a directory but i only want to see the latest files. i need this so that i can see what's the latest file i have based on their number.

for example, i have 3 files: file-1.txt,file-2.txt and file-3.txt
how do i make it so that only file-3.txt is shown?

i found that i could use dircache.listdir(path) but that just lists the files. hope my question is not too confusing.

Re: list files in a directory

Wed, 28 May 2003 10:01:34 -0700

: hi, i just started with python. i want to get a list of files in a directory but i only want to see the latest files. i need this so that i can see what's the latest file i have based on their number.
:
: for example, i have 3 files: file-1.txt,file-2.txt and file-3.txt
: how do i make it so that only file-3.txt is shown?
:
: i found that i could use dircache.listdir(path) but that just lists the files. hope my question is not too confusing.

That's a tough one. I'll let you know if I find anything. Please post here if you find something yourself. Thanks!

infidel

Re: list files in a directory

Wed, 28 May 2003 11:07:32 -0700

This message was edited by rul at 2003-5-28 11:8:38

ok i got it partly working, here's the code

import dircache
list = dircache.listdir('test')
i = 0
check = len(list[0])
temp = []
count = len(list)

while count != 0:
if len(list[i]) != check:
temp.append(list[i-1])
check = len(list[i])
else:
i = i + 1
count = count - 1

print temp

------------------------------------------------------------------
inside the test directory, i have these files:
abc - 10.txt
abc - 11.txt
abc - 12.txt
xyz - 201.txt
xyz - 202.txt
xyz - 203.txt

however, the output is only abc - 12.txt
what i want is for the program to output abc - 12.txt and xyz - 203.txt

I got it

Wed, 28 May 2003 13:26:33 -0700

Very simple once I stopped to actually think about it.

import os

highestnumbers = {}
highestfiles = {}

for filename in os.listdir("C:\\TEMP\\py"):
    basename, extension = filename.split('.')
    prefix, number = basename.split(' - ')
    if number > highestnumbers.get(prefix, 0):
        highestnumbers[prefix] = number
        highestfiles[prefix] = filename

prefixes = highestfiles.keys()
prefixes.sort()
for prefix in prefixes:
    print 'Latest file for %s prefix is: %s' % (prefix, highestfiles[prefix])

This, of course, depends on consistently applied naming conventions. If the file names change patterns at all then it won't work. Should get you started, though.

infidel

Re: I got it

Wed, 28 May 2003 21:02:00 -0700

: Very simple once I stopped to actually think about it.
:
:

: import os
: 
: highestnumbers = {}
: highestfiles = {}
: 
: for filename in os.listdir("C:\\TEMP\\py"):
:     basename, extension = filename.split('.')
:     prefix, number = basename.split(' - ')
:     if number > highestnumbers.get(prefix, 0):
:         highestnumbers[prefix] = number
:         highestfiles[prefix] = filename
: 
: prefixes = highestfiles.keys()
: prefixes.sort()
: for prefix in prefixes:
:     print 'Latest file for %s prefix is: %s' % (prefix, highestfiles[prefix])
:

:
: This, of course, depends on consistently applied naming conventions. If the file names change patterns at all then it won't work. Should get you started, though.
:
:
: infidel
:
:

it works great! thanks a lot!

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Sat, 30 Jan 2010 03:13:19 -0700

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Sat, 30 Jan 2010 03:16:36 -0700

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