parsing directory for certain filetypes

[royG]

hi
i wrote a function to parse a given directory and make a sorted list
of files with .txt,.doc extensions .it works,but i want to know if it
is too bloated..can this be rewritten in more efficient manner?

here it is...

from string import split
from os.path import isdir,join,normpath
from os import listdir

def parsefolder(dirname):
filenms=[]
folder=dirname
isadr=isdir(folder)
if (isadr):
dirlist=listdir(folder)
filenm=""
for x in dirlist:
filenm=x
if(filenm.endswith(("txt","doc"))):
nmparts=[]
nmparts=split(filenm,'.' )
if((nmparts[1]=='txt') or (nmparts[1]=='doc')):
filenms.append(filenm)
filenms.sort()
filenameslist=[]
filenameslist=[normpath(join(folder,y)) for y in filenms]
numifiles=len(filenameslist)
print filenameslist
return filenameslist
folder='F:/mysys/code/tstfolder'
parsefolder(folder)
thanks,
RG

[sam]

royG napisał(a):

i wrote a function to parse a given directory and make a sorted list
of files with .txt,.doc extensions .it works,but i want to know if it
is too bloated..can this be rewritten in more efficient manner?

Probably this should be rewriten and should be very compact. Maybe you should
grab string:

find $dirname -type f -a \( -name '*.txt' -o -name '*.doc' \)

and split by "\n"?
--
UFO Occupation
www.totalizm.org

[jay graves]

On Mar 10, 8:57 am, royG <[e-mail]:

i wrote a function to parse a given directory and make a sorted list
of files with .txt,.doc extensions .it works,but i want to know if it
is too bloated..can this be rewritten in more efficient manner?

Try the 'glob' module.

....
Jay

[sam]

Robert Bossy napisał(a):

I leave you the exercice to add .doc files. But I must say (whoever's
listening) that I was a bit disappointed that glob('*.{txt,doc}') didn't
work.

"{" and "}" are bash invention and not POSIX standard unfortunately

--
UFO Occupation
www.totalizm.org

[jay graves]

On Mar 10, 9:28 am, Robert Bossy <[e-mail]:

Personally, I'd use glob.glob:

import os.path
import glob

def parsefolder(folder):
path = os.path.normpath(os.path.join(folder, '*.py'))
lst = [ fn for fn in glob.glob(path) ]
lst.sort()
return lst

Why the 'no-op' list comprehension? Typo?

....
Jay

[Tim Chase]

i wrote a function to parse a given directory and make a sorted list

of files with .txt,.doc extensions .it works,but i want to know if it
is too bloated..can this be rewritten in more efficient manner?

here it is...

from string import split
from os.path import isdir,join,normpath
from os import listdir

def parsefolder(dirname):
filenms=[]
folder=dirname
isadr=isdir(folder)
if (isadr):
dirlist=listdir(folder)
filenm=""
for x in dirlist:
filenm=x
if(filenm.endswith(("txt","doc"))):
nmparts=[]
nmparts=split(filenm,'.' )
if((nmparts[1]=='txt') or (nmparts[1]=='doc')):
filenms.append(filenm)
filenms.sort()
filenameslist=[]
filenameslist=[normpath(join(folder,y)) for y in filenms]
numifiles=len(filenameslist)
print filenameslist
return filenameslist
folder='F:/mysys/code/tstfolder'
parsefolder(folder)

It seems to me that this is awfully baroque with many unneeded
superfluous variables. Is this not the same functionality (minus
prints, unused result-counting, NOPs, and belt-and-suspenders
extension-checking) as

def parsefolder(dirname):
if not isdir(dirname): return
return sorted([
normpath(join(dirname, fname))
for fname in listdir(dirname)
if fname.lower().endswith('.txt')
or fname.lower().endswith('.doc')
])

In Python2.5 (or 2.4 if you implement the any() function, ripped
from the docs[1]), this could be rewritten to be a little more
flexible...something like this (untested):

def parsefolder(dirname, types=['.doc', '.txt']):
if not isdir(dirname): return
return sorted([
normpath(join(dirname, fname))
for fname in listdir(dirname)
if any(
fname.lower().endswith(s)
for s in types)
])

which would allow you to do both

parsefolder('/path/to/wherever/')

and

parsefolder('/path/to/wherever/', ['.xls', '.ppt', '.htm'])

In both cases, you don't define the case where isdir(dirname)
fails. Caveat Implementor.

-tkc
[1] http://docs.python.org/lib/built-in-funcs.html

[Robert Bossy]

jay graves wrote:

On Mar 10, 9:28 am, Robert Bossy <[e-mail]:

>Personally, I'd use glob.glob:

import os.path
import glob

def parsefolder(folder):
path = os.path.normpath(os.path.join(folder, '*.py'))
lst = [ fn for fn in glob.glob(path) ]
lst.sort()
return lst

Why the 'no-op' list comprehension? Typo?

My mistake, it is:

import os.path
import glob

def parsefolder(folder):
path = os.path.normpath(os.path.join(folder, '*.py'))
lst = glob.glob(path)
lst.sort()
return lst

[royG]

On Mar 10, 8:03 pm, Tim Chase wrote:

In Python2.5 (or 2.4 if you implement the any() function, ripped
from the docs[1]), this could be rewritten to be a little more
flexible...something like this (untested):

that was quite a good lesson for a beginner like me..
thanks guys

in the version using glob()

>path = os.path.normpath(os.path.join(folder, '*.txt'))
lst = glob.glob(path)

is it possible to check for more than one file extension? here i will
have to create two path variables like
path1 = os.path.normpath(os.path.join(folder, '*.txt'))
path2 = os.path.normpath(os.path.join(folder, '*.doc'))

and then use glob separately..
or is there another way?

RG

[Gerard Flanagan]

On Mar 11, 6:21 am, royG <[e-mail]:

On Mar 10, 8:03 pm, Tim Chase wrote:

In Python2.5 (or 2.4 if you implement the any() function, ripped
from the docs[1]), this could be rewritten to be a little more
flexible...something like this (untested):

that was quite a good lesson for a beginner like me..
thanks guys

in the version using glob()

path = os.path.normpath(os.path.join(folder, '*.txt'))
lst = glob.glob(path)

is it possible to check for more than one file extension? here i will
have to create two path variables like
path1 = os.path.normpath(os.path.join(folder, '*.txt'))
path2 = os.path.normpath(os.path.join(folder, '*.doc'))

and then use glob separately..
or is there another way?

I don't think you can match multiple patterns directly with glob, but
`fnmatch` - the module used by glob to do check for matches - has a
`translate` function which will convert a glob pattern to a regular
expression (string). So you can do something along the lines of the
following:

---------------------------------------------

import os
from fnmatch import translate
import re

d = '/tmp'
patt1 = '*.log'
patt2 = '*.ini'
patterns = [patt1, patt2]

rx = '|'.join(translate(p) for p in patterns)
patt = re.compile(rx)

for f in os.listdir(d):
if patt.match(f):
print f

---------------------------------------------

hth

Gerard

[jay graves]

On Mar 11, 12:21 am, royG <[e-mail]:

On Mar 10, 8:03 pm, Tim Chase wrote:
in the version using glob()

path = os.path.normpath(os.path.join(folder, '*.txt'))
lst = glob.glob(path)

is it possible to check for more than one file extension? here i will
have to create two path variables like
path1 = os.path.normpath(os.path.join(folder, '*.txt'))
path2 = os.path.normpath(os.path.join(folder, '*.doc'))

and then use glob separately..
or is there another way?

use a loop. (untested)

def parsefolder(folder):
lst = []
for pattern in ('*.txt','*.doc'):
path = os.path.normpath(os.path.join(folder, pattern))
lst.extend(glob.glob(path))
lst.sort()
return lst

[Tim Chase]

royG wrote:

On Mar 10, 8:03 pm, Tim Chase wrote:

>In Python2.5 (or 2.4 if you implement the any() function, ripped
from the docs[1]), this could be rewritten to be a little more
flexible...something like this (untested):

that was quite a good lesson for a beginner like me..
thanks guys

in the version using glob()

>path = os.path.normpath(os.path.join(folder, '*.txt'))
lst = glob.glob(path)

is it possible to check for more than one file extension? here i will
have to create two path variables like
path1 = os.path.normpath(os.path.join(folder, '*.txt'))
path2 = os.path.normpath(os.path.join(folder, '*.doc'))

and then use glob separately..

Though it doesn't use glob, the 2nd solution I gave (the one that
uses the any() function you quoted) should be able to handle an
arbitrary number of extensions...

-tkc