count files in a directory

**James Stroud** · Jul 19 '05, 01:28 AM

Re: count files in a directory

On Friday 20 May 2005 07:12 pm, rbt wrote:[color=blue]
> I assume that there's a better way than this to count the files in a
> directory recursively. Is there???
>
> def count_em(valid_ path):
> x = 0
> for root, dirs, files in os.walk(valid_p ath):
> for f in files:
> x = x+1
> print "There are", x, "files in this directory."
> return x
>
> rbt[/color]

def count_em(valid_ path):
root, dirs, files = os.walk(valid_p ath)
return len(files)

--
James Stroud
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095

http://www.jamesstroud.com/

**James Stroud** · Jul 19 '05, 01:28 AM

Re: count files in a directory

Come to think of it

file_count = len(os.walk(val id_path)[2])

--
James Stroud
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095

http://www.jamesstroud.com/

**Scott David Daniels** · Jul 19 '05, 01:28 AM

Re: count files in a directory

James Stroud wrote:[color=blue]
> Come to think of it
>
> file_count = len(os.walk(val id_path)[2])
>
> --
> James Stroud
> UCLA-DOE Institute for Genomics and Proteomics
> Box 951570
> Los Angeles, CA 90095
>
> http://www.jamesstroud.com/[/color]

Somee possible answers are:

# files directly in path
file_count = len(os.walk(pat h)[2])

# files and dirs directly in path
file_count = len(os.listdir( path))

# files in or below path
file_count = 0
for root, dirs, files in os.walk(path):
file_count += len(files)

# files and dirs in or below path
file_count = 0
for root, dirs, files in os.walk(path):
file_count += len(files) + len(dirs)

--Scott David Daniels
Scott.Daniels@A cm.Org

**rbt** · Jul 19 '05, 01:28 AM

Re: count files in a directory

James Stroud wrote:[color=blue]
> Come to think of it
>
> file_count = len(os.walk(val id_path)[2])[/color]

I get this traceback:

PythonWin 2.4 (#60, Nov 30 2004, 09:34:21) [MSC v.1310 32 bit (Intel)] on win32.
Portions Copyright 1994-2004 Mark Hammond (mhammond@skipp inet.com.au) - see
'Help/About PythonWin' for further copyright information.
Traceback (most recent call last):
File "C:\Program
Files\Python24\ Lib\site-packages\python win\pywin\frame work\scriptutil s.py", line 310,
in RunScript
exec codeObject in __main__.__dict __
File "C:\Documen ts and Settings\rbt\De sktop\newa\repl icate.py", line 57, in ?
A = count_em(X)
File "C:\Documen ts and Settings\rbt\De sktop\newa\repl icate.py", line 51, in count_em
count = len(os.walk(val id_path)[2])
TypeError: unsubscriptable object

**rbt** · Jul 19 '05, 01:28 AM

Re: count files in a directory

James Stroud wrote:[color=blue]
> def count_em(valid_ path):
> root, dirs, files = os.walk(valid_p ath)
> return len(files)[/color]

Here's another Tback:
[color=blue][color=green][color=darkred]
>>> Traceback (most recent call last):[/color][/color][/color]
File "C:\Program
Files\Python24\ Lib\site-packages\python win\pywin\frame work\scriptutil s.py", line 310,
in RunScript
exec codeObject in __main__.__dict __
File "C:\Documen ts and Settings\rbt\De sktop\newa\repl icate.py", line 62, in ?
A = count_em(X)
File "C:\Documen ts and Settings\rbt\De sktop\newa\repl icate.py", line 56, in count_em
root, dirs, files = os.walk(valid_p ath)
ValueError: need more than 2 values to unpack

**James Stroud** · Jul 19 '05, 01:28 AM

Re: count files in a directory

Sorry, I've never used os.walk and didn't realize that it is a generator.

This will work for your purposes (and seems pretty fast compared to the
alternative):

file_count = len(os.walk(val id_path).next()[2])

The alternative is:

import os
import os.path

file_count = len([f for f in os.listdir('.') if os.path.isfile( f)])

On Friday 20 May 2005 08:08 pm, rbt wrote:[color=blue]
> James Stroud wrote:[color=green]
> > def count_em(valid_ path):
> > root, dirs, files = os.walk(valid_p ath)
> > return len(files)[/color]
>
> Here's another Tback:[color=green][color=darkred]
> >>> Traceback (most recent call last):[/color][/color]
>
> File "C:\Program
> Files\Python24\ Lib\site-packages\python win\pywin\frame work\scriptutil s.py",
> line 310, in RunScript
> exec codeObject in __main__.__dict __
> File "C:\Documen ts and Settings\rbt\De sktop\newa\repl icate.py", line 62,
> in ? A = count_em(X)
> File "C:\Documen ts and Settings\rbt\De sktop\newa\repl icate.py", line 56,
> in count_em root, dirs, files = os.walk(valid_p ath)
> ValueError: need more than 2 values to unpack[/color]

--
James Stroud
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095

http://www.jamesstroud.com/

**Heiko Wundram** · Jul 19 '05, 01:28 AM

Re: count files in a directory

Am Samstag, 21. Mai 2005 06:25 schrieb James Stroud:[color=blue]
> This will work for your purposes (and seems pretty fast compared to the
> alternative):
>
> file_count = len(os.walk(val id_path).next()[2])[/color]

But will only work when you're just scanning a single directory with no
subdirectories. ..!

The alternative (which will work regardless of subdirectories) is something
like the following:

heiko@heiko ~ $ python
Python 2.4 (#1, Apr 3 2005, 00:49:51)
[GCC 3.4.3-20050110 (Gentoo Linux 3.4.3.20050110-r1, ssp-3.4.3.20050110-0,
pie- on linux2
Type "help", "copyright" , "credits" or "license" for more information.[color=blue][color=green][color=darkred]
>>> import os
>>> path = "/home/heiko"
>>> file_count = sum((len(f) for _, _, f in os.walk(path)))
>>> file_count[/color][/color][/color]
55579[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

HTH!

--
--- Heiko.
see you at: http://www.stud.mh-hannover.de/~hwundram/wordpress/

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**rbt** · Jul 19 '05, 01:29 AM

Re: count files in a directory

Heiko Wundram wrote:[color=blue]
> Am Samstag, 21. Mai 2005 06:25 schrieb James Stroud:
>[color=green]
>>This will work for your purposes (and seems pretty fast compared to the
>>alternative ):
>>
>>file_count = len(os.walk(val id_path).next()[2])[/color]
>
>
> But will only work when you're just scanning a single directory with no
> subdirectories. ..!
>
> The alternative (which will work regardless of subdirectories) is something
> like the following:
>
> heiko@heiko ~ $ python
> Python 2.4 (#1, Apr 3 2005, 00:49:51)
> [GCC 3.4.3-20050110 (Gentoo Linux 3.4.3.20050110-r1, ssp-3.4.3.20050110-0,
> pie- on linux2
> Type "help", "copyright" , "credits" or "license" for more information.
>[color=green][color=darkred]
>>>>import os
>>>>path = "/home/heiko"
>>>>file_coun t = sum((len(f) for _, _, f in os.walk(path)))
>>>>file_coun t[/color][/color]
>
> 55579
>
>
> HTH!
>[/color]

Thanks! that works great... is there any significance to the underscores that you
used? I've always used root, dirs, files when using os.walk() do the underscores make
it faster... or more efficient?

**rbt** · Jul 19 '05, 01:29 AM

Re: count files in a directory

James Stroud wrote:[color=blue]
> Sorry, I've never used os.walk and didn't realize that it is a generator.
>
> This will work for your purposes (and seems pretty fast compared to the
> alternative):
>
> file_count = len(os.walk(val id_path).next()[2])[/color]

Thanks James... this works *really* well for times when I only need to count files in
the current directory (no recursion). I think others will find it useful as well.

**Steven Bethard** · Jul 19 '05, 01:29 AM

Re: count files in a directory

rbt wrote:[color=blue]
> Heiko Wundram wrote:[color=green][color=darkred]
>>>>> import os
>>>>> path = "/home/heiko"
>>>>> file_count = sum((len(f) for _, _, f in os.walk(path)))
>>>>> file_count[/color][/color]
>
> Thanks! that works great... is there any significance to the underscores
> that you used? I've always used root, dirs, files when using os.walk()
> do the underscores make it faster... or more efficient?[/color]

No, they don't make any semantic difference; they work just like any
other identifier. It's just that, by convention, single underscores
indicate that we don't care what values these names are bound to. So in
the example above, Heiko indicates that we don't care about what you
would normally call 'root' and 'dirs'.

HTH,

STeVe

count files in a directory

count files in a directory

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