On 6/22/2015 9:32 PM, Paul Rubin wrote:
Travis Griggs travisgri...@gmail.com writes:
The following seems to obtuse/clever for its own good:
return sum(1 for _ in self.path.iterdir())
I disagree. For one who understands counting and Python, this is a
direct way to define the count of a
Travis Griggs wrote:
Subject nearly says it all.
If i’m using pathlib, what’s the simplest/idiomatic way to simply count
how many files are in a given directory?
I was surprised (at first) when
len(self.path.iterdir())
I don’t say anything on the in the .stat() object that helps
I use len(list(self.path.iterdir()))
You get an extra list created in there. Do you care?
Laura
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I should proof my posts before I send them, sorry
Subject nearly says it all.
If i’m using pathlib, what’s the simplest/idiomatic way to simply count how
many files are in a given directory?
I was surprised (at first) when
len(self.path.iterdir())
didn’t work.
I don’t see anything in the
On Mon, Jun 22, 2015 at 4:33 PM, Travis Griggs travisgri...@gmail.com wrote:
I should proof my posts before I send them, sorry
Subject nearly says it all.
If i’m using pathlib, what’s the simplest/idiomatic way to simply count how
many files are in a given directory?
I was surprised (at
Subject nearly says it all.
If i’m using pathlib, what’s the simplest/idiomatic way to simply count how
many files are in a given directory?
I was surprised (at first) when
len(self.path.iterdir())
I don’t say anything on the in the .stat() object that helps me.
I could of course do the
Travis Griggs travisgri...@gmail.com writes:
The following seems to obtuse/clever for its own good:
return sum(1 for _ in self.path.iterdir())
I've generally done something like that. I suppose it could be added to
itertools.
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Am Samstag, 21. Mai 2005 06:25 schrieb James Stroud:
This will work for your purposes (and seems pretty fast compared to the
alternative):
file_count = len(os.walk(valid_path).next()[2])
But will only work when you're just scanning a single directory with no
subdirectories...!
The
Heiko Wundram wrote:
Am Samstag, 21. Mai 2005 06:25 schrieb James Stroud:
This will work for your purposes (and seems pretty fast compared to the
alternative):
file_count = len(os.walk(valid_path).next()[2])
But will only work when you're just scanning a single directory with no
only need to count
files in
the current directory (no recursion). I think others will find it useful as
well.
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rbt wrote:
Heiko Wundram wrote:
import os
path = /home/heiko
file_count = sum((len(f) for _, _, f in os.walk(path)))
file_count
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
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_path):
for f in files:
x = x+1
print There are, x, files in this directory
On Friday 20 May 2005 07:12 pm, rbt wrote:
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_path):
for f in files:
x = x+1
Come to think of it
file_count = len(os.walk(valid_path)[2])
--
James Stroud
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095
http://www.jamesstroud.com/
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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(valid_path).next()[2])
The alternative is:
import os
import os.path
file_count = len([f for f in
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