On Tue, Mar 17, 2009 at 6:34 AM, A.T.Hofkamp wrote:
>> http://personalpages.tds.net/~kent37/kk/00012.html
>
> Nice web-page!
Thanks!
> You can do the above statements also iteratively of course
>
> for i in ...
> s = read()
> # write s
>
> but since the loop does nothing with either s or read
Kent Johnson wrote:
On Mon, Mar 16, 2009 at 12:30 PM, A.T.Hofkamp wrote:
I don't know what code is executed in an assignment exactly, but
**possibly**, first the 'read()' is executed (thus loading a very big string
into memory), before assigning the value to the variable (which releases the
pr
On Mon, Mar 16, 2009 at 12:30 PM, A.T.Hofkamp wrote:
> I don't know what code is executed in an assignment exactly, but
> **possibly**, first the 'read()' is executed (thus loading a very big string
> into memory), before assigning the value to the variable (which releases the
> previous value of
Cheetah1000 wrote:
I can't speak for Python 2.6, but using Jython 2.1 (Python 2.1 for Java),
the code only looks at the file you are trying to extract()/read(). Near
the end of the zip archive is a directory of all the files in the archive,
with the start position and length of each file. Jytho
Alan Gauld wrote:
>
>
> "Sander Sweers" wrote
>
>> ... I would expect zf.read(zfilename) to only read the
>> requested file in the zipfile.
>
> That's a dangerous assumption. You might be right but I'd want to
> do some tests first to see. But if even one zipped file was big the
> same wou
Sander Sweers wrote:
2009/3/10 Alan Gauld :
newFile.write(zf.read(zfilename))
Remember you are reading the file into memory and then writing it
out again in a single operation, that will use twice the space of the
uncompressed files - plus some extra for overhead.
Question, Do you m
"Sander Sweers" wrote
out again in a single operation, that will use twice the space of
the
uncompressed files - plus some extra for overhead.
Question, Do you mean the file in the zipfile (zfilename) or the
whole
zipfile (zf)? I would expect zf.read(zfilename) to only read the
requested
2009/3/10 Alan Gauld :
>> newFile.write(zf.read(zfilename))
>
> Remember you are reading the file into memory and then writing it
> out again in a single operation, that will use twice the space of the
> uncompressed files - plus some extra for overhead.
Question, Do you mean the file in
"Harris, Sarah L" wrote
However I still have a memory error when I try to run it on three
or more files that are over 100 MB?
And this surprises you? :-)
How much memory do you have free on your computer when you run this?
newFile.write(zf.read(zfilename))
Remember you are re
On Tue, Mar 10, 2009 at 8:45 AM, Harris, Sarah L
wrote:
> That looks better, thank you.
> However I still have a memory error when I try to run it on three or more
> files that are over 100 MB?
How big are files in the zip file?
It seems that in this line
newFile.write(zf.read(zfilename))
the
That looks better, thank you.
However I still have a memory error when I try to run it on three or more files
that are over 100 MB?
import zipfile, glob, os
from os.path import isfile
zipnames=filter(isfile, glob.glob('*.zip'))
for zipname in zipnames:
zf=zipfile.ZipFile(zipname, 'r')
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