>>> I'm dealing with several large items that have been zipped up to
>>> get quite impressive compression. However, uncompressed, they're
>>> large enough to thrash my memory to swap and in general do bad
>>> performance-related things. I'm trying to figure out how to
>>> produce a file-like iterator out of the contents of such an item.
>>
>> The Time Machine in action again - that's already done, but in SVN. You
>> want the new ZipFile.open(filename) method, which returns a file-like
>> object.
>
> Thanks! I'll give the 2.6 version a try.
Just to follow up on this, I dropped the the 2.6 version of
zipfile.py in my project folder (where the machine is currently
running Python2.4), used the ZipFile.open() and it worked fine.
I was able to successfully extract a 960 meg MDB file from the
zip-file. The one thing that did throw me off is that it
rejected specifying that the file be opened as binary:
z = ZipFile('foo.zip')
f = z.open('path/to/file.mdb', 'rb') #failed
f = z.open('path/to/file.mdb') # worked
but just opening with no type specification did allow me to
extract the file successfully. I don't know if I just struck it
lucky with newline/EOF translations, or if it really does do
binary file handling and you don't get a choice of non-binary
file handling.
Anyways, thanks to Gabriel (and all the authors of Python
zipfile.py library) for the solution.
-tkc
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