How do you unzip very large files in python?

Using python 2.4 and the built-in ZipFile library, I cannot read very large zip files (greater than 1 or 2 GB) because it wants to store the entire contents of the uncompressed file in memory. Is there another way to do this (either with a third-party library or some other hack), or must I "shell out" and unzip it that way (which isn't as cross-platform, obviously).

From stackoverflow

• Have a look at http://stackoverflow.com/questions/297345/create-a-zip-file-from-a-generator-in-python which discusses a similar probem.

  Marc Novakowski : Thanks but unfortunately they just discuss zipping a file, not unzipping. If you look at the source code in the zipfile.py library, it uses zlib to decompress a file into a string, which is what's using all the memory.

• Here's an outline of decompression of large files.

  import zipfile
  import zlib
  import os
  
  src = open( doc, "rb" )
  zf = zipfile.ZipFile( src )
  for m in  zf.infolist():
  
      # Examine the header
      print m.filename, m.header_offset, m.compress_size, repr(m.extra), repr(m.comment)
      src.seek( m.header_offset )
      src.read( 30 ) # Good to use struct to unpack this.
      nm= src.read( len(m.filename) )
      if len(m.extra) > 0: ex= src.read( len(m.extra) )
      if len(m.comment) > 0: cm= src.read( len(m.comment) ) 
  
      # Build a decompression object
      decomp= zlib.decompressobj(-15)
  
      # This can be done with a loop reading blocks
      out= open( m.filename, "wb" )
      result= decomp.decompress( src.read( m.compress_size ) )
      out.write( result )
      result = decomp.flush()
      out.write( result )
      # end of the loop
      out.close()
  
  zf.close()
  src.close()
  

  Marc Novakowski : This is exactly what I was looking for - thanks!