How do you unzip very large files in python?

Here's an outline of decompression of large files.

import zipfileimport zlibimport ossrc = 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()

As of Python 2.6, you can use ZipFile.open() to open a file handle on a file, and copy contents efficiently to a target file of your choosing:

import errnoimport osimport shutilimport zipfileTARGETDIR = '/foo/bar/baz'with open(doc, "rb") as zipsrc:    zfile = zipfile.ZipFile(zipsrc)    for member in zfile.infolist():       target_path = os.path.join(TARGETDIR, member.filename)       if target_path.endswith('/'):  # folder entry, create           try:               os.makedirs(target_path)           except (OSError, IOError) as err:               # Windows may complain if the folders already exist               if err.errno != errno.EEXIST:                   raise           continue       with open(target_path, 'wb') as outfile, zfile.open(member) as infile:           shutil.copyfileobj(infile, outfile)

This uses shutil.copyfileobj() to efficiently read data from the open zipfile object, copying it over to the output file.