Instead of using extract or extractall, just get the data and write it to any file you like. Here's a code sample that does what you want:
import os, sys, time
import zipfile
ENC = 'cp437'
outdir = unicode(os.path.abspath('.'))
outzip = 'c:/1temp/timbersales.zip'
zf = zipfile.ZipFile(outzip, 'r')
for info in zf.infolist():
fn, dtz = info.filename, info.date_time # , info.file_size
# some zips have dirs listed as files. Catch
# and bypass those.
name = os.path.basename(fn)
if not name:
continue
# get our filename converted from bytes to unicode
fn_uni = fn.decode(ENC, 'replace')
bn_uni = os.path.basename(fn_uni)
# this method is about 15% faster than extractall, and
# preserves modify and access dates
c = zf.open(fn)
outfile = os.path.join(outdir, bn_uni)
# try/except to avoid problems with locked files, etc
# do in chunks to avoid memory problems
chunk = 2**16
try:
with open(outfile, 'wb') as f:
s = c.read(chunk)
f.write(s)
while not len(s) < chunk:
s = c.read(chunk)
f.write(s)
c.close()
# set modify and access dates to that inside the zip
dtout = time.mktime(dtz + (0, 0, -1))
os.utime(outfile, (dtout, dtout))
except IOError:
c.close()
This example does all the files in the zip, but you can easily add a couple of lines to check for specific files. It also will overwrite any files in the work dir with the same name as the one being extracted.