python 解压zip 慢_Python和zlib：解压缩级联流的速度非常慢

I've been supplied with a zipped file containing multiple individual streams of compressed XML. The compressed file is 833 mb.

If I try to decompress it as a single object, I only get the first stream (about 19 kb).

I've modified the following code supplied as a answer to an older question to decompress each stream and write it to a file:

import zlib

outfile = open('output.xml', 'w')

def zipstreams(filename):

"""Return all zip streams and their positions in file."""

with open(filename, 'rb') as fh:

data = fh.read()

i = 0

print "got it"

while i < len(data):

try:

zo = zlib.decompressobj()

dat =zo.decompress(data[i:])

outfile.write(dat)

zo.flush()

i += len(data[i:]) - len(zo.unused_data)

except zlib.error:

i += 1

outfile.close()

zipstreams('payload')

infile.close()

This code runs and produces the desired result (all the XML data decompressed to a single file). The problem is that it takes several days to work!

Even though there are tens of thousands of streams in the compressed file, it still seems like this should be a much faster process. Roughly 8 days to decompress 833mb (estimated 3gb raw) suggests that I'm doing something very wrong.

Is there another way to do this more efficiently, or is the slow speed the result of a read-decompress-write---repeat bottleneck that I'm stuck with?

Thanks for any pointers or suggestions you have!

解决方案

It's hard to say very much without more specific knowledge of the file format you're actually dealing with, but it's clear that your algorithm's handling of substrings is quadratic-- not a good thing when you've got tens of thousands of them. So let's see what we know:

You say that the vendor states that they are

using the standard zlib compression library.These are the same compression routines on which the gzip utilities are built.

From this we can conclude that the component streams are in raw zlib format, and are not encapsulated in a gzip wrapper (or a PKZIP archive, or whatever). The authoritative documentation on the ZLIB format is here: http://tools.ietf.org/html/rfc1950

So let's assume that your file is exactly as you describe: A 32-byte header, followed by raw ZLIB streams concatenated together, without any other stuff in between. (Edit: That's not the case, after all).

Python's zlib documentation provides a Decompress class that is actually pretty well suited to churning through your file. It includes an attribute unused_data whose documentation states clearly that:

The only way to determine where a string of compressed data ends is by actually decompressing it. This means that when compressed data is contained part of a larger file, you can only find the end of it by reading data and feeding it followed by some non-empty string into a decompression object’s decompress() method until the unused_data attribute is no longer the empty string.

So, this is what you can do: Write a loop that reads through data, say, one block at a time (no need to even read the entire 800MB file into memory). Push each block to the Decompress object, and check the unused_data attribute. When it becomes non-empty, you've got a complete object. Write it to disk, create a new decompress object and initialize iw with the unused_data from the last one. This just might work (untested, so check for correctness).

Edit: Since you do have other data in your data stream, I've added a routine that aligns to the next ZLIB start. You'll need to find and fill in the two-byte sequence that identifies a ZLIB stream in your data. (Feel free to use your old code to discover it.) While there's no fixed ZLIB header in general, it should be the same for each stream since it consists of protocol options and flags, which are presumably the same for the entire run.

import zlib

# FILL IN: ZHEAD is two bytes with the actual ZLIB settings in the input

ZHEAD = CMF+FLG

def findstart(header, buf, source):

"""Find `header` in str `buf`, reading more from `source` if necessary"""

while buf.find(header) == -1:

more = source.read(2**12)

if len(more) == 0: # EOF without finding the header

return ''

buf += more

offset = buf.find(header)

return buf[offset:]

You can then advance to the start of the next stream. I've added a try/except pair since the same byte sequence might occur outside a stream:

source = open(datafile, 'rb')

skip_ = source.read(32) # Skip non-zlib header

buf = ''

while True:

decomp = zlib.decompressobj()

# Find the start of the next stream

buf = findstart(ZHEAD, buf, source)

try:

stream = decomp.decompress(buf)

except zlib.error:

print "Spurious match(?) at output offset %d." % outfile.tell(),

print "Skipping 2 bytes"

buf = buf[2:]

continue

# Read until zlib decides it's seen a complete file

while decomp.unused_data == '':

block = source.read(2**12)

if len(block) > 0:

stream += decomp.decompress(block)

else:

break # We've reached EOF

outfile.write(stream)

buf = decomp.unused_data # Save for the next stream

if len(block) == 0:

break # EOF

outfile.close()

PS 1. If I were you I'd write each XML stream into a separate file.

PS 2. You can test whatever you do on the first MB of your file, till you get adequate performance.