python: Open file from zip without temporary extracting it

问题

How can I open files from a zip archive without extracting them first?

I\'m using pygame. To save disk space, I have all the images zipped up. Is it possible to load a given image directly from the zip file? For example: pygame.image.load(\'zipFile/img_01\')

回答1:

Vincent Povirk's answer won't work completely;

import zipfile
archive = zipfile.ZipFile('images.zip', 'r')
imgfile = archive.open('img_01.png')
...

You have to change it in:

import zipfile
archive = zipfile.ZipFile('images.zip', 'r')
imgdata = archive.read('img_01.png')
...

For details read the ZipFile docs here.

回答2:

import io, pygame, zipfile
archive = zipfile.ZipFile('images.zip', 'r')

# read bytes from archive
img_data = archive.read('img_01.png')

# create a pygame-compatible file-like object from the bytes
bytes_io = io.BytesIO(img_data)

img = pygame.image.load(bytes_io)

I was trying to figure this out for myself just now and thought this might be useful for anyone who comes across this question in the future.

回答3:

In theory, yes, it's just a matter of plugging things in. Zipfile can give you a file-like object for a file in a zip archive, and image.load will accept a file-like object. So something like this should work:

import zipfile
archive = zipfile.ZipFile('images.zip', 'r')
imgfile = archive.open('img_01.png')
try:
    image = pygame.image.load(imgfile, 'img_01.png')
finally:
    imgfile.close()

来源：https://stackoverflow.com/questions/19371860/python-open-file-from-zip-without-temporary-extracting-it

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python

zipfile