Question

我想解压缩不同路径中不同目录中的文件。代码如下，错误是无效的数据流。请帮帮我。非常感谢你。

import sys
import os
import bz2
from bz2 import decompress

path = "Dir"
for(dirpath,dirnames,files)in os.walk(path):
   for file in files:
       filepath = os.path.join(dirpath,filename)
       newfile = bz2.decompress(file)
       newfilepath = os.path.join(dirpath,newfile)

Answer 1

bz2.compress / decompress使用二进制数据：

>>> import bz2
>>> compressed = bz2.compress(b'test_string')
>>> compressed
b'BZh91AY&SYJ|i\x05\x00\x00\x04\x83\x80\x00\x00\x82\xa1\x1c\x00 \x00"\x03h\x840"
P\xdf\x04\x99\xe2\xeeH\xa7\n\x12\tO\x8d \xa0'
>>> bz2.decompress(compressed)
b'test_string'

简而言之 - 您需要手动处理文件内容。如果你有非常大的文件，你应该更喜欢使用bz2.BZ2Decompressor到bz2.decompress，因为后者要求你将整个文件存储在一个字节数组中。

for filename in files:
    filepath = os.path.join(dirpath, filename)
    newfilepath = os.path.join(dirpath,filename + '.decompressed')
    with open(newfilepath, 'wb') as new_file, open(filepath, 'rb') as file:
        decompressor = BZ2Decompressor()
        for data in iter(lambda : file.read(100 * 1024), b''):
            new_file.write(decompressor.decompress(data))

您还可以使用bz2.BZ2File使这更简单：

for filename in files:
    filepath = os.path.join(dirpath, filename)
    newfilepath = os.path.join(dirpath, filename + '.decompressed')
    with open(newfilepath, 'wb') as new_file, bz2.BZ2File(filepath, 'rb') as file:
        for data in iter(lambda : file.read(100 * 1024), b''):
            new_file.write(data)

Answer 2

bz2.decompress获取压缩的数据并对其进行充气。您传递文件名，而不是文件中的数据！

请改为：

zipfile = bz2.BZ2File(filepath) # open the file
data = zipfile.read() # get the decompressed data
newfilepath = filepath[:-4] # assuming the filepath ends with .bz2
open(newfilepath, 'wb').write(data) # write a uncompressed file

Answer 3

这应该有效

for file in files:
    archive_path = os.path.join(dirpath,filename)
    outfile_path = os.path.join(dirpath, filename[:-4])
    with open(archive_path, 'rb') as source, open(outfile_path, 'wb') as dest:
        dest.write(bz2.decompress(source.read()))

解压缩bz2文件

3 个答案: