I would like to decompress the files in different directories which are in different routes. And codes as below and the error is invalid data stream. Please help me out. Tha
bz2.decompress
takes compressed data and inflates it. You pass a filename, not the data in the file!
Do this instead:
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
bz2.compress/decompress work with binary data:
>>> 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'
In short - you need to process file contents manually. In case you have very large files you should prefer using bz2.BZ2Decompressor
to bz2.decompress
, because the latter requires that you store the entire file in a byte array.
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))
You can also use bz2.BZ2File
to make this even simpler:
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)
This should work
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()))