Develop Reference

how to upload chunks of a string longer than 2147483647 bytes?

I am trying to upload a file around ~5GB size as below but, it throws the error string longer than 2147483647 bytes. It sounds like there is a limit of 2 GB to upload. Is there a way to upload data in chunks? Can anyone provide guidance?
logger.debug(attachment_path)
currdir = os.path.abspath(os.getcwd())
os.chdir(os.path.dirname(attachment_path))
headers = self._headers
headers['Content-Type'] = content_type
headers['X-Override-File'] = 'true'
if not os.path.exists(attachment_path):
raise Exception, "File path was invalid, no file found at the path %s" % attachment_path
filesize = os.path.getsize(attachment_path)
fileToUpload = open(attachment_path, 'rb').read()
logger.info(filesize)
logger.debug(headers)
r = requests.put(self._baseurl + 'problems/' + problemID + "/" + attachment_type + "/" + urllib.quote(os.path.basename(attachment_path)),
headers=headers,data=fileToUpload,timeout=300)
ERROR:
string longer than 2147483647 bytes
UPDATE:
def read_in_chunks(file_object,chunk_size=30720*30720):
"""Lazy function (generator) to read a file piece by piece.
Default chunk size: 1k."""
while True:
data = file_object.read(chunk_size)
if not data:
break
yield data
f = open(attachment_path)
for piece in read_in_chunks(f):
r = requests.put(self._baseurl + 'problems/' + problemID + "/" + attachment_type + "/" + urllib.quote(os.path.basename(attachment_path)),
headers=headers,data=piece,timeout=300)

Your question has been asked on the requests bug tracker; their suggestion is to use streaming upload. If that doesn't work, you might see if a chunk-encoded request works.
[edit]
Example based on the original code:
# Using `with` here will handle closing the file implicitly
with open(attachment_path, 'rb') as file_to_upload:
r = requests.put(
"{base}problems/{pid}/{atype}/{path}".format(
base=self._baseurl,
# It's better to use consistent naming; search PEP-8 for standard Python conventions.
pid=problem_id,
atype=attachment_type,
path=urllib.quote(os.path.basename(attachment_path)),
),
headers=headers,
# Note that you're passing the file object, NOT the contents of the file:
data=file_to_upload,
# Hard to say whether this is a good idea with a large file upload
timeout=300,
)
I can't guarantee this would run as-is, since I can't realistically test it, but it should be close. The bug tracker comments I linked to also mention that sending multiple headers may cause issues, so if the headers you're specifying are actually necessary, this may not work.
Regarding chunk encoding: This should be your second choice. Your code was not specifying 'rb' as the mode for open(...), so changing that should probably make the code above work. If not, you could try this.
def read_in_chunks():
# If you're going to chunk anyway, doesn't it seem like smaller ones than this would be a good idea?
chunk_size = 30720 * 30720
# I don't know how correct this is; if it doesn't work as expected, you'll need to debug
with open(attachment_path, 'rb') as file_object:
while True:
data = file_object.read(chunk_size)
if not data:
break
yield data
# Same request as above, just using the function to chunk explicitly; see the `data` param
r = requests.put(
"{base}problems/{pid}/{atype}/{path}".format(
base=self._baseurl,
pid=problem_id,
atype=attachment_type,
path=urllib.quote(os.path.basename(attachment_path)),
),
headers=headers,
# Call the chunk function here and the request will be chunked as you specify
data=read_in_chunks(),
timeout=300,
)

How to print the content of zipped gzip'd files

Ok, so I have a zip file that contains gz files (unix gzip).
Here's what I do --
def parseSTS(file):
import zipfile, re, io, gzip
with zipfile.ZipFile(file, 'r') as zfile:
for name in zfile.namelist():
if re.search(r'\.gz$', name) != None:
zfiledata = zfile.open(name)
print("start for file ", name)
with gzip.open(zfiledata,'r') as gzfile:
print("done opening")
filecontent = gzfile.read()
print("done reading")
print(filecontent)
This gives the following result --
>>>
start for file XXXXXX.gz
done opening
done reading
Then stays like that forever until it crashes ...
What can I do with filecontent?
Edit : this is not a duplicate since my gzipped files are in a zipped file and i'm trying to avoid extracting that zip file to disk. It works with zip files in a zip file as per How to read from a zip file within zip file in Python? .

I created a zip file containing a gzip'ed PDF file I grabbed from the web.
I ran this code (with two small changes):
1) Fixed indenting of everything under the def statement (which I also corrected in your Question because I'm sure that it's right on your end or it wouldn't get to the problem you have).
2) I changed:
zfiledata = zfile.open(name)
print("start for file ", name)
with gzip.open(zfiledata,'r') as gzfile:
print("done opening")
filecontent = gzfile.read()
print("done reading")
print(filecontent)
to:
print("start for file ", name)
with gzip.open(name,'rb') as gzfile:
print("done opening")
filecontent = gzfile.read()
print("done reading")
print(filecontent)
Because you were passing a file object to gzip.open instead of a string. I have no idea how your code is executing without that change, but it was crashing for me until I fixed it.
EDIT: Adding link to GZIP docs from James R's answer --
Also, see here for further documentation:
http://docs.python.org/2/library/gzip.html#examples-of-usage
END EDIT
Now, since my gzip'ed file is small, the behavior I observe is that is pauses for about 3 seconds after printing done reading, then outputs what is in filecontent.
I would suggest adding the following debugging line after your print "done reading" -- print len(filecontent). If this number is very, very large, consider not printing the entire file contents in one shot.
I would also suggest reading this for more insight into what I expect is your problem: Why is printing to stdout so slow? Can it be sped up?
EDIT 2 - an alternative if your system does not handle file io on zip files, causing no such file errors in the above:
def parseSTS(afile):
import zipfile
import zlib
import gzip
import io
with zipfile.ZipFile(afile, 'r') as archive:
for name in archive.namelist():
if name.endswith('.gz'):
bfn = archive.read(name)
bfi = io.BytesIO(bfn)
g = gzip.GzipFile(fileobj=bfi,mode='rb')
qqq = g.read()
print qqq
parseSTS('t.zip')

Most likely your problem lies here:
if name.endswith(".gz"): #as goncalopp said in the comments, use endswith
#zfiledata = zfile.open(name) #don't do this
#print("start for file ", name)
with gzip.open(name,'rb') as gzfile: #gz compressed files should be read in binary and gzip opens the files directly
#print("done opening") #trust in your program, luke
filecontent = gzfile.read()
#print("done reading")
print(filecontent)
See here for further documentation:
http://docs.python.org/2/library/gzip.html#examples-of-usage

How to replace duplicate files with hard links using python?

I'm a photographer and doing many backups. Over the years I found myself with a lot of hard drives. Now I bought a NAS and copied all my pictures on one 3TB raid 1 using rsync. According to my script about 1TB of those files are duplicates. That comes from doing multiple backups before deleting files on my laptop and being very messy. I do have a backup of all those files on the old hard drives, but it would be a pain if my script messes things up. Can you please have a look at my duplicate finder script and tell me if you think I can run it or not? I tried it on a test folder and it seems ok, but I don't want to mess things up on the NAS.
The script has three steps in three files. In this First part I find all image and metadata files and put them into a shelve database (datenbank) with their size as key.
import os
import shelve
datenbank = shelve.open(os.path.join(os.path.dirname(__file__),"shelve_step1"), flag='c', protocol=None, writeback=False)
#path_to_search = os.path.join(os.path.dirname(__file__),"test")
path_to_search = "/volume1/backup_2tb_wd/"
file_exts = ["xmp", "jpg", "JPG", "XMP", "cr2", "CR2", "PNG", "png", "tiff", "TIFF"]
walker = os.walk(path_to_search)
counter = 0
for dirpath, dirnames, filenames in walker:
if filenames:
for filename in filenames:
counter += 1
print str(counter)
for file_ext in file_exts:
if file_ext in filename:
filepath = os.path.join(dirpath, filename)
filesize = str(os.path.getsize(filepath))
if not filesize in datenbank:
datenbank[filesize] = []
tmp = datenbank[filesize]
if filepath not in tmp:
tmp.append(filepath)
datenbank[filesize] = tmp
datenbank.sync()
print "done"
datenbank.close()
The second part. Now I drop all file sizes which only have one file in their list and create another shelve database with the md5 hash as key and a list of files as value.
import os
import shelve
import hashlib
datenbank = shelve.open(os.path.join(os.path.dirname(__file__),"shelve_step1"), flag='c', protocol=None, writeback=False)
datenbank_step2 = shelve.open(os.path.join(os.path.dirname(__file__),"shelve_step2"), flag='c', protocol=None, writeback=False)
counter = 0
space = 0
def md5Checksum(filePath):
with open(filePath, 'rb') as fh:
m = hashlib.md5()
while True:
data = fh.read(8192)
if not data:
break
m.update(data)
return m.hexdigest()
for filesize in datenbank:
filepaths = datenbank[filesize]
filepath_count = len(filepaths)
if filepath_count > 1:
counter += filepath_count -1
space += (filepath_count -1) * int(filesize)
for filepath in filepaths:
print counter
checksum = md5Checksum(filepath)
if checksum not in datenbank_step2:
datenbank_step2[checksum] = []
temp = datenbank_step2[checksum]
if filepath not in temp:
temp.append(filepath)
datenbank_step2[checksum] = temp
print counter
print str(space)
datenbank_step2.sync()
datenbank_step2.close()
print "done"
And finally the most dangerous part. For evrey md5 key i retrieve the file list and do an additional sha1. If it matches I delete every file in that list execept the first one and create a hard link to replace the deleted files.
import os
import shelve
import hashlib
datenbank = shelve.open(os.path.join(os.path.dirname(__file__),"shelve_step2"), flag='c', protocol=None, writeback=False)
def sha1Checksum(filePath):
with open(filePath, 'rb') as fh:
m = hashlib.sha1()
while True:
data = fh.read(8192)
if not data:
break
m.update(data)
return m.hexdigest()
for hashvalue in datenbank:
switch = True
for path in datenbank[hashvalue]:
if switch:
original = path
original_checksum = sha1Checksum(path)
switch = False
else:
if sha1Checksum(path) == original_checksum:
os.unlink(path)
os.link(original, path)
print "delete: ", path
print "done"
What do you think?
Thank you very much.
*if that's somehow important: It's a synology 713+ and has an ext3 or ext4 filesystem.

This looked good, and after sanitizing a bit (to make it work with python 3.4), I ran this on my NAS. While I had hardlinks for files that had not been modified between backups, files that had moved were being duplicated. This recovered that lost disk space for me.
A minor nitpick is that files that are already hardlinks are deleted and relinked. This does not affect the end result anyway.
I did slightly alter the third file ("3.py"):
if sha1Checksum(path) == original_checksum:
tmp_filename = path + ".deleteme"
os.rename(path, tmp_filename)
os.link(original, path)
os.unlink(tmp_filename)
print("Deleted {} ".format(path))
This makes sure that in case of a power-failure or some other similar error, no files are lost, though a trailing "deleteme" is left behind. A recovery script should be quite trivial.

Why not compare the files byte for byte instead of the second checksum? One in a billion two checksums might accidentally match, but direct comparison shouldn't fail. It shouldn't be slower, and might even be faster. Maybe it could be slower when there are more than two files and you have to read the original file for each other. If you really wanted you could get around that by comparing blocks of all the files at once.
EDIT:
I don't think it would require more code, just different. Something like this for the loop body:
data1 = fh1.read(8192)
data2 = fh2.read(8192)
if data1 != data2: return False

Note: If you're not wedded to Python, there are exsting tools to do the heavy lifting for you:
https://unix.stackexchange.com/questions/3037/is-there-an-easy-way-to-replace-duplicate-files-with-hardlinks

How do you create a hard link.
In linux you do
sudo ln sourcefile linkfile
Sometimes this can fail (for me it fails sometimes). Also your python script needs to run in sudo mode.
So I use symbolic links:
ln -s sourcefile linkfile
I can check for them with os.path.islink
You can call the commands like this in Python:
os.system("ln -s sourcefile linkfile")
or like this using subprocess:
import subprocess
subprocess.call(["ln", "-s", sourcefile, linkfile], shell = True)
Have a look at execution from command line and hard vs. soft links
When it works, could you post your whole code? I would like to use it, too.

Write files to disk with python 3.x

Using BottlePy, I use the following code to upload a file and write it to disk :
upload = request.files.get('upload')
raw = upload.file.read()
filename = upload.filename
with open(filename, 'w') as f:
f.write(raw)
return "You uploaded %s (%d bytes)." % (filename, len(raw))
It returns the proper amount of bytes every single time.
The upload works fine for file like .txt, .php, .css ...
But it results in a corrupted file for other files like .jpg, .png, .pdf, .xls ...
I tried to change the open() function
with open(filename, 'wb') as f:
It returns the following error:
TypeError('must be bytes or buffer, not str',)
I guess its an issue related to binary files ?
Is there something to install on top of Python to run upload for any file type ?
Update
Just to be sure, as pointed out by #thkang I tried to code this using the dev version of bottlepy and the built-in method .save()
upload = request.files.get('upload')
upload.save(upload.filename)
It returns the exact same Exception error
TypeError('must be bytes or buffer, not str',)
Update 2
Here the final code which "works" (and dont pop the error TypeError('must be bytes or buffer, not str',)
upload = request.files.get('upload')
raw = upload.file.read().encode()
filename = upload.filename
with open(filename, 'wb') as f:
f.write(raw)
Unfortunately, the result is the same : every .txt file works fine, but other files like .jpg, .pdf ... are corrupted
I've also noticed that those file (the corrupted one) have a larger size than the orginal (before upload)
This binary thing must be the issue with Python 3x
Note :
I use python 3.1.3
I use BottlePy 0.11.6 (raw bottle.py file, no 2to3 on it or anything)

Try this:
upload = request.files.get('upload')
with open(upload.file, "rb") as f1:
raw = f1.read()
filename = upload.filename
with open(filename, 'wb') as f:
f.write(raw)
return "You uploaded %s (%d bytes)." % (filename, len(raw))
Update
Try value:
# Get a cgi.FieldStorage object
upload = request.files.get('upload')
# Get the data
raw = upload.value;
# Write to file
filename = upload.filename
with open(filename, 'wb') as f:
f.write(raw)
return "You uploaded %s (%d bytes)." % (filename, len(raw))
Update 2
See this thread, it seems to do same as what you are trying...
# Test if the file was uploaded
if fileitem.filename:
# strip leading path from file name to avoid directory traversal attacks
fn = os.path.basename(fileitem.filename)
open('files/' + fn, 'wb').write(fileitem.file.read())
message = 'The file "' + fn + '" was uploaded successfully'
else:
message = 'No file was uploaded'

In Python 3x all strings are now unicode, so you need to convert the read() function used in this file upload code.
The read() function returns a unicode string aswell, which you can convert into proper bytes via encode() function
Use the code contained in my first question, and replace the line
raw = upload.file.read()
with
raw = upload.file.read().encode('ISO-8859-1')
That's all ;)
Further reading : http://python3porting.com/problems.html

unzipping file results in "BadZipFile: File is not a zip file"

I have two zip files, both of them open well with Windows Explorer and 7-zip.
However when i open them with Python's zipfile module [ zipfile.ZipFile("filex.zip") ], one of them gets opened but the other one gives error "BadZipfile: File is not a zip file".
I've made sure that the latter one is a valid Zip File by opening it with 7-Zip and looking at its properties (says 7Zip.ZIP). When I open the file with a text editor, the first two characters are "PK", showing that it is indeed a zip file.
I'm using Python 2.5 and really don't have any clue how to go about for this. I've tried it both with Windows as well as Ubuntu and problem exists on both platforms.
Update: Traceback from Python 2.5.4 on Windows:
Traceback (most recent call last):
File "<module1>", line 5, in <module>
zipfile.ZipFile("c:/temp/test.zip")
File "C:\Python25\lib\zipfile.py", line 346, in init
self._GetContents()
File "C:\Python25\lib\zipfile.py", line 366, in _GetContents
self._RealGetContents()
File "C:\Python25\lib\zipfile.py", line 378, in _RealGetContents
raise BadZipfile, "File is not a zip file"
BadZipfile: File is not a zip file
Basically when the _EndRecData function is called for getting data from End of Central Directory" record, the comment length checkout fails [ endrec[7] == len(comment) ].
The values of locals in the _EndRecData function are as following:
END_BLOCK: 4096,
comment: '\x00',
data: '\xd6\xf6\x03\x00\x88,N8?<e\xf0q\xa8\x1cwK\x87\x0c(\x82a\xee\xc61N\'1qN\x0b\x16K-\x9d\xd57w\x0f\xa31n\xf3dN\x9e\xb1s\xffu\xd1\.....', (truncated)
endrec: ['PK\x05\x06', 0, 0, 4, 4, 268, 199515, 0],
filesize: 199806L,
fpin: <open file 'c:/temp/test.zip', mode 'rb' at 0x045D4F98>,
start: 4073

files named file can confuse python - try naming it something else. if it STILL wont work, try this code:
def fixBadZipfile(zipFile):
f = open(zipFile, 'r+b')
data = f.read()
pos = data.find('\x50\x4b\x05\x06') # End of central directory signature
if (pos > 0):
self._log("Trancating file at location " + str(pos + 22)+ ".")
f.seek(pos + 22) # size of 'ZIP end of central directory record'
f.truncate()
f.close()
else:
# raise error, file is truncated

I run into the same issue. My problem was that it was a gzip instead of a zip file. I switched to the class gzip.GzipFile and it worked like a charm.

astronautlevel's solution works for most cases, but the compressed data and CRCs in the Zip can also contain the same 4 bytes. You should do an rfind (not find), seek to pos+20 and then add write \x00\x00 to the end of the file (tell zip applications that the length of the 'comments' section is 0 bytes long).
# HACK: See http://bugs.python.org/issue10694
# The zip file generated is correct, but because of extra data after the 'central directory' section,
# Some version of python (and some zip applications) can't read the file. By removing the extra data,
# we ensure that all applications can read the zip without issue.
# The ZIP format: http://www.pkware.com/documents/APPNOTE/APPNOTE-6.3.0.TXT
# Finding the end of the central directory:
# http://stackoverflow.com/questions/8593904/how-to-find-the-position-of-central-directory-in-a-zip-file
# http://stackoverflow.com/questions/20276105/why-cant-python-execute-a-zip-archive-passed-via-stdin
# This second link is only losely related, but echos the first, "processing a ZIP archive often requires backwards seeking"
content = zipFileContainer.read()
pos = content.rfind('\x50\x4b\x05\x06') # reverse find: this string of bytes is the end of the zip's central directory.
if pos>0:
zipFileContainer.seek(pos+20) # +20: see secion V.I in 'ZIP format' link above.
zipFileContainer.truncate()
zipFileContainer.write('\x00\x00') # Zip file comment length: 0 byte length; tell zip applications to stop reading.
zipFileContainer.seek(0)
return zipFileContainer

I had the same problem and was able to solve this issue for my files, see my answer at
zipfile cant handle some type of zip data?

I'm very new at python and i was facing the exact same issue, none of the previous methods were working.
Trying to print the 'corrupted' file just before unzipping it returned an empty byte object.
Turned out, I was trying to unzip the file right after writing it to disk, without closing the file handler.
with open(path, 'wb') as outFile:
outFile.write(data)
outFile.close() # was missing this
with zipfile.ZipFile(path, 'r') as zip:
zip.extractall(destination)
Closing the file stream then unzipping the file resolved my issue.

Sometime there are zip file which contain corrupted files and upon unzipping the zip gives badzipfile error. but there are tools like 7zip winrar which ignores these errors and successfully unzip the zip file. you can create a sub process and use this code to unzip your zip file without getting BadZipFile Error.
import subprocess
ziploc = "C:/Program Files/7-Zip/7z.exe" #location where 7zip is installed
cmd = [ziploc, 'e',your_Zip_file.zip ,'-o'+ OutputDirectory ,'-r' ]
sp = subprocess.Popen(cmd, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)

Show the full traceback that you got from Python -- this may give a hint as to what the specific problem is. Unanswered: What software produced the bad file, and on what platform?
Update: Traceback indicates having problem detecting the "End of Central Directory" record in the file -- see function _EndRecData starting at line 128 of C:\Python25\Lib\zipfile.py
Suggestions:
(1) Trace through the above function
(2) Try it on the latest Python
(3) Answer the question above.
(4) Read this and anything else found by google("BadZipfile: File is not a zip file") that appears to be relevant

I faced this problem and was looking for a good and clean solution; But there was no solution until I found this answer. I had the same problem that #marsl (among the answers) had. It was a gzipfile instead of a zipfile in my case.
I could unarchive and decompress my gzipfile with this approach:
with tarfile.open(archive_path, "r:gz") as gzip_file:
gzip_file.extractall()

Have you tried a newer python, or if that is too much trouble, simply a newer zipfile.py? I have successfully used a copy of zipfile.py from Python 2.6.2 (latest at the time) with Python 2.5 in order to open some zip files that weren't supported by Py2.5s zipfile module.

In some cases, you have to confirm if the zip file is actually in gzip format. this was the case for me and i solved it by :
import requests
import tarfile
url = ".tar.gz link"
response = requests.get(url, stream=True)
file = tarfile.open(fileobj=response.raw, mode="r|gz")
file.extractall(path=".")

for this this happened when the file wasn't downloaded fully I think. So I just delete it in my download code.
def download_and_extract(url: str,
path_used_for_zip: Path = Path('~/data/'),
path_used_for_dataset: Path = Path('~/data/tmp/'),
rm_zip_file_after_extraction: bool = True,
force_rewrite_data_from_url_to_file: bool = False,
clean_old_zip_file: bool = False,
gdrive_file_id: Optional[str] = None,
gdrive_filename: Optional[str] = None,
):
"""
Downloads data and tries to extract it according to different protocols/file types.
note:
- to force a download do:
force_rewrite_data_from_url_to_file = True
clean_old_zip_file = True
- to NOT remove file after extraction:
rm_zip_file_after_extraction = False
Tested with:
- zip files, yes!
Later:
- todo: tar, gz, gdrive
force_rewrite_data_from_url_to_file = remvoes the data from url (likely a zip file) and redownloads the zip file.
"""
path_used_for_zip: Path = expanduser(path_used_for_zip)
path_used_for_zip.mkdir(parents=True, exist_ok=True)
path_used_for_dataset: Path = expanduser(path_used_for_dataset)
path_used_for_dataset.mkdir(parents=True, exist_ok=True)
# - download data from url
if gdrive_filename is None: # get data from url, not using gdrive
import ssl
ctx = ssl.create_default_context()
ctx.check_hostname = False
ctx.verify_mode = ssl.CERT_NONE
print("downloading data from url: ", url)
import urllib
import http
response: http.client.HTTPResponse = urllib.request.urlopen(url, context=ctx)
print(f'{type(response)=}')
data = response
# save zipfile like data to path given
filename = url.rpartition('/')[2]
path2file: Path = path_used_for_zip / filename
else: # gdrive case
from torchvision.datasets.utils import download_file_from_google_drive
# if zip not there re-download it or force get the data
path2file: Path = path_used_for_zip / gdrive_filename
if not path2file.exists():
download_file_from_google_drive(gdrive_file_id, path_used_for_zip, gdrive_filename)
filename = gdrive_filename
# -- write downloaded data from the url to a file
print(f'{path2file=}')
print(f'{filename=}')
if clean_old_zip_file:
path2file.unlink(missing_ok=True)
if filename.endswith('.zip') or filename.endswith('.pkl'):
# if path to file does not exist or force to write down the data
if not path2file.exists() or force_rewrite_data_from_url_to_file:
# delete file if there is one if your going to force a rewrite
path2file.unlink(missing_ok=True) if force_rewrite_data_from_url_to_file else None
print(f'about to write downloaded data from url to: {path2file=}')
# wb+ is used sinze the zip file was in bytes, otherwise w+ is fine if the data is a string
with open(path2file, 'wb+') as f:
# with open(path2file, 'w+') as f:
print(f'{f=}')
print(f'{f.name=}')
f.write(data.read())
print(f'done writing downloaded from url to: {path2file=}')
elif filename.endswith('.gz'):
pass # the download of the data doesn't seem to be explicitly handled by me, that is done in the extract step by a magic function tarfile.open
# elif is_tar_file(filename):
# os.system(f'tar -xvzf {path_2_zip_with_filename} -C {path_2_dataset}/')
else:
raise ValueError(f'File type {filename=} not supported.')
# - unzip data written in the file
extract_to = path_used_for_dataset
print(f'about to extract: {path2file=}')
print(f'extract to target: {extract_to=}')
if filename.endswith('.zip'):
import zipfile # this one is for zip files, inspired from l2l
zip_ref = zipfile.ZipFile(path2file, 'r')
zip_ref.extractall(extract_to)
zip_ref.close()
if rm_zip_file_after_extraction:
path2file.unlink(missing_ok=True)
elif filename.endswith('.gz'):
import tarfile
file = tarfile.open(fileobj=response, mode="r|gz")
file.extractall(path=extract_to)
file.close()
elif filename.endswith('.pkl'):
# no need to extract it, but when you use the data make sure you torch.load it or pickle.load it.
print(f'about to test torch.load of: {path2file=}')
data = torch.load(path2file) # just to test
assert data is not None
print(f'{data=}')
pass
else:
raise ValueError(f'File type {filename=} not supported, edit code to support it.')
# path_2_zip_with_filename = path_2_ziplike / filename
# os.system(f'tar -xvzf {path_2_zip_with_filename} -C {path_2_dataset}/')
# if rm_zip_file:
# path_2_zip_with_filename.unlink(missing_ok=True)
# # raise ValueError(f'File type {filename=} not supported.')
print(f'done extracting: {path2file=}')
print(f'extracted at location: {path_used_for_dataset=}')
print(f'-->Succes downloading & extracting dataset at location: {path_used_for_dataset=}')
you can use my code with pip install ultimate-utils for the most up to date version.

In the other case, this warning showing up when the ml/dl model has different format.
For the example:
you want to open pickle, but the model format is .sav
Solution:
you need to change the format to original format
pickle --> .pkl
tensorflow --> .h5
etc.

In my case, the zip file itself was missing from that directory - thus when I tried to unzip it, I got the error "BadZipFile: File is not a zip file". It got resolved after I moved the .zip file to the directory. Please confirm that the file is indeed present in your directory before running the python script.

In my case, the zip file was corrupted. I was trying to download the zip file with urllib.request.urlretrieve but the file wouldn't completely download for some reason.
I connected to a VPN, the file downloaded just fine, and I was able to open the file.