I'm trying to create a download function for my streamlit app. But what I currently have allows me to download a zip file via a button on my streamlit app but unfortunately it also saves it to my local folder. I don't want it to save to my local folder. The problem is when I initialize the file_zip object. I want the zip file in a specific name ideally the same name of the file that the user upload with a '.zip' extension (i.e datafile that contains the string file name as a parameter in the function). But everytime I do that it keeps saving the zip file in my local folder. Is there an alternative to this? BTW I'm trying to save list of pandas dataframe into one zip file.
def downloader(list_df, datafile, file_type):
file = datafile.name.split(".")[0]
#create zip file
with zipfile.ZipFile("{}.zip".format(file), 'w', zipfile.ZIP_DEFLATED) as file_zip:
for i in range(len(list_df)):
file_zip.writestr(file+"_group_{}".format(i)+".csv", pd.DataFrame(list_df[i]).to_csv())
file_zip.close()
#pass it to front end for download
zip_name = "{}.zip".format(file)
with open(zip_name, "rb") as f:
bytes=f.read()
b64 = base64.b64encode(bytes).decode()
href = f'Click Here To Download'
st.markdown(href, unsafe_allow_html=True)
It sounds like you want to create the zip file in memory and use it later to build a base64 encoding. You can use an io.BytesIO() object with ZipFile, rewind it, and read the data back for base64 encoding.
import io
def downloader(list_df, datafile, file_type):
file = datafile.name.split(".")[0]
#create zip file
zip_buf = io.BytesIO()
with zipfile.ZipFile(zip_buf, 'w', zipfile.ZIP_DEFLATED) as file_zip:
for i in range(len(list_df)):
file_zip.writestr(file+"_group_{}".format(i)+".csv", pd.DataFrame(list_df[i]).to_csv())
zip_buf.seek(0)
#pass it to front end for download
zip_name = "{}.zip".format(file)
b64 = base64.b64encode(zip_buf.read()).decode()
del zip_buf
href = f'Click Here To download'
st.markdown(href, unsafe_allow_html=True)
I'm trying to zip several images in a ZIP file.
Although It's a Django app, files are not stored in the same app. I want to fetch them from a url list.
I get the following error: FileNotFoundError [Errno 2] No such file or directory: 'image1.jpg'
def download_image(request):
fotos = ['https://storage.googleapis.com/some/image1.jpg', 'https://storage.googleapis.com/some/image2.jpg']
f = StringIO()
zip = ZipFile(f, 'w')
for foto in fotos:
url = urlopen(foto)
filename = str(foto).split('/')[-1]
zip.write(filename, url.read())
zip.close()
response = HttpResponse(f.getvalue(), content_type="application/zip")
response['Content-Disposition'] = 'attachment; filename=image-test.zip'
return response
I reviewed several posts and finaly followed this one how can I export multiple images using zipfile and urllib2 - django
But I can't get this working. Any clues welcome. Thanks in advance.
Per the zip documentation, write() is used to write an existing file into the filesystem (by path). I think that you are looking for writestr(), which can be passed the actual file contents.
I had uploaded zip file in my azure account as a blob in azure container.
Zip file contains .csv, .ascii files and many other formats.
I need to read specific file, lets say ascii file data containing in zip file. I am using python for this case.
How to read particular file data from this zip file without downloading it on local? I would like to handle this process in memory only.
I am also trying with jypyter notebook provided by azure for ML functionality
I am using ZipFile python package for this case.
Request you to assist in this matter to read the file
Please find following code snippet.
blob_service=BlockBlobService(account_name=ACCOUNT_NAME,account_key=ACCOUNT_KEY)
blob_list=blob_service.list_blobs(CONTAINER_NAME)
allBlobs = []
for blob in blob_list:
allBlobs.append(blob.name)
sampleZipFile = allBlobs[0]
print(sampleZipFile)
The below code should work. This example accesses an Azure Container using an Account URL and Key combination.
from azure.storage.blob import BlobServiceClient
from io import BytesIO
from zipfile import ZipFile
key = r'my_key'
service = BlobServiceClient(account_url="my_account_url",
credential=key
)
container_client = service.get_container_client('container_name')
zipfilename = 'myzipfile.zip'
blob_data = container_client.download_blob(zipfilename)
blob_bytes = blob_data.content_as_bytes()
inmem = BytesIO(blob_bytes)
myzip = ZipFile(inmem)
otherfilename = 'mycontainedfile.csv'
filetoread = BytesIO(myzip.read(otherfilename))
Now all you have to do is pass filetoread into whatever method you would normally use to read a local file (eg. pandas.read_csv())
you could use below code for reading file inside .zip file without extracting in python
import zipfile
archive = zipfile.ZipFile('images.zip', 'r')
imgdata = archive.read('img_01.png')
For details , you can refer to ZipFile docs here
Alternatively, you can do something like this
-- coding: utf-8 --
"""
Created on Mon Apr 1 11:14:56 2019
#author: moverm
"""
import zipfile
zfile = zipfile.ZipFile('C:\\LAB\Pyt\sample.zip')
for finfo in zfile.infolist():
ifile = zfile.open(finfo)
line_list = ifile.readlines()
print(line_list)
Here is the output for the same
Hope it helps.
I have some problem with excel(xlsx) file.I want to just open and save operation using python code.I have tried with python but couldn't found
cursor = context.cursor()
s3 = boto3.resource('s3')
bucket = s3.Bucket('bucket')
objects = bucket.objects.all()
for obj in objects:
if obj.key.startswith('path/filename'):
filename=obj.key
openok=open(obj)
readok = openok.readlines()
readok.close()
print ('file open and close sucessfully')```
You can't read/interact with files directly on s3 as far as I know.
I'd recommend downloading it locally, and then opening it. You can use the builtin tempfile module if you want to save it to a temporary path.
with tempfile.TemporaryDirectory() as tmpdir:
local_file_path = os.path.join(tmpdir, "tmpfile")
bucket.download_file(obj.key, local_file_path)
openok=open(local_file_path)
readok = openok.readlines()
readok.close()
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.