I'm currently trying to open an .xlsx file with zipfile on Python, finding all files with namelist(), then using .count() to find all images in .png format within the archive.
My problem is currently, the list returned by namelist() function returns only 1680 elements.
After saving the xlsx file as an html, I am able to view all images contained in the excel spreadsheet and the total file count is 3,352 files.
I checked documentation for zipfile and exhausted the best Google searches I could muster. I appreciate any hints or advice!
Here's the snippet of code I'm using:
import zipfile as zf
xlsx = 'myfile.xlsx'
xlsx_file = zf.ZipFile(xlsx)
fileList = xlsx_file.namelist()
maybe convert it to a wheel file? wheel works good to me
Related
I am trying to extract zip files using the zipfile module's extractall method.
My code snippet is
import zipfile
file_path = '/something/airway.zip'
dir_path = 'something/'
with zipfile.ZipFile(file_path, "r") as zip_ref:
zip_ref.extractall(dir_path)
I have two zip files named, test (1.1 mb) and airway (520 mb).
For test.zip the folder contains all the files but for airway.zip, it creates another folder inside my target folder named Airway, and then extracts all the files there. Even after renaming the airway.zip to any garbage name, the result was same.
Is there some workaround to get only the files extracted in my target folder? It is critical for me as I'm doing this extraction automated from django
Python version: 3.9.6;
Django version: 2.2
I ran your code and it seems to be only a problem of the zipfile itself. If you create a zipfile by selecting only the elements you get the result you got with test.zip. If you create it by selecting a folder holding the elements the folder will be there if you extract it again, no matter what you name your zip file.
I have two articles related to this:
https://www.kite.com/python/docs/zipfile.ZipFile.extractall
https://www.geeksforgeeks.org/working-zip-files-python/
Even if both of these articles do not solve your problem then I think that instead of zipping the files in the folder you just zipped the folder itself so try by zipping the files inside the folder.
I am working on a project that manipulates with a document's xml file. My approach is like the following. First convert the DOCX document into a zip archive, then extract the contents of that archive in order to have access to the document.xml file, and finally convert the XML to a txt in order to work with it.
So i did all the above on a document, and everything worked perfectly, but when i decided to use a different document, the Zipfile library doesnt extract the content of the new ZIP archive, however it somehow extracts the contents of the old document that i processed before, and converts the document.xml file into document.txt without even me even running that block of code that converts the XML into txt.
The worst part is the old document is not even in the directory anymore, so i have no idea how Zipfile is extracting the content of that particular document when its not even in the path.
This is the code I am using in Jupyter notebook.
import shutil
import zipfile
# Convert the DOCX to ZIP
shutil.copyfile('data/docx/input.docx', 'data/zip/document.zip')
# Extract the ZIP
with zipfile.ZipFile('zip/document.zip', 'r') as zip_ref:
zip_ref.extractall('data/extracted/')
# Convert "document.xml" to txt
os.rename('extracted/word/document.xml', 'extracted/word/document.txt')
# Read the txt file
with open('extracted/word/document.txt') as intxt:
data = intxt.read()
This is the directory tree for the extracted zip archive for the first document.
data -
1-docx
2-zip
3-extracted/-
1-customXml/
2-docProps/
3-_rels
4-[Content_Types].xml
5-word/-document.txt
The 2nd document's directory tree should be as following
data -
1-docx
2-zip
3-extracted/-
1-customXml/
2-docProps/
3-_rels
4-[Content_Types].xml
5-word/-document.xml
But Zipfile is extracting the contents of the first document even when the DOCX file is not in the directory.I am also using Ubuntu 20.04 so i am not sure if it has to do with my OS.
I suspect that you are having issues with relative paths, as unzipping any Word document will create the same file/directory structure. I'd suggest using absolute paths to avoid this. What you may also want to do is, after you are done manipulating and processing the extracted files and directories, delete them. That way you won't encounter any issues with lingering files.
I'm working with zipped files in python for the first time, and I'm stumped.
I read the documentation for zipfile, but I'm not sure what would be the best way to do what I'm trying to do. I have a zipped folder with CSV files inside, and I'd like to be able to open the zip file, and retrieve certain values from the csv files inside.
Do I use zipfile.extract(file name here) to bring it to the current working directory? And if I do that, do I just use the file name to work with the file, or does this index or list them differently?
Currently, I manually extract all files in the zipped folder to the current working directory for my project, and then use the csv module to read them. All I'm really trying to do is remove that step.
Any and all help would be greatly appreciated!
You are looking to avoid extracting to disk, in the zip docs for python there is ZipFile.open() which gives you a file-like object. That is an object that mostly behaves like a regular file on disk, but it is in memory. It gives a bytes array when read, at least in py3.
Something like this...
from zipfile import ZipFile
import csv
with ZipFile('abc.zip') as myzip:
print(myzip.filelist)
for mf in myzip.filelist:
with myzip.open(mf.filename) as myfile:
mc = myfile.read()
c = csv.StringIO(mc.decode())
for row in c:
print(row)
The documentation of Python is actually quite good once one has learned how to find things as well as some of the basic programming terms/descriptions used in the documentation.
For some reason csv.BytesIO is not implemented, hence the extra step via csv.StringIO.
I'm trying to use python to search within a .zip file without actually extracting the zipped file. I understand re.search can do searches within files, but will it do the same for files that have yet to be extracted?
The way to do this is using the zipfile module, it allows reading the name list and other meta info from the zip file prior to extracting the content.
import zipfile
zf = zipfile.ZipFile('example.zip', 'r')
print zf.namelist()
You can read more here about the Zipfile Library
This is my problem.
I need to combine text, picture and video (any codec) into one file.
I know there is binary files. How would I go about packaging and reading the file.
For example, In the one file I store the text, then the png and then the video.
In another Python file I extract the files again and display as I please.
Regards,
Renier Engelbrecht
You could use the zipfile module - it creates a single file from arbitrary components.
Sample usage (Python 3):
import zipfile
# Write zip file
with zipfile.ZipFile("combined_file.zip", mode='w', compression=zipfile.ZIP_STORED) as archive:
archive.write("file_1.ext")
archive.write("file_2.ext")
# Extract contents later
with zipfile.ZipFile("combined_file.zip", mode='r') as archive:
archive.extractall()