I'm just wondering how can i create a .pyd file using python3.5 on win10.By the way I just want to know how to create it using command line like how you create .pyc file.
Something like
f = open("NAMEOFFILE.pyd", "r")
f.close()
should do the trick. this should work with other arbitrary file types. .txt, .json, .py, .whateverfileextensionyouwant
Related
I was trying to create a file without a name in python (only filetype)
I tried this -
open(".gitignore","w+").close()
But it does not work.
edit - it does work real issue is in getting file through glob.glob
classify_folder_name = #path of the folder which contain .gitignore file
rel_paths = glob.glob(classify_folder_name + '/**', recursive=True)
for local_file in rel_paths:
print(local_file)
it does not print .gitignore file.
Any help will be appreciated.
Note -: don't want to use os.listdir()
There are few things that you might check:
files with dot at the beginning are hidden so whatever OS you are using, make sure you have hidden files visibility enabled
It might be saved in different directory
open(".gitignore","w+").close()
It would be better if you do this:
To create a file:
with open('.gitignore', 'w') as fp:
pass
I'll start by mentioning that I've no knowledge in Python but read online that it could help me with my situation.
I'd like to do a few things using (I believe?) a Python script.
I have a bunch of .yml files that I want to transfer the contents into one main .yml file (let's call it Main.yml). However, I'd also like to be able to take the name of each individual .yml and add it before it's content into Main.yml as "##Name". If possible, the script would look like each file in a directory, instead of having to list every .yml file I want it to look for (my directory in question only contains .yml files). Not sure if I need to specify, but just in case: I want to append the contents of all files into Main.yml & keep the indentation (spacing). P.S. I'm on Windows
Example of what I want:
File: Apes.yml
Contents:
Documentation:
"Apes":
year: 2009
img: 'link'
After running the script, my Main.yml would like like:
##Apes.yml
Documentation:
"Apes":
year: 2009
img: 'link'
I'm just starting out in Python too so this was a great opportunity to see if my newly learned skills work!
I think you want to use the os.walk function to go through all of the files and folders in the directory.
This code should work - it assumes your files are stored in a folder called "Folder" which is a subfolder of where your Python script is stored
# This ensures that you have the correct library available
import os
# Open a new file to write to
output_file = open('output.txt','w+')
# This starts the 'walk' through the directory
for folder , sub_folders , files in os.walk("Folder"):
# For each file...
for f in files:
# create the current path using the folder variable plus the file variable
current_path = folder+"\\"+f
# write the filename & path to the current open file
output_file.write(current_path)
# Open the file to read the contents
current_file = open(current_path, 'r')
# read each line one at a time and then write them to your file
for line in current_file:
output_file.write(line)
# close the file
current_file.close()
#close your output file
output_file.close()
I am trying to extract a file name with this format--> filename.tar.gz10
I have tried mutpile wayd but for all of them, I get the error that is unknow format. it works fine for files ends with tar.gz00. I tried to change the name but still does not work.
Here are what I have tried,
import tarfile
file = tarfile.open('filename.tar.gz10')
file.extractall('./extracted_path')
file.close()
Another way is,
shutil.unpack_archive('./filename.tar.gz10', './extracted_path', 'tar.gz17')
Thanks for your help in advance.
This coule be because the archive was split into smaller chunks, on linux you could do so using the split -b command so one big file is actually multiple smaller ones now, and they are named like
file.tar.gz01
file.tar.gz02
file.tar.gz03
file.tar.gz04
etc...
you wont be able to decompress these file individually, so you have to concatenate them first into one file then decompress.
To verify whther it was split or not, run file {filename} and if does not recognize it as a gzip compressed archive then it is propably split (this is why you get unknown format error)
You can try to do the following:
from glob import glob
import os
path = '/path/to/' # location of your files
list_of_files = glob(path + '*.tar.gz*') # list all gzip files
bash_command = 'gzip -dk filename.tar.gz' + ' '.join(list_of_files) # create bash command to concatenate the files
os.system(bash_command)
Does any one know why i got an error of "FileNotFoundError: [Errno 2] No such file or directory: 'bcs.xlsx'" when i'm loading this file of size 2mb it has around 60,000 rows and 4 columns.
i tried using csv instead of xlsx but i get the same error and i've checked hundreds times that the script and the file are at he same directory.
This is because Python does not find your file, errors are not lying.
But there's a misunderstanding in your question, you checked that the file is in the same directory as your script, but that's not the check you have to do. You have to check the file is in the current working directory of your python script.
To see your current working directory, use:
import os
print(os.getcwd())
And as we're at it you can list this directory:
print(os.listdir())
I don't know how you execute your script, but if you're using a terminal emulator, a typical way to give a file name to a program is by argument, not hardcoding its name, like by using argparse. And if you do this way, your shell completion may help you naming your file properly, like:
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('file', type=argparse.FileType('r'))
args = parser.parse_args()
print(args.file.read())
Now on a shell if you type:
python3 ./thescript.py ./th[TAB]
your shell will autocomplete "./th" to "./thescript.py" (if and only if it exists), highly reducing the probablity of having a typo. Typically if there's a space in the filename like "the script.py", your shell should properly autocomplete the\ script.py.
Also if you use argparse with the argparse.FileType as I did, you'll have a verbose error in case the file does not exist:
thescript.py: error: argument file: can't open 'foo': [Errno 2] No such file or directory: 'foo'
But… you already have a verbose error.
I am opening some VTU files from Directory X and there are other output files in that directory (for example log.txt) that I want to open via a plugin. If I do a os.getcwd() I end up in ParaViews installation directory. What I want is the directory of the VTU files I loaded BEFORE applying the plugin... So basically the start Point of the Pipline.
You could do something like this to get the reader
myreader = FindSource('MyReader')
then get the file name via the FileName attribute
myreader.FileName