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
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()
Upon looping a directory to delete txt files ONLY - a message is returned indicating The System cannot find the file specified: 'File.txt'.
I've made sure the txt files that I'm attempting to delete exist in the directory I'm looping. I've also checked my code and to make sure it can see my files by printing them in a list with the print command.
import os
fileLoc = 'c:\\temp\\files'
for files in os.listdir(fileLoc):
if files.endswith('.txt'):
os.unlink(files)
Upon initial execution, I expected to see all txt files deleted except for other non-txt files. The actual result was an error message "FileNotFoundError: [WinError 2] The system cannot find the file specified: 'File.txt'.
Not sure what I'm doing wrong, any help would be appreciated.
It isn't found because the the path you intended to unlink is relative to fileLoc. In fact with your code, the effect is to unlink the file relative to the current working directory. If there were *.txt files
in the cwd then the code would have unfortunate side-effects.
Another way to look at it:
Essentially, by analogy, in the shell what you're trying to do is equivalent to this:
# first the setup
$ mkdir foo
$ touch foo/a.txt
# now your code is equvalent to:
$ rm *.txt
# won't work as intended because it removes the *.txt files in the
# current directory. In fact the bug is also that your code would unlink
# any *.txt files in the current working directory unintentionally.
# what you intended was:
$ rm foo/*.txt
The missing piece was the path to the file in question.
I'll add some editorial: The Old Bard taught us to "when in doubt, print variables". In other words, debug it. I don't see from the OP an attempt to do that. Just a thing to keep in mind.
Anyway the new code:
Revised:
import os
fileLoc = 'c:\\temp\\files'
for file in os.listdir(fileLoc):
if file.endswith('.txt'):
os.unlink(os.path.join(fileLoc,file))
The fix: os.path.join() builds a path for you from parts. One part is the directory (path) where the file exists, aka: fileLoc. The other part is the filename, aka file.
os.path.join() makes a whole valid path from them using whatever OS directory separator is appropriate for your platform.
Also, might want to glance through:
https://docs.python.org/2/library/os.path.html
How can I use Python to create new folders relative to my current working directory?
For example, my path is C:/Documents/Code with no folders within and just has my Python file. How do I store some data within C:/Documents/Code/Data without hard coding the absolute path?
This is what I've been trying:
path = "/Data/file.txt"
file = open(path, "w")
This gives me the error of "No such file or directory".
Thanks for any assistance!
Prepend the path with a single dot ., which denotes your current working directory:
path = "./Data/file.txt"
# ^
As per the code below I am having issues with the zipping a directory using the python 3 shutil.make_archive function. The .testdir will be zipped but it is being zipped in /home/pi, instead of /home/pi/Backups.
zip_loc = '/home/pi/.testdir'
zip_dest = '/home/pi/Backups/'
shutil.make_archive(zip_loc, 'zip', zip_dest)
Could anyone explain what I am doing wrong?
Reading the docs here I came up with:
zip_loc = '/home/pi/.testdir'
zip_dest = '/home/pi/Backups/'
shutil.make_archive(base_dir=zip_loc, root_dir=zip_loc, format='zip', base_name=zip_dest)
From the docs:
base_name is the name of the file to create, including the path, minus any format-specific extension.
root_dir is a directory that will be the root directory of the archive; for example, we typically chdir into root_dir before creating the archive.
base_dir is the directory where we start archiving from; i.e. base_dir will be the common prefix of all files and directories in the archive.
root_dir and base_dir both default to the current directory.
Before to write the archive, move to the good directory :
old_path = os.getcwd()
os.chdir(path)
-> write the archive
After writing the archive move back to old directory :
os.chdir(old_path)