Develop Reference

Why os.path.realpath doesn't work properly

i just downloaded a file called "N_PR_8705_004A_.doc" in my "Downloads" folder and i want to put it into my "Stage NLP" folder using os. I know how to do it without os but i'd like that shit to work it's faster and it simply doesnt. First i tried to get the path of my file doing this:
import os
os.path.dirname(os.path.abspath("N_PR_8705_004A_.doc"))
# or os.path.realpath it's the same
and the result i get is:
'C:\\Users\\f002722\\Stage NLP'
whereas when i do list all the files in this folder doing:
os.listdir("C:\\Users\\f002722\\Stage NLP")
you clearly see it is simply not there:
['.ipynb_checkpoints',
'ADR service study - D2 (1st part).pdf',
'basetal.py',
'Codes test',
'Cours NLP.ipynb',
'e Deorbit',
'edot CDF study.pdf',
'edot_v5.pdf',
'Entrainement.ipynb',
'ESA edot workshop May 6th 2014 - Summary.msg',
'ESA_edotWorkshop-_Envisat_attitude-Copy1',
'ESA_edotWorkshop-_Envisat_attitude.pdf',
'ESA_edotWorkshop_GNC_.pdf',
'ESA_INNOCENTI_Challenges.pdf',
'ESA_Robin_Biesbroek_edot.pdf',
'GMV_edot_Symposium.pdf',
'JOP_edotWorkshop.pdf',
'KT_HAARMANN_Edot.pdf',
'MDA_edot_Symposium_-_Robotic_Capture.pdf',
'MDA_eDot_Symposium_-_Robotic_Capture.pdf.kx2zd5w.partial',
'Note_Ariane_NLP.ipynb',
'Note_Ariane_NLP_2.ipynb',
'Note_Ariane_NLP_3.ipynb',
'OHB_eDotWorkshop_ADRM.pdf',
'OHB_Sweden_eDotWorkshop_PRISMA_and_IRIDES.pdf',
'SKA_Polska_eDotWorkshop_Net_Simulator.pdf',
'TAS_Carole_Billot_edot.pdf',
'Test.ipynb',
'Text_clustering_v3_2.py',
'Webinar_OOSandADR_7May2020.pdf',
'__pycache__']
So what the hell is going on i'm out of ideas here.
Thx in advance

I think I have a possible answer to your question. Neither realpath nor abspath require their arguments to name existing files. In particular, the documentation for abspath() says: "On most platforms, this is equivalent to calling the function normpath() as follows: normpath(join(os.getcwd(), path))."
This means that if you have a Python script that has a line like,
foo = os.path.dirname(os.path.abspath("doesnotexist"))
then the value of foo will be the current working directory of the script. Since "doesnotexist" isn't the name of a file in this directory, it won't show up if you do os.listdir(foo).
I notice that you wrote that "N_PR_8705_004A_.doc" was in your "Downloads" directory, which is obviously not the same as 'C:\\Users\\f002722\\Stage NLP'. If 'C:\\Users\\f002722\\Stage NLP' is the working directory for your Python script, then running os.path.dirname(os.path.abspath("N_PR_8705_004A_.doc")) is just like writing os.path.dirname(os.path.abspath("doesnotexist")), for the reasons that I just gave.
Python can't automatically figure out the path of a file just by giving it a relative file name. For example, there could be many files named README.txt on a system, each in different directories, so there's no way for os.path.abspath('README.txt') to know which of those directories you want.
To move the file "N_PR_8705_004A_.doc" from the "Downloads" directory to 'C:\\Users\\f002722\\Stage NLP', you'd probably need to do something like this:
import shutil
shutil.move('C:\\Users\\f002722\\Downloads\\N_PR_8705_004A_.doc',
'C:\\Users\\f002722\\Stage NLP')
presuming, of course, that the "Downloads" directory was inside 'C:\\Users\\f002722'.

Can i shorten a file location to the .py files location

first post here so sorry if it's hard to understand. Is it possible to shorten the directory in python to the location of the .py file?. For example, if I wanted to grab something from the directory "C:\Users\Person\Desktop\Code\Data\test.txt", and if the .py was located in the Code folder, could I shorten it to "\data\test.txt". I'm new to python so sorry if this is something really basic and I just didn't understand it correctly.
I forgot to add i plan to use this with multiple files, for example: "\data\test.txt" and \data\test2.txt

import os
CUR_FILE = os.path.abspath(__file__)
TARGET_FILE = "./data/test.txt"
print(os.path.join(CUR_FILE, TARGET_FILE))
With this, you can move around your Code directory anywhere and not have to worry about getting the full path to the file.
Also, you can run the script from anywhere and it will work (you don't have to move to Code's location to run the script.

You can import os and get current working directory ,this will give you the location of python file and then you can add the location of folder data and the file stored in that ,code is given below
import os
path=os.getcwd()
full_path1=path+"\data\test.txt"
full_path2=path+"\data\test2.txt"
print(full_path1)
print(full_path2)
I think this will work for your case and if it doesn't work then add a comment

vscode - read file from current folder where .py file is

I'm very new to programming, and to vscode.
I'm learning Python and currently I am learning about working with files.
The path looks like this: /home/anewuser/learning/chapter10.
The problem: completely basic "read file in python" lesson does not work in vscode because no such file or directory error raises when running my .py file, located in ~/learning/chapter10. But vscode wants that my .txt file I am supposed to open in python, to be in ~/learning directory, then it works. I don't like this behaviour.
All I want is to be able to read file placed in the directory where the .py file is. How to do this?

Because in your case ~/learning is the default cwd (current working directory), VSCode looks for pi_digits.txt in that location. If you put pi_digits.txt beside file_reader.py (which is located at ~/learning/chapter10), you'll have to specify the path (by prepending chapter10/ to the .txt file).
So you should do this:
with open('chapter10/pi_digits.txt') as file_object:
contents = file_object.read()
print(contents)
If you want to change the default current working directory (for example you want to change it to ~/learning/chapter10), you'll have to do the following:
~/learning/chapter10/file_reader.py
import os # first you need to import the module 'os'
# set the cwd to 'chapter10'
os.chdir('chapter10')
# now 'file_reader.py' and 'pi_digits.txt' are both in the cwd
with open('pi_digits.txt') as file_object:
contents = file_object.read()
print(contents)
With os.chdir('chapter10') you've set chapter10 as the default cwd, in which VSCode now will look for pi_digits.txt.
For detailed information about os.chdir() you can read through the official documentation or take a look at this post on stackoverflow.

In "User Settings", use the search bar to look for "python.terminal.executeInFileDir" and set (=) its value to "true" instead of "false".
I took this answer from here
How to run python interactive in current file's directory in Visual Studio Code?
this is my first time putting an answer on StackOverflow
so I apologize if I didn't do it the right way

Python libtorrent, get file list names

I am using libtorrent for python 3.6 . I just want to get any file names that downloaded with a session, e.g. the folder name, the files name etc.
I searched around the web didn't come across anything. I am using the follow example:
https://www.libtorrent.org/python_binding.html
So when the download progress finish, i want to know what files this session downloaded. How can achieve that? Thanks in advance!

Finally found the answer, the code is:
handle = libtorrent.add_magnet_uri(session, magnetLink,params)
session.start_dht()
while not handle.has_metadata():
time.sleep(1)
torinfo = handle.get_torrent_info()
for x in range(torinfo.files().num_files()):
print(torinfo.files().file_path(x))
The code above prints the file names that came with the magnet file.

Python - extract and modify a file path in all files in a directory in linux

I have files .sh files and .json files in which there are file paths given to point to a specific directory, but I should keep on changing the file path, depending on where my python scipt is run.
eg:content of one of my .sh file is
"cd /home/aswany/BotStudioInstallation/databricks/platform/databricksastro"
and I should change the file path via python code where the following path
"/home/aswany/BotStudioInstallation/" keep on changing depending on where databicks is located,
I tried the following code:
replaceAll(str(self.currentdirectory)+
"/databricks/platform/devsettings.json",
"/home/holmes/BotStudioInstallation",self.currentdirectory)
and function replaceAll is:
def replaceAll(self,file,searchExp,replaceExp):
for line in fileinput.input(file, inplace=1):
if searchExp in line:
line = line.replace(searchExp,replaceExp)
sys.stdout.write(line)
but above code only replaces a line
"home/holmes/BotStudioInstallation" to the current directory I am logged in,bt it cannot be sure that "home/holmes/BotStudioInstallation" is the only possibility it keep on changing like "home/aswany/BotStudioInstallation","home/dev3/BotStudioInstallation" etc ,I thought of regular expression for this.
please help me

Not sure I 100% understood your issue, but maybe I can help nonetheless.
As pointed out by J.F. Sebastian, you can use relative paths and remove the base part of the path. Using ./databricks/platform/devsettings.json might be enough. This is by far the most elegant solution.
If for any reason it is not, you can keep the directory you need to access, then append it to the base directory whenever you need it. That should allow you to deal with changes in the base directory. Though in the case the files will be used by other applications than your own, that might not be an option.
dir = get_dir_from_json()
dir_with_base = self.currentdirectory + dir
Alternatively, not an elegant solution though, without using regex you can use a "pattern" to always replace.
{
"directory": "<<_replace_me_>>/databricks/platform"
}
Then you know you can always replace "<<_replace_me_>>" with the base directory.