Develop Reference

How can I find out which path os.path points to?

i am a web developer (php, js, css and ...).
i order a python script for remove image background. it worked in cmd very well but when running it from php script, it dosnt work.
i look at the script for find problem and i realized that the script stops at this line:
net.load_state_dict(self.torch.load(os.path.join("../library/removeBG/models/", name, name + '.pth'), map_location="cpu"))
I guess the problem with the script is that it can't find the file, and probably the problem is caused by the path that os.path points to.
Is it possible to print the path that os .path points to?
If not, do you have a solution to this problem?

This should be enough:
name = 'name'
p = os.path.join("../library/removeBG/models/", name, name + '.pth')
print(p)
This is what i get:
>>> ../library/removeBG/models/name/name.pth

The problem here is that the php script might be in different directory so while executing the python script via php script, the os.path points to the directory from where it is being executed i.e. the location of php script.
TLDR; Try using absolute path.

Selenium how to save screenshot to specific directory in Python

I would like to take a screenshot for my selenium driver and save it to a specific directory. Right now, I can run:
driver.save_screenshot('1.png')
and it saves the screenshot within the same directory as my python script. However, I would like to save it within a subdirectory of my script.
I tried the following for each attempt, I have no idea where the screenshot was saved on my machine:
path = os.path.join(os.getcwd(), 'Screenshots', '1.png')
driver.save_screenshot(path)
driver.save_screenshot('./Screenshots/1.png')
driver.save_screenshot('Screenshots/1.png')

Here's a kinda hacky way, but it ought to work for your end result...
driver.save_screenshot('1.png')
os.system("mv 1.png /directory/you/want/")
You might need to use the absolute path for your file and/or directory in the command above, not 100% sure on that.

You can parse the file path you want to the save_screenshot function.
As your doing this already a good thing to check is that os.getcwd is the same as the location of the script (may be different if your calling it from somewhere else) and that the directory exists, this can be created via os.makedirs.
import os
from os import path
file_dir = path.join(os.getcwd(), "screenshots")
os.makedirs(file_dir, exist_ok=True)
file_path = path.join(file_dir, "screenshot_one.png")
driver.save_screenshot(file_path)
If os.getcwd is not the right location, the following will get the directory of the current script..
from os import path
file_dir = path.dirname(path.realpath(__file__))

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'.

Python searching for image in wrong directory

I am trying to load some assets onto my program that I have them in a folder called 'Graphics', which is inside the folder 'Snake', which is inside the folder 'Projects', which is inside the folder 'Python'. However, also inside that folder 'Python' is another folder named 'HelloWorld'.
I am trying to load some assets in a program that I am running in 'Snake' and Python is searching for the assets in the 'HelloWorld' folder (which is where I used to keep my python files).
I get the error:
FileNotFoundError: No file 'Projects/Snake/Graphics/apple.png' found in working directory 'C:\Users\35192\OneDrive - WC\Desktop\Python\HelloWorld'
I believe that for this I have to change the default directory for vs code. I have changed the default directory for the command prompt and it did nothing. Perhaps this is because the python that I am running in the command prompt is different from the one in vs code (?)
How do I fix this?
Thank you in advance.
Edit:
This is how I am currently loading the image:
apple = pygame.image.load('Projects\Snake\Graphics\apple.png').convert_alpha()

Use pathlib to construct the path to your images. You wil have to add import pathlib to your code.
pathlib.Path(__file__) will give you the path to the current file. pathlib.Path(__file__).parent will give you the folder. Now you can construct the path with the / operator.
Try the following code and check the output.
import pathlib
print(pathlib.Path(__file__))
print(pathlib.Path(__file__).parent)
print(pathlib.Path(__file__).parent / 'Grahics' / 'apple.png')
Now you will be able to move the full project to a totally different folder without having to adjust any code.
Your code example looks like this: apple = pygame.image.load('Projects\Snake\Graphics\apple.png').convert_alpha()
If you import pathlib you can replace that with the dynamic approach:
path_to_image= pathlib.Path(__file__).parent / 'Grahics' / 'apple.png'
apple = pygame.image.load(path_to_image).convert_alpha()
I'm quite sure that pygame can work with a path from pathlib. If not then you have to convert the path to a string manually
apple = pygame.image.load(str(path_to_image)).convert_alpha()

You don't need to change the default directory. Just load from the full directory. That should look something like: "C:\Users\...\Python\Snake\Graphics\apple.png".

I think the simplest way is to first see your active directory by simply typing in
pwd, and then you could simply change the directory by cd("C:/path/to/location"), remember you have to use the backslash, or just use the following library:
import os
os.chdir("C:/path/to/location")
As pydragon posted, you could also import it by just giving the import function a path.

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