Develop Reference

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

Call all files from a menu

I've got a task to do that is crushing my head. I have five .py documents and I want to make a menu in another .py so I can run any of them by introducing a string inside an input() but don't really see the way to do that and I don't know if there is somehow I can.
I have tried import every file to the 6th file but I don't even know how to start.
I would like it just to be seen as simple as it can sound, but yet I find it really hard.

If you just want to run them, then try this:-
import os
file_path = input("Enter the path of your file = ")
os.system(file_path)
If the file that you are trying to execute is not in the current
directory, i.e. doesn't exist in the same folder as the currently
executing python file, then you have to provide it's full path.
Path Format:-. C:\Users\lmYoona\OneDrive\Desktop\example.py
If the python file you are trying to execute is in the same directory as
the currently executing python file, then abstract name will also
work
Path Format:- example.py
P.S.:- I would only recommend this method if all you want is just to execute the other python file, rather then importing stuff from it.

How to open any program in Python?

Well I searched a lot and found different ways to open program in python,
For example:-
import os
os.startfile(path) # I have to give a whole path that is not possible to give a full path for every program/software in my case.
The second one that I'm currently using
import os
os.system(fileName+'.exe')
In second example problem is:-
If I want to open calculator so its .exe file name is calc.exe and this happen for any other programs too (And i dont know about all the .exe file names of every program).
And assume If I wrote every program name hard coded so, what if user installed any new program. (my program wont able to open that program?)
If there is no other way to open programs in python so Is that possible to get the list of all install program in user's computer.
and there .exe file names (like:- calculator is calc.exe you got the point).
If you want to take a look at code
Note: I want generic solution.

There's always:
from subprocess import call
call(["calc.exe"])
This should allow you to use a dict or list or set to hold your program names and call them at will. This is covered also in this answer by David Cournapeau and chobok.

You can try with os.walk :
import os
exe_list=[]
for root, dirs, files in os.walk("."):
#print (dirs)
for j in dirs:
for i in files:
if i.endswith('.exe'):
#p=os.getcwd()+'/'+j+'/'+i
p=root+'/'+j+'/'+i
#print(p)
exe_list.append(p)
for i in exe_list :
print('index : {} file :{}'.format(exe_list.index(i),i.split('/')[-1]))
ip=int(input('Enter index of file :'))
print('executing {}...'.format(exe_list[ip]))
os.system(exe_list[ip])
os.getcwd()+'/'+i prepends the path of file to the exe file starting from root.
exe_list.index(i),i.split('/')[-1] fetches just the filename.exe
exe_list stores the whole path of an exe file at each index

Can be done with winapps
First install winapps by typing:
pip install winapps
After that use the library:
# This will give you list of installed applications along with some information
import winapps
for app in winapps.list_installed():
print(app)
If you want to search for an app you can simple do:
application = 'chrome'
for app in winapps.search_installed(application):
print(app)

How to access file in parent directory using python?

I am trying to access a text file in the parent directory,
Eg : python script is in codeSrc & the text file is in mainFolder.
script path:
G:\mainFolder\codeSrc\fun.py
desired file path:
G:\mainFolder\foo.txt
I am currently using this syntax with python 2.7x,
import os
filename = os.path.dirname(os.getcwd())+"\\foo.txt"
Although this works fine, is there a better (prettier :P) way to do this?

While your example works, it is maybe not the nicest, neither is mine, probably. Anyhow, os.path.dirname() is probably meant for strings where the final part is already a filename. It uses os.path.split(), which provides an empty string if the path end with a slash. So this potentially can go wrong. Moreover, as you are already using os.path, I'd also use it to join paths, which then becomes even platform independent. I'd write
os.path.join( os.getcwd(), '..', 'foo.txt' )
...and concerning the readability of the code, here (as in the post using the environ module) it becomes evident immediately that you go one level up.

To get a path to a file in the parent directory of the current script you can do:
import os
file_path = os.path.join(os.path.dirname(os.path.dirname(__file__)), 'foo.txt')

You can try this
import environ
environ.Path() - 1 + 'foo.txt'

to get the parent dir the below code will help you:
import os
os.path.abspath(os.path.join('..', os.getcwd()))