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Python on RPi: Errno 2 No such file or directory: 'config.json' but config.json is in the same folder as the main.py script

Hey all
I am trying to run a python script from GitHub, https://github.com/dudisgit/gmod_toolgun_prop for a project with a functioning screen and I put a command at the end of the .bashrc file
python3 /home/pi/gmod_toolgun_prop-main/main.py
so that the code executes as soon as the RPi powers up. When running the script in Thonny's Python IDE on my RPi 2B it executes no problem and the screen works. However when I open terminal I get an error message from the code running in the .bashrc file:
Traceback (most recent call last):
File "/home/pi/gmod_toolgun_prop-main/main.py", line 381, in <module>
main()
File "/home/pi/gmod_toolgun_prop-main/main.py", line 357, in main
with open(args.config) as config_file:
FileNotFoundError: [Errno 2] No such file or directory: 'config.json'
However, the config.json file is in the same folder as the main.py file as shown below:
Screenshot of file explorer showing config.json is in the same folder as main.py
And here's a screenshot of the error message
And here's the code that is refenced in the error message as line 357:
with open(args.config) as config_file:
config = json.load(config_file)
The entirety of the main.py script is in the Github link as attached in the first paragraph as well as the config.json file and other relevant files.
I am fairly new to the Python programming space so I don't understand what could be causing this error nor how this script handles opening the config.json file.
I have tried creating a custom service but it spits out the same error. Crontab and the local.bashrc file just doesn't work straight up for this. This is the furthest I have got with it attempting to execute on boot.

Maybe it helps to use the absolute path of config.json, i.e.(according to your description)
/home/pi/gmod_toolgun_prop-main/config.json
instead of the simple filename config.json.

If Python says the file is not there, then the file is not there. The only question is, where is there?
The filename in this case is config.json. Since there's no / (no directory name), the name is taken to be relative to the current working directory. That might or might not be the same as the directory of the main Python module, here /home/pi/gmod_toolgun_prop-main/main.py.
You can verify that by printing the current working directory just before opening the file. You can use os.getcwd to do that. Or, use strace(1) to show the interpreter's attempts to open config.json.

Convert python file into exe using auto-py-to-exe

I have just finished a project where I have made a connect 4 game and am trying to convert it to an exe file using auto-py-to-exe.
I want to use the one-file option, however every time it finishes and I run it, it would come up with an error:
Failed to execute script 'main' due to unhandled exception: No file 'Assets/icon.png' found in working directory '...'
Then in the box it says:
Traceback (most recent call last):
File "main.py", line 32, in <module>
FileNotFoundError: No file 'Assets/window-icon.png' found in working directory '...'.
I've tried quite a few alterations, e.g. not using the image, but then it would come up with the same error but for a different added file.
How can I fix this?
EDIT: I've tried it again by using the os module and giving the full directories to all the files in main.py, but that hasn't changed anything.

I was also facing this problem then I got the solutions from analysing other threads.
You have to update your script by adding this function
import sys
import os
def resource_path(relative_path):
""" Get absolute path to resource, works for dev and for PyInstaller """
base_path=getattr(sys,'_MEIPASS',os.path.dirname(os.path.abspath(__file__)))
return os.path.join(base_path, relative_path)
and changing address of every additional files used in your script
for example
icon = resource_path("game.icon")
You just have to use auto-py-to-exe just like shown below
Showing the options you can just use
Add all the files used in the Additional File section of it and make sure that the period(.) is there in the destination
That's it !

Opening a file in Python gives me errors

I am quite new to Python and I am having problems opening a file in Python.
I want to open a text file called 'coolStuff' in a Folder on my Desktop and this is how I type in the command but I still get an error message. The file exists and so I do not understand why I get that error message.
open("coolStuff.txt","r")
Traceback (most recent call last):
File "<pyshell#11>", line 1, in <module>
open("coolStuff.txt","r")
FileNotFoundError: [Errno 2] No such file or directory: 'coolStuff.txt'

If you want to simply supply a filename like coolStuff.txt without also providing a full directory, then you have to make sure that Python is running in the same directory as the file. If you aren't sure what directory Python is running in, try this:
import os
print(os.getcwd())

You have two options:
let's say your file is in C:\path\to\dir\coolStuff.txt
1.
open(r'C:\path\to\dir\coolStuff.txt','r')
2.
import os
os.chdir(r'c:\path\to\dir')
open('coolStuff.txt', 'r')

Because the file you want to open is not in the current directory.You can find the file 'coolStuff.txt' in the terminal and launch your python environment at the same directory.

Error opening megawarc archive from Python

I've found myself having to use a python script to access a webarchive.
What I have is a 'megawarc' web archive file from http://archive.org/details/archiveteam-fanfiction-warc-11. I need to un-megawarc this, using the python script found at https://github.com/alard/megawarc.
I'm trying to run the 'restore' command, and I have the three files needed (FILE.warc.gz,
FILE.tar, and FILE.json.gz) from the first link.
I have both python 2.7 and 3.3 installed.
--------------update--------------
I've ran both this method..
python megawarc restore FILE
and this method..
Make sure you have the files megawarc and ordereddict.py in the same directory, with the files you want to convert.
Rename the file megawarc to megawarc.py
Open a python console in this directory
Type the following code (line by line) :
import sys
sys.argv = ['megawarc','restore','FILE']
import megawarc
megawarc.main()
using python 2.7, and this is what I get..
c:\Python27>python megawarc restore FILE
Traceback (most recent call last):
File "megawarc", line 563, in <module>
main()
File "megawarc", line 552, in main
mwr.process()
File "megawarc", line 460, in process
self.process_entry(entry, tar_out)
File "megawarc", line 478, in process_entry
entry["target"]["offset"], entry["target"]["size"])
File "megawarc", line 128, in copy_to_stream
raise Exception("End of file: %d bytes expected, but %d bytes read." % (buf_size, l))
Exception: End of file: 4096 bytes expected, but 236 bytes read.
Is there something else i'm missing?
I have the following files all in
c:\python27
FILE.megawarc.json.gz
FILE.megawarc.tar
FILE.megawarc.warc.gz
megawarc
ordereddict.py
Is this some type of corrupt file error? Is there something i'm missing?

On the second link you provided, there are two important files :
megawarc
ordereddict.py
The executable script is megawarc. To run it, you have to launch it in a shell with
python megawarc restore FILE
Alternatively, if you're using a UNIX-based system. You can do
chmod +x megawarc
To give megawarc script executable property and then run it with
./megawarc restore FILE
Here, FILE is the actual name you should type if the 3 files you have are FILE.warc.gz, FILE.tar, and FILE.json.gz. You have to change this parameter by the common prefix to your 3 input files if needed.
EDIT :
Okay, i found an alternative that would work if you don't have a standard shell to start the script in command line.
What you have to do is :
Make sure you have the files megawarc and ordereddict.py in the same directory, with the files you want to convert.
Rename the file megawarc to megawarc.py
Open a python console in this directory
Type the following code (line by line) :
import sys
sys.argv = ['megawarc','restore','FILE']
import megawarc
megawarc.main()
This should work, i've just tried it.
Hope it will help.

Python I/O cannot find file, but path seems OK

I am with a python script. I want to open a file to retrieve data inside. I add the right path to sys.path:
sys.path.append('F:\WORK\SIMILITUDE\ALGOCODE')
sys.path.append('F:\WORK\SIMILITUDE\ALGOCODE\DTW')
More precisely, the file file.txt I will open is in DTW folder, and I also add upper folder ALGOCODE. Then, I have command
inputASTM170512 = open("file.txt","r")
I have this present:
Traceback (most recent call last):
File "<pyshell#24>", line 1, in <module>
inputASTM170512 = open("ASTM-170512.txt","r")
IOError: [Errno 2] No such file or directory: 'ASTM-170512.txt'
Why? Do you have any idea?

open() checks only the current working directory and does not traverse your system path looking for the file. Only import works with that mechanism.
You will either need to change your working directory before you open the file, with os.chdir(PATH) or to include the entire path when trying to open it.

When you try to open file with open, for example:
open("ASTM-170512.txt","r")
you will try to open a file in the current directory.
It does not depend on sys.path. The sys.path variable is used when you try to import modules, but not when you open files.
You need to specify the full path to file in the open or change the current directory to the correspondent place (I think that the former is better).