Python 3.7

Is it possible to update my python version to 3.7? I'd like to run my app but it depends on 3.7 since I the libraries I'm using aren't available for 2.7

Your account already supports Python 3.7 -- you just need to use the right command to start it. Because there are many versions of Python installed, you need to use python3.7 to start Python 3.7, python3.8 to start 3.8, and so on. Each version has its own version of pip, for example pip3.7 for 3.7, pip3.8 for 3.8.

The commands python and pip are aliases for python2.7 and pip2.7 respectively, because the operating system requires it.

Ah, thank you so much. Now I'm facing a different problem. I'm trying to run my code in Python 3.7

I've installed a module in the bash with pip3.7 install module --user Now I'm trying to use the following code:

print("hello from python 3.7")
import module

Using this, I still do get the error message that the module was not found, however installing it was succesfull. Any ideas of what I've done wrong? Thank you!

what's module? do you mean to put in the actual module name?

Yeah, I'm trying to import the Discord module. I've already installed it using pip3.7 but can't import it.

Any idea of how to resolve this problem?

that sounds suspiciously like you either did not install into python3.7 or you are not running python3.7. or that you installed into or you are running in a virtualenv, vs you running/installing outside of the virtualenv.

I did use:

pip3.7 install discord --user

And my code starts with:

print("hello from python 3.7")
import discord

If I'm correct, that should do the trick, but it doesn't. I have no idea what else I could do. If needed, feel free to take a look at my code / files.

It looks like you're running the code from the editor, and your account is configured to use Python 3.8 when you do that. If you want to use Python 3.7, this help page explains how to specify the Python version in the editor. If you're happy to use Python 3.8, then you should use pip3.8 instead of pip3.7 to install modules.