Setting up and integrating Jupyter Notebook development environment with VSC

[enoren5 (Timon Fisher)]

@aivarpaalberg

Have you selected/set base i.e. which Python version/environment should VS Code code use to run cell/notebook?

When I client the “Select Kernel” button at the top right in the screenshot below, it reveals an empty menu pictured at the top:

Here is “Python v3.8.5 (conda)” showing as selected at the bottom and slightly to the left:

This exhaustive question on Stack Overflow with the same issue as mine includes many creative potential solutions: python - jupyter server : not started, no kernel in vs code - Stack Overflow

One Stack Overflow member recommends:

How I solved it:

1. Press Command+Shift+P to open a new command pallete
2. Type >Python: Select Intepreter to start jupyter notebook server
3. Open the notebook again

Someone else suggests:

1. Press Command+Shift+P on mac, Ctrl+Shift+p on windows
2. Type Jupyter: Select Interpreter to start Jupyterserver
3. It would show you a dropdown of python versions installed.
4. I chose python 3.7.5 and it worked for me. you can choose the python version installed on your machine.

Another member says:

I have seen all possible solutions but not work, finally I just upgrade jupyter, notebook, and jupyterlab, like pip install -U jupyterlab, and I can choose the kernel in VScode!

None of the above resolved my original issue.

Here is a new twist: After installing jupyterlab or some other package (or set of packages), after restarting VSC, I I encountered this weird error:

It’s a condition box. I clicked: “Yes”. Jupyter is still not integrated properly.