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Error message when opening a ipython console : TypeError: ChannelABC() takes no arguments
Solution that I found : downgrad jupyter_client to version <6 : jupyter/jupyter_client#531 ipython/ipykernel#489
Error message when opening a ipython console : TypeError: ChannelABC() takes no arguments
Solution that I found : downgrad jupyter_client to version <6 :
jupyter/jupyter_client#531
ipython/ipykernel#489