I am new to programming and i decided to learn Python first, so;
I installed Python, latest version 3.4. and I am trying to open Python IDLE(GUI) mode, so when I open I
My problem was that the .py file wasn't on my local machine. It was on a shared directory.
After moving the file to my local machine, I quit getting the error.
Just to note my particular issue, this happens for me when my RAM gets full and my CPU gets busy.
The problem is because of a network socket timeout on the IPC pipes between the RPC subprocess.
It's a poor design (insecure and prone to failure) that's commonly used for IPC instead of process pipes.
The fix is to clear out some RAM and CPU usage and wait a minute before trying again.
And for developers, the fix is to stop using sockets for IPC and use proper process pipes.
Yes, it's the same exact socket timeout issue you experience with your browser, though on modern browsers, the page just stops loading instead of displaying a timeout error screen.
(note this assumes the case of a good WAN connection, with a local timeout)
The fix is here .
Open task manager, kill pythonw.exe, if you find 2 instances or more. Should work after that.
Delete all newely created .py files in the directory with Python. for example random.py, end.py - that was my problem that caused the same notification window. Reason in filename conflicts.
I fixed it, I needed to run IDLE with admin privileges. (I am using Windows 7 x64). Hope this helps.
Go to C:/Users/[your user]/AppData/Local/Programs/Python/Python35-32 and delete or rename every *.py file in this directory which is named after a certain method, function, module or library. Then run IDLE. Should work.
Hope I could help