lipd.readLipd.py needs improvement -- here's some simple things! #37
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chrismheiser
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Mar 12, 2018
Previously, to use readLipd with a file in your current working directory you would have to call readLipd(“./file.lpd”). Now, you can drop the prefix and call readLipd(“file.lpd”). This will NOT work for directories, as providing an empty parameter will trigger the GUI browse. You must use readLipd(“./”) to read the whole current directory for files.
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I fixed the issue with the "./" You don't need it to call a file in your current working directory anymore. We decided to keep the current use of the GUI as the fallback when the function is called with an empty "usr_path" parameter. If someone wants to open a file directly, then they can provide pass that path to the function call which circumvents the GUI. |
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@ericsteig writes:
The default for lipd.readLipdshould be to expect a filename, not open a GUI.
*The path should by default assume './'.
Details:
The default for lipd.readLipd is to open a GUI (which I would never want). Also, the GUI doesn't open on my machine. I don't care since I'll never use it!
I have a lipd file called GISP2.lpd.
If I'm already in the right directory, I should be able to read it with:
D = lipd.readLipd('GISP2.lpd')
but that doesn't work. I have to say:
D = lipd.readLipd('./GISP2.lpd')
which just is silly.
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