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wjy-yy avatar wjy-yy commented on August 17, 2024

I guess the expression might be:

results_df['median_rmse'] = [compute_median_segment_error(row) for row in results]

Is it right?

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mdr223 avatar mdr223 commented on August 17, 2024

Hi there!

If my recollection is correct, this dataframe has one row per-trial, and for each trial we store the RMSE for each of the N segments (N=5) in its own column. We want to compute the median of these N=5 RMSE's for the trial. Thus, we execute this lambda with axis=1, so it's input is an entire row of the dataframe (I.e. one trial) and its output is the median of the per-segment RMSEs for that trial. Please let me know if this isn't clear, as I'd be happy to provide more pointers into the code once I'm at my laptop!

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wjy-yy avatar wjy-yy commented on August 17, 2024

Thanks for your prompt reply. I think it is clear for me. I will recheck it. If the expression could be rewritten, would you mind my PR for that part?

I will continue to understand your great code. Looking for further discussion with you.

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mdr223 avatar mdr223 commented on August 17, 2024

My apologies for the delayed response! I'd be happy to consider a PR / revision as long as it logically computes the same outcome.

I might also suggest / encourage creating a fork of this repository if you'd like to do some more custom work of your own, as this repository is mostly meant to remain in a static state for future readers of our paper to be able to reproduce our results.

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