Comments (4)
I guess the expression might be:
results_df['median_rmse'] = [compute_median_segment_error(row) for row in results]
Is it right?
from inquest.
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!
from inquest.
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.
from inquest.
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.
from inquest.
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from inquest.