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norm_functional_contrib- Error or misunderstaning #206
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Hey @ragavishn, Yes that should be what that column corresponds to for each sample and function. Would you mind sharing the table where you are noticing this problem? If you don't mind sharing it, you can email it to me here: Thanks, Gavin |
Hi Gavin, Thank you so much for taking your time to look into this. Please find below the values I get when i sum the columns norm_func_contrib by function. I have emailed you my EC_metagenome_out/pred_metagenome_contrib.tsv file as well. Please let me know if I have missed something. Thank you. function norm_taxon_function_contrib |
Thanks for sending these files. That column is supposed to sum to one per sample / function, so that’s definitely concerning. Would you mind also sending the original input files and the exact command you used to generate the stratified table (e.g. through a google drive or dropbox link)? I’d just like to confirm that I can reproduce the same problem from the full output file and double-check that the output file was written correctly. Thanks, Gavin |
I will try to upload the original input file. Last column (norm_func_contrib) alone was not calculated past 100 rows. I am afraid its more of an pandas.groupby fn issue as it works for small datasets and your test data. |
My apologies @ragavishn, I just realized that I had totally forgotten about this... I thought it had been resolved, but it seems not be the case. I'm looking into it now and I believe this is a bug unfortunately, at least for larger tables. |
Thanks again - this is now fixed in the latest version (v2.5.0): https://github.com/picrust/picrust2/releases, which should soon be available through conda. |
I have run picrust2 pipeline for my smaples and I notice that the sum of 'norm_taxon_func_contrib' for a given function is not equal to 1. am I missing something here? Can you please clarify how this is being calculated? Our understanding is
'norm_taxon_func_contrib' = taxon_fn_abun / (sum of taxon_fn abun for the given fn)
Thank you.
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