Just a comment that Python supports sets which make this more efficient and easier to read: a = set(list1)b = set(list2)intersection = a & bunion = a | b len(intersection) / float(union) On Tue, 2019-05-21 at 15:15 +0200, Bakary N'tji Diallo wrote: > Hello there, > > Hope you are doing very well. > > > > I would like to compute tanimoto similarity using obspectrophore > fingerprint. I have found a formula on the net for the tanimoto > similarity: > def tanimoto(list1, list2): > #list1 and list2 #spectrophore fingerprint > intersection = [common_item for common_item in list1 if common_item > in list2] > return float(len(intersection))/(len(list1) + len(list2) - > len(intersection)) > > > For now, I tested this using the same compound as input and the > result is 1. > Just wanted to confirm this is the right formula for Tanimoto > coefficient for spectrophore. > > Thanks > > Best regards > -- > Bakary N’tji DIALLO > PhD Student (Bioinformatics), Research Unit in Bioinformatics (RUBi) > Mail: diallobaka...@gmail.com | Skype: diallobakary4 > Tel: +27798233845 | +223 74 56 57 22 | +223 97 39 77 14 > > > > > _______________________________________________OpenBabel-discuss > mailing listopenbabel-disc...@lists.sourceforge.net > https://lists.sourceforge.net/lists/listinfo/openbabel-discuss
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