Steven D'Aprano <steve+pyt...@pearwood.info> added the comment:
> In the spirit of "perfect is the enemy of good", would it be > reasonable to start with a simple, fast implementation using > exp-mean-log? Then if someone wants to make it more accurate later, > they can do so. I think that is a reasonable idea. On the basis that something is better than nothing, go ahead. We can discuss accuracy and speed issues later. Getting some tricky cases down for reference: # older (removed) implementation py> geometric_mean([7]*2) 7.0 py> geometric_mean([7]*15) 7.0 # Raymond's newer (faster) implementation py> exp(fmean(map(log, [7]*2))) 6.999999999999999 py> exp(fmean(map(log, [7]*15))) 6.999999999999999 py> geometric_mean([3,27]) 9.0 py> geometric_mean([3,27]*5) 9.0 py> exp(fmean(map(log, [3,27]))) 9.000000000000002 py> exp(fmean(map(log, [3,27]*5))) 8.999999999999998 py> x = 2.5e15 py> geometric_mean([x]*100) 2500000000000000.0 py> exp(fmean(map(log, [x]*100))) 2499999999999999.5 On the other hand, sometimes rounding errors work in our favour: py> geometric_mean([1e50, 1e-50]) # people might expect 1.0 0.9999999999999998 py> 1e-50 == 1/(1e50) # even though they aren't quite inverses False py> exp(fmean(map(log, [1e50, 1e-50]))) 1.0 ---------- _______________________________________ Python tracker <rep...@bugs.python.org> <https://bugs.python.org/issue27181> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
