Python: geometric mean a**1/n if a too large

Question

i have large list of values:

[23.22, 50.44 .... 32.53]

after i should get next value:

reduce(operator.mul, [Decimal(i) for i in list])

print value - 4.248649022193430909459625077E+583 this value is too large but i have next action:

value**1/len(list)

I can not get the value if the list is very large - Is it possible to get the geometric mean in that case?

`value**1` is `value`, so `value**1/len(list)` is just `value/len(list)`. That is probably not what you meant to write, but that is what you have written. — khelwood, Oct 05 '17 at 08:13
See this answer: https://stackoverflow.com/a/43099751/107660 — Duncan, Oct 05 '17 at 08:15

score 6 · Accepted Answer · answered Oct 05 '17 at 08:14

6

Instead of multiplication of the items, sum their logarithms:

geometric_mean = (a1 * ... * an) ** (1/n) =
               = exp((log(a1) + .. + log(an)) / n)

answered Oct 05 '17 at 08:14

Dmitry Bychenko

1 Answers1