You mean quicker as in faster or as in easier? Using indexing is probably
easiest but not fastest:
julia> M = randn(5,5)
5x5 Array{Float64,2}:
-1.51134 1.2331 -0.186083 0.412282 -1.13114
-0.0411132 0.124684 0.377426 0.622427 -0.278162
0.788182 -0.834092 1.09506 1.45196 -0.620717
1.45856 -0.889277 -0.768049 -0.95953 -0.340098
1.69666 0.695566 0.706905 0.63089 0.958474
julia> mean(M[M .> 0])
0.8751552351448673
If you want the fastest, loop through the matrix and keep totals and counts:
julia> function meanpos(a)
t = zero(eltype(a))
c = 0
for x in a
x > 0 || continue
t += x
c += 1
end
return t/c
end
meanpos (generic function with 1 method)
julia> meanpos(M)
0.8751552351448673
This should be fairly efficient for any non-sparse collection.
On Tue, Jul 28, 2015 at 11:50 AM, Jude <[email protected]> wrote:
> Hi!
>
> Probably a quite basic question but was wondering if someone knows a quick
> way to get the average of theh positive elements of a matrix, eg, If I have
> an m*n matrix and want to get the average for each m so have m*1 but only
> want to work with non-zero elements and ignore the zero elements and
> negative elements
>
> I guess I could using indexing and pull out the indexes corresponding to
> the positive elements but just wondering if anyone knew quicker ways!
>
> Thanks!
>
