| Re: averaging over same index [message #45292 is a reply to message #45291] |
Tue, 23 August 2005 09:45  |
btt
Messages: 345
Registered: December 2000
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Senior Member |
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andi.walther@web.de wrote:
> Hello,
>
> I have an array V(alues) and an array S(ubscripts of a target array)
> and I want to
> extract the mean of all values with the same index in order to put them
> into the new array.
>
> As a simple example:
> V = [ 3, 7, 99, 5, 2 , 10]
> S = [ 1, 3, 3, 2 , 0 ,1]
>
> new vector should be --> new = [ 2 , 6.5 , 5, 53 ]
>
> slow way would be: for n = 0 , max(S)-1 do new[n]=mean(v[where(S eq
> n)])
>
> a bit faster without WHERE in the loop:
>
> VSorted = v[sort(S)]
> SSorted = S[sort(S)]
> uu = uniq(iSorted)
> for n = 0 , n_elements(uu)-1 do new[iSorted[uu[n]]] =
> mean(vSorted[n:(n-1)>0])
>
> But is there a way to do what I want without resorting to
> a loop? In my real-world problem I have vectors with 500000 elements
> and the number of the occurence of indices
> are quite irragular and can exceed 30 times.
>
> Thanks Andi
>
Hello,
You might take a look at histogram. Something like the following quick
knock-off might work.
PRO testmean
V = [ 3, 7, 99, 5, 2 , 10]
S = [ 1, 3, 3, 2 , 0 ,1]
h = HISTOGRAM(s, rev = r, loc = loc)
nH = n_elements(h)
newV = FltArr(nH)
for i = 0, nH-1 Do begin
if h[i] GT 0 then $
newV[i] = MEAN(V[r[r[i] : r[i+1]-1]])
endfor
print, newV
end
IDL> testmean
% Compiled module: TESTMEAN.
% Compiled module: MEAN.
% Compiled module: MOMENT.
2.00000 6.50000 5.00000 53.0000
Cheers,
Ben
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