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Re: I would like to average the first n columns based on duplicate values of the n+1th column [message #93709 is a reply to message #93708] Tue, 04 October 2016 04:17 Go to previous messageGo to previous message
belkaraza is currently offline  belkaraza
Messages: 6
Registered: June 2016
Junior Member
Am Dienstag, 4. Oktober 2016 12:32:48 UTC+2 schrieb Markus Schmassmann:
> On 10/03/2016 11:05 PM, belkaraza@web.de wrote:
>> Can Someone help me solve this problem in IDL:
>> "I have a matrix with duplicate numbers in one of the columns. I
>> would
>> like to average the rows with duplicate numbers. For example, I have
>> duplicate values in a matrix A in column 3:
>> A =
>> 1 2 1
>> 4 4 2
>> 5 4 2
>> 4 5 2
>> 5 5 3
>> 10 3 3
>>
>>
>> B =
>> 1 2 1
>> 4.3333 4.3333 2.0000
>> 7.5000 4.0000 3.0000
>>
>> where each row is the average values of the duplicate rows of column 3.
>>
>> Can anyone help?"
>>
>> found here:
>> http://stackoverflow.com/questions/15270019/i-would-like-to- average-the-first-n-columns-based-on-duplicate-values-of-the -n1
>
> if isa(A,/integer) then begin
> h=histogram(A[2,*],reverse_indices=ri)
> idx=where(h ne 0,n)
> B=fltarr(3,n)
> for i=0,n-1 do begin
> if ri[idx[i]] eq ri[idx[i]+1]-1 then $
> B[0,i]=A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]] else $
> B[0,i]=mean(A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]],dim=2)
> endfor
> endif else
> values=A[2,uniq(A[2,*],sort(A[2,*]))]
> ; if A[2,*] is already sorted, A[2,uniq(A[2,*])] is sufficient there
> n=n_elements(values)
> B=fltarr(3,n)
> for i=0,n-1 do begin
> w=where(A[2,*] eq values[i],cnt)
> if w cnt 1 then B[0,i]=A[*,where(A[2,*] eq values[i])] else $
> B[0,i]=mean(A[*,where(A[2,*] eq values[i])],dim=2,/nan)
> endfor
> endelse
>
>
> hope that does it, Markus


Hey, thanks for the answer. The last if loop is bugged. if w cnt 1 then B[0,i]
Can't see how to fix that
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