a = [[4,2,0,5],[9,0,1,5],[0,4,2,1],[1,2,3,4]]
b = sort(a) ;sort the whole array
sizeX = 4 ;(use DIM instead)
sizeY = 4
nbElements = n_elements(a)
c = b/sizeX ;tells you, for each element of b, which line of A the
;index correspond to
useless = histogram(c, reverse_indices = ri)
;get the reverse indices of the histogram. It tells you, for each line
of a (print useless, you will get 4 4 4 4), the indices of b that
correspond to it.
;get the sorted indices, resorted by lines. You have, for sure, 4 bin
here, so the first indice of b is located at ri[5].
d = b[ri[sizeX+1:sizeX+nbElements]]
minus = indgen(sizeY) * sizeX
minusBIG = transpose(rebin(minus,sizeX,sizeY))
;you want the indices on each line and not on the whole array (if that's
what you want, d is fine for you then)
result = reform(d-minusBIG, sizeX,sizeY)
IDL> print, result
2 1 0 3
1 2 3 0
0 3 2 1
0 1 2 3
I suggest you to learn to use the histogram... it's wonderful what you
can do with it!
Jean
humphreymurray@[EMAIL PROTECTED]
wrote:
> Hi,
>
> In IDL, is there a way to independly sort the columns of a 2d matrix
> without looping through and sorting each row individually?
>
> Currenly I'm using:
>
> for x = long(0), x_size - 1 do begin
> sorted_indexs[0,x] = sort(matrix[*,x])
> endfor
>
> For example, if the matrix contained the following values:
>
> 4 2 0 5
> 9 0 1 5
> 0 4 2 1
> 1 2 3 4
>
> I want the result matrix to contain index's like:
>
> 2 1 0 3
> 1 2 3 0
> 0 3 2 1
> 0 1 2 3
>
> Here, each column is sorted as if it's an independant vector.
>
> Cheers.
>
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