| Re: Technique to find maximum in 100x100 element moving box [message #93758 is a reply to message #93753] |
Thu, 13 October 2016 03:36   |
Markus Schmassmann
Messages: 129
Registered: April 2016
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Senior Member |
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On 10/12/2016 11:26 PM, Samantha Tushaus wrote:
> For reference, this is my current code in which I essentially use a truncated edge method.
>
> FOR i = 0, nx-1 DO BEGIN
> FOR j = 0, ny-1 DO BEGIN
> IF (i-50) LT 0 THEN low_ind_i = 0 ELSE low_ind_i = i-50
> IF (j-50) LT 0 THEN low_ind_j = 0 ELSE low_ind_j = j-50
> IF (i+50) GT (nx-1) THEN hi_ind_i = nx-1 ELSE hi_ind_i = i+50
> IF (j+50) GT (ny-1) THEN hi_ind_j = ny-1 ELSE hi_ind_j = j+50
> data_max[i,j] = max(data[low_ind_i:hi_ind_i,low_ind_j:hi_ind_j])
> ENDFOR
> ENDFOR
still looping, but faster without the ifs:
data_max=make_array(size(data,/dim),type=size(data,/type))
FOR i = 0, nx-1 DO FOR j = 0, ny-1 DO data_max[i,j] = $
max(data[(i-50)>0:(i+50)<nx-1,(j-50)>0:(j+50)<ny-1])
There might be a way with histogram and it's reverse index (possibly
using ORD beforehand if data is float), but depending on the data
(mostly how many different values end up being in data_max and how many
values larger than it's smallest one exist) it might not be faster.
You can also try looping twice over -50 to 50, but i doubt that would be
faster.
Good luck, Markus
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