| Re: A new puzzle for histogram [message #50143 is a reply to message #50142] |
Fri, 15 September 2006 15:38   |
Jean H.
Messages: 472
Registered: July 2006
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Senior Member |
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Hi,
If I corectly understand your problem, you might want to look at the
rebin function:
a = indgen(4) ;could be your angle
b = transpose(indgen(4)) ;could be your range
print, rebin(a,4,4) * rebin(b,4,4)
>> 0 0 0 0
0 1 2 3
0 2 4 6
0 3 6 9
hope that helps...
Jean
gknoke wrote:
> So, I've got this piece of code which is horribly horribly inefficient.
> I know the solution lies in a clever application of the histogram
> function, but being Friday afternoon my brain isn't seeing it. Anyone
> else have any insight on how I might approach it?
>
> This particular routine is mapping a piece of data from polar to
> cartesian coordinates. Currently the code generates sin/cos angle
> tables and calculates the x,y coordinates in meters for each
> range/angle, and then converts that to an x,y coordinate in terms of
> pixels from the center. I realize the calculation of the x,y
> coordinates can be replaced with a simple vector operation, but I can't
> see how to turn the separate resulting arrays for x and y into a single
> array I can use the histogram function to match to the mapped grid.
>
> ;Setup the output grid
> map = fltarr(xysize, xysize)
> count = intarr(xysize, xysize)
>
> cos_table = cos(angles)
> sin_table = sin(angles)
>
> for j_theta = 0, n_elements(angles)-1 do begin
> for i_range = 0, n_range-1 do begin
> ;Calculate x and y in meters
> x = r_pts(i_range)*cos_table(j_theta)
> y = r_pts(i_range)*sin_table(j_theta)
>
> ;Find corresponding pixel on mapped grid
> ix = round((x-x0)/cellsize)+xysize/2
> jy = round((y-y0)/cellsize)+xysize/2
>
> ;If the pixel coord is inside the image put the data point there
> if(ix ge 0 and ix le xysize-1) then begin
> if(jy ge 0 and jy le xysize-1) then begin
> map(ix,jy)=map(ix,jy) + data(i_range,j_theta)
> count(ix,jy)=count(ix,jy)+1
> endif
> endif
> endfor
> endfor ;End of nearest neighbor loops
>
> Thanks,
>
> --Greg
>
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