| Re: array operations [message #55807] |
Wed, 12 September 2007 08:11 |
Conor
Messages: 138
Registered: February 2007
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Senior Member |
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The other way to calculate the mean is by using rebin. When rebin
shrinks an array, it takes a mean of values. So rather than using:
the general way of doing it is;
mean_image = rebin(myarray,ncols,nrows)
or in your case:
mean_image = rebin(myarray,10,10)
rebin shrinks the array down to new dimensions of size 10x10. It
averages along the third dimension when making the new array.
On Sep 12, 6:59 am, Steve <f...@k.e> wrote:
> Since this sounds something like some of the processing I do, have you
> considered a different "average" - the median, idl's median function
> does have a dimension parameter, so:-
>
> medpixel=median(myarray,dimension=3,/even)
>
> Steve
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| Re: array operations [message #55813 is a reply to message #55807] |
Wed, 12 September 2007 03:59  |
Steve[5]
Messages: 10
Registered: September 2007
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Junior Member |
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Since this sounds something like some of the processing I do, have you
considered a different "average" - the median, idl's median function
does have a dimension parameter, so:-
medpixel=median(myarray,dimension=3,/even)
Steve
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| Re: array operations [message #55814 is a reply to message #55813] |
Wed, 12 September 2007 03:18 |
Craig Markwardt
Messages: 1869
Registered: November 1996
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Senior Member |
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payon@gmx.de writes:
> Hello,
> I have a small question to an array operation.
>
> I have a 3-dimensional array. 10x10x100 ... the first 2 dimensions
> 10x10 are spatial dimensions (so one image with 10 width with and 10
> pix height). The third dimension is the time dimension. So every
> images was acquired 100 times.
>
> What I wanted to do is now to compute the the mean of every pixel in
> time dimension.
>
> So for one special pixel (e.g. [3,3]) i would write
>
> meanpixel = mean(myarray[3,3,*])
>
> but how is it, if i would like to do this operation for every pixel in
> the spatial dimensions? I just saw a possibility with a for loop.
>
If you are just computing the mean value, then using TOTAL is the
fastest. You can sum over any axis you wish, and then divide by the
number of pixels in the sum.
For example,
mean_image = total(myarray,3) / 100
would give the mean image. The "3" tells TOTAL to sum over the third
dimension.
Good luck,
Craig
--
------------------------------------------------------------ --------------
Craig B. Markwardt, Ph.D. EMAIL: craigmnet@REMOVEcow.physics.wisc.edu
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
------------------------------------------------------------ --------------
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| Re: array operations [message #55815 is a reply to message #55814] |
Wed, 12 September 2007 02:32 |
Spon
Messages: 178
Registered: September 2007
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Senior Member |
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On Sep 12, 9:13 am, pa...@gmx.de wrote:
> Hello,
> I have a small question to an array operation.
>
> I have a 3-dimensional array. 10x10x100 ... the first 2 dimensions
> 10x10 are spatial dimensions (so one image with 10 width with and 10
> pix height). The third dimension is the time dimension. So every
> images was acquired 100 times.
>
> What I wanted to do is now to compute the the mean of every pixel in
> time dimension.
>
> So for one special pixel (e.g. [3,3]) i would write
>
> meanpixel = mean(myarray[3,3,*])
>
> but how is it, if i would like to do this operation for every pixel in
> the spatial dimensions? I just saw a possibility with a for loop.
>
> like
> FOR j = 0, 9 DO BEGIN
>
> FOR k = 0, 9 DO BEGIN
>
> meanarray[i,j]=mean(myarray[i,j,*])
>
> ENDFOR
>
> ENDFOR
>
> This isn't a very fast possibility especially when the array goes big,
> and i have to do that operation for a full image frame which is
> 1024x1024 pixel.
>
> Is there maybe another way how to act with it? I know that IDL is very
> strong with array operations, so maybe there is any another solution,
> which maybe doesn't need the two loops. Thanks a lot for your kind
> responses.
>
> greetings
>
> martin
Hello Martin, hello list!
Apologies in advance for any glaring errors, this is my first post...
You may wish to make use of the fact that, while the 'MEAN' function
has no [DIMENSIONS] keyword, the 'TOTAL' function does. Why? I'm not
really sure.
e.g.:
s=size(myarray)
meandata=(TOTAL(myarray,3,/NAN) / s(3))
I'm not sure this is any faster, but I suspect so. Another thing you
might consider if you've got large datasets of floating points, is to
bytescale (BYTSCL) your data first; I guess it depends how important
your raw absolute data are to you, or if you're looking at relative
rather than absolute values and can afford to shrink down your dataset
before you begin juggling with it.
Also, not directly related, but worth a careful read or 16: J. D.
Smith's amazing dimension juggling tutorials:
http://www.dfanning.com/tips/rebin_magic.html
http://www.dfanning.com/tips/array_concatenation.html
plus, of course, anything and everything written by the brilliant Dave
Fanning without whom I'd still be trapped in a FOR loop myself :-)
Hope this helps,
Chris
PS I tried playing around with the PRESERVE_TYPE keyword, but just
like the helpfile predicted I got wild overflows and complete
gibberish in the output (my raw data is currently single-precision
float).
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