| Re: Multi-dimensions without for loop ? [message #38739] |
Fri, 26 March 2004 07:01  |
Emmanuel Christophe
Messages: 11
Registered: March 2004
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Junior Member |
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ok, I guess my example using 'mean' was not well choosen. I understand
how to make it using 'total' now and it's working because this is a sum.
But in this case:
Let define a simple array (dimension 2x2 to explain, but i'd like to
make it work on bigger array):
|a00 a10|
|a01 a11|
Is it possible to get in one step without using 'for' loops
dimension (1x2)
| a00.a10 |
| a01.a11 |
or
| max([a00,a10]) |
| max([a01,a11]) |
Thanks,
Emmanuel
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| Re: Multi-dimensions without for loop ? [message #38762 is a reply to message #38739] |
Thu, 25 March 2004 06:22  |
Andrea Pitacco
Messages: 3
Registered: March 2004
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Junior Member |
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On Thu, 25 Mar 2004 11:47:31 +0100, Emmanuel Christophe
<melaneum555@yahoo.fr> wrote:
> Hi,
>
> I'm trying to optimize some IDL code, removing for loops.
>
> I'm using the mean function to get the average of each line (this is for
> the example, could be another function). From a 2 dimensional array, I
> want to remove the mean of each column.
>
> Here is a sample using a loop:
> --------------------------------
> for j=0,size-1 do begin
> vect=data[j,*]
> datac[j,*]=vect-mean(vect)
> endfor
> --------------------------------
>
> How to do it in one instruction: if i'm using something like
> 'mean(data)', i'll get the average for the whole array, and not line by
> line.
Hi Emmanuel,
I have not understood clearly if you like to have the column mean
being removed from the data column, or the row (line?) mean removed
from the data row. 'Seems more likely the first. Anyway:
In the first case you may want to try:
datac = data - Total(data,2) / (Size(data))[2] #
Replicate(1D,(Size(data))[2])
In the second:
datac = data - Total(data,1) / (Size(data))[1] ##
Replicate(1D,(Size(data))[1])
Regards, Andrea
Andrea Pitacco
University of Padova, Italy
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| Re: Multi-dimensions without for loop ? [message #38763 is a reply to message #38762] |
Thu, 25 March 2004 06:09 |
btt
Messages: 345
Registered: December 2000
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Senior Member |
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Emmanuel Christophe wrote:
> Hi,
>
> I'm trying to optimize some IDL code, removing for loops.
>
> I'm using the mean function to get the average of each line (this is for
> the example, could be another function). From a 2 dimensional array, I
> want to remove the mean of each column.
>
> Here is a sample using a loop:
> --------------------------------
> for j=0,size-1 do begin
> vect=data[j,*]
> datac[j,*]=vect-mean(vect)
> endfor
> --------------------------------
>
> How to do it in one instruction: if i'm using something like
> 'mean(data)', i'll get the average for the whole array, and not line by
> line.
>
> the instruction 'total' give me something similar to what I want:
> 'total(data,1)' will make the sum in only one direction.
>
> How to get that with ordinary function ? and is it possible ? I'm
> thinking of something like 'data[0:size-1,*]'...
>
Hello,
You can do it without a loop, but unfortunately not with the built-in
statistics routines.
IDL> data = Findgen(4,6)
IDL> colTotal = TOTAL(data,2)
IDL> dim = Size(data, /dim)
IDL> colMean = Rebin(colTotal/dim[1], dim)
IDL> newData = data-colMean
But you could write your own routine to add the neat dimension handling
to MEAN, MOMENT, etc that you already get with TOTAL, MAX, etc.
Ben
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| Re: Multi-dimensions without for loop ? [message #38817 is a reply to message #38739] |
Fri, 26 March 2004 15:36 |
JD Smith
Messages: 850
Registered: December 1999
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Senior Member |
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On Fri, 26 Mar 2004 16:01:07 +0100, Emmanuel Christophe wrote:
> ok, I guess my example using 'mean' was not well choosen. I understand
> how to make it using 'total' now and it's working because this is a sum.
>
> But in this case:
>
> Let define a simple array (dimension 2x2 to explain, but i'd like to
> make it work on bigger array):
> |a00 a10|
> |a01 a11|
>
> Is it possible to get in one step without using 'for' loops
>
> dimension (1x2)
>
> | a00.a10 |
> | a01.a11 |
p=product(a,1)
> or
>
> | max([a00,a10]) |
> | max([a01,a11]) |
>
m=max(a,DIMENSION=1)
A pleasingly IDL-esque lack of symmetry there. Probably requires
IDL>5.6.
JD
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