| A more efficient way of multiplying this [message #91581] |
Sat, 01 August 2015 19:44  |
stefano.rgc
Messages: 2
Registered: August 2015
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Junior Member |
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Hello, I have learnt a lot in this site. Now I will ask my first question.
Imagine you have a vector [x,y,z] and an array [[A1,A2,A3,...,An],[B1,B2,B3,...,Bn],[C1,C2,C3,...,Cn]]. What I need is multiply x by the vector [A], y by vector [B] and z by vector [C]. Currently I do this with a "for loop" (just 3 loops in this example) but this is very inefficient when the dimensions get bigger.
Some of you know a way to do this without resorting to a "for loop"?
I will appreciate any reply. Thanks.
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| Re: A more efficient way of multiplying this [message #91582 is a reply to message #91581] |
Sun, 02 August 2015 06:14  |
Heinz Stege
Messages: 189
Registered: January 2003
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Senior Member |
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Hello,
I modified Nikola's Code a little (see below). First I excluded the
creation of the arrays a and b from the time measurement. Second I
added the Reform/Rebin method. You can find a lot of discussions in
other threads of this group and in the internet. (Thanks to JD Smith
and David Fanning.)
Here are my results:
IDL> .run multiply_arrays_example
% Compiled module: $MAIN$.
Slow Loop: 7.8099990
Fast Loop: 0.48500013
"# operator": 0.31200004
reform/rebin: 0.14099979
and here is the file multiply_arrays_example.pro:
a = randomu(seed,3)
b = RANDOMU(seed, 150, 3)
seconds0 = SYSTIME(1)
for k=0,9999 do begin ; "Slow Loop" (scaled by 10)
c1 = FLTARR(150, 3)
FOR i = 0, 2 DO FOR j = 0, 149 DO c1[j, i] = a[i]*b[j, i]
end
seconds1 = SYSTIME(1)
for k=0,99999 do begin ; "Fast Loop"
c2 = FLTARR(150, 3)
FOR i = 0, 2 DO c2[*, i] = a[i]*b[*, i]
end
seconds2= SYSTIME(1)
for k=0,99999 do begin ; "# operator"
aa = (FLTARR(150)+1)#a
c3 = aa*b
end
seconds3 = SYSTIME(1)
for k=0,99999 do begin ; "reform/rebin"
c4 = rebin(reform(a,1,3),150,3,/sample)*b
end
seconds4 = SYSTIME(1)
print, 'Slow Loop: ', (seconds1-seconds0)*10
if array_equal(c2,c1) then print, 'Fast Loop: ', seconds2-seconds1
if array_equal(c3,c1) then print, '"# operator":', seconds3-seconds2
if array_equal(c4,c1) then print, 'reform/rebin:', seconds4-seconds3
end ; end of file multiply_arrays_example.pro
Cheers, Heinz
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| Re: A more efficient way of multiplying this [message #91584 is a reply to message #91581] |
Sun, 02 August 2015 19:00 |
stefano.rgc
Messages: 2
Registered: August 2015
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Junior Member |
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Thanks for the quick replies.
@karo03de. What you write is basically the same I wrote: "Currently I do this with a "for loop" (just 3 loops in this example)"
@Nikola Vitas. Thank you for the "# operator" way. It was very instructive.
@Heinz Stege. As you did, I also had figured the way with rebin/reform but I was afraid of how big the "rebin(reform(a,1,3),150,3,/sample)" variable could have been; that because the actual dimensions were 7000 and 5000 instead of 3 and 150 (as you used). But now I realized the amount of memory is only 3 MB so it is such not a big deal. Also thanks for the /sample keyword; I didn't know it.
I will manage it with the rebin/reform way as @Heinz Stege suggested it. But I thought that there was a way to do this without the need of resorting to "copy" something many times, as it does rebin or the line "aa = FINDGEN(100)#(FLTARR(150)+1)" at @Nikola Vitas code.
Thanks again.
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