| Re: Matrix Multiplication in IDL [message #35994] |
Mon, 04 August 2003 14:06 |
James Kuyper
Messages: 425
Registered: March 2000
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Senior Member |
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K Banerjee wrote:
>
> Folks,
>
> In IDL:
>
> VB> a = indgen(2, 4) - 2
> VB> print, a ; Actually prints transpose(a)
> -2 -1
> 0 1
> 2 3
> 4 5
Not quite. You've defined a as being a 2x4 matrix, and that's exactly
what it prints. Specifically, you'll find that the elements are printed
as
a(0,0) a(1,0)
a(0,1) a(1,1)
a(0,2) a(1,2)
a(0,3) a(1,3)
If you interpret that as transpose(a), then you haven't adapted your
thinking yet to IDL's way of handling arrays. To print transpose(a),
type:
print, transpose(a)
> Keeping the above points in mind, I am trying to understand the
> IDL command:
>
> transpose(G)##G
>
> (In my case, G is (296 x 4).)
The fundamental confusing thing is that IDL's array operators use the
opposite convention from IDL itself, as to the meaning of rows and
columns. Rather than getting hung up on the defintions of rows, columns,
and transposes, let's just investigate how it works.
a = INTARR(l,m)
b = INTARR(n,o)
c = a # b
d = b ## a
The matching rule is:
m eq n
Both c and d are (l,o) arrays. The contents of c[i,j] are the same as
the contents of d[i,j]. Both of them are
total( transpose( a[i,*])*b[*,j] ))
> In IDL, the above matrix product turns out to be (4 x 4). However,
> I was expecting the matrix product to be (296 x 296) since I
> interpret the above IDL command as carrying out the matrix
> multiplication:
To get a (296,269) array out of g, you must type either
h = g # TRANSPOSE(g)
or
h = TRANSPOSE(g) ## g
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