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From: Doug Schwarz on 7 Apr 2010 18:14
In article <hpip5f$iij$1(a)fred.mathworks.com>,
"Matt Fig" <spamanon(a)yahoo.com> wrote:

> But Doug, by the example given:
>
>
> M = magic(2)*pi;
> V = [sqrt(2);sqrt(3)];
>
> % Juliette's technique:
> P(:,1)=M(:,1).*V;
> P(:,2)=M(:,2).*V
>
> % BSXFUN
> P2 = bsxfun(@times,M,V)
>
>
> >>isequal(P,P2)
> ans =
> 1

You're right, of course -- I must be getting old. :-(

--
Doug Schwarz
dmschwarz&ieee,org
Make obvious changes to get real email address.
From: Juliette Salexa on 7 Apr 2010 18:16
Thanks Matt and Steve,
I always look up command documentation on the MW website rather than typing
the help command , because I find that it's incredibly faster. When I type:
help bsxfun

the command window says "busy" for about 25 seconds before the help page comes up.
And the website has more functionality.

but in the future I know where to look if I'm wondering the etymology of a command!
From: Matt J on 7 Apr 2010 19:23
"Juliette Salexa" <juliette.physicist(a)gmail.com> wrote in message <hpikta$bjd$1(a)fred.mathworks.com>...

>
> I am currently doing:
>
> P(:,1)=M(:,1).*V
> P(:,2)=M(:,2).*V
>
>
> The only way I can think of generalizing this to larger dimensions would be using loops.
===============

Note, that for-loops have been found to be more optimal than bsxfun when one the dimensions of M is very small. For example, if M is 10000x10 and
V is 10000x1 it would probably be better to loop over the 10 columns of M as you've proposed. Similarly, if there are a small number of rows, you would loop over those.

Here is a function which I use in situations like these (and which I was forced to use before bsxfun came along). It analyzes the optimum looping strategy based on the dimensions of M:

function M=matmvec(M,v)
%"Matrix times vector". Multiplies a vector into columns or rows of matrix.
%
% M_out=mattvec(M,v)
%
%The input M is a matrix.
%
%If v is a row vector, it will be multiplied into every row of M.
%
%If v is a column vector it will be multiplied into every column of M.
%
%It might be faster to do M*diag(v) or similar, but this route
%conserves more memory.


[Mrows,Mcols]=size(M);
vlen=length(v);
mindim=min([Mrows,Mcols]);


if size(v,2)==1 %v is a column vector


if vlen~=Mrows; error 'Sizes not compatible'; end
if Mcols==mindim, %loop over columns

for ii=1:Mcols
M(:,ii)=M(:,ii).*v;
end

else %loop over rows

for ii=1:vlen
M(ii,:)=M(ii,:).*v(ii);
end

end


else %v is a row vector

if vlen~=Mcols; error 'Sizes not compatible'; end

if Mcols==mindim %loop over columns

for ii=1:Mcols
M(:,ii)=M(:,ii).*v(ii);
end

else %loop over rows

for ii=1:Mrows
M(ii,:)=M(ii,:).*v;
end

end

end
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