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Re: Apply function to vector
From: |
Mike Miller |
Subject: |
Re: Apply function to vector |
Date: |
Tue, 21 Jun 2011 02:15:53 -0500 (CDT) |
User-agent: |
Alpine 2.00 (DEB 1167 2008-08-23) |
On Mon, 20 Jun 2011, Jordi GutiƩrrez Hermoso wrote:
On 20 June 2011 04:59, Fritz Fischer <address@hidden> wrote:
For convenient reasons to plot functions I would like to apply a function to
a vector:
e.g.
x = 0:1:2*pi
y = sin(x)
plot(x,y)
So far, everything works fine.
But if I want to apply one of my functions the same way,
e.g.
x = 0:1:40
y = rayleigh_optical_density(x) %contains just some simple calculations
I get the following message:
error: for A^b, A must be square
I read about element-wise operations, but I guess this is of no use here,
isn't it?
It depends how you wrote rayleigh_optical_density. Did you vectorise
it? What functions does it call? You need to either rewrite it to use
elementwise calculations or to call arrayfun on it:
arrayfun(@rayleigh_optical_density, x)
For example, if you have scalars w, x, y and z, You could do this:
(w*x/y)^z
And if all four variables became vectors of, say, length four, you might
want to do something like this:
output=zeros(4,1); # initialize output vector
for i=1:4, output(i) = (w(i)*x(i)/y(i))^z(i) ; end
But that is slow and awkward and unnecessary. This does the same thing
(assuming these are column vectors):
output = (w.*x./y).^z ;
If they were column vectors, the output would be a column vector. The
"for" loop above will produce a column vector as output no matter which
kinds of vectors were used as input.
Another plus: When all four vectors are scalars, (w*x/y)^z and
(w.*x./y).^z provide the same output. Therefore, when writing code,
consider the possibility that you might use vectors or matrices instead of
scalars and write the code accordingly.
Mike
--
Michael B. Miller, Ph.D.
Minnesota Center for Twin and Family Research
Department of Psychology
University of Minnesota