Should I be using arrayfun in Matlab like this?

Question

I started learning a little Matlab a couple days ago.

I wanted to plot a Dirac-comb-like approximation, so given that I knew some functional programming, and that I have been told "you shouldn't need for loops in Matlab", I ended up with this:

M = 50
dx = 0.1
r = 20
x = -r/dx:r/dx
y = arrayfun(@(k) dx .* sum(exp(-2j * pi * dx * k * (-M:dx:M))), x)

But I feel arrayfun isn't a good way to do this -- it just feels awkward/overkill in Matlab.

Or maybe it's just me, I don't know.
Is there a better way to plotting this graph without resorting to arrayfun, or is this the best way?

Rody Oldenhuis · Accepted Answer · 2012-09-20 06:26:31Z

2

First, use semicolons behind commands to suppress output, it really makes a difference in performance:

M  = 50;
dx = 0.1;
r  = 20;
x  = -r/dx : r/dx;

Then, dot-operators (.*, ./, etc.) are for element-wise operations. The multiplication you do inside the arrayfun (dx .* sum(exp(...))) is a scalar times a vector. In this case element-wise and normal multiplications are the same. It is a good habit to keep normal multiplications for scalar*vector; it helps prevent bugs.

Then, the arrayfun is unneccesary. You can accomplish the same like so:

y = dx * sum( exp(-2j*pi*dx * (-M:dx:M).'*x) );

The product -2j*pi*dx is a product between all scalars. The product (-M:dx:M).'*x however is a product between matrices. Since sum sums down the columns (dimension 1) by default, the results are the same. This solution has larger memory overhead, but is a lot faster than arrayfun.

Note that I've used .' for transpose. In Matlab, the notation A' means conjugate transpose, and A.' means normal transpose. Especially in the context of complex math like you have, this is very important. Learn the difference, and remember it well.

edited Sep 20, 2012 at 6:26

answered Sep 20, 2012 at 4:47

Rody Oldenhuis

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3 Comments

user541686 Over a year ago

+1 this is an awesome answer/advice, thank you! Just to clarify, when you say exp((-M:dx:M).' * x) -- is that supposed to be a matrix exponential, or is it doing something else? I've never used matrix exponentials for this!

Rody Oldenhuis Over a year ago

@Mehrdad: No, expm() is the matrix exponential. exp( [some matrix] ) is just the element-wise exponential.

user541686 Over a year ago

Ahhhh that makes more sense, I was wondering how to do that! Thank you! :)

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