Creating few columns of the Vandermonde matrix without using vander

Hi,

I want to create the first n columns of the Vandermonde matrix (I need this matrix in order to fit a curve through given points with the lsqlin function). Of course the commands:

A = vander(x);
A = A(1:n);

would do the work, but since the vector x might have over million values and n could be very small, 10 for example, it would be very inefficient.

I already did it with a for loop:

A = ones(length(x), n);

for i = 2 : n
    A(:, i) = (x.^(i - 1));
end

but I think it can be done more elegant without a loop. Is there a function that does this? Or can someone give me an idea how to do it without a loop? I am thinking that repmat or reshape might be useful.

Answer 1

[Jan]

You calculate the Vandermonde matrix from right to left. Is this intented? Do you want to get the n leftmost columns (smallest values)?

The power operation consumes a lot of time.

function A = ShortVander(x, n)
x = x(:);  % Column vector
A = ones(length(x), n);
for i = 2:n
  A(:, i) = x .* A(:, i-1);  % [EDITED, Thanks Steven]
end

Look into the source code of VANDER to find a very equal approach...

[Aleksandar]

Ah yes, this should be even faster. Thanks!

[Steven Morrow]

The multiplication should be compnenet-wise, i.e. the line in the for loop should be A(:,i) = x.*A(:,i-1)

[Are Mjaavatten]

This is nice, but why not let the built-in vander function take the number of columns as an optional second argument?

Answer 2

[Aleksandar]

I just found a function that does this without any loops, just as I was looking for, and it is using repmat and cumprod:

http://www.mathworks.com/matlabcentral/fileexchange/20896-vandermonde-matrix/content/vanderm.m

[Jan]

Please compare the times with my posted FOR loop. Under Matlab 2009a and for x = rand(1, 1000) the CUMPROD method needs >50% more time. I've tried this version before I've posted the loop:

function A = ShortVander(x, n)
v = v(:);
A = v(:, ones(1, n));
A(:, 1) = 1;
A = cumprod(A);

But this is also about 50% slower than the loop. If you work with large arrays it has the additional problem, that it needs a lot of temporary memory.

Answer 3

[Steven Lord]

You could do this pretty easily if you know the powers to which you want to raise the data used to create the Vandermonde matrix.

data = 1:6;
powers = [5 3 1];
V = vander(data)

V = 6×6

1 1 1 1 1 1 32 16 8 4 2 1 243 81 27 9 3 1 1024 256 64 16 4 1 3125 625 125 25 5 1 7776 1296 216 36 6 1

V2 = reshape(data, [], 1).^powers

V2 = 6×3

1 1 1 32 8 2 243 27 3 1024 64 4 3125 125 5 7776 216 6

The columns of V2 are the first, third, and fifth columns of V corresponding to data^5, data^3, and data^1.