Chap vector to matrix with special structure.

I have a collumn vector consisting of n elemtents (size nx1):

P=[P1 ..... Pn]'

What I which to do, is transform this into a matrix with a certain number of rows. This is best explained by an example e.g. k=3 rows:

M=
P1 P2 ... Pn-2
P2 P3 ... Pn-1
P3 P4 ... Pn

Or k=4 rows
M=
P1 P2 ... Pn-3
P2 P3 ... Pn-2
P3 P4 ... Pn-1
P4 P5 ... Pn

In general
M=
P1 P2 ...  Pn-k+1
P2 P3
.  .
.  .
Pk Pk+1 ... Pn

I hope my point is clear and you can help me. I'm performing this operation such that I can perform a multiplication more efficiently.

Thank you guys in advance!

Answer 1

[Martijn]

My own answer:

dum=hankel(C)
M=dum(1:k,1:n-k+1)

[Martijn]

Elapsed time is 0.000184 seconds.

[Sean de Wolski]

What is the complete code for yours? i.e. n,k

[Martijn]

clear all; clc;

%test timing

P = [1 2 pi 5 exp(1)];

tic

k=3 %free choice variable

dum=hankel(P);

PP1 =dum(1:k,1:length(P)-k+1)

toc

tic

dum=hankel(P);

PP2 =dum(1:3,1:3)

toc

tic

idx = 1:ceil(numel(P)./2);

PP3 = P(bsxfun(@plus,idx',idx-1))

toc

PP1 =

1.0000 2.0000 3.1416

2.0000 3.1416 5.0000

3.1416 5.0000 2.7183

Elapsed time is 0.007529 seconds.

PP2 =

1.0000 2.0000 3.1416

2.0000 3.1416 5.0000

3.1416 5.0000 2.7183

Elapsed time is 0.000640 seconds.

PP3 =

1.0000 2.0000 3.1416

2.0000 3.1416 5.0000

3.1416 5.0000 2.7183

Elapsed time is 0.000752 seconds.

[Martijn]

Strangely enough it looks like PP is calculated slower when using indices k and n to specify which part to 'select' from the hankel matrix.

Answer 2

[Sean de Wolski]

P = [1 2 pi 5 exp(1)];
idx = 1:ceil(numel(P)./2);
PP = P(bsxfun(@plus,idx',idx-1))

[Martijn]

Elapsed time is 0.004266 seconds.

[Martijn]

How does your work for a vector P of length n and, number of row in PP of k? I'm looking for a generic solution.

[Martijn]

Done it:

tic

idx = 1:k

idx2= 1:length(P)-k+1

PP3 = P(bsxfun(@plus,idx',idx2-1))

toc

Elapsed time is 0.001055 seconds.

While my generic method is : Elapsed time is 0.007312 seconds.