split a vecor into several parts (of different sizes) to later fill a matrix.

Hi!

I have a vector V and I want to split it in several parts of different known sizes (in the example below, the sizes would be: [3,6,5,...]), to build a matrix like this one (example):

M = [ V(1)  V(4)  V(10) ....
      V(2)  V(5)  V(11) ....
      V(3)  V(6)  V(12)
       0    V(7)  V(13)
       0    V(8)  V(14)
       0    V(9)    0
       0     0      0   ]

so each column is padded with zeros until a certain number of rows is reached (fixed a priori)

Any idea of how to do that WITHOUT using a loop (for ... end)? i.e. doing it in just one step (like matricial operation) because this has to be done a lot of times and with a loop the required time would be too much.

Thanks in advance!

Merce

[Davide Ferraro]

I can't understand how your matrix should be structured. Can you please clarify?

[Sean de Wolski]

My guess is the loop is the fastest way to do this.

[Tom]

I agree with Sean, you already know the size M has to be so you can preallocate M as a matrix of zeros and then fill it in a column at a time in a loop

Answer 1

[Andrei Bobrov]

try this is code

V = 10:10:140; % eg
s =[3     6     5];
b =  cumsum(s) - s + 1;

i0 = bsxfun(@plus,b,(0:max(diff(b))-1)');
t = bsxfun(@lt,i0,[b(2:end) numel(V)+1]);
i0(~t) = 1;
out = V(i0).*t;

OR

V = 10:10:140;
s =[3     6     5];
a = cellfun(...
       @(x,y)[x(:);zeros(max(s)-y,1)],mat2cell(V(:)',1,s),num2cell(s(:)'),'un',0);
out = [a{:}];

[David T_]

Nice code, but you have an extra row at the bottom filled with zeros. Just a very small change needed:

                                   here
  V=(1:15)/pi; % e.g.                |
  b = [1 4 10 12];                   V
  i0 = bsxfun(@plus,b,(0:max(diff(b))-1)');
  t = bsxfun(@lt,i0,[b(2:end) numel(V)+1]);
  i0(~t) = 1;
  out = V(i0).*t;

[Andrei Bobrov]

Hello David, thank you for reply, I corrected.

Answer 2

[Jos (10584)]

Take a look at MAT2CELL and PADCAT

% data
V = 1:15 ;
sz = [3 5 6 1] ; % lengths
% engine
C = mat2cell(V(:),sz,1) ;
[M,tf] = padcat(C{:}) ;
M(~tf) = 0 ; % replace NaNs by 0, per request

PADCAT can be found here: http://www.mathworks.com/matlabcentral/fileexchange/22909

Answer 3

[Sean de Wolski]

I have a simple for-loop showing a 2.7x and a 9.5x speed up over Andrei's solutions for V,S being this size. In general, a for-loop will smoke conversions so cell arrays any time and with the JIT accelerator it can generally hold up to complicated vectorized methods. Consider Andrei's dataset:

function looptest

%Data: same data all
V = 10:10:140; % eg
s =[3     6     5];
times = zeros(1,3);

for ii = 1:100;
  %Andrei's first method
  tic
  b =  cumsum(s) - s + 1;
  i0 = bsxfun(@plus,b,(0:max(diff(b))-1)');
  t = bsxfun(@lt,i0,[b(2:end) numel(V)+1]);
  i0(~t) = 1;
  out = V(i0).*t;
  times(1) = times(1)+toc;

    %Andrei's second method
    tic;
    a = cellfun(...
        @(x,y)[x(:);zeros(max(s)-y,1)],mat2cell(V(:)',1,s),num2cell(s(:)'),'un',0);
    out2 = [a{:}];
    times(2) = times(2)+toc;

    %trivial for-loop
    tic
    ns = numel(s);
    out3 = zeros(max(s),ns);
    sel = cumsum([1 s(1:end-1)]); %start element
    eel = sel+s-1; %end element
    for jj = 1:ns
        out3(1:s(jj),jj) = V(sel(jj):eel(jj));
    end
    times(3) = times(3)+toc;
end
disp(times./times(3));

My System: 2.7907 9.5219 1.0000

Note if we blow V and S up a fair amount, Andrei's bsxfun solution does surpass the for-loop for at least a little while.

Answer 4

[Merce Casas Prat]

Thank you for all the comments!

I actually compared the for loop option with mat2cell+padcat and the loop is faster as some of you mentioned. But thanks anyway!