split a vecor into several parts (of different sizes) to later fill a matrix.
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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
3 个评论
Davide Ferraro
2012-8-30
I can't understand how your matrix should be structured. Can you please clarify?
Sean de Wolski
2012-8-30
My guess is the loop is the fastest way to do this.
Tom
2012-8-30
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
回答（4 个）
Andrei Bobrov
2012-8-30
编辑：Andrei Bobrov
2012-8-30
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{:}];
2 个评论
David T_
2012-8-30
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
2012-8-30
Hello David, thank you for reply, I corrected.
Jos (10584)
2012-8-30
编辑：Andrei Bobrov
2012-8-30
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
Sean de Wolski
2012-8-30
编辑：Sean de Wolski
2012-8-30
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.
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2012-8-30
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