Generate specific matrix from an array

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Pierre
Pierre el 1 de Oct. de 2013
Comentada: Laurent el 1 de Oct. de 2013
I have an array A=[-2 -1 0 1 3 5] which has 6 values and maximum value is 5 and minimum value is -2. I wanna generate a 6 by 10(=5+abs(-2)+3) matrix B. B=[1 1 1 0 0 0 0 0 0 0; 0 1 1 1 0 0 0 0 0 0; 0 0 1 1 1 0 0 0 0 0; 0 0 0 1 1 1 0 0 0 0; 0 0 0 0 0 1 1 1 0 0; 0 0 0 0 0 0 0 1 1 1 ]. We can assume the virtual number of column in matrix B is [-3 -2 -1 0 1 2 3 4 5 6]. Each row data in matrix B is from each number in array A. For example, -2 in array A means we should set (-2+1=-1), (-2) and (-2-1=-3) for 1 on the first row in matrix B, so the value should be [1 1 1 0 0 0 0 0 0 0]. -1 in array A means we should set (-1+1=0), (-1) and (-1-1=-2) for 1 on the second row in matrix B, so the value should be [0 1 1 1 0 0 0 0 0 0]. In this example, the size of array A is 1 by 6, Actuallly, it could be 1 by n. Please tell the faster way to generate matrix B. Thanks

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Pierre
Pierre el 1 de Oct. de 2013
Sorry, I posted MATLAB code again. Is there any person to tell me whether there is other faster mwthod to generate matrix B? Thanks.
A=[-2 -1 0 1 3 5 ] ;
[size_r, size_c]=size(A);
max_value=max(A);
min_value=abs(min(A));
B=zeros(size_c, max_value+min_value+3);
row_num=[1:1:size_c,1:1:size_c,1:1:size_c];
col_num=[A+min_value+1,A+min_value+2,A+min_value+3];
idx=sub2ind(size(B),row_num,col_num);
B(idx)=1;
  1 comentario
Laurent
Laurent el 1 de Oct. de 2013
Did you measure the time? On my machine my code is about 5 to 6 times faster than your code. How big will your n be in the end?

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Más respuestas (2)

Laurent
Laurent el 1 de Oct. de 2013
Editada: Laurent el 1 de Oct. de 2013
Below I show one way to do it. For big arrays there might be a faster way if you don't use for-loops, but for small n there is no need for this.
A=[-2 -1 0 1 3 5];
B=zeros(length(A),10);
tempA=A+3;
for ii=1:length(A)
B(ii,tempA(ii):tempA(ii)+2)=1;
end
If you want the size of B to depend on the values in A you could replace line 2 with:
B=zeros(length(A), max(A)+5);
  1 comentario
Pierre
Pierre el 1 de Oct. de 2013
if possible, is there any way to generate this matrix without for loop?

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Pierre
Pierre el 1 de Oct. de 2013
I found this way to generate it as following. Is there any person to tell me whether there is other faster mwthod to generate matrix B? Thanks.
A=[-2 -1 0 1 3 5 ] [size_r, size_c]=size(A); max_value=max(A); min_value=abs(min(A)) B=zeros(size_c, max_value+min_value+3); row_num=[1:1:size_c,1:1:size_c,1:1:size_c];
col_num=[A+min_value+1,A+min_value+2,A+min_value+3]; idx=sub2ind(size(B),row_num,col_num); B(idx)=1;
  1 comentario
Jan
Jan el 1 de Oct. de 2013
Please follow the "? Help" link on this page to learn how to format code in the forum. Posting not readable code is useless.

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