How can I convert a numeric matrix(a) to a 0 and 1 matrix(b)?

Question

Moe 2014-4-25

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评论： Moe 2014-4-25

采纳的回答： Geoff Hayes

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Hi everyone,

suppose I have a matrix:

a = [3;
     1;
     4;
     2]

Then I want it to be:

 b = [0 0 1 0;
0 0 0;
0 0 1;
1 0 0]

Explanation of first row in matrix "b" (how it's created, manual):

if a(1)=1
     b(1,1)=1
elseif b(1,1)=0
end
if a(1)=2
     b(1,2)=1
elseif b(1,2)=0
end
if a(1)=3
     b(1,3)=1
elseif b(1,3)=0
end
if a(1)=4
     b(1,4)=4
elseif b(1,4)=0
end

and so on for rows 2,3 and 4 in matrix a.

I'm looking for a automatic loop or function in the Matlab, because real size of matrix a is 1000 rows.

Please help me, Thanks.

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Answer 1

Geoff Hayes 2014-4-25

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Hi Mohammed,

Why not just use a simple loop?

r=size(a,1);     % get the number of rows in a (first dimension,1)
b=zeros(r,r);    % initialize the b matrix to all zeros
for i=1:r
   b(i,a(i))=1;  % set the ith row of b with the a(i) column set to one
end

Geoff

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Geoff Hayes 2014-4-25

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Hi Mohammed,

So this isn't much different from before - your b now has 8 rows (since a has 4 rows) and all rows of b are paired. The first row of a either populates row 1 or row 2 of b, the second row of a populates either row 3 or row 4 of b, etc. with the ith row of a populating either row 2i-1 or 2i of b:

for i=1:r
   if a(i,2)==1
        b(2*i-1,a(i,1))=1; 
   else 
        b(2*i,a(i,1))=1; 
   end
end

That should do it!

Geoff

Moe 2014-4-25

Thanks Geoff, It was great and worked very well.

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Answer 2

Jos (10584) 2014-4-25

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No need for for-loops. This shows the power of linear indexing:

    a = [3 1 4 2] % these specify column indices
    b = zeros(numel(a), max(a))
    idx = sub2ind(size(b), 1:numel(a), a(:).') % linear indices
    b(idx) = 1

As for your second question:

    a = [3 2; 1 1; 4 2; 2 1]
    b = zeros(2*size(a,1), max(a(:,1)))
    colix = a(:,1).'
    rowix = 2*(1:size(a,1)) - (a(:,2)==1).' 
    idx = sub2ind(size(b), rowix, colix) 
    b(idx) = 1