Converting location of a 2x3 vector into a matrix with value 1

Suppose I have

a = 2×3

1 3

2 4

7 8

Now I want to create a matrix of dimension 10 x 10 where the entries 1,3 and 2,4 and 7,8 are equal to one.

Z = zeros(10) % 10 x 10 matrix containing only zeros

Z(a(1,1),a(1,2))=1 % now entry 1,3 is equal to 1

This is an illustrative example and I could for sure just code the second line three times. However for a large matrix a this will be tedious. I have tried to solve this problem with some for loops but without any positve result.

[Arif Hoq]

a = [1     3;2     4;7     8];
Z = zeros(10) ;% 10 x 10 matrix containing only zeros
Z(a(1,1),a(1,1))=1; % now entry 1,3 is equal to 1
Z(a(1,1),a(1,2))=1;
Z(a(2,1),a(2,1))=1;
Z(a(2,2),a(2,2))=1;
Z(a(3,1),a(3,1))=1;
Z(a(3,2),a(3,2))=1

Z = 10×10

1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

is that your expected output ?

[Wietze Zijpp]

Thank you for your response. It is more than the output I expect. I would like to only obtain the output

Z(a(1,1),a(1,2))=1; % so entry 1.3 =1

Z(a(2,1),a(2,2))=1; % entry 2,4 = 1

Z(a(3,1),a(3,2))=1; % entry 7,8 = 1

[Arif Hoq]

or this one ?

a = [1     3;2     4;7     8];
Z = zeros(10) ;% 10 x 10 matrix containing only zeros
Z(a(1,1),a(1,2))=1; % now entry 1,3 is equal to 1
Z(a(2,1),a(2,2))=1;
Z(a(3,1),a(3,2))=1

Z = 10×10

0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Answer 1

[Voss]

a = [1     3     
     2     4     
     7     8];
Z = zeros(10,10);
Z(sub2ind(size(Z),a(:,1),a(:,2))) = 1

Z = 10×10

0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Answer 2

[Arif Hoq]

if 2nd one is your expecter output , then

a = [1     3;2     4;7     8];
Z = zeros(10) ;% 10 x 10 matrix containing only zeros
row=size(a,1);
j=1:2;
for i=1:row
        Z(a(i,j(1)),a(i,j(2)))=1;
end

Answer 3

[Stephen23]

"However for a large matrix a this will be tedious."

If you have a large matrix it may be better if it were a sparse array (which can make operations using it more efficient), in which case this task is very easy:

a = [1,3;2,4;7,8];
m = sparse(a(:,1),a(:,2),1,10,10)

m = 
   (1,3)        1
   (2,4)        1
   (7,8)        1

full(m) % checking

ans = 10×10

0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Answer 4

[Bruno Luong]

a = [1,3;2,4;7,8]; % assumed there is no repeated indexes
A = accumarray(a,1,[10,10])

A = 10×10

0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0