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

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Wietze Zijpp
Wietze Zijpp 2022-3-22
编辑： Bruno Luong 2022-3-23
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.
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Wietze Zijpp
Wietze Zijpp 2022-3-23
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
Arif Hoq 2022-3-23
Ran in:
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

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采纳的回答

Voss
Voss 2022-3-23
编辑：Voss 2022-3-23
Ran in:
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

更多回答（3 个）

Arif Hoq
Arif Hoq 2022-3-23
编辑：Arif Hoq 2022-3-23
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
Stephen23
Stephen23 2022-3-23
编辑：Stephen23 2022-3-23
Ran in:
"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
Bruno Luong
Bruno Luong 2022-3-23
编辑：Bruno Luong 2022-3-23
Ran in:
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

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