Finding different values of a matrix in a single loop
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Bruno Baptista
el 1 de Dic. de 2020
Comentada: Ameer Hamza
el 1 de Dic. de 2020
I have this matrix
4 5
2 8
3 4
5 6
1 3
2 4
2 7
i need to find a number in the matrix wich has to be a variable. For example if i need the number 3 then matlab has to get the first row that contains 3 ( in this case 3 4) then it has to find the first row with the number of the other collumn and do this for the rest of the matrix and has to works for any matrix (n,2)
Starting with the number 3 i should get this :
3 4
2 4
2 7
and then put it on a vector : 3 4 2 7
Was only able to get this till now
cid= [4 5;
2 8;
3 4;
5 6;
1 3;
2 4;
2 7]
z=zeros(1,3);
v=zeros(1,3);
x=3
y=0;
for i=1:linhas
for j=1:colunas
if cid( i,j) == x
v(1,i) = cid(i,2)
z(1,i) = cid(i,1)
z(1:x) = 0;
x = v(i) ;
end
end
end
6 comentarios
Respuesta aceptada
Ameer Hamza
el 1 de Dic. de 2020
Editada: Ameer Hamza
el 1 de Dic. de 2020
Edit: This is modified to follows the rule as you described in your comments.
A = [
4 5
2 8
3 4
5 6
1 3
2 4
2 7];
a = 3;
M = [];
m = a;
count = 1;
col = 1;
while true
[idx, ~] = find(A==a, 1);
if isempty(idx)
break;
end
M(count, :) = A(idx, :);
count = count + 1;
a = setdiff(M(end, :), a);
m(count) = a;
A(1:idx, :) = [];
end
Result
>> M
M =
3 4
2 4
2 7
>> m
m =
3 4 2 7
9 comentarios
Más respuestas (1)
KSSV
el 1 de Dic. de 2020
Read about ismember.
A = [ 4 5
2 8
3 4
5 6
1 3
2 4
2 7] ;
xi = 2 ;
[c,ia] = ismember(A(:,1),xi) ;
iwant = A(c,:)
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