Finding different values of a matrix in a single loop

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
         

[Timo Dietz]

You wrote: "number of the other collumn"

Which other coulmn do you mean in case you have a (n,3) - matrix?

[KALYAN ACHARJYA]

"then it has to find the first row with the number of the other collumn"

What number, as the input is 3? Also how did you get it?

2     4
2     7

[dpb]

OK, I see [3 4] and then the [2 4]. By what rule did you obtain the [2 7] ?

[Bruno Baptista]

sorry, its not (n,3), its (n,2).

if i start with 3 then it has to find the first row with the number 3 wich is [3 4]. So now that it found the row with 3 i want it to search the first row with the number 4( its number 4 cause it was on the other collumn of the row with 3).

[Bruno Baptista]

it has to work for every matrix (n,2); i written 3 for mistake.

last result isnt [2 8] because i want it to start over from the previous row.

[Bruno Baptista]

i obtain obtain the [2 7] because after it gets [2 4] then it has to search the number in the other collumn wich is 2

Answer 1

[Ameer Hamza]

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

[Setsuna Yuuki.]

Awesome!

[Bruno Baptista]

for it to be correct m should be giving [ 3 4 2 7 ]

[Ameer Hamza]

Try the updated code.

[Bruno Baptista]

seems to work only when 3 is the origin number, when its 4 for example its shows an incomplete vector

[Ameer Hamza]

It gives output m = [4 5] if you set a = 4. What is the expected output? There is no other 5 in second column.

[Bruno Baptista]

i expected m = [4 5 6] , theres still one 5 in the first column

[Ameer Hamza]

The rule is still a bit confusing. See the updated code in my answer. Is that what you want?

[Bruno Baptista]

i think it works now, thks for the effort

[Ameer Hamza]

I am glad to be of help!

Answer 2

[KSSV]

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,:)

iwant = 3×2

2 8 2 4 2 7

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