Bringing arrays to the same length

There are three arrays (two vectors and a matrix), I need to bring them to the same length

a=[ 1.2,1.3,1.4]; % by array "a" i need to sort the values ​​in array "b" and "m"
b=[ 1.2,0.9,1.3,2.6,1.4];
m=[1,2;26,90;1,2;120,0;1,0];
% what do i need to get
ba=[ 1.2,1.3,1.4];
ma=[1,2;1,2;1,0];

I will be glad to any advice

[DGM]

At first glance, I thought you're looking for the intersections:

a=[1.2,1.3,1.4]; % by array "a" i need to sort the values in array "b" and "m"
b=[1.2,0.9,1.3,2.6,1.4];
m=[1,2;26,90;1,2;120,0;1,0];

ba = intersect(b,a)

ba = 1×3

1.2000 1.3000 1.4000

ma = intersect(m,a)

ma =

  0×1 empty double column vector

... but there's no intersection between m and a. I don't know what rules you use to get the results you describe. Is ma a function of the indices of the intersection of b and a?

[ba idx] = intersect(b,a);
ma = m(idx,:)

ma = 3×2

1 2 1 2 1 0

[Benjamin Kraus]

I'm having some trouble understanding your question.

If I understand correctly, you are trying to calculate the values in ba and ma based somehow on the values of a, b, and m, but I don't see how you calculate the values in ma from any of the vectors, and ba appears to be equal to a. Can you give more description (in words) of the algorithm you are hoping to use to derive ba and ma based on a, b, and m?

Answer 1

[Rik]

There are two issues with the solution by Arif:

• It uses find on a logical array, but that is only used to index an array
• It assume the the first column in m will be 1 for all the rows we're looking for

The edits below solve both.

a=[ 1.2,1.3,1.4];
b=[ 1.2,0.9,1.3,2.6,1.4];
m=[1,2;26,90;1,2;120,0;1,0];

[L,idx]=ismember(a,b); % find the element of a in b
ba=a(L)

ba = 1×3

1.2000 1.3000 1.4000

ma=m(idx(L),:)

ma = 3×2

1 2 1 2 1 0

Answer 2

[Arif Hoq]

try this:

a=[ 1.2,1.3,1.4]; % by array "a" i need to sort the values in array "b" and "m"
b=[ 1.2,0.9,1.3,2.6,1.4];
m=[1,2;26,90;1,2;120,0;1,0];

[idx]=find(ismember(b,a)); % find the element of a in b
ba=b(idx)

ba = 1×3

1.2000 1.3000 1.4000

ma=m((m==1),:) % find the first element=1

ma = 3×2

1 2 1 2 1 0