Merge two cells into a single cell?

Hello World,

I'm basically trying to find the most frequent column vector of the vertically concatenated matrix of

x=[2 3 4 3 5 4 4]
y=[7 9 8 9 7 3 7]

I tried doing

z=[x;y]
mode(z,2)

But this only gives me the most frequent value for each row. I need the frequent combination ( in my case, 3;9 not 4;7)

So, right now I'm trying to find a way to make a matrix such that I'll get the following using x and y:

w=[27 39 48 39 57 43 47]

How could I do that?

[TUD]

So, since my real goal is to find the most frequent combination, and not to merge them, I used this code to finally get it. There probably is an easier way.

clear all
clc

a=zeros(1,7);

x=[25 3 4 3 5 4 4]
y=[7 9 8 9 7 3 7]

z=[x;y]

for i=1:7
    for j=1:7
        if (z(1,j)==z(1,i))&&(z(2,j)==z(2,i))
            a(i)=a(i)+1;
        end
    end
end

a

maxim=max(a)

[locate]=find(a==maxim)

freq=z(:,locate(1))

Answer 1

[Mischa Kim]

TUD, use

 w = str2num(num2str([x' y'], '%-d'))'

[TUD]

Thanks for responding Mischa. When I run the code you gave, I still get a 2x7 matrix. I want it as a 1x7 matrix.

[Mischa Kim]

TUD, this should do the job:

 x = [25 3 4 3 5 4 4];
 y = [ 7 9 8 9 7 3 7];
 z = str2num(strcat(num2str(x'),num2str(y')))
 z =
   257
    39
    48
    39
    57
    43
    47

Answer 2

[Jos (10584)]

w = 10*x + y

[TUD]

Thanks Jos. This does work, but it depends on the size of the values, that is, if the y is double or single digit, etc. I want this for decimals too.

Answer 3

[Jos (10584)]

Another option:

x=[2 3 4 3 5 4 4]
y=[7 9 8 9 7 3 7]  

[xy,~,ix] = unique([x(:) y(:)],'rows') ;
[~,ix2] = mode(ix)
MyMode = xy(ix2,:).'

[TUD]

Jos,

This is the result I get:

x =

       2     3     4     3     5     4     4

y =

       7     9     8     9     7     3     7

ix2 =

       2

MyMode =

       3

While I need

Mymode= [3; 9]

I realized that I don't really need to find the best combination for my problem. But anyway, I still have added an edit that gets the best combination if anyone needs it.

Thanks for you help!