## How to group rows of data into a cell of a cell array?

### John Hunt (view profile)

on 17 Oct 2017
Latest activity Answered by Walter Roberson

on 17 Oct 2017

### Cedric Wannaz (view profile)

I have a data that has xyz information as well as three numbers that group each point together. As an example a row might have [ -1 2 -4 1 2 15] So that atom is at point (-1,2,-4) and it is defined by 1,2,and 15. There is an unknown amount of atoms that have different locations but have the same defined numbers. I want to group all of the atoms that have the same 3 defining numbers in a matrix that is in a cell array (or into structure if that is easier) QN. I have the following code but I am not adding more data to one cell. This is what I have
j=1;
for k = 1:(length(TypesQN)-1)
for p = 1:(length(locQN)-1)
if TypesQN(k,1:4)==locQN(p,4:7)
qn=locQN(p,:);
QN{j}=qn;
j=j+1;
end
end
end
Where the amount of defined numbers is know (TypesQN) and the total dataset length is known (locQN).

### Cedric Wannaz (view profile)

on 17 Oct 2017
Edited by Cedric Wannaz

### Cedric Wannaz (view profile)

on 17 Oct 2017

Here is one way. We start by building a fake data set:
>> A = [rand(10,3), randi(2,10,3)]
A =
0.7513 0.8407 0.3517 1.0000 1.0000 1.0000
0.2551 0.2543 0.8308 1.0000 2.0000 1.0000
0.5060 0.8143 0.5853 2.0000 1.0000 1.0000
0.6991 0.2435 0.5497 2.0000 2.0000 2.0000
0.8909 0.9293 0.9172 2.0000 1.0000 1.0000
0.9593 0.3500 0.2858 1.0000 2.0000 2.0000
0.5472 0.1966 0.7572 2.0000 1.0000 2.0000
0.1386 0.2511 0.7537 1.0000 2.0000 2.0000
0.1493 0.6160 0.3804 1.0000 2.0000 1.0000
0.2575 0.4733 0.5678 1.0000 2.0000 1.0000
Then we find group IDs per unique combination of 3 "ID" numbers:
>> [~,~,gId] = unique( A(:,4:end), 'rows' )
gId =
1
2
4
6
4
3
5
3
2
2
and we use these to split the array:
>> groups = splitapply( @(x){x}, A, gId )
groups =
6×1 cell array
{1×6 double}
{3×6 double}
{2×6 double}
{2×6 double}
{1×6 double}
{1×6 double}
where you can see that the 3 numbers are the same for e.g. group 2:
>> groups{2}
ans =
0.2551 0.2543 0.8308 1.0000 2.0000 1.0000
0.1493 0.6160 0.3804 1.0000 2.0000 1.0000
0.2575 0.4733 0.5678 1.0000 2.0000 1.0000
Note that if you don't want to save the ID numbers in the groups, you can split as follows:
>> groups = splitapply( @(x){x(:,1:3)}, A, gId )
groups =
6×1 cell array
{1×3 double}
{3×3 double}
{2×3 double}
{2×3 double}
{1×3 double}
{1×3 double}
>> groups{2}
ans =
0.2551 0.2543 0.8308
0.1493 0.6160 0.3804
0.2575 0.4733 0.5678