Keeping track of the Index while using "cell2mat" function

hey guys, I have a Cell Array a and am using cell2mat function for further clustering analysis, lets say:

a = {[1 2 3; 7 8 9], [4 5 6;10 11 12; 13 14 15]};
b = cell2mat(a(:));

rng(2)
k=2;
[idx] = kmeans(b,k);

which_in_cluster = cell(k,1);
for j = 1:k
    which_in_cluster{j} = find(idx == j);
end

Cluster_1 = b(which_in_cluster{1,1},:); % 1st Cluster
Cluster_2 = b(which_in_cluster{2,1},:); % 2nd Cluster

Cluster_1 =

     7     8     9
    10    11    12
    13    14    15


Cluster_2 =

     1     2     3
     4     5     6

I am now looking for a way to know in which Cell number and row these clusters have originally been in a. So in this example:

     Cluster_1 =

     7     8     9   % This has been in Cell number 1, row index 2 of a
    10    11    12   % This has been in Cell number 2, row index 2 of a
    13    14    15   % This has been in Cell number 2, row index 3 of a
    

Any help is greatly appreciated.

[Ko Fa]

If anyone got an idea or can send me to the right direction, I'd really appreciate that.

Answer 1

[Walter Roberson]

key = cell2mat((arrayfun(@(IDX) [repmat(IDX, size(a{IDX},1), 1), (1:size(a{IDX},1).']) , 1:numel(a), 'uniform', 0));

Now index rows of key by the cluster number to get the cell number and row number.

[Ko Fa]

Thanks for ur answer Walter, when trying to use ur code on the given example I get the error

Invalid Expression. When calling a function or indexing a variable, use parantheses. Otherwise, check for missmatched delimiters.

Putting this in:

key = cell2mat((arrayfun(@(idx) [repmat(idx, size(a{idx},1), 1), (1:size(a{idx},1).')] , 1:numel(a), 'uniform', 0)));

Gives me:

Error using horzcat

Dimensions of arrays being concatenated are not consistent

[Walter Roberson]

a = {[1 2 3; 7 8 9], [4 5 6;10 11 12; 13 14 15]};
key = arrayfun(@(idx) [repmat(idx,size(a{idx},1),1), (1:size(a{idx},1)).'], 1:numel(a), 'uniform', 0);
celldisp(key)

 
key{1} =
 
     1     1
     1     2

 
 
key{2} =
 
     2     1
     2     2
     2     3

 

[Ko Fa]

Ty, I'm still not quite sure how to read the output given by key.

key{1} is a 2x2 matrix. Cluster_1 is a 3x3 matrix. Not quite sure how the cell number and row number are extracted from key.

[Walter Roberson]

a = {[1 2 3; 7 8 9], [4 5 6;10 11 12; 13 14 15]};
b = cell2mat(a(:));

key = cell2mat(arrayfun(@(idx) [repmat(idx,size(a{idx},1),1), (1:size(a{idx},1)).'], (1:numel(a)).', 'uniform', 0))

key = 5×2

1 1 1 2 2 1 2 2 2 3

rng(2)
k=2;
[idx] = kmeans(b,k);

B = [b,key];

Cluster = cell(k,1);
for j = 1:k
    which_in_cluster{j} = find(idx == j);
    Cluster{j} = B(idx == j,:);
end
celldisp(Cluster)

 
Cluster{1} =
 
     7     8     9     1     2
    10    11    12     2     2
    13    14    15     2     3

 
 
Cluster{2} =
 
     1     2     3     1     1
     4     5     6     2     1

 

Last two columns are cell index in a, then row number within that cell entry.

[Ko Fa]

Thank you Walter, my original question has been answered but I do have a follow up question on this if you don't mind. I can post this in a new question if you wish so.

I exctracted the last 2 Columns from the first Cell of Cluster

c = Cluster{1};
IDX2 = c(:,end-1:end)

IDX2 =
     1     2    % Cell 1 - Row 2
     2     2    % Cell 2 - Row 2
     2     3    % Cell 2 - Row 3

I now have another Cell Array, with the same size as a (only consisting of nx1 matrices). For example:

f = {[1;2],[2;3;4]}
>> celldisp(f)
f{1} =
     0.5
     0.2

f{2} =
     0.7
     0.9
     0.8

How can I now extract the corresponding numbers from f (they represent the frequencies in my analysis) from the information represented in IDX2?

so, that the output would look something like this:

F =
    0.2000
    0.9000
    0.8000

[Walter Roberson]

F = arrayfun(@(R) f{IDX2(R,1)}(IDX2(R,2)), (1:size(IDX2,1).'))

This is probably one of the cases where it is just clearer to write a loop, and probably more efficient:

nrow = size(IDX2,1);
F = zeros(nrow,1);
for K = 1 : nrow
    F(K) = f{IDX2(K,1)}(IDX2(K,2));
end

[Ko Fa]

A pleasure! Thank you