Unable to perform assignment because the left and right sides have a different number of elements.

I have a problem with the attached script sectrion. anr is a 46x46 double Matrix, with calculated values. I cant find my error but it probably is a very simple one.

Thx for the Help in advance.

[Stephen23]

@Jonas Jeskulke: please show us the complete error message. This means all of the red text.

[Jonas Jeskulke]

Unable to perform assignment because the left and right sides have a different number of
elements.

Error in DendroClusterMatrix (line 66)
           k(i) = find(irow(i) == irow(is));

thats the whole Error message.

[Stephen23]

If there are zero matches or more than one match your code will fail.

The LHS uses indexing to refer to one array element, which thus requires the RHS to be a scalar. But find will return non-scalar outputs if it detects zero/multiple matches. There seems to be nothing in your code preventing or handling non-scalar outputs from find.

[Jonas Jeskulke]

So do understand you correctly? It shloud work? if nothing helps i can upload the whole script maybe there is a problem within it that is causing that?

[Guillaume]

As Stephen explained, find can return from 0 to many elements, whereas

k(i) = ...

requires a single element to be valid.

More than one element can easily be worked around by limiting find to return the first match

k(i) = find(irow(i) == irow(is), 1);

But this will still error if there's no match.

Unfortunately, you haven't explained what the code is supposed to be doing. Most likely, you don't need the loops.

Note that the line:

[irow,icol,ind] = find(anr>0 & anr<=10);

is a bit mislieading. The variable ind is not an index, it's the actual values found in the input of find, which in this case is guaranteed to be all 1.

[Jonas Jeskulke]

with the whole script i am first calculating the angle between 2 vektors (XYZ) and the next step is to check if there are angles between >0 and <=10 i am using the find for that.

What i want to do the is saving all the values of the first row , then check if the the next and so on. BUT if for example in the first row the second value is 6 i want to save all the values from the second row in the same file/Matrix as the first one.

i hope this is understandable.

[Guillaume]

Sorry, no it's not clear. Rather than explaining in term of what the code should do, please explain what result you want. Also what is the size of the input(s)? Ideally give an example of inputs and desired output.

[Jonas Jeskulke]

Ok sure thing. as final output i want to have "clusters" of data. as example: i have a matrix looking like this as input (anr):

0 6.8 6.2 12.4 96.3

0 0 4.9 12.3 89.9

0 0 0 11.1 90.6

0 0 0 0 7.5

0 0 0 0 0

so from this input i want ot get Matrixes that look like this

1 2

1 3

2 3

and a second on with

4 5

so i have a cluster in the first matrix a cluster in the second but they are not intertwinde witch each other. does this make it a bit clearer?

[Guillaume]

What defines a cluster? Adjacent values (orthogonal? diagonal?) between 0 and 10? If so, have you got the image processing toolbox?

[Jonas Jeskulke]

i dont have any toobox s far. I will try to explain: a cluster is formed if the values match ceartain criteria:

1. have to be between >0 and <=10

2. if it is in the same row

3. if it is in another row, but the row it is in is a a collum in an allready exsiting

"cluster matrix" (example: in the first row the second value meets the criteria.

so the values from the second row will be written in the same matrix)

Answer 1

[Guillaume]

Your criteria for the cluster connectivity is a bit strange in my opinion since it doesn't actually require adjacency. If that's what you want I think the easiest may be to build a graph of connectivity and let matlab split the graph into its components. No toolbox needed for that.

I've tested with the following matrix:

inrange = [1 0 0 1 0 0 1 0 0;
           1 0 1 0 0 0 0 0 0;
           0 1 0 0 1 1 0 0 0;
           0 0 0 0 0 0 1 1 0;
           0 0 0 0 0 0 0 0 1;
           0 1 0 0 1 0 0 0 0];

which according to your criteria has 3 clusters (elements of rows 1, 2, 4 are one cluster, elements of rows 3 and 5 another, and the element of row 6 is another). In your code,

irange = anr > 0 & anr < 10;

With that:

[row, col] = find(inrange);  %locate elements in range
%now build connectivity graph. First iterate over the points to find which points share the same column or row
connectto =  arrayfun(@(idx) find(row(idx) == row | col(idx) == col), 1:numel(row), 'UniformOutput', false);  %Note that this will create self-loop as a point obviously has the same column and row as itself. Self-loops don't matter for connectivity
%convert the above in a 2 column matrix of start and end nodes
connectivity = [repelem((1:numel(row))', cellfun(@numel, connectto)), vertcat(connectto{:})];
%build graph
g = graph(connectivity(:, 1), connectivity(:, 2));
%optionally plot it
plot(g);
%get connected components of the graph, i.e. a cluster by the given criteria of sharing a row or column
clusters = conncomp(g, 'OutputForm', 'cell');
%replace node indices by [row, col] coordinates
clusters = cellfun(@(idx) [row(idx), col(idx)], clusters, 'UniformOutput', false);
%for display:
celldisp(clusters)

[Jonas Jeskulke]

Thank you a lot thihs is working like a charm. i was stuck with that problem for a long time.

I appreciate the work you put into this.

Best regards Jonas J.