How to find the find the pattern in each row of Matrix?

Hallo, Thanks for reading!

The array that I am working with consist of 8 column.

1       2       3       4       5       6       7       8
==========================================================
0       3.2     0       3.3     0       2       19      0
3.2     0       3.2     0       3.3     0       2       0  
0       0       6.2     0       0       8       21      0   
3.2     9       3.2     0       0       0       2       0  
...     

The goals is to find the column number which are non-zeros. For example: row1 = 2468, row2 = 1357, row3= 367 so on. With these I would like finally find out what is the most frequent pattern.

The current code is

for i=1:length(data)
    k(i,:)= find(data(i,:)~=0);
end

However, my array is inconsistent how to solve this problem? Returning the following error:

Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.

Answer 1

[Kazi Alam]

Already found a better option!

Here is the code.

for i=1:length(outerwall)
 k(:,i)= str2double(sprintf('%d',find(outerwall(i,:)~=0)));
end

Edit:

for more info

https://de.mathworks.com/matlabcentral/answers/483625-how-to-combine-all-columns-of-an-array-into-one-column

[Adam Danz]

This solution is bad for two reasons.

1. Is very inefficient. It converts vectors to strings, then back to a numbers, and uses a loop, all of which can be avoided.
2. Most importantly, this method will fail if there are more than 9 columns. Example: let's say the pattern in row j is [1 3] and the pattern in row k is [13]. That produces the same output "13" even though the 2 patterns have no overlap at all. Another example: what columns are represented by '123'? [1 2 3] or [1, 23] or [123]?

dpb's answer is much more efficient, cleaner, quicker, and robust.

Answer 2

[dpb]

A=[A;A(3,:)];                                   % make sure one row has same pattern
% the engine
B=(A~=0);                                       % convert to logical array for pattern, independent of value
[~,ib]=ismember(B,unique(B,'rows'),'rows');     % locations of each pattern
N=histcounts(ib);                               % count number of each pattern located

For the above augmented, array, this returns

>> N
N =
     2     1     1     1
>> 

which shows the first row was duplicated; all rest were unique, only occurring once.

[Adam Danz]

Good one!

[Kazi Alam]

Thanks! I have found a better option. I will post it below.