Select matrices with nonzero rows from a bigger matrix ?

Hello guys!

I am struggling with this problem:

I have a big matrix that is made of few zeros raws and many nonzero raws.

This is a fake simple example to explain, because in reality I am working with big data :(:

B= [0 0 0 0;
    0 0 0 0;
    0 0 0 0;
    0 0 0 0 ;
    1 1 2 3;
    0 0 0 3;
    1 1 1 0;
    4 4 4 4;
    0 0 0 0 ;
    0 0 0 0;
    0 0 0 0 ;
    0 0 0 8;
    0 0 1 0;
    0 1 1 0;
    0 0 0 0;
    0 0 0 0 ];

The zero rows in this matrix, indicate that the signal stopped and I just need to analyse the non zero raws in chunks.

Therefore would like to obtain:

chunk_1=[1 1 2 3;0 0 0 3;1 1 1 0;4 4 4 4]
chunk_2=[0 0 0 8;0 0 1 0;0 1 1 0;]

As you can see the chunks don't have to have the same amount of raws, since I analyse them individually, and the zero chunks also don't have the same number of rows.

I hope you guys have more knowledge than ve and could help me!

Thanks in advance !

Answer 1

[Ameer Hamza]

This nice little function from the image processing toolbox (bwconncomp) will be very useful here.

B = [0 0 0 0;0 0 0 0;0 0 0 0;0 0 0 0 ;1 1 2 3;0 0 0 3;1 1 1 0;4 4 4 4; 0 0 0 0 ;0 0 0 0;0 0 0 0 ;0 0 0 8;0 0 1 0;0 1 1 0;0 0 0 0;0 0 0 0 ];

mask = sum(B, 2)~=0;
comps = bwconncomp(mask);
compMaskList = comps.PixelIdxList;

B_components = cell(numel(compMaskList), 1);
for i=1:numel(compMaskList)
    B_components{i} = B(compMaskList{i}, :);
end

This shorter version is equivalent to the above code

comps = bwconncomp(sum(B, 2)~=0);
B_comps = cellfun(@(x) {B(x, :)}, comps.PixelIdxList);

[Geraldine Ejimadu]

woow, thank you. I didn't expect such a fast response :)

If you don't mind, what does the function bwconncomp do exactly? I am still at a basic level so I don't understand the code so well.

Thank You

[Ameer Hamza]

bwconncomp is actually an image processing function. It is used to find out connected components in a black-white image (binary). So we can use it to find out connected non-zero rows by creating a mask with this term 'sum(B, 2)~=0'.

[Geraldine Ejimadu]

Ok, got it!

What do you mean by "connected"/connectivity referred to the non zero rows?

[Ameer Hamza]

If you run 'sum(B, 2)~=0', you get

>> sum(B, 2)~=0
ans =
  16×1 logical array
   0
   0
   0
   0
   1
   1
   1
   1
   0
   0
   0
   1
   1
   1
   0
   0

So bwconncomp detects that there are two groups of 1s and gives back their row number.

[Geraldine Ejimadu]

It works, thank you so much!!!

Answer 2

[dpb]

Any number of these kinds of Q? on Answers -- under "runs" or any number of other terms...but, it's pretty simple to code from scratch--

d=diff([false;any(B,2)]);                                       % locations with any observations--T
ix=[find(d==1) find(d==-1)];                                    % find start, end of sections 0->1, 1->0
C=arrayfun(@(i) B(ix(i,1):ix(i,2),:),1:size(ix,1),'UniformOutput',false);  % cell array of contents by group

[Geraldine Ejimadu]

Thank You so much! I am trying to understand line 2 and 3, so that I can make it work for my data.

Could you please explain them to me? Line 2, and the use of arrayfun and the indexing in line 3?

[dpb]

Try it with your simple example data...start from inside out--first what is "d"?---

>> [B d]
ans =
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     1     1     2     3     1
     0     0     0     3     0
     1     1     1     0     0
     4     4     4     4     0
     0     0     0     0    -1
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     8     1
     0     0     1     0     0
     0     1     1     0     0
     0     0     0     0    -1
     0     0     0     0     0
>> 

any(x,2) gave us a true everywhere there was any nonzero row; diff() on that is only +/-1 when the value changes -- an initial start, then stop, ...

Next, simply extract those start:stop locations to an array of two vectors.

arrayfun is just a for...end loop on one line -- it uses the anonymous function @(i) B(ix(i,1):ix(i,2),:) as its action -- that's just a colon operation of B(i1:i2) for rows and : for all columns.

The next term is the indices over which to loop: 1:size(ix,1) which is just the number of row in our indexing array we built. Could write the function to use two input arguments and pass ix(:,1), ix(:,2); I just used the one index to the array and put the ix array inside.

Lastly, since returns a variable number of outputs, arrayfun needs to know to return results as a cell array or 'uniformoutput', false Otherwise, it'll fail as the output would have to be able to be concatenated into a single ouptut. But, you don't really want that here,anyway; you get the cell array of the data you're looking for for each case in a cell containing the array of doubles.

[Geraldine Ejimadu]

Thank you!! This is really helpful!

[dpb]

Very common type problem and typical use of diff for pattern-matching. "Trick" worth knowing; can adapt to many situations with thought.

May not always have Image Processing TB (as I don't) so need to be able to do oneself...