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How extract sub matrix without zeros from a big matrix

Asked by gianluca scutiero on 3 Nov 2017
Latest activity Edited by Cedric Wannaz
on 3 Nov 2017
Hello everybody ... I have a big matrix [N] as shown in Figure. I need to extract sub-matrix from [N]: [A],[B],[C],[D], [E] .... The end of each row can contain some zeros (red part). So I need to extract these sub-matrix. If [A] has the same number of columns of [E], I have to combine them in a unique matrix ([F]=[A];[E]). [N] does not contain zeros in the white part. Could you help me with this problem? Thank you very much for your time

  2 Comments

what do you mean by big matrix? be specific? what's the size? How do you identify A,B,C,D and E within N? How are they distributed across N's rows?
big matrix is about 30000x16. Because the zeros are at the end of each row. I have attached an example of the matrix. For example here A can be made by the first 3 rows.

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3 Answers

Answer by Cedric Wannaz
on 3 Nov 2017
Edited by Cedric Wannaz
on 3 Nov 2017
 Accepted Answer

Here is an example; we first build a test data set:
>> N = randi( 10, 10, 4 ) ;
>> for k = 1 : 10, N(k,1+randi(3,1):end) = 0 ; end
>> N
N =
8 0 0 0
3 3 9 0
6 0 0 0
7 3 0 0
9 0 0 0
10 4 0 0
6 0 0 0
2 3 0 0
2 7 4 0
3 5 6 0
Then sorting/grouping can be achieved as follows:
>> gId = sum( N == 0, 2 ) ;
>> groups = splitapply( @(x){x}, N, gId ) ;
With that you get:
>> groups
groups =
3×1 cell array
{3×4 double}
{3×4 double}
{4×4 double}
>> groups{1}
ans =
3 3 9 0
2 7 4 0
3 5 6 0
>> groups{2}
ans =
7 3 0 0
10 4 0 0
2 3 0 0
>> groups{3}
ans =
8 0 0 0
6 0 0 0
9 0 0 0
6 0 0 0
This assumes that there is no zero aside from the trailing ones on each row. We can work releasing this requirement if there can be zeros elsewhere, and on truncation to the non-zero part if you really need it.
EDIT : Here are the few extra steps if you wanted to deal with situations with zeros in the middle of non-zeros, and if you needed truncation: I start by adding a zeros in N(9,2) to test that it is working:
>> N(9,2) = 0 ;
Then
>> [r, c] = find( N ) ;
last_nzc = splitapply( @max, c, r ) ;
gId = findgroups( size(N, 2) - last_nzc + 1 ) ;
groups = splitapply( @(x,c){x(:,1:c(1))}, N, last_nzc, gId ) ;
With that we get:
>> groups{1}
ans =
3 3 9
2 0 4
3 5 6
>> groups{2}
ans =
7 3
10 4
2 3
>> groups{3}
ans =
8
6
9
6

  1 Comment

Note that I initially wrote x(:,1:c) in the call to SPLITAPPLY, which went through. That is surprising(!) This lead me to evaluate the following:
>> 1 : [4,5,6]
ans =
1 2 3 4
which is interesting ...

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Answer by Guillaume
on 3 Nov 2017

If I understood correctly:
%N: your big matrix
rowswithnozeros = cellfun(@(row) nonzeros(row).', num2cell(N, 2), 'UniformOutput', false);
rowlength = cellfun(@numel, rowswithnozeros);
[~, ~, subs] = unique(rowlength, 'stable'); %s'stable' optional
matriceswithnozeros = accumarray(subs, (1:numel(rowswithnozeros))', [], @(rows) {vertcat(rowswithnozeros{rows})});

  3 Comments

Also your code works well Guillaume and it is good because remove also the zeros. But is there the possibility to join the rows generated with the same numbers of columns ?
"But is there the possibility to join the rows generated with the same numbers of columns ?"
That's the whole purpose of the last three lines.
Sorry !!!! I appreciated very much your help ... thank you :)

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Answer by gianluca scutiero on 3 Nov 2017

Thank you very much, Cedric Wannaz. It works very well. :)

  1 Comment

My pleasure, but please see my last edits: the extra operations truncate zeros and manage cases with zeros inside the non-zero blocks!

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