How extract sub matrix without zeros from a big matrix
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gianluca scutiero
el 3 de Nov. de 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 comentarios
KL
el 3 de Nov. de 2017
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?
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Cedric
el 3 de Nov. de 2017
Editada: Cedric
el 3 de Nov. de 2017
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
Más respuestas (2)
Guillaume
el 3 de Nov. de 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 comentarios
Guillaume
el 3 de Nov. de 2017
"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.
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