# How extract sub matrix without zeros from a big matrix

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gianluca scutiero 2017년 11월 3일
편집: Cedric Wannaz 2017년 11월 3일
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 ##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
gianluca scutiero 2017년 11월 3일
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. 댓글을 달려면 로그인하십시오.

### 채택된 답변

Cedric Wannaz 2017년 11월 3일
편집: Cedric Wannaz 2017년 11월 3일
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
##### 댓글 수: 1표시숨기기 없음
Cedric Wannaz 2017년 11월 3일
편집: Cedric Wannaz 2017년 11월 3일
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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### 추가 답변(2개)

Guillaume 2017년 11월 3일
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})});
##### 댓글 수: 3표시숨기기 이전 댓글 수: 2
gianluca scutiero 2017년 11월 3일
Sorry !!!! I appreciated very much your help ... thank you :)

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gianluca scutiero 2017년 11월 3일
Thank you very much, Cedric Wannaz. It works very well. :)
##### 댓글 수: 1표시숨기기 없음
Cedric Wannaz 2017년 11월 3일
편집: Cedric Wannaz 2017년 11월 3일
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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