필터 지우기
필터 지우기

Finding consecutive data with non zero in array

조회 수: 11 (최근 30일)
edward kabanyas
edward kabanyas 2019년 9월 15일
편집: edward kabanyas 2019년 10월 7일
Hi all;
I want to find consecutive data with non zero value in array. However, as long as the data with 0 value is less than 3 points, it is still considered as part of the previous data set. The data with the same length is classified into same array.
For example:
xx = [1 2 3 0 1 2 0 0 0 1 2 3 4 0 1 0 0 0 1 1 1];
The expected result
xx1 =[1 2 3 0 1 2; 1 2 3 4 0 1];
xx2=[1 1 1];
I try to find it using looping procedure but it takes only onsecutive data with non zero value and I can not find the solution for this criteria "as long as the data with 0 value is less than 3 points".
Hope some help.
Thank you.
Edward
  댓글 수: 2
Rik
Rik 2019년 9월 15일
Numbered variables are a bad idea. Also, it should be feasible to adapt your code to also work for the shorter runs of zeros. Can you share your code here? Attach it in an m file if it is more than about 20 lines long.
Otherwise, you could try the RunLength FEX submission by Jan to find the run lengths and continue from there.
Rik
Rik 2019년 9월 16일
Comment posted as answer by edward kabanyas:
Thank you Rik. The following is my code:
threshold = 0;
transitions = diff([0, xx > threshold, 0]);
runstarts = find(transitions == 1);
runends = find(transitions == -1) - 1;
blocks = arrayfun(@(s, e) xx(s:e), runstarts, runends, 'UniformOutput', false);
celldisp(blocks)
However, it takes only onsecutive data with non zero value and I can not find the solution for the criteria "as long as the data with 0 value is less than 3 points". Then, the cell with same size in blocks is not merged into one cell as I need above.

댓글을 달려면 로그인하십시오.

채택된 답변

Andrei Bobrov
Andrei Bobrov 2019년 9월 16일
편집: Andrei Bobrov 2019년 10월 5일
for R2013a (and for your data from all_data.txt)
f = fopen('all_data.txt');
Data = textscan(f,'%f %f %f %f %f %f %f','CollectOutput',1);
fclose(f);
Data = Data{:};
lo = any(Data(:,6:7) ~= 0,2);
d = find(lo);
ii = [true;diff(d) > 3];
i2 = ii([2:end,1]);
i = zeros(size(Data,1),1);
i(d(ii)) = 1;
i(d(i2) + 1) = -1;
j = cumsum(i);
jj = cumsum([false;diff(j) == 1]).*j;
out = accumarray(jj+1,(1:size(Data,1))',[],@(x){Data(sort(x),:)});
out = out(2:end);
  댓글 수: 10
Andrei Bobrov
Andrei Bobrov 2019년 10월 5일
I'm fix.
edward kabanyas
edward kabanyas 2019년 10월 5일
편집: edward kabanyas 2019년 10월 7일
Thank you very much Andrei. It is perfect, thank you again.

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Rik
Rik 2019년 9월 16일
I used the mfile version of Jan's FEX submission RunLength, because I can't get my compiler to work right now. With the runlegth determined it is relatively easy to find the groups that need to be merged. Then you can still use the code you proposed to find the transitions.
xx = [1 2 3 0 1 2 0 0 0 1 2 3 4 0 1 0 0 0 1 1 1];
[a,b]=RunLength_M(xx~=0);
L= a(:)==false & b<3;%to be merged
%checks for egde cases should be perfomed here:
%-first group is a small group of zeros
%-last group is a small group of zeros
L_prev=[L(2:end);false];
L_next=[false;L(1:(end-1))];
b(L_prev)=b(L_prev)+b(L)+b(L_next);
a(L | L_next)=[];%remove merged parts
b(L | L_next)=[];%remove merged parts
x=RunLength_M(a,b);
transitions = diff([0, RunLength_M(a,b), 0]);
runstarts = find(transitions == 1);
runends = find(transitions == -1) - 1;
blocks = arrayfun(@(s, e) xx(s:e), runstarts, runends, 'UniformOutput', false);
celldisp(blocks)
  댓글 수: 5
Rik
Rik 2019년 10월 3일
What I meant to ask is if you were using a matrix or a vector a input. Attaching the data as a mat file would probably work best.
Rik
Rik 2019년 10월 5일
편집: Rik 2019년 10월 5일
It turns out you also need to use (:) for b if your input is a column vector. I just tested this on R2011a, so it should also works on ancient releases. I'll edit my answer.
Edit: except that your example is integer only and your data isn't, which seems to be causing problems.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 String Parsing에 대해 자세히 알아보기

제품


릴리스

R2013a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by