Divide an array in areas based on its values

조회 수: 10 (최근 30일)
Giorgos Papakonstantinou
Giorgos Papakonstantinou 2013년 5월 28일
Suppose that I want find the edges of ones and of not ones such that I can divide an in areas the array. Suppose the array is as the example below:
a=[1 1 1 1 1 1 12 10 1 1 1 1 1 11 12 1 1 1 2 3]
a(1:6) -->area1 of ones
a(7:8) -->area1 of not ones
a(9:a13) -->area2 of ones
a(14:15)-->area2 of not ones
a(16:19)-->area3 of ones
a(20) -->area3 of not ones
and so on..
  댓글 수: 2
Giorgos Papakonstantinou
Giorgos Papakonstantinou 2013년 5월 28일
Obviously its duplicated. I did because the comments are sometimes neglected. :-)

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채택된 답변

Azzi Abdelmalek
Azzi Abdelmalek 2013년 5월 28일
a=[1 1 1 1 1 1 12 10 1 1 1 1 1 11 12 1 1 1 2 3 ]
idx=find(a==1)
e=[diff(idx) 100]
f=find(e~=1)
ff=idx(f)
ii1=[idx(1) idx(f(1:end-1)+1)]
ii2=idx(f)
idx=find(a~=1)
e=[diff(idx) 100]
f=find(e~=1)
ff=idx(f)
jj1=[idx(1) idx(f(1:end-1)+1)]
jj2=idx(f)
for k=1:numel(ii1)
area1(k)={ii1(k):ii2(k)};
area0(k)={jj1(k):jj2(k)};
end

추가 답변 (2개)

Andrei Bobrov
Andrei Bobrov 2013년 5월 28일
a=[1 1 1 1 1 1 12 10 1 1 1 1 1 11 12 1 1 1 2 3 ];
a1 = a == 1;
ii = [true, diff(a1)~=0];
idx = cumsum(ii);
out = accumarray(idx(:),a(:),[],@(x){x});

Giorgos Papakonstantinou
Giorgos Papakonstantinou 2013년 5월 28일
I amazed by the simplicity of the codes both of you provided. Although I have some difficulty to understand them they solve my problem.
I would be grateful if yo provided some explanation Thank you very much.
  댓글 수: 3
Giorgos Papakonstantinou
Giorgos Papakonstantinou 2013년 5월 28일
This answer is common for both of the answers that I got above considering the level of my knowledge in matlab. Therefore this answer corresponds to Azzi's and Andrei's answer and this the reason that I posted as an answer and not as a comment. I would accept both answers if this was possible. I accepted Azzi's because it is easier to comprehend.
Andrei Bobrov
Andrei Bobrov 2013년 5월 29일
Please read the description of the three functions in the MATLAB documentation: diff, cumsum, accumarray.

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