The longest consecutive values in a vector and the position at which it starts and ends

조회 수: 2 (최근 30일)
I have a large matrix where I want to find the value that has been repeated the most. Then define its starting and ending indexes. For example
Thanks for the help in advanse!
A = [35, 25, 40, 20, 20, 30, 30, 30, 30, 30, 9, 20, 30, 10, 30]
The solution should be as below
The_Answer = 30
Starting_index = 6;
Ending_index = 10
  댓글 수: 2
Geoff Hayes
Geoff Hayes 2021년 10월 13일
@Yaser Khojah - is this homework? What have you tried so far? What are the dimensions of the large matrix?
Yaser Khojah
Yaser Khojah 2021년 10월 13일
It is a large project, however, I have the code in a different computer which I do not have an access to it. I have tried using diff.

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채택된 답변

Star Strider
Star Strider 2021년 10월 13일
One approach —
A = [35, 25, 40, 20, 20, 30, 30, 30, 30, 30, 9, 20, 30, 10, 30];
[Au,~,ix] = unique(A, 'stable');
Tally = accumarray(ix,1);
HiFreq = Au(Tally==max(Tally));
Lv = false(size(A));
Lv(A==HiFreq) = true;
Start = strfind(Lv, [0 1])+1;
End = [strfind(Lv,[1 0]) numel(A)];
Len = End - Start;
[~,Idx] = max(Len);
Desired_Answer = HiFreq
Desired_Answer = 30
Desired_Start = Start(Idx)
Desired_Start = 6
Desired_End = End(Idx)
Desired_End = 10
.
  댓글 수: 8
Yaser Khojah
Yaser Khojah 2021년 10월 14일
Thanks so much. Take your time and hopefully it will work :)
Star Strider
Star Strider 2021년 10월 14일
This appears to work correctly for all of them, and with only minor changes in my original code.
To test it, un-comment (remove the ‘%’) from the ‘A’ vector to test , then run the code. (Keep the ‘%’ for the others not being tested. I included my original ‘Test’ vector as well in the ‘A Library’ of test vectors. The fprintf call allowed me to keep track of the loop iterations easily. I’m leaving it in, although commented so it won’t execute.)
A = [35, 25, 40, 20, 20, 30, 30, 30, 30, 30, 9, 20, 30, 10, 30];
% A = [30, 30, 30, 30, 30, 30, 30, 30, 35, 25, 40, 20, 20, 30, 30, 30, 30, 30, 9, 20, 30, 10, 30]; % Test
% A = [9 9 8 9 8 8 8 7 2 1];
% A = [9 9 8 8 8 7 2 9 3];
% A = [9 9 9 8 8 8 7 2 9 3];
% A = [9 8 7 6 8 7 2 9 3];
[Au,~,ix] = unique(A, 'stable');
Tally = accumarray(ix,1);
HiFreq = Au(Tally==max(Tally));
% Lv = false(size(A));
for k = 1:numel(HiFreq)
Element = HiFreq(k);
Lv = false(size(A));
Lv(A==HiFreq(k)) = true;
Lv = [false Lv];
Start = strfind(Lv, [0 1]);
End = unique([strfind(Lv,[1 0]) numel(A)]-1);
minidx = min(numel(Start),numel(End));
EndStt = [End(1:minidx); Start(1:minidx)];
Len = End(1:minidx) - Start(1:minidx);
[~,Idx(k)] = max(Len);
EndStart(:,k) = EndStt(:,Idx(k));
% fprintf('-------------------------\n')
end
HiFreqv = [];
Startv = [];
Endv = [];
Check = -diff(EndStart);
if all(Check)
[~,IxES] = max(-diff(EndStart));
HiFreqv = HiFreq(IxES);
Startv = EndStart(2,IxES);
Endv = EndStart(1,IxES);
end
Desired_Answer = HiFreqv
Desired_Answer = 30
Desired_Start = Startv
Desired_Start = 6
Desired_End = Endv
Desired_End = 10
Definitely an interesting problem!
.

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추가 답변 (2개)

Matt J
Matt J 2021년 10월 13일
편집: Matt J 2021년 10월 13일
Using "Tools for Processing Consecutive Repetitions in Vectors",
A = [35, 25, 40, 20, 20, 30, 30, 30, 30, 30, 9, 20, 30, 10, 30];
[starts,stops,lengths]=groupLims(groupConsec(A),1);
[~,i]=max(lengths);
The_Answer = A(starts(i))
The_Answer = 30
Starting_index = starts(i)
Starting_index = 6
Ending_index = stops(i)
Ending_index = 10
  댓글 수: 3
Matt J
Matt J 2021년 10월 14일
Really? It doesn't look like anyone has downloaded it recently.
Yaser Khojah
Yaser Khojah 2021년 10월 14일
I want to use it but I cant dowload this code in my project machine.

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Image Analyst
Image Analyst 2021년 10월 14일
If you have the Image Processing Toolbox (like most people do), you can use bwareafilt() to extract the longest run. Then the code becomes simply:
A = [35, 25, 40, 20, 20, 30, 30, 30, 30, 30, 9, 20, 30, 10, 30]
da = bwareafilt([0, diff(A)] == 0, 1)
startingIndex = max([1, find(da, 1, 'first')-1])
endingIndex = find(da, 1, 'last')
You see
A =
35 25 40 20 20 30 30 30 30 30 9 20 30 10 30
da =
1×15 logical array
0 0 0 0 0 0 1 1 1 1 0 0 0 0 0
startingIndex =
6
endingIndex =
10

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