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

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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
  2 Comments
Yaser Khojah
Yaser Khojah on 13 Oct 2021
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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Accepted Answer

Star Strider
Star Strider on 13 Oct 2021
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
.
  8 Comments
Star Strider
Star Strider on 14 Oct 2021
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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More Answers (2)

Matt J
Matt J on 13 Oct 2021
Edited: Matt J on 13 Oct 2021
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
  3 Comments

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Image Analyst
Image Analyst on 14 Oct 2021
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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