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

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Yaser Khojah2021-10-13

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 个评论显示隐藏 1更早的评论
Yaser Khojah 2021-10-13
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.

### 采纳的回答

Star Strider 2021-10-13
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 个评论显示隐藏 7更早的评论
Star Strider 2021-10-14
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!
.

### 更多回答（2 个）

Matt J 2021-10-13

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 个评论显示隐藏 2更早的评论
Yaser Khojah 2021-10-14
I want to use it but I cant dowload this code in my project machine.

Image Analyst 2021-10-14
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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