Finding number(s) that is(are) repeated consecutively most often

Given this array for example:

a = [1 2 2 2 1 3 2 1 4 4 4 5 1]

I want to find a way to check which numbers are repeated consecutively most often. In this example, the output should be [2 4] since both 2 and 4 are repeated three times consecutively.

Another example:

a = [1 1 2 3 1 1 5]

This should return [1 1] because there are separate instances of 1 being repeated twice.

Answer 1

[Paras Gupta]

Hi,

It is my understanding that you intend to find all the numbers for which consective occurence is maximum. The following code illustrates how to achieve the same.

A = [1 2 2 2 1 3 2 1 4 4 4 5 1];

% B is a logical array with true(1) at indices where the difference between adjacent 
% elements is not zero
B = diff(A)~=0

B = 1×12 logical array
   1   0   0   1   1   1   1   1   0   0   1   1

% Append 1 at the start and end so that the first and last elements of A
% are also considered during the computation
C = [1, B, 1]

C = 1×14

1 1 0 0 1 1 1 1 1 0 0 1 1 1

% D gives us the indices in A where current element is different than the previous element 
% (that is no consective occcurence with the previous element)
D = find(C)

D = 1×10

1 2 5 6 7 8 9 12 13 14

% E gives us the count of consecutive occurences for all elements in A. We can verify the sum 
% of elements in E is equal to the length of A
E = diff(D)

E = 1×9

1 3 1 1 1 1 3 1 1

% Find the maximum consecutive occurence
maxCount = max(E)

maxCount = 3

% There can be multiple consective occurences withcount same as the maximum
idx = find(E == maxCount)

idx = 1×2

2 7

% D(idx) gives us the indices in A where maximum consective occurences start
% array m gives us the numbers repeated consecutively most often 
values = A(D(idx))

values = 1×2

2 4

A compact way to write down the above code, provided for reference.

B = [1 1 2 3 1 1 5];
x = find([1, diff(B)~=0, 1]);
y = diff(x);
maxCount = max(y);
indices = find(y == maxCount);
values = B(x(indices))

values = 1×2

1 1

You can refer to the linked documentations for diff, find, and max functions for a better understanding of how the above code works.

Hope this helps!

Answer 2

[Matt J]

Using,

https://www.mathworks.com/matlabcentral/fileexchange/78008-tools-for-processing-consecutive-repetitions-in-vectors

>> a = [1 2 2 2 1 3 2 1 4 4 4 5 1];

>> [starts,~,runLengths]=groupLims(a);

>> a(starts(runLengths==max(runLengths)))

ans =

2 4