Determine the number of elements in succession in a vector that are equal in a succinct way

I have a sequence, say:

x = [4 4 4 4 4 1 1 1 1 1 1 2 2 2 2 2 2 2 4 4 4 4 4 4 4];

I need to get the 4's at the end of the sequence, and count how many of them occur in a row. I can't simply do the sum of 4's that are occurring in x, because there are 4's earlier on in the sequence. Is there a function, or succinct way of doing this that doesn't require looping through the whole vector?

[David Haydock]

edited the question to say "in succession" instead of "in a row" to not confuse

Answer 1

[Bruno Luong]

You can use many runlength in filesubmission. I use here my own;

x = [4 4 4 4 4 1 1 1 1 1 1 2 2 2 2 2 2 2 4 4 4 4 4 4 4]

x = 1×25

4 4 4 4 4 1 1 1 1 1 1 2 2 2 2 2 2 2 4 4 4 4 4 4 4

[len, v] = runlengthencoder(x);
len(find(v==4,1,'last'))

ans = 7

function [len, v, gr, subidx] = runlengthencoder(X)
% [len, v, gr, subidx] = runlengthencoder(X)
% Run-length encoder
%
% INPUT
%   X is (1 x n) row vector, column is also allowed
% OUTPUTS:
%   len: integer arrays (1 x m)
%   v: (1 x m) ordering subset of X, such that two adjadcent elements are differents
%       and X = replelem(v, len)
%   gr: (1 x n) integer, group number (value in 1:m)
%   subidx: (1 x n) integer, interior indexes of X with in the group
%
% See also: runlengthdecoder
if ~isrow(X)
    X = reshape(X, 1, []);
end
n = size(X,2);
if n > 0
    b = [true, diff(X)~=0];
    ij = find([b, true]);
    len = diff(ij);
    v = X(b);
    if nargout >= 3
        gr = repelem(1:length(len),len);
        if nargout >= 4
            subidx = ones(1,n);
            subidx(ij(2:end-1)) = 1-len(1:end-1);
            subidx = cumsum(subidx, 2);
        end
    end
else
    [len, v, gr, subidx] = deal([]);
end
end % runlengthencoder

[David Haydock]

You have no idea how much help this function is to me. It makes all of my code so much more streamlined. Thank you so much!