I want to get how many times a value sequence is repeated in a column and the average number of times it repeated in a sequence.

suppose i have a column

[0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 ]

i want to get , that the repetative sequence of 1 is repeated 5 times in the column. (So here answer should be 5)

and the average number of times it repeated in a sequence is (4+3+3+4+3)/5 = 3.4 (So here answer should be 3.4)

Sorry for my english!

Answer 1

[Walter Roberson]

C = [0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 1  0 0 0 0 0 0 0 1 1 1  0 0 0 0 1 1  1 1 0 0 0 0 0 0 1 1 1 0 0 0 ]

C = 1×53

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

%for this purpose, C MUST be a row vector
starts = strfind([0 C], [0 1]);
stops = strfind([C 0], [1 0]);
number_of_repeats = length(starts)

number_of_repeats = 5

average_repeats = mean(stops - starts + 1)

average_repeats = 3.4000

%alternative, needs Image Processing Toolbox
%for this purpose C could be a row or column vector
info = regionprops(logical(C), 'Area');
number_of_repeats = length(info)

number_of_repeats = 5

average_repeats = mean([info.Area])

average_repeats = 3.4000

Answer 2

[Voss]

You can use strfind() to find a pattern in a vector like this. Use pattern [0 1] to find where the 1's start, and use pattern [1 0] to find where the 1's end. Prepend and append a 0 to x to correctly handle the cases when x starts/ends with a 1.

x = [0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 1  0 0 0 0 0 0 0 1 1 1  0 0 0 0 1 1  1 1 0 0 0 0 0 0 1 1 1 0 0 0].';

sequence_start = strfind([0 x(:).' 0],[0 1])

sequence_start = 1×5

10 21 31 38 48

sequence_end = strfind([0 x(:).' 0],[1 0])

sequence_end = 1×5

14 24 34 42 51

sequence_length = sequence_end-sequence_start

sequence_length = 1×5

4 3 3 4 3

n_sequences = numel(sequence_length)

n_sequences = 5

avg_sequence_length = mean(sequence_length)

avg_sequence_length = 3.4000

Answer 3

[Stephen23]

"suppose i have a column"

V = [0;0;0;0;0;0;0;0;0;1;1;1;1;0;0;0;0;0;0;0;1;1;1;0;0;0;0;0;0;0;1;1;1;0;0;0;0;1;1;1;1;0;0;0;0;0;0;1;1;1;0;0;0];
D = find(diff([0;V(:);0]));
M = mean(D(2:2:end)-D(1:2:end))

M = 3.4000

N = numel(D)/2

N = 5