How to find indices in a vector satisfying a condition for a minimum consecutive number of entries

I have some time series

>> y = [2.1,2.5,4.6,2.2,3.4,2.1,3.3,4.4,2.4,5.6,4.3,4.4,2.3,3.0,5.7,3.4,3.8,3.9,1.6,2.0];

What I want to know is where in my time series I have consecutive numbers that satisfy some logical, lets say the logical is

condition = y < 4 & y > 1

Of course I know I can do

>> y(condition)
ans =

    2.1000    2.5000    3.6000    2.2000    3.4000    2.1000    3.3000    2.4000    2.3000    3.0000    3.4000    3.8000    3.9000    1.6000    2.0000

But that is not what I want. I want to know where in the vector y this condition is satisfied but only when it is satisfied n times in a row. So if n = 3 then I would like to construct the logical below

condition2 = [0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1]

I have true where the elements of y satisfy the condition as part of 3 or more consecutive numbers, and false everywhere else. Even though the first 2 elements of y satisfy the logical condition, I don't want them because they aren't part of 3 or more values that satisfy the logical. For the solution I'd like to be able to vary n.

Answer 1

[Jon]

Here's one approach.

You could shorten this up a little by combining some of the individual statements into more complicated ones, but I expanded it out here to makeit easier to follow

y = [2.1,2.5,4.6,2.2,3.4,2.1,3.3,4.4,2.4,5.6,4.3,4.4,2.3,3.0,5.7,3.4,3.8,3.9,1.6,2.0];

% check where condition applies
idl = y<4 & y>1

% mark begin and end of each group where condition is met with a 1 or -1
% resp
ijmp = diff([0,idl]); 

% just keep marks where groups begin
ibeg = ijmp>0;

% assign a group number to each group of ones
igroup = cumsum(ibeg).*idl

% find number of elements in each group, and where they come from
[count,edges,bin] = histcounts(igroup)

% assign number of repeats to each element according to which group it is
% in
numrep = count(bin)

% make a logical vector set true for each element that is in a group with a
% sufficient number of repeats
irep = numrep >= 3 & idl % don't include elements that are not in groups

[Jon]

You could also do it like this, counting the repetitions a little more directly like this without using histcounts:

y = [2.1,2.5,4.6,2.2,3.4,2.1,3.3,4.4,2.4,5.6,4.3,4.4,2.3,3.0,5.7,3.4,3.8,3.9,1.6,2.0];

% check where condition applies
idl = y<4 & y>1

% mark begin and end of each group where condition is met with a 1 or -1
% resp
ijmp = diff([0,idl,0]); 

% find where where groups begin
ibeg = ijmp(1:end-1)>0; 

% assign a group number to each group of ones
igroup = cumsum(ibeg).*idl

% find number of elements in each group
counts = find(ijmp==-1) - find(ijmp==1)

% assign number of repeats to each element according to which group it is
% in
numrep = zeros(size(igroup))
numrep(igroup>0) = counts(igroup(igroup>0))

% make a logical vector set true for each element that is in a group with a
% sufficient number of repeats
irep = numrep >= 3

[brandon burkholder]

Works perfectly. Thanks

[Jon]

Here's a further simplified approach

y = [2.1,2.5,4.6,2.2,3.4,2.1,3.3,4.4,2.4,5.6,4.3,4.4,2.3,3.0,5.7,3.4,3.8,3.9,1.6,2.0];

% check where condition applies
idl = y<4 & y>1

% mark begin and end of each group where condition is met with a 1 or -1
% resp
ijmp = diff([0,idl,0]); 

% find number of elements in each group
itot = cumsum(cumsum(ijmp));
counts = itot(ijmp < 0) - itot(ijmp > 0) + 1;

% assign number of repeats to each element according to which group it is
% in
ijmp(ijmp>0) = counts;
ijmp(ijmp<0) = -counts;
numrep = cumsum(ijmp(1:end-1));

% make a logical vector set true for each element that is in a group with a
% sufficient number of repeats
irep = numrep >= 3