find value every time counter increase by 2.

hello community kindly let me ask your support:

I have 2 vectors:

a = [0 0 0 0 1 1 2 2 3 3 3 3 4]'; %Counter

b = [0 1 2 3 4 5 6 7 8 9 10 11 12]'; %value to be found

c = [0 0 0 0 0 1 1 0 0 0 0 0 1]'; %flag when condition is met

what I'm trying to do is to get a value from b every time a increases its value by 2 and when c = 1

for j=1:length(time)

if a(j) == a(j) ++2 && c == 1

d(j,1) = b(j);

end

I'm trying to get something like:

d = [ 0 0 0 0 0 0 6 0 0 0 0 12]

can not find the solution......

any advice will be highly appreciated.

[John D'Errico]

Your example does not match what you asked.

a = [0 0 0 0 1 1 2 2 3 3 3 3 4]

a NEVER increases its value by 2 from the previous element in that vector.

Are you asking to find when a has increased by 2 from the previous case you found? So there the 6th element is 2 greater than 0, the start of the vector. But what are you asking?

[A-Rod]

Thank you John to take time to look at my question.

I didn't meant to say previous element in that vector a.

I meant to say everytime counter a increases its value by two from the start of the vector which is 0

when a = 2 and then a = 4 and so on.....

Answer 1

[Voss]

a = [0 0 0 0 1 1 2 2 3 3 3 3 4]';    %Counter 
b = [0 1 2 3 4 5 6 7 8 9 10 11 12]'; %value to be found
c = [0 0 0 0 0 1 1 0 0 0 0 0 1]';    %flag when condition is met

Indices of a that are 0,2,4,6,8,... more than the first element of a, and where c is not zero:

idx = mod(a-a(1),2) == 0 & c;

Or indices of a that are 0,2,4,6,8,... more than the first element of a, and where c is not zero, and it's the first time hitting that value in a (assuming a never decreases from one element to the next):

idx = mod(a-a(1),2) == 0 & c & [false; diff(a) > 0];

idx is the same in both cases, for the example posted.

Construct d:

d = zeros(size(b));
d(idx) = b(idx);
disp(d)

     0
     0
     0
     0
     0
     0
     6
     0
     0
     0
     0
     0
    12

Another example; this time the choice of method makes a difference because c(8) is 1, so you have two consecutive 2's in a, with c being 1 at both locations (Do you want to have both in d or just the first one?):

a = [0 0 0 0 1 1 2 2 3 3 3 3 4]';    %Counter 
b = [0 1 2 3 4 5 6 7 8 9 10 11 12]'; %value to be found
c = [0 0 0 0 0 1 1 1 0 0 0 0 1]';    %flag when condition is met

idx1 = mod(a-a(1),2) == 0 & c;
d1 = zeros(size(b));
d1(idx1) = b(idx1);
disp(d1)

     0
     0
     0
     0
     0
     0
     6
     7
     0
     0
     0
     0
    12

idx2 = mod(a-a(1),2) == 0 & c & [false; diff(a) > 0];
d2 = zeros(size(b));
d2(idx2) = b(idx2);
disp(d2)

     0
     0
     0
     0
     0
     0
     6
     0
     0
     0
     0
     0
    12

[A-Rod]

It works, thanks a lot for sharing it

[Voss]

You're welcome!