find value every time counter increase by 2.

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A-Rod
A-Rod 2024-3-8
评论： Voss 2024-3-11
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.
  2 个评论
John D'Errico
John D'Errico 2024-3-8
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
A-Rod 2024-3-8
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.....

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Voss
Voss 2024-3-8
Ran in:
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

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