How to find the first element in ascending numbers that repeat?

I want to find the index of the first instance of a number only when the next number is ascending.

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

b = find(A==1)

Actual Output:

     1     6     8    10    12    17    19    21    22

Desired Output:

1 12 22

Answer 1

[DGM]

You said you want the first instance where the next number is larger, but the desired output you gave [1 12 22] is all instances where the next number is larger.

A = [0 5 1 0 1 2 3 4 0 1 0 1 0 1 0 1 2 3 4 0 1 0 1 0 1 1 2 3 4]
numtofind = 1;

% find all instances
% ignore the last element, since it has no following element
b = find(A(1:end-1)==numtofind)

% pick whichever type of result you need
%b = b(find(A(b)<A(b+1),1)) % the first instance where the next number is larger
b = b(A(b)<A(b+1)) % all instances where the next number is larger

Just pick whichever one you need.

[Mirthand]

Thank you! I needed (% all instances where the next number is larger).

Is there a way to change it from the next number is larger to the next two numbers are larger?

For example if I wanted to use numtofind = 0;

I want to ignore 010101

but keep instances where 012

[DGM]

Sure

A = [0 5 1 0 1 2 3 4 0 1 0 1 2 1 0 1 2 3 4 0 1 0 1 0 1 1 2 3 4 1]
numtofind = 1;

b = find(A(1:end-2)==numtofind) % find all instances
b = b(A(b)<A(b+1) & A(b)<A(b+2)) % all instances where the next two numbers are larger

Answer 2

[Bruno Luong]

A = [1 2 3 4 0 1 0 1 0 1 0 1 2 3 4 0 1 0 1 0 1 1 2 3 4]
b = find(A(1:end-1)==1 & diff(A)>0)