How to find the first number, ignore subsequent until a greater number repeats.

A = [ 0 3 0 3 0 3 0 3 0 4 0 4 0 4 0 4 0 5 0 5 0 5 0 5 0 5 0 3 0 3 0 3]

b = find(A==3)
   2     4     6     8    28    30    32

Desired Output:

   2 28

My attempt is below:

c = diff(b)
d = unique(c)

Which then gives 2 20 but not sure how to go back to the index since this is the diff.

[Stephen23]

Where in your code do you check the "until a greater number repeats" requirement?

[dpb]

What would be desired result if

A = [ 0 3 0 3 0 3 0 3 0 4 0 3 0 4 0 4 0 5 0 5 0 5 0 5 0 5 0 3 0 3 0 3]

so

b = find(A==3)
   2     4     6     8    12    28    30    32

?

[Mirthand]

2 28

[Mirthand]

Where in your code do you check the "until a greater number repeats" requirement?

That's true, I think it doesn't nececessarily need to follow that.

[dpb]

Both solutions so far for the hypothesized slightly different input vector.

A = [ 0 3 0 3 0 3 0 3 0 4 0 3 0 4 0 4 0 5 0 5 0 5 0 5 0 5 0 3 0 3 0 3]

return

[2 12]

The Q? raised by S Cobeldick seems pertinent if the correct answer is, indeed, to be the one with 28 and not 12 unless it is able to be assured the input pattern must follow that of the first example precisely in not having one of the magic numbers possibly being repeated after the intervening value.

Answer 1

[Bruno Luong]

A = [ 0 3 0 3 0 3 0 3 0 4 0 4 0 4 0 4 0 5 0 5 0 5 0 5 0 5 0 3 0 3 0 3]:
b = find(A==3);
c = diff(b);
b([1 find(c>c(1),1,'first')+1])

you'll get

ans =

     2    28

[Mirthand]

Can you explain this line?

b([1 find(c>c(1),1,'first')+1])

[Bruno Luong]

b is the indices in A of 3s

c is the distance between 2 indices,

so

j = find(c>c(1),1,'first')

returns in jthe place where the distance beween 2 indices is larger than the first distance c(1) ("a greater number").

Finally

b([1 j+1])

is just for purpose of getting back indices in A of the 3 and what you call "a greater number repeat"

Answer 2

[Matt J]

Is this what you want?

A = [ 0 3 0 3 0 3 0 3 0 4 0 4 0 4 0 4 0 5 0 5 0 5 0 5 0 5 0 3 0 3 0 3];

cA=cummax(A);

b1=find(A==3,1);
b2=find(A==3 & cA>3 ,1);

b=[b1,b2]

b = 1×2

2 28