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

2 visualizzazioni (ultimi 30 giorni)
Mirthand il 7 Apr 2021
Commentato: Bruno Luong il 8 Apr 2021
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.
##### 5 CommentiMostra 3 commenti meno recentiNascondi 3 commenti meno recenti
Mirthand il 7 Apr 2021
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 il 7 Apr 2021
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.

Accedi per commentare.

### Risposta accettata

Bruno Luong il 7 Apr 2021
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
##### 2 CommentiMostra NessunoNascondi Nessuno
Mirthand il 8 Apr 2021
Can you explain this line?
b([1 find(c>c(1),1,'first')+1])
Bruno Luong il 8 Apr 2021
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"

Accedi per commentare.

### Più risposte (1)

Matt J il 7 Apr 2021
Modificato: Matt J il 7 Apr 2021
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
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

Accedi per commentare.

### Categorie

Scopri di più su OFDM in Help Center e File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by