Azzera filtri
Azzera filtri

Find series of maxima of array (and matrix) within blocks of size n

1 visualizzazione (ultimi 30 giorni)
I have a simple problem, and it's, perhaps, less simple extension:
  1. I have an array A=[1 2 5 1 0 0 5 0 1 2 4 0 5 3 2 0 1 0 0 5] (all positive). How do I produce a vector B, which contains the maxima of, say, blocks of n=5 within A?
Namely, I want to produce B=[5 5 5 5].
2. I have a matrix: A=[1 2 5 1 0 0 5 0 1 2 4 0 5 3 2 0 1 0 0 5; 1 2 5 1 0 0 7 0 1 2 4 0 7 3 2 0 1 0 0 5; 1 2 5 1 0 0 5 0 1 2 4 0 5 3 2 0 1 0 0 5;] . Now, I want to get the matrix: C=[5 5 5 5; 5 7 7 5; 5 5 5 5] .
Can these two tasks be achomplished with a single procedure, without using loops?
Thanks!

Risposta accettata

Ted Shultz
Ted Shultz il 30 Ago 2019
Modificato: Ted Shultz il 30 Ago 2019
You could use rehsape and max:
[reshape(max(reshape(A',5,[])), [], size(A,1))]'
Test:
A=[1 2 5 1 0 0 5 0 1 2 4 0 5 3 2 0 1 0 0 5]
[reshape(max(reshape(A',5,[])), [], size(A,1))]'
A=[1 2 5 1 0 0 5 0 1 2 4 0 5 3 2 0 1 0 0 5; 1 2 5 1 0 0 7 0 1 2 4 0 7 3 2 0 1 0 0 5; 1 2 5 1 0 0 5 0 1 2 4 0 5 3 2 0 1 0 0 5;]
[reshape(max(reshape(A',5,[])), [], size(A,1))]'
ans =
5 5 5 5
ans =
5 5 5 5
5 7 7 5
5 5 5 5

Più risposte (1)

Bruno Luong
Bruno Luong il 30 Ago 2019
Modificato: Bruno Luong il 30 Ago 2019
maxfun = @(A) squeeze(max(reshape(A,size(A,1),5,[]),[],2));
A = [1 2 5 1 0 0 5 0 1 2 4 0 5 3 2 0 1 0 0 5]
B = maxfun(A)
A = [1 2 5 1 0 0 5 0 1 2 4 0 5 3 2 0 1 0 0 5; 1 2 5 1 0 0 7 0 1 2 4 0 7 3 2 0 1 0 0 5; 1 2 5 1 0 0 5 0 1 2 4 0 5 3 2 0 1 0 0 5]
C = maxfun(A)
  1 Commento
Erez
Erez il 30 Ago 2019
Awsome solution! I can't formally accept two solutions on this website, so I'll "accept" the first (arbitrarily). But this solution of yours works just as great. Thanks!

Accedi per commentare.

Categorie

Scopri di più su Loops and Conditional Statements 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