Azzera filtri
Azzera filtri

How to repeat a vector over and over?

3 visualizzazioni (ultimi 30 giorni)
I have a vector with a length of 1023 elements (1 by 1023 elements). I want this vector to be 1 by 50000, in other words, change it up to 50000 elements by copying the 1023 elements over and over and over until it stretches to 50000 elements. The thing is, sometimes the original vector's length and the desired one are not compatiable, like this example I gave. 50000/1023 is not necessarily an integer, so the code will have to copy the 1023 elements 48 times and then there will be 896 spaces left, so they will have to be filled with the first 896 elements of the 1023 original elements. I hope someone can help me do this efficiently because I have to do this process like a million time for some analysis I am running :)

Risposta accettata

John D'Errico
John D'Errico il 6 Apr 2023
Modificato: John D'Errico il 6 Apr 2023
So write a function that does it for you.
vec = 1:4;
repvec(vec,7)
ans = 1×7
1 2 3 4 1 2 3
vec = [0;1];
repvec(vec,8)
ans = 8×1
0 1 0 1 0 1 0 1
function newvec = repvec(vec,finallength)
% replicates elements of a vector to have a final length
nvec = numel(vec);
reps = ceil(finallength/nvec);
newvec = repmat(vec(:),reps,1);
newvec(finallength+1:end) = [];
if isrow(vec), newvec = newvec.'; end
end
repvec will insure the resulting vector is a row or column vector, depending on what came in.

Più risposte (2)

Torsten
Torsten il 6 Apr 2023
Spostato: Torsten il 6 Apr 2023
a = rand(1,1023);
a = repmat(a,1,ceil(50000/1023));
a = a(1:50000)

the cyclist
the cyclist il 6 Apr 2023
Modificato: the cyclist il 6 Apr 2023
Here is one way:
% Inputs (for example small enough to easily see the result)
N = 7;
v = [2 3 5];
% Algorithm
vlen = numel(v);
b = rem(N,vlen);
out = [repmat(v,1,(N-b)/vlen) v(1:b)]
out = 1×7
2 3 5 2 3 5 2

Categorie

Scopri di più su Structures in Help Center e File Exchange

Prodotti


Release

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by