Search through a vector and replace numbers at specific indices with the index
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
So this question is really hard to ask probably because it is so easy to answer. But I could really use some help. I have a 100 by 1 vector A. For any number in A I want to make sure the number located at its index is that actual number. If it's a different, I would like to replace it with its index.
For example at index 2, we have 1, but if we look at vector A, 2 appears at index 8, I want to replace the 1 at index 2, with 2. Also at index 6, we have 1 but if we scan through vector A, 6 appears at index 11, I would like to replace the 1 at index 6 with 6 and so forth. Thank you in advance!
A=[1 1 3 4 5 1 7 2 9 5 6 12 1 4 3 16 1 9 1 16 3 4 23 10 4 21 18 5 6 12 11 4 27 5 12 12 4 5 3 12 12 12 9 12 34 1 12 34 31 3 8 9 3 54 4 27 3 11 1 60 31 9 9 24 9 12 9 7 23 4 5 1 5 12 1 3 5 1 23 1 24 3 4 9 18 26 7 21 5 9 1 3 5 11 25 4 7 14 23 21]
The answer looks something like this
answer=[1 2 3 4 5 6 7 8 9 10 11 12 1 14 3 16 1 18 1 16 21 4 23 24 25 26 27 5 6 12 31 4 27 34 12 12 4 5 3 12 12 12 9 12 34 1 12 34 31 3 8 9 54 54 4 27 3 11 60 60 31 9 9 24 9 12 9 7 23 4 5 1 5 12 1 3 5 1 23 1 24 3 4 9 18 26 7 21 5 9 1 3 5 11 25 4 7 14 23 21];
The code should be something like this I think, but this is not giving me the answer am looking for.
B=(1:100 ); for v=1:n if B(v)==any(A(v)) A(v)=B(v) end end
0 commentaires
Réponses (1)
KSSV
le 6 Déc 2017
Modifié(e) : KSSV
le 6 Déc 2017
Instead of replacing, you can striaght away generate a row vector from 1 to 100? This is your output eh? Still, you can do like below:
A=[1 1 3 4 5 1 7 2 9 5 6 12 1 4 3 16 1 9 1 16 3 4 23 10 4 21 18 5 6 12 11 4 27 5 12 12 4 5 3 12 12 12 9 12 34 1 12 34 31 3 8 9 3 54 4 27 3 11 1 60 31 9 9 24 9 12 9 7 23 4 5 1 5 12 1 3 5 1 23 1 24 3 4 9 18 26 7 21 5 9 1 3 5 11 25 4 7 14 23 21];
pos = 1:length(A) ;
% get indices equal positions
idx = A==pos ;
%%If not replace
A(~idx) = pos(~idx) ;
I hope, I got your question correct. :)
1 commentaire
Voir également
Catégories
En savoir plus sur Matrix Indexing dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!