How can I rewrite this in a simple short code
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Jorge Arturo Clares Pastrana
le 21 Jan 2022
Commenté : Walter Roberson
le 21 Jan 2022
MID is a 1 x 15 Matrix with numerical val
ea1 = 1;
ea2 = ((MID(1,2)-MID(1,1))/(MID(1,2)));ea3 = ((MID(1,3)-MID(1,2))/(MID(1,3)));
ea4 = ((MID(1,4)-MID(1,3))/(MID(1,4)));ea5 = ((MID(1,5)-MID(1,4))/(MID(1,5)));
ea6 = ((MID(1,6)-MID(1,5))/(MID(1,6)));ea7 = ((MID(1,7)-MID(1,6))/(MID(1,7)));
ea8 = ((MID(1,8)-MID(1,7))/(MID(1,8)));ea9 = ((MID(1,9)-MID(1,8))/(MID(1,9)));ea10 = ((MID(1,10)-MID(1,9))/(MID(1,10)));
ea11 = ((MID(1,11)-MID(1,10))/(MID(1,11)));ea12 = ((MID(1,12)-MID(1,11))/(MID(1,12)));ea13 = ((MID(1,13)-MID(1,12))/(MID(1,13)));ea14 = ((MID(1,14)-MID(1,13))/(MID(1,14)));ea15 = ((MID(1,15)-MID(1,14))/(MID(1,15)));
EA = abs([ea1;ea2;ea3;ea4;ea5;ea6;ea7;ea8;ea9;ea10;ea11;ea12;ea13;ea14;ea15]);
0 commentaires
Réponse acceptée
Voss
le 21 Jan 2022
EA = abs([1 (MID(2:end)-MID(1:end-1))./MID(2:end)].');
3 commentaires
Walter Roberson
le 21 Jan 2022
(MID(2:end)-MID(1:end-1))
looks as if it could be done more compactly as diff(MID)
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Interpolation 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!