How can I rewrite this in a simple short code
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Jorge Arturo Clares Pastrana
am 21 Jan. 2022
Kommentiert: Walter Roberson
am 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 Kommentare
Akzeptierte Antwort
Voss
am 21 Jan. 2022
EA = abs([1 (MID(2:end)-MID(1:end-1))./MID(2:end)].');
3 Kommentare
Walter Roberson
am 21 Jan. 2022
(MID(2:end)-MID(1:end-1))
looks as if it could be done more compactly as diff(MID)
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Interpolation finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!