How can I rewrite this in a simple short code
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Jorge Arturo Clares Pastrana
2022 年 1 月 21 日
コメント済み: Walter Roberson
2022 年 1 月 21 日
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]);
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採用された回答
Voss
2022 年 1 月 21 日
EA = abs([1 (MID(2:end)-MID(1:end-1))./MID(2:end)].');
3 件のコメント
Walter Roberson
2022 年 1 月 21 日
(MID(2:end)-MID(1:end-1))
looks as if it could be done more compactly as diff(MID)
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