How to replace some Number in row matrix with NaN

4 vues (au cours des 30 derniers jours)
Yared Daniel
Yared Daniel le 15 Juin 2021
I have a row vector like this
Y = [5 7 18 16 40 18 9 32 15 4 9 60 40 24 10 13];
What I want is to replace a number Y(i) with NaN if and only if its abs diff with Y(i-1) and Y(i+1) is greater than 10. if not the number remains as it was.
For instance abs(Y(4)-Y(5))=24 which is > 10 & abs(Y(5)-Y(6))=22 >10 therefore Y(5) = 40 has to be replaced with NaN
Y_new = [5 7 18 16 NaN 18 9 NaN NaN 4 9 NaN NaN Nan 10 13];
I have tried the follwing code and id didn't worked
clear
clc
Y = [5 7 18 16 40 18 9 32 15 4 9 60 40 24 10 13];
len = numel(Y);
A=Y;
start = 1;
for jj = 2 :len -1
if abs(Y(start)- Y(jj))<10
start=jj;
elseif abs(Y(start)- Y(jj))>10 && abs(Y(jj)- Y(jj+1))<10
Y(start)=NaN;
start = jj+1;
elseif abs(Y(start)- Y(jj))>10 && abs(Y(jj)- Y(jj+1))>10
Y(jj) = NaN;
start = jj+1;
else
end
end

Réponse acceptée

Yared Daniel
Yared Daniel le 17 Juin 2021
I have solved it!
clear
Y = [5 7 18 16 40 18 9 32 15 4 9 60 40 24 10 13];
idx = find(abs(diff(Y))>10);
L = numel(idx);
Y_new=Y;
for i=1:L-1
if abs(idx(i)-idx(i+1))==1
Y_new(idx(i)+1) =NaN;
else
end
end

Plus de réponses (1)

KSSV
KSSV le 15 Juin 2021
Y = [5 7 18 16 40 18 9 32 15 4 9 60 40 24 10 13];
Y_new = [5 7 18 16 NaN 18 9 NaN NaN 4 9 NaN NaN NaN 10 13];
dY = diff([0 Y]) ;
idx = dY>10 ;
Y1 = Y ;
Y1(idx) = NaN ;
  3 commentaires
KSSV
KSSV le 15 Juin 2021
Successive difference is considered. Your explanation is confusing. Try to extend the same logic to your case.
Yared Daniel
Yared Daniel le 15 Juin 2021
I have corrected the explanation to make clear

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