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How to replace some Number in row matrix with NaN

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Yared Daniel
Yared Daniel on 15 Jun 2021
Answered: Yared Daniel on 17 Jun 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

Accepted Answer

Yared Daniel
Yared Daniel on 17 Jun 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

More Answers (1)

KSSV
KSSV on 15 Jun 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 Comments
Yared Daniel
Yared Daniel on 15 Jun 2021
I have corrected the explanation to make clear

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