How to replace a number in a row vector with NaN in certain condition

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I have raw vector like this
x = [15 13 9 20 12 15 2 25 17 3 2 5 18 4 12 30 32 40 1 19];
What I need is to replace a number with NaN when the following condition happen.
Whenever the abs difference between x(i) and x(i+1) is >10 replace x(i+1) with NaN , and continue next computation between x(i) and x(i+2) by skipping the value that is assigned as ‘NaN’. If the abs diff between x(i) and x(i+2) is <10, Then the next computation has to be between x(i+2) and x(i+3). But if the abs diff between x(i) and x(i+2) was >10, x(i+2) would be replace by ‘Nan” and next computation would be between x(i) and x(i+3).
In similar way computation continue and returned result should be as follow
x_new = [15 13 9 NaN 12 15 NaN NaN 17 NaN NaN NaN 18 NaN 12 NaN NaN NaN NaN 19]

Accepted Answer

per isakson
per isakson on 29 May 2021
Edited: per isakson on 29 May 2021
abs(x(6)-x(8)) is equal to 10. Accourding to x_new, x(8) should be replaced by NaN. Thus I have use ">=".
%%
vec = [15 13 9 20 12 15 2 25 17 3 2 5 18 4 12 30 32 40 1 19];
len = numel( vec );
new = vec;
pos = 1;
for jj = 2 : len
if abs(vec(pos)-vec(jj)) >= 10 % Notice ">="
new(jj) = nan;
else
pos = jj;
end
end
%%
x_new = [15 13 9 NaN 12 15 NaN NaN 17 NaN NaN NaN 18 NaN 12 NaN NaN NaN NaN 19];
disp( x_new ), disp( new )
15 13 9 NaN 12 15 NaN NaN 17 NaN NaN NaN 18 NaN 12 NaN NaN NaN NaN 19 15 13 9 NaN 12 15 NaN NaN 17 NaN NaN NaN 18 NaN 12 NaN NaN NaN NaN 19
  1 Comment
Yared Daniel
Yared Daniel on 30 May 2021
Thank you so much for understanding my question and solving exactly what I need

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