Entering a vector in a function with if else statement

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Ida Lunde Hygum
Ida Lunde Hygum el 12 de Mzo. de 2018
Comentada: KL el 12 de Mzo. de 2018
Hi Matlab community.
I made this function that should sort out data that differs more than +/- 20 % giving it "NaN" if it does. If the previous measurement varied more than 20 % it should should not do so, which i tried with the command: "(sorted_data(i-1)==NaN) && (i>1)" but when I use the debugger it does not enter this statement even though it is true. How come it won't enter this? Thanks for the help!
Kind Regards
function sorted_data=sort_for_outliers(data)
sorted_data=zeros(length(data),1);
for i=1:length(data)
if i==1
sorted_data(i)=[data(i)];
elseif (sorted_data(i-1)==NaN) && (i>1)
sorted_data(i)=[data(i)];
elseif data(i)>1.20.*data(i-1)
sorted_data(i)=[NaN];
elseif data(i)<0.80.*data(i-1)
sorted_data(i)=[NaN];
else
sorted_data(i)=[data(i)];
end
end
end
  1 comentario
Torsten
Torsten el 12 de Mzo. de 2018
At the beginning of your function, you set "sorted_data" to zero. Thus sorted_data(i-1) will not be NaN.
Best wishes
Torsten.

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Respuestas (1)

KL
KL el 12 de Mzo. de 2018
Editada: KL el 12 de Mzo. de 2018
I'm afraid your usage of sorted_data(i-1)==NaN would always return false. The right way to find if a value is nan is to use isnan
P.S: It's always good to pay attention to the warnings on the right hand vertical bar on your editor window.
  2 comentarios
Ida Lunde Hygum
Ida Lunde Hygum el 12 de Mzo. de 2018
Hi! I tried this and it worked, thank you!
KL
KL el 12 de Mzo. de 2018
You're welcome!

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