Creating a column vector form data in table to meet specific requirements.
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Hassen Mohamed Yousif Abdelatif
le 7 Déc 2021
Commenté : Stephen23
le 8 Déc 2021
I tried so many times to write a vector that meet specific requirements but it gave me scalar.
The requirements are:
If their age is greater than 75 or their BMI is greater than 50 I want the variable risk to a value of 3, signifying high risk. If their age is greater than 50 and less than or equal to 75 and their BMI is greater than 40 and less than or equal to 50, then I want the variable risk for that subject to 2 (medium risk) and if they don't fit any of the above criteria I would like their risk to 1, signifying low risk.
Notice both (BMI and age are vector of 20x1 double)
My code is
for index=1:length(age)
for index_1=1:length(BMI)
if age>75
risk(high_risk) = 3;
elseif (age>50 & age<=75)
risk(medium_risk) = 2;
elseif (BMI>50)
risk(high_risk) =3;
elseif ( BMI>40 & BMI<=50)
risk(medium_risk)=2
else
risk=1
end
end
end
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David Hill
le 7 Déc 2021
risk=ones(size(age));
risk(age>75|BMI>50)=3;
risk(age>50&age<=75&BMI>40&BMI<=50)=2;
8 commentaires
Stephen23
le 8 Déc 2021
Apparently because only ten participants meet those requirements.
If you have a 20x5 table and select 10 rows from it then you will get a 10x5 table.
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