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For loop for a function

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susman on 15 Feb 2021
Commented: susman on 15 Feb 2021
I have the following function and I want to run it for age 30 to 100. I think I need to develop a for loop here. Can anyone please help me out? I know if I change the value of age step by step like 30, 31 .... then I can do it but I need to run it as a for loop
function hf = myfun(age)
age = [30:100]
hf = zeros(5,5);
hf(1,2) = exp(-0.0625.*age-0.0134); % exp(age effect+time effect)
hf(1,5) = exp(-9.65573+0.01844+0.08218*age+0.02246); % exp(intercept+ age effect+time effect)
hf(2,3) = exp(-1.6660-0.1116.*age-0.0025); % exp(intercept+ age effect+time effect)
hf(2,4) = exp(-8.96236+0.07691.*age + 0.00978); % assuming the death rate of male of same age(Hubener et al.)
hf(2,5) = exp(-9.65573+0.08218.*age+0.02246); % self-mortality
hf(3,2) = exp(-0.0625.*age-0.0134+0.0676); %exp(intercept+ age effect+time effect+marriage once before)
hf(3,5) = exp(-9.65573+0.08218.*age+0.02246-0.11853);
hf(4,2) = exp(-0.4176-0.0625-0.0134.*age);
hf(4,5) = exp(-9.65573+0.08218.*age+0.02246-0.00415);
hf(1,1) = -(hf(1,2)+hf(1,5))
hf(2,2) = -(hf(2,3)+hf(2,4)+hf(2,5))
hf(3,3) = -(hf(3,2)+hf(3,5))
hf(4,4) = -(hf(4,2)+hf(4,5))

Accepted Answer

Jakob B. Nielsen
Jakob B. Nielsen on 15 Feb 2021
You construct the for loop like this:
for i=1:size(age,2) %by referencing your age array, you can change the age values and it will still work
hf(i,:,:) = zeros(5,5);
hf(i,1,2) = exp(-0.0625.*age(i)-0.0134); %index every step of hf to rely on i, and use the i'th value of age for the evaluations of the whole way
In the end, you will have hf(1,:,:) contain your results from age(1) in this case age=30, hf(2,:,:) age = 31 and so on.
susman on 15 Feb 2021
instead of getting hf(1,:,:) I am getting hf(:,:,1), hf(:,:,2) and so on.
Is there any syntax problem?

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