MATLAB: Evaluate function handle (determine the value of function handle using the code

determine value of functionhandleevalue function handle

Please tell me how to determine the value of a function handle in a quick way. 
hf12 = @(age) exp(-0.0625.*age-0.0134) % exp(age effect+time effect) 
hf13 = 0
hf14 = 0
hf15 =  @(age) exp(-9.65573+0.01844+0.08218*age+0.02246) % exp(intercept+ age effect+time effect) 

hf11 = 0
hf21 = 0
hf23 = @(age) exp(-1.6660-0.1116.*age-0.0025) % exp(intercept+ age effect+time effect) 
hf24 = @(age)exp(-8.96236+0.07691.*age + 0.00978) % assuming the death rate of male of same age(Hubener et al.)
hf25 = @(age)exp(-9.65573+0.08218.*age+0.02246) % self-mortality 
hf22 = 0
hf31 = 0
hf32 = @(age) exp(-0.0625.*age-0.0134+0.0676) %exp(intercept+ age effect+time effect+marriage once before)
hf34 = 0
hf35 =  @(age) exp(-9.65573+0.08218.*age+0.02246-0.11853)
hf33 = 0
hf41 = 0
hf42 = @(age) exp(-0.4176-0.0625-0.0134.*age)
hf43 = 0
hf45 = @(age) exp(-9.65573+0.08218.*age+0.02246-0.00415) %exp(intercept+ age effect+time effect+widowhood effect)
hf44 = 0
hf51 = 0
hf52 = 0
hf53 = 0
hf55 = 0
hf54 = 0
age =30
% Evaluate matrix Q at age=30
Q30 = [-(hf15(30)+hf12(30)), hf12(30), hf13, hf14, hf15(30); ...
          hf21, -(hf23(30)+hf25(30)), hf23(30), hf24(30), hf25(30); ...
          hf31, hf32(30), -(hf32(30)+hf35(30)), hf34, hf35(30); ...
          hf41, hf42(30), hf43, -(hf42(30)+hf45(30)), hf45(30);  ...
          hf51, hf52, hf53, hf54, hf55]
It takes so long if I have to determine Q matrix for each age from 20 to 100 years. I want to figure out, how can I do this in some loop or some other way like:
% put all function handles in a matrix like 
      Q = [h11, hf12, hf13, hf14, hf15; ...
          hf21, h22, hf23, hf24, hf25; ...
          hf31, hf32, h33, hf34, hf35; ...
          hf41, hf42, hf43, h44, hf45;  ...
          hf51, hf52, hf53, hf54, hf55]
% and then determine Q for each age
% Determine the value of "Q" for ages 20 to 100.

Best Answer

Defining lots and lots of function handles is not an efficient approach. One function is much simpler:
myfun(30)
ans = 5×5
         0    0.1513         0         0    0.0008
         0         0    0.0066    0.0013    0.0008
         0    0.1619         0         0    0.0007
         0    0.4139         0         0    0.0008
         0         0         0         0         0

function hf = myfun(age)
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);
end
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