Is it possible to creare a matrix of functions?

Hello, i am working on a code for my university. Is it possible to create a vector of 229 functions in a way that if i select an element, for exemple tc(3), as in the image, i can work on the 3st function of my vector. Thank you very much!

[Guillaume]

In your image, what is the input variable of each function? Are you sure that you actually mean function?

[Stephen23]

It is certainly possible to create a cell array of function handles, but I very much doubt that such a thing would be an efficient way to code what you show. Most likely some basic vectorized code would suffice, with some simple indexing afterwards.

[Walter Roberson]

A key word in what Stephen said is "cell array". It is not possible to create a plain array of function handles.

F = [@(x) x.^2, @(x) sin(x)]      %this is an error
F = {@(x) x.^2, @(x) sin(x)}      %this is fine

[Riccardo Bartoloni]

If you want, you have the entire part of the code at this post: https://it.mathworks.com/matlabcentral/answers/459500-how-can-i-find-the-optimal-parameters-that-minimize-sse?s_tid=prof_contriblnk

For Guillaume

Yes, i want to have different functions in my vector. I have to minimize SSE finding 15 parameters. x(4) in this example should be the fourth parameter.

For Stephen Cobeldick

My scope is to minimize the last line of my code (that you find in the first link), so i want to write this last line as a function. Do you think that this is not the correct method?

For Walter Roberson

I tried to modify the initial part of my code with something like that:

clc
clear
t=0.33333: 0.333333: 76.333333; %length(t)=229
load('DUM.mat'); %Matrice 4x229
t2=t;
t2(length(t2))=[ ]
t2=[0 t2];
tc=[];
for i=1:229
summ= @(x) t(i).^x(4)-t2(i).^x(4);
tc={tc summ}; %1x229-1x229=1x229 
end

but as a result for tc i don't have a vector of function (1x229) but only a 1x2 cell.

[Walter Roberson]

tc{i} = summ;

[Riccardo Bartoloni]

Thank you Walter Roberson, now i think i am nearest from the solution. But i would like to give value to each t(i) present in vector tc, assigning the correspandant value of vector t (1x229). This is my code now. As a solution i have 1x229 functions but each element is the same general function[@(x)t(i)^x(4)-t2(i)^x(4)]. On the contrary i would like to have for example for i=1: @(x) 0.333^x(4)-0^x(4), for i=2: @(x) 0.666^x(4)-0.333^x(4)

clc
clear
t=0.33333: 0.333333: 76.333333; %length(t)=229
load('DUM.mat'); %Matrice 4x229
expbeta=@(x)exp(x(5:8)*DUM); %1x4*4x229=1x229
t2=t;
t2(length(t2))=[ ]
t2=[0 t2];
tc=[];
for i=1:229
summ= @(x) t(i)^x(4)-t2(i)^x(4);
tc{i}=summ; %1x229-1x229=1x229 
end

[John D'Errico]

Do you mean like this?

f = @(t1,t2,x) t1.^x(4) - t2.^x(4);
t = 0:0.2:1;
x0 = [2 3 5 7];

f(t(2:end), t(1:end-1),x0)
ans =
     1.28e-05    0.0016256     0.026355      0.18172      0.79028

I think you just need to learn to work with function handles, with functions in general, and to learn some more MATLAB, learning how to vectorize your code, etc..

[Stephen23]

"Do you think that this is not the correct method?"

I very much doubt that you need to create so many function handles (doing so entirely defeats the point of functions). Your code certainly does not make use of any of MATLAB vectorization abilities, which means you are avoiding one of MATLAB'S most powerful features.

Most likely you could use some of the standard optimization tools:

https://www.mathworks.com/products/optimization.html

You might like to read this

https://www.mathworks.com/matlabcentral/answers/228557-experts-of-matlab-how-did-you-learn-any-advice-for-beginner-intermediate-users

[Riccardo Bartoloni]

Thank you very much Stephen Cobeldick! Thanks to your suggestions, using vectorization, i was able to delete a lot of double "for" loop, and now my code is more or less 20 times faster.....Sorry but i am a beginner at using Matlab... :D :D

I paste my new code. My fundamental question is: how can i do to find the optimal parameters (vector Param) that minimize SSE (in my code called "SQ_TOT"), without knowing these parameters?

clc
clear
t=0.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333:0.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333:76.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333;
Param=[0.5254010 11.4402640 29861.1830810 2.0010250 -0.051606000 -0.057373000 0.036517000 -0.099394000 1.647586000 81.195773000 1.422753000 -0.078617000 0.036363000 0.106596000 0.128553000]
load('DUM.mat'); %4x229
DUMqA=[Param(5:8)];
expbeta=exp(DUMqA*DUM);
t1=t.^Param(4);
t2=t.^Param(4);
t2(length(t))=[];
t2=[0 t2];
tc=t1-t2;
lt=length(t);
cohort=zeros(1,lt);

for i=1:length(t1);
  summ=[zeros(1,i-1) expbeta(i:lt).*tc(1:lt+1-i)];
  cohort=[cohort;summ];
end;
cohort(1,:)=[];   

HEY_A=cumsum(cohort,2);

F_A=(1-(Param(3)./(Param(3)+HEY_A)).^Param(2)).*(1-Param(1));
%%%%Retention Model%%%%%
DUMqR=[Param(12:15)];

expbeta_R=exp(DUMqR*DUM);

t1R=t.^Param(11);
t2R=t.^Param(11);
t2R(length(t))=[];
t2R=[0 t2R];
tc_R=t1R-t2R;


cohort_R=[];
%cohort_R= matrice riga 1000
for i=1:length(t1);
  summ=[zeros(1,i-1) expbeta_R(i:length(t1)).*tc_R(1:length(t)+1-i)];
  cohort_R=[cohort_R;summ];
end;
cohort_R(1,:)=[];

HEY_R=cumsum(cohort_R,2);

F_R=(1-(Param(10)./(Param(10)+HEY_R)).^Param(9));

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%Addition%%%%
load('housed.mat'); %1x229
HouseCha1=housed;
%HouseCha1= riga 19 
housed(229)=[];
HouseCha2=[0 housed];
ChaninPop=HouseCha1-HouseCha2;
 
F_AZ=[zeros(lt,1) F_A];

EXP_A=[];
EXP_AA=[];

EC_RVERO=zeros(1,lt);

EC=[];
%EC= riga 976 oppure colonna IA e riga 1712 
F_RZ=[zeros(228,1) F_R];
F_AZ(:,end)=[];
F_RZ(:,end)=[];
ChaninPopR=ChaninPop(1:end-1);
EC_RVERO=[0];
for i=1:lt
EXP_AA=(F_A-F_AZ).*(ChaninPop+EC_RVERO)';
EC=sum(EXP_AA);
EC(lt)=[];
EXP_RR=(F_R-F_RZ).*(EC)';
EC_RVERO=sum(EXP_RR);
EC_RVERO=[0 EC_RVERO];
EC_RVERO(end)=[];
end
EC=[EC sum(EXP_AA(:,end))];
EC_RVERO(1)=[];
EC_RVERO=[EC_RVERO sum(EXP_RR(:,end))];  

EC3m_R=[0 0];
for i=3:lt;
    summ=sum(EC_RVERO(i-2:i));
    EC3m_R=[EC3m_R summ];
end;

EC3m=[EC(1) sum(EC(1:2))];
for i=3:lt;
    summ=sum(EC(i-2:i));
    EC3m=[EC3m summ];
end;

load('DishADD.mat'); %1x229

DIV_=ADD~=0;
DIVEC3=DIV_.*EC3m;
DIV_A=(DIVEC3-ADD);
DIV_A2=(DIV_A).^2;
SQ_ADD=sum(DIV_A2(1,:));

load('LOSS.mat'); %1x229
DIV=LOSS~=0;
DIV_L=DIV.*EC3m_R;
DIV_LOSS=(DIV_L-LOSS)
DIV_LOSS2=(DIV_LOSS).^2
SQ_LOSS=sum(DIV_LOSS2(1,:));


EXP_E=EC-EC_RVERO;
EXP_END=[];    
for i=1:lt;
    summ=sum(EXP_E(1:i));
    EXP_END=[EXP_END summ];
end;

load('END.mat');
DIV1=END~=0;
DIV2=DIV1-DIV_;
DIV_E=DIV2.*EXP_END;
END_DIV=DIV2.*END;

DIV_E(1)=EXP_END(1)

SQ_EN=[(EXP_END(1))^2];
pos=find(DIV2==1);

for i=4:3:pos(length(pos));
DIV_END=((DIV_E(i)-DIV_E(i-3))-(END_DIV(i)-END_DIV(i-3)))^2;
SQ_EN=[SQ_EN DIV_END];
end;
SQ_END=sum(SQ_EN(1,:));
SQ_END=SQ_END+(END(length(END))-EXP_END(length(EXP_END)))^2;

SQ_TOT= SQ_LOSS+SQ_END+SQ_ADD

[Guillaume]

Note that there's never any point(*) to write numbers with that many digits:

t=0.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333:0.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333:76.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333;

A number of class double can only store about 16 different digits and matlab will ignore the extra digits.

t = 0.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333
%and
t = 0.3333333333333333
%and the simpler
t = 1/3

will all result in the exact same value stored in memory (which if decoded in decimal by James Tursa's excellent num2strexact is 0.333333333333333314829616256247390992939472198486328125, the best approximation to 1/3 you can get with double)

The best way to write your expression is thus:

t = 1/3 : 1/3 : 76+1/3;

------

* others than for messing up the formatting of activity feed on this forum, it turns out!

Answer 1

[John D'Errico]

Do you want to create function handles? For example,

fmat = {@(x) sin(x), @(x) cos(x), @(x) exp(x), @(x) x^2 + 1};

fmat{1}(2)
ans =
       0.9093

fi = fmat{2}
fi =
  function_handle with value:
    @(x)cos(x)
    

Or you could create an array of symbolic objects.

syms x
fmat = [x, x^2+2, (x+1) -2*x];

then you could use tham as functions too.

matlabFunction(fmat(1))
ans =
  function_handle with value:
    @(x)x

A virtue of an array of symbolic objects is you can do many things with them all.

int(fmat,x)
ans =
[ x^2/2, (x*(x^2 + 6))/3, (x*(x + 2))/2, -x^2]

And you can even use them all at once, in one call.

fall = matlabFunction(fmat)
fall =
  function_handle with value:
    @(x)[x,x.^2+2.0,x+1.0,x.*-2.0]
>> fall(3)
ans =
     3    11     4    -6

[Riccardo Bartoloni]

Thank you John D'Errico! And how i have to do if i want to create this vector with a for loop?

tc={@(x)t(1)^x(4)-t2(1)^x(4), @(x)t(2)^x(4)-t2(2)^x(4),...
    @(x)t(229)^x(4)-t2(229)^x(4)}

[John D'Errico]

I think you are making it more complex than it need be.

It would seem easiest if you made it a function of three inputs. So now you would need only one function.

tc = @(x,t1,t2) t1.^x(4) - t2.^x(4);
tc([1 3 5 4],1:3,2:2:6)
ans =
         -15        -240       -1215      

So, you can use tc as a vectorized function handle. No need to have 229 function handles, because they all are just the same function, with different values of t.

[Riccardo Bartoloni]

But in my case i don't know the value of vector x. Because at the end i have to find the optimal vector x that minimize SSE. Maybe looking at this post it should be clearest https://it.mathworks.com/matlabcentral/answers/459500-how-can-i-find-the-optimal-parameters-that-minimize-sse?s_tid=prof_contriblnk

[Riccardo Bartoloni]

I am arrived here. How i have to do if i want to create a new function that is the multiplication between the second function of expbeta (expbeta{2}) and the first function of tc (tc{1})?

clc
clear
t=0.33333: 0.333333: 76.333333; %length(t)=229
load('DUM.mat'); %Matrice 4x229
t2=t;
t2(length(t2))=[ ]
t2=[0 t2];
lt=length(t);
for i=1:lt;
summ= @(x) t(i)^x(4)-t2(i)^x(4);
tc{i}=summ; %1x229-1x229=1x229 
end
for i=1:lt;
summ=@(x)exp(x(5:8)*DUM(1:4,i));
expbeta{i}=summ;
end