How to calculate number of function evaluation used?

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Hi guys, can some one help me to solve my problem?
i've be facing difficulty to find the number of function evaluation. let's say that i have f(x)=x^3+x^2-2 with initial value of x(0)=1.5
i want to calculate x(n+1)=x(n)-(f(x(n)/fprime(x(n))) until n=5
thus, i want to calculate how many time of f(x(n)) have been use until n=5. but if the first iteration has use f(x(1)) then the second iteration also use f(x(1)), it is considered as we only use f(x(1)) one time only.
can someone help me

Risposta accettata

Steven Lord
Steven Lord il 20 Giu 2017
If you're using release R2017a or later you could use memoizedFunction. Once you've used memoizedFunction to solve the problem, call stats on it to determine how many times and with which inputs it was called.
  3 Commenti
Steven Lord
Steven Lord il 21 Giu 2017
Yes, that's how I would do it. The first time you call the function with a new set of inputs, that increments TotalMisses. Each subsequent time you call the function with inputs that are already in the cache, that increments TotalHits.
Tamas Kis
Tamas Kis il 23 Dic 2022
Modificato: Tamas Kis il 23 Dic 2022
Found a way to do this using a handle class: https://www.mathworks.com/matlabcentral/fileexchange/122422-counted-function-countedfunction. In an example I included it is about 5 times faster than using a memoized function.
If you have a function handle f, you can simply do
[f,count] = count_function_calls(f)
and keep using f in the same way as before. When you want to know the number of function calls, you just use
count()

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Più risposte (2)

Torsten
Torsten il 20 Giu 2017
When n=5, f and fprime have been called 5 times each, namely to evaluate f(x1),f(x2),...,f(x5), fprime(x1),fprime(x2),...,fprime(x5).
Best wishes
Torsten.
  4 Commenti
Torsten
Torsten il 20 Giu 2017
Modificato: Torsten il 20 Giu 2017
f=@(x)x^3+x^2-2;
fprime=@(x)3*x^2+2*x;
x=4;
for n=1:5
x=x-f(x)/fprime(x);
end
Best wishes
Torsten.
NUR ZAFIRAH MOHD SIDEK
NUR ZAFIRAH MOHD SIDEK il 20 Giu 2017

@Jan Simon: ops..sorry..because this is my first time.

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Jan
Jan il 20 Giu 2017
function fx = f(x)
persistent Count
if isempty(Count)
Count = 0;
end
if nargin == 0 % Reply counter and reset it
fx = Count;
Count = 0;
return;
end
fx = x^3+x^2-2;
Count = Count + 1;
end
Now:
f();
x = f(1) + f(2) + f(3);
Count = f()
  2 Commenti
NUR ZAFIRAH MOHD SIDEK
NUR ZAFIRAH MOHD SIDEK il 21 Giu 2017
and is it the same process if i want to find more than one function at the same time?
Jan
Jan il 21 Giu 2017
I do not understand you question. This example shows how to insert and request a counter ina function. Of course you can do exactly the same with other functions also.

Accedi per commentare.

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