I am creating two functions f(i) and f(j) and I want to find the value of i and j simultaneously such that the difference f(i) -f(j) is minimized.
However on running the code below I am getting an error.
I have two functions with parameter i and j as below
bu1<- function(j){
sum(linkinc_lev1$gdp*(1/(1+ (linkinc_lev1$use_gro*(1+j/100))))
}
bu1<- function(j){
sum(linkinc_lev2$gdp*(1/(1+ (linkinc_lev2$use_gro*(1+i/100))))
}
Now I need to find the value of i and j simultaneously such that difference of above functions minimized.
I was trying like
f1<- function(j,i) abs(bu1(j)-td1(i))
ans_lev1<-optimize(f1, lower=-100, upper=100),
but getting error Error in td1(i) : argument "i" is missing, with no default
Is there any way in R to minimize functions based on two parameters?
Yes, there is, almost all optimizers work on parameter vectors. You should modify your function to something like
f1<- function(param) abs(bu1(param[2])-td1(param[1]))
i.e. the function takes a single argument "param", and inside the function you fetch the values of interest out of it.
A note: if you use abs() you end up with a non-differentiable objective function. You have to select an optimizer that can handle it (SANN and Nelder-Mead can, for instance). I would rather do
f1 <- function(param) (bu1(param[2])-td1(param[1]))^2
still the same solution but now differentiable, and you can feed it to most optimizers.
Related
I have a function that I am optimizing using the optimx function in R (I'm also open to using optim, since I'm not sure it will make a difference for what I'm trying to do). I have a gradient that I am passing to optimx for (hopefully) faster convergence compared to not using a gradient. Both the function and the gradient use many of the same quantities that are computed from each new parameter set. One of these quantities in particular is very computationally costly, and it's redundant to have to compute this quantity twice for each iteration - once for the function, and again for the gradient. I'm trying to find a way to compute this quantity once, then pass it to the function and the gradient.
So here is what I am doing. So far this works, but it is inefficient:
optfunc<-function(paramvec){
quant1<-costlyfunction(paramvec)
#costlyfunction is a separate function that takes a while to run
loglikelihood<-sum(quant1)**2
#not really squared, but the log likelihood uses quant1 in its calculation
return(loglikelihood)
}
optgr<-function(paramvec){
quant1<-costlyfunction(paramvec)
mygrad<-sum(quant1) #again not the real formula, just for illustration
return(mygrad)
}
optimx(par=paramvec,fn=optfunc,gr=optgr,method="BFGS")
I am trying to find a way to calculate quant1 only once with each iteration of optimx. It seems the first step would be to combine fn and gr into a single function. I thought the answer to this question may help me, and so I recoded the optimization as:
optfngr<-function(){
quant1<-costlyfunction(paramvec)
optfunc<-function(paramvec){
loglikelihood<-sum(quant1)**2
return(loglikelihood)
}
optgr<-function(paramvec){
mygrad<-sum(quant1)
return(mygrad)
}
return(list(fn = optfunc, gr = optgr))
}
do.call(optimx, c(list(par=paramvec,method="BFGS",optfngr() )))
Here, I receive the error: "Error in optimx.check(par, optcfg$ufn, optcfg$ugr, optcfg$uhess, lower, : Cannot evaluate function at initial parameters." Of course, there are obvious problems with my code here. So, I'm thinking answering any or all of the following questions may shed some light:
I passed paramvec as the only arguments to optfunc and optgr so that optimx knows that paramvec is what needs to be iterated over. However, I don't know how to pass quant1 to optfunc and optgr. Is it true that if I try to pass quant1, then optimx will not properly identify the parameter vector?
I wrapped optfunc and optgr into one function, so that the quantity quant1 will exist in the same function space as both functions. Perhaps I can avoid this if I can find a way to return quant1 from optfunc, and then pass it to optgr. Is this possible? I'm thinking it's not, since the documentation for optimx is pretty clear that the function needs to return a scalar.
I'm aware that I might be able to use the dots arguments to optimx as extra parameter arguments, but I understand that these are for fixed parameters, and not arguments that will change with each iteration. Unless there is also a way to manipulate this?
Thanks in advance!
Your approach is close to what you want, but not quite right. You want to call costlyfunction(paramvec) from within optfn(paramvec) or optgr(paramvec), but only when paramvec has changed. Then you want to save its value in the enclosing frame, as well as the value of paramvec that was used to do it. That is, something like this:
optfngr<-function(){
quant1 <- NULL
prevparam <- NULL
updatecostly <- function(paramvec) {
if (!identical(paramvec, prevparam)) {
quant1 <<- costlyfunction(paramvec)
prevparam <<- paramvec
}
}
optfunc<-function(paramvec){
updatecostly(paramvec)
loglikelihood<-sum(quant1)**2
return(loglikelihood)
}
optgr<-function(paramvec){
updatecostly(paramvec)
mygrad<-sum(quant1)
return(mygrad)
}
return(list(fn = optfunc, gr = optgr))
}
do.call(optimx, c(list(par=paramvec,method="BFGS"),optfngr() ))
I used <<- to make assignments to the enclosing frame, and fixed up your do.call second argument.
Doing this is called "memoization" (or "memoisation" in some locales; see http://en.wikipedia.org/wiki/Memoization), and there's a package called memoise that does it. It keeps track of lots of (or all of?) the previous results of calls to costlyfunction, so would be especially good if paramvec only takes on a small number of values. But I think it won't be so good in your situation because you'll likely only make a small number of repeated calls to costlyfunction and then never use the same paramvec again.
I have written a function that takes two arguments, a number between 0:16 and a vector which contains four parameter values.
The output of the function does change if I change the parameters in the vector, but it does not change if I change the number between 0:16.
I can add, that the function I'm having troubles with, includes another function (called 'pi') which takes the same arguments.
I have checked that the 'pi' function does actually change values if I change the value from 0:16 (and it does also change if I change the values of the parameters).
Firstly, here is my code;
pterm_ny <- function(x, theta){
(1-sum(theta[1:2]))*(theta[4]^(x))*exp((-1)*theta[4])/pi(x, theta)
}
pi <- function(x, theta){
theta[1]*1*(x==0)+theta[2]*(theta[3]^(x))*exp((-1)*(theta[3]))+(1-
sum(theta[1:2]))*(theta[4]^(x))*exp((-1)*(theta[4]))
}
Which returns 0.75 for pterm_ny(i,c(0.2,0.2,2,2)), were i = 1,...,16 and 0.2634 for i = 0, which tells me that the indicator function part in 'pi' does work.
With respect to raising a number to a certain power, I have been told that one should wrap the wished number in a 'I', as an example it would be like;
x^I(2)
I have tried to do that in my code, but that didn't help either.
I can't remember the argument for doing it, but I expect that it's to ensure that the number in parentheses is interpreted as an integer.
My end goal is to get 17 different values of the 'pterm' and to accomplish that, I was thinking of using the sapply function like this;
sapply(c(0:16),pterm_ny,theta = c(0.2,0.2,2,2))
I really hope that someone can point out what I'm missing here.
In advance, thank you!
You have a theta[4]^x term both in your main expression and in your pi() function; these are cancelling out, leaving the result invariant to changes in x ...
Also:
you might want to avoid using pi as your function name, as it's also a built-in variable (3.14159...) - this can sometimes cause confusion
the advice about using the "as is" function I() to protect powers is only relevant within formulas, e.g. as used in lm() (linear regression). (It would be used as I(x^2), not x^I(2)
when we create a function why we use -(minus) in return, in R program
see this example
f <- function(pars) {
L <- (n*log(pars[1]))+(n*pars[1]*log(T1))+(n*pars[1]*log(pars[2]))-
(n*log((T1^pars[1])-(pars[2]^pars[1])))- ((pars[1]+1)*sum(log(pars[2]+x)))
return(-L)
}
here why we use (-L) in return what if I use return (L)
The sourse of example Error in f(x, ...) : argument "x" is missing, with no default in nlm
Usually they do this with the aim of maximization. When optimizing functions in R the default set up of the optimizing functions is usually to minimize. A Short cut to this is just to use a - sign in the log likelihood function which on being minimized will in turn be maximized. Although not necessary. You can be able to use the controls within the optimizing functions to indicate whether you need to maximize or minimize your log likelihood function
In this case, you use a minus sign because you a create a function which returns -L. The minus sign is not mandatory in R functions in general: you could have also written
L = -L
return(L)
I'd like to save computation time,
by avoiding running the same function with the same arguments multiple times.
given the following code:
f1 <- function(a,b) return(a+b)
f2 <- function(c,d,f) return(c*d*f)
x <- 3
y <- 4
f2(1,2,f1(x,y))
let's assume that the computation of 'f' function argument is hard,
and I'd like to cash the result somehow, so that I'd know if it had ever been executed before.
here is my main question:
I assume I can generate a key myself for f1(3,4),
example: key <- paste('f1',x,y), do my own bookkeeping and avoid running it again.
however, is it possible for f2 to generate such a key from f automatically and return it to me?
(for any function with any arguments)
if not / alternatively, before I pass f1(x,y) can I generate such a key in a generic manner,
that would work for any function with any arguments?
thanks much
Interesting question. I never thought about this.
A quick google search found this package: R.cache.
The function addMemoization takes a function as argument, and returns a function that should cache its results.
I haven't used this package myself, so I don't know how well it works, but it seems to fit what you are looking for.
Suppose you have a function f<- function(x,y,z) { ... }. How would you go about passing a constant to one argument, but letting the other ones vary? In other words, I would like to do something like this:
output <- outer(x,y,f(x,y,z=2))
This code doesn't evaluate, but is there a way to do this?
outer(x, y, f, z=2)
The arguments after the function are additional arguments to it, see ... in ?outer. This syntax is very common in R, the whole apply family works the same for instance.
Update:
I can't tell exactly what you want to accomplish in your follow up question, but think a solution on this form is probably what you should use.
outer(sigma_int, theta_int, function(s,t)
dmvnorm(y, rep(0, n), y_mat(n, lambda, t, s)))
This calculates a variance matrix for each combination of the values in sigma_int and theta_int, uses that matrix to define a dennsity and evaluates it in the point(s) defined in y. I haven't been able to test it though since I don't know the types and dimensions of the variables involved.
outer (along with the apply family of functions and others) will pass along extra arguments to the functions which they call. However, if you are dealing with a case where this is not supported (optim being one example), then you can use the more general approach of currying. To curry a function is to create a new function which has (some of) the variables fixed and therefore has fewer parameters.
library("functional")
output <- outer(x,y,Curry(f,z=2))