Can I pass more arguments to optimize() function in R

Question

The optimize function is only designed to handle one function argument as I understand it. But sometimes a function can depend on different input ranges of variables and parameters. for example: ``` x.r<-rnorm(100) y<-function(x,p) -x^2+p*1/exp(x^3) optimize(y,upper=range(x.r)[2],lower=range(x.r)[1]) ``` x.r. the domain of x being passed to y can change from sample to sample. Instead of using a generic x, I want to be able to draw from many x.r domains of values to run y over. Is there some way to modify the optimize function so that I could pass both y, the function, and x it's domain variables as well as the optimize parameter(s) p? With the above I get an error: ``` > optimize(y,upper=range(x.r)[2],lower=range(x.r)[1]) Error in p * 1 : 'p' is missing ``` edit: per suggestion below (which makes sense... but) I tried. ``` x.r<-rnorm(100) p<-seq(-5,5,1) y<-function(x,p) -x^2+p*1/exp(x^3) optimize(y,upper=range(x.r)[2],lower=range(x.r)[1],p) ``` results was: Error in p * 1 : 'p' is mis…

Accepted Answer

I call `optimise` over p

```
x.r <- rnorm(100)
p   <- seq(-5,5,1)
y<-function(x,p) -x^2+ p/exp(x^3)
ll <- lapply(p,function(i)
         optimize(y,upper=range(x.r)[2],lower=range(x.r)[1],p=i))

ll[[which.min(sapply(ll,'[[','minimum'))]]

$minimum
[1] -2.14628

$objective
[1] -98357.67
```

Of course if `maximum = TRUE` the last becomes

```
ll[[which.max(sapply(ll,'[[','maximum'))]]
```

Can I pass more arguments to optimize() function in R

Solution

Problem