konrad1991 / dfdr Goto Github PK

If an expression contains a call to a function, then we need to use the chain rule. That means we need to compute the derivative of this function. We need to find it and do this, but we can't do it in the rewriting when we compute a derivative, we need to put that code into the function we create, otherwise, we are evaluating df/dx in a context that might be different from when we later call the derivative.

The issue is similar to how global variables are handled. Now, we just leave them be in expressions, and that works because df/dx is evaluated in the same environment as f. We can't do the same thing with derived functions. If we refer to a function g, we can't compute dg/dx when we compute df/dx, we need to compute it when we call df/dx.

Hello,

I'm using R-4.2.2 with Linux Mint. When I enclose the body of a function with curly braces {}, I can't apply the dfdr functions to this function; I get the error "function { is not supported". That sounds strange because there are such examples in the documentation.

Could the approach in the function 'd' be used for more complicated functions. For instance:

f <- function(y) {
y[1] = a + y[1]*b
y[2] = y[2]*y[1]
return(y)
}

Is it actually automatic differentiation in forward or reverse mode?

Right now, I don't handle parentheses at all. This is a bit silly, but that is how it is. I just didn't include ( in the parsing of calls.

Code such as

dfdx <- d(f, x)

looks better than

dfdx <- d(f, "x")

It is simple enough to handle the former with substitute(), but then it is harder to use programmatically. With tidy-eval, though, it should be easy enough to implement it.

It should be easy enough to figure out how to simplify constant expressions. There is a little more involved in doing partial evaluation, having to do with references to scopes around f that makes expressions mutable.

It should be pretty straightforward to automatically compute the gradient and hessian of a function if we can compute the derivative with respect to any individual parameter.

Clean up the code a little and test it, and then make a release.

Hello,

Thank you for this amazing package.

I have the function f below. I'm wondering whether we can add the function d to the list of supported functions in order to compute the gradient of f, in the same way we can do it for derivatives.

d <- function(x, y ,z) 
  sqrt((-sin(x) * sin(y) + cos(x) * cos(z))^2 + (-sin(y) * sin(z) + cos(y) * cos(x))^2 + (-sin(z) * sin(x) + cos(z) * cos(y))^2)

f <- function(x, y, z)
  (
    cos(
      x - (-sin(x) * sin(y) + cos(x) * cos(z)) / d(x, y, z)
    )
    * sin(
      y - (-sin(y) * sin(z) + cos(y) * cos(x)) / d(x, y, z)
    )
    + cos(
      y - (-sin(y) * sin(z) + cos(y) * cos(x)) / d(x, y, z)
    )
    * sin(
      z - (-sin(z) * sin(x) + cos(z) * cos(y)) / d(x, y, z)
    )
    + cos(
      z - (-sin(z) * sin(x) + cos(z) * cos(y)) / d(x, y, z)
    )
    * sin(
      x - (-sin(x) * sin(y) + cos(x) * cos(z)) / d(x, y, z)
    )
  ) * (
    (
      cos(
        x + (-sin(x) * sin(y) + cos(x) * cos(z)) / d(x, y, z)
      )
      * sin(
        y + (-sin(y) * sin(z) + cos(y) * cos(x)) / d(x, y, z)
      )
      + cos(
        y + (-sin(y) * sin(z) + cos(y) * cos(x)) / d(x, y, z)
      )
      * sin(
        z + (-sin(z) * sin(x) + cos(z) * cos(y)) / d(x, y, z)
      )
      + cos(
        z + (-sin(z) * sin(x) + cos(z) * cos(y)) / d(x, y, z)
      )
      * sin(
        x + (-sin(x) * sin(y) + cos(x) * cos(z)) / d(x, y, z)
      )
    )
  )