devinbost - 8 months ago 35

R Question

I am able to compute a symbolic derivative in R for the logit function with this statement:

`deriv(quote(exp(-9.3 + 0.0146*x)/(1 + exp(-9.3 + 0.0146*x))),"x")`

The result is an expression:

`expression({`

.expr4 <- exp(-9.3 + 0.0146 * x)

.expr5 <- 1 + .expr4

.expr7 <- .expr4 * 0.0146

.value <- .expr4/.expr5

.grad <- array(0, c(length(.value), 1L), list(NULL, c("x")))

.grad[, "x"] <- .expr7/.expr5 - .expr4 * .expr7/.expr5^2

attr(.value, "gradient") <- .grad

.value

})

However, when I try to return the expression in a function, such as like this:

`DerivLogit <- function(x){`

deriv(expression(exp(-9.3 + 0.0146*x)/(1 + exp(-9.3 + 0.0146*x))),"x")

}

evaluating DerivLogit(x) of course doesn't substitute the parameter x for the variable x in my expression. And so, for example, DerivLogit(1) equals DerivLogit(2), both of which simply return the expression without any parameter substitution.

Is there a way to convert the derivative expression into a function that I can evaluate where the parameter (e.g. x) will substitute so that I can see the numerical outcome for the given value of x?

And if so, how do I do so in R?

Answer Source

You can use `eval`

to evaluate the expression

```
DerivLogit <- function(x){
eval(deriv(expression(exp(-9.3 + 0.0146*x)/(1 + exp(-9.3 + 0.0146*x))),"x"))
}
```