Kristian Nielsen - 3 years ago 163

R Question

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!

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Answer Source

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)`

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