Can't you just do this with apply? Something like this:
f = (x, y, z) -> x + y + z^2
let x=3, y=4, z=5
apply(f, x, y, z)
end
On Sunday, February 2, 2014 5:47:36 PM UTC+2, Walking Sparrow wrote:
>
> Let me clarify a little bit. My question is actually the following:
>
> In R, one can do something like
>
> > f <- function(x1, x2, x3, x4, x5, x6) { some expressions that you like
> to use }
> > evaluate.at <- list(x1 = 2, x2 = 2.3, x3 = 2, x4 = 1.2, x5 = 3.4, x6 =
> 5.6)
> > do.call(f, evaluate.at) # get the value
>
> "do.call" can accept any valid function, and a list of variable values.
> There is no restriction to the function or the number of variables.
> Sometimes this is very useful.
>
> How do we do similar things in Julia? Does Julia have a function or macro
> similar to "do.call"?
>
> On Sunday, February 2, 2014 4:09:05 AM UTC-8, Mauro wrote:
>>
>> I don't quite comprehend your problem, so maybe this doesn't help. But
>> as far as I can tell, there is no need for macros:
>>
>> Just define your function as a normal function, which can be evaluated
>> for any (x,z):
>>
>> julia> f(x,z) = x + x*z + log(sin(x) + 2) * log(cos(z) + 2)
>> f (generic function with 1 method)
>> julia> x0 = 2.3; z0 = 1.4;
>> julia> f(x0,z0)
>> 6.302488546391614
>>
>> For the first step: use automatic differentiation on the function in
>> question. The package https://github.com/scidom/DualNumbers.jl can do
>> this:
>>
>> julia> using DualNumbers
>> julia> xdu = dual(x0,1)
>> 2.3 + 1.0du
>>
>> # this gives (f(x0,z0), \partial f / \partial x at (x0,z0)) :
>> julia> f(xdu,z)
>> 6.302488546391614 + 2.2120074784516013du
>>
>> julia> zdu = dual(z,1)
>> 1.4 + 1.0du
>> # this gives (f(x0,z0), \partial f / \partial z at (x0,z0)) :
>> julia> f(x,zdu)
>> 6.302488546391614 + 1.8413102782559223du
>>
>> (double check that the derivatives are right but I think that is how it
>> should work)
>>
>> On Sat, 2014-02-01 at 23:42, [email protected] wrote:
>> > You are right about that I have an R background. What I am trying to do
>> is
>> > to evaluate a function given by the user. For example,
>> >
>> > I want to write a function that can compute the marginal effects of a
>> > linear or logistic model. For simplicity, let's just use linear
>> regression.
>> > If the user did a linear regression using the following model (I am
>> using
>> > the formula syntax from R)
>> >
>> > y ~ x + z + sin(x) * sin(z) for the data set my_data, which has three
>> > columns x, y, and z
>> >
>> > Then the marginal effects at the mean are computed like this: First,
>> > compute the first derivative of 1+ x + z + sin(x) * sin(z). This can be
>> > done in R using the function "deriv" to get the expression of the first
>> > derivative. In the second step, I need to substitute the mean values of
>> x
>> > and z into the result of the first step. An example of this would be
>> the
>> > "margins" function in the R package "PivotalR"
>> > (http://cran.r-project.org/web/packages/PivotalR/ and
>> > https://github.com/gopivotal/PivotalR)
>> >
>> > Right now, I have no idea how to do the first step in Julia. But that
>> is
>> > OK, because I just started learning Julia.
>> >
>> > Now my question is in the second step. The user can use any complex
>> > expressions in the linear regression like y ~ x + x*z + log(sin(x) + 2)
>> *
>> > log(cos(z) + 2), and the data set my_data and formula can have any
>> number
>> > of variables like x1, x2, ...., x1000. So when you write the code for
>> the
>> > value substitution in the second step, you cannot know which function
>> and
>> > what variables you will have.
>> >
>> > So in Julia or R, I need a function or macro F(f, [....]) that does
>> this:
>> > given a function f, whose format is the input from the user, and a set
>> of
>> > variable values [...], whose number and names are also the input from
>> the
>> > user, F(f, [...]) returns the value of f evaluated at the values [...].
>> For
>> > example, the user inputs
>> >
>> > f = 1 + z + cos(x)*log(2+cos(z))/(2+sin(x))
>> >
>> > and [x = 2.3, z = 1.4],
>> >
>> > F should return the value of f evaluated at x = 2.3 and z = 1.4.
>> >
>> > This can be done in R, see "margins" function in PivotalR, which
>> actually
>> > does big data computation in-database. The problem is how to do the
>> same
>> > thing in Julia?
>> >
>> > Hope my explanation makes my question clearer.
>> >
>> > On Saturday, February 1, 2014 2:35:38 PM UTC-8, Jameson wrote:
>> >>
>> >> You need to provide more detail on what you are trying to do with
>> this.
>> >> You seem
>> >> to be confusing several concepts involving the usage of expressions,
>> >> macros, and functions. I can't tell if you are trying to write special
>> >> syntax, or are just unaware of anonymous functions:
>> >>
>> >> Mostly, why is :(sin(x) + cos(y) * sin(z)) an expression, and not a
>> >> function? It seems like you perhaps have an R background?
>> >>
>> >> f(x,y,z) = (sin(x) + cos(y) * sin(z))
>> >> f(1,2,3)
>> >>
>> >> On Sat, Feb 1, 2014 at 12:04 PM, Walking Sparrow
>> >> <[email protected]<javascript:>>
>>
>> >> wrote:
>> >> > So the real question is how to generate a code block like this
>> >> >
>> >> > quote
>> >> > x = 2
>> >> > y = 3
>> >> > .....
>> >> > x + y + ....
>> >> > end
>> >> >
>> >> > Need to embed a for loop inside the macro definition?
>> >> >
>> >> >
>> >> >
>> >> > On Saturday, February 1, 2014 8:52:30 AM UTC-8, Walking Sparrow
>> wrote:
>> >> >>
>> >> >> Please forgive me if this is a stupid question. Suppose I have an
>> >> >> expression
>> >> >>
>> >> >> :(sin(x) + cos(y) * sin(z))
>> >> >>
>> >> >> and the values of x, y, z.
>> >> >>
>> >> >> How can I write a macro that can substitute the values of x, y, z
>> into
>> >> the
>> >> >> above expression? The number of values that I want to substitute
>> >> depends on
>> >> >> the actual use cases and thus is unknown.
>> >> >>
>> >> >> I wrote a function that can do this
>> >> >>
>> >> >> function substitute(expr::Expr, vals::Array{Expr,1})
>> >> >> for i = 1:length(vals)
>> >> >> @eval $(vals[i])
>> >> >> end
>> >> >> @eval $expr
>> >> >> end
>> >> >>
>> >> >> x = 10
>> >> >> y = 23
>> >> >>
>> >> >> substitute(:(x+y), [:(x = 2), :(y = 3)])
>> >> >>
>> >> >> x
>> >> >> y
>> >> >>
>> >> >> But if you run the above code, you will see that the values of
>> global x
>> >> >> and y are changed, which is not what I intend to do. This is
>> because
>> >> "eval"
>> >> >> does the evaluation in the global scope. Besides, I think it is a
>> bad
>> >> coding
>> >> >> pattern to use eval and it is slow.
>> >> >>
>> >> >> It would be better if this can be done using macro. But I have no
>> idea
>> >> >> about how to do this.
>> >>
>>
>> --
>> Sent with my mu4e
>>
>>
