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] <javascript:> 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
>
>
