John, thank you for the suggestions. I learned a lot by reading your blogs.
On Saturday, February 1, 2014 4:31:47 PM UTC-8, John Myles White wrote: > > If you want to do this, the easiest way is to define your own > implementation of the @~ macro that the latest version Julia uses to parse > expressions that look like R’s formulas. > > That will give you access to the quoted expressions you’d need to > manipulate to do your analysis. > > Given those quoted expressions, you’ll need to define a symbolic > differentiation tool that’s rich enough to handle the inputs you want to > process. The Calculus package handles symbolic differentiation for a good > chunk of functions, but you may need to extend it to your use case. > > It may be worth noting that your example makes very heavy usage of R’s > non-standard evaluation functionality, which is something that the Julia > community has not invested much time into developing yet. Most Julia > programmers tend to avoid operating on symbolic expressions. > > — John > > On Feb 1, 2014, at 3:42 PM, Walking Sparrow > <[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]> >> 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. >> > >
