[R] Symbolic derivation using D in package stats - how do I properly convert the returned call into a character string?

[Gabor Grothendieck]

You could try using Ryacas (home page at http://code.google.com/p/ryacas/):

> library(Ryacas)
>
> u1 <- Sym("u1"); u2 <- Sym("u2"); u3 <- Sym("u3"); u4 <- Sym("u4")
> th1 <- Sym("th1"); th2 <- Sym("th2"); th3 <- Sym("th3"); th4 <- Sym("th4")
>
> C1 <- ((1 - 2 + u1^(-th1)) + (((1 - 2 + u2^(-th2)) + (((1 -
+    2 + u3^(-th3)) + u4^(-th3))^(-1/th3))^(-th2))^(-1/th2))^(-th1))^(-1/th1)
> C2 <- ((1 - 2 + u1^(-th1)) + (((1 - 2 + u2^(-th2)) +
+    u3^(-th2))^(-1/th2))^(-th1))^(-1/th1)
>
> dC1 <- deriv(C1, list("u1", "u2", "u3", "u4"))
> dC2 <- deriv(C2, list("u1", "u2", "u3", "u4"))
>
> dC1 / dC2
expression((u1^-th1 - 1 + ((u2^-th2 - 1 + ((u3^-th3 - 1 +
u4^-th3)^(-1/th3))^-th2)^(-1/th2))^-th1)^-(1/th1 +
    1) * (((u2^-th2 - 1 + ((u3^-th3 - 1 +
u4^-th3)^(-1/th3))^-th2)^(-1/th2))^-(th1 +
    1) * (th1 * ((u2^-th2 - 1 + ((u3^-th3 - 1 +
u4^-th3)^(-1/th3))^-th2)^-(1/th2 +
    1) * (th2 * u2^-(th2 + 1))))) * (th1 * th2)^2/(((u1^-th1 -
    1 + ((u2^-th2 - 1 + ((u3^-th3 - 1 +
u4^-th3)^(-1/th3))^-th2)^(-1/th2))^-th1)^-(1/th1 +
    1) * (((u2^-th2 - 1 + ((u3^-th3 - 1 +
u4^-th3)^(-1/th3))^-th2)^(-1/th2))^-(th1 +
    1) * (th1 * ((1/th2 + 1) * (th2 * u2^-(th2 + 1)) * (u2^-th2 -
    1 + ((u3^-th3 - 1 + u4^-th3)^(-1/th3))^-th2)^-(1/th2 + 2) *
    (th2 * u2^-(th2 + 1)) - (u2^-th2 - 1 + ((u3^-th3 - 1 +
u4^-th3)^(-1/th3))^-th2)^-(1/th2 +
    1) * (th2 * ((th2 + 1) * u2^-(th2 + 2))))) - th1 * ((u2^-th2 -
    1 + ((u3^-th3 - 1 + u4^-th3)^(-1/th3))^-th2)^-(1/th2 + 1) *
    (th2 * u2^-(th2 + 1))) * (((u2^-th2 - 1 + ((u3^-th3 - 1 +
    u4^-th3)^(-1/th3))^-th2)^(-1/th2))^-(th1 + 2) * ((th1 + 1) *
    ((u2^-th2 - 1 + ((u3^-th3 - 1 + u4^-th3)^(-1/th3))^-th2)^-(1/th2 +
        1) * (th2 * u2^-(th2 + 1)))))/th2) + ((u2^-th2 - 1 +
    ((u3^-th3 - 1 + u4^-th3)^(-1/th3))^-th2)^(-1/th2))^-(th1 +
    1) * (th1 * ((u2^-th2 - 1 + ((u3^-th3 - 1 +
u4^-th3)^(-1/th3))^-th2)^-(1/th2 +
    1) * (th2 * u2^-(th2 + 1)))) * ((u1^-th1 - 1 + ((u2^-th2 -
    1 + ((u3^-th3 - 1 + u4^-th3)^(-1/th3))^-th2)^(-1/th2))^-th1)^-(1/th1 +
    2) * ((1/th1 + 1) * (((u2^-th2 - 1 + ((u3^-th3 - 1 +
u4^-th3)^(-1/th3))^-th2)^(-1/th2))^-(th1 +
    1) * (th1 * ((u2^-th2 - 1 + ((u3^-th3 - 1 +
u4^-th3)^(-1/th3))^-th2)^-(1/th2 +
    1) * (th2 * u2^-(th2 + 1)))))))/th2) * (th1 * th2)^2))
> Simplify("%")
expression((u1^-th1 - 1 + ((u2^-th2 - 1 + ((u3^-th3 - 1 +
u4^-th3)^(-1/th3))^-th2)^(-1/th2))^-th1)^-(1/th1 +
    1) * (th1 * u1^-(th1 + 1)) * th1/(th1 * ((u1^-th1 - 1 + ((u2^-th2 -
    1 + u3^-th2)^(-1/th2))^-th1)^-(1/th1 + 1) * (th1 * u1^-(th1 +
    1)))))

On 11/21/06, Daniel Berg <daniel at danielberg.no> wrote:
> Dear all,
>
> I am using the function 'D' in the 'stats' package to perform symbolic
> derivation.
> This works very well and it is much faster than e.g. Mathematica (at
> least for my purposes).
> First, I would like to thank the development team for this excellent
> function.
>
> However, I run into trouble in some cases, particularly when I am to do
> some operations on long expressions obtained from using D.
> The problem seems to occur when I convert a 'call' or an 'expression' to
> a 'character', then some parts of the original 'expression' is lost.
> Is there some way of transforming a call or an expression into a
> character string that I do not know of or alternatively, can I merge two
> calls or two expressions by a numerical operation, e.g. c = a / b where
> both a and b are symbolic expressions and I wish for c to be a symbolic
> expression as well?
>
> Take e.g. the following example:
> #---------------------------------
> # First, I try with d=2, now it works:
> d<-2; u<-rep("u1",d); th<-rep("th1",d)
> for(i in 1:d) { u[i] <- paste("u",i,sep=""); th[i] <- paste("th",i,sep="") }
> dC1 <- expression((((1 - 2 + u1^(-th1)) + u2^(-th1))^(-1/th1)))
> dC2 <- expression(u1)
> #These expressions were returned from a combination of several
> expressions of mine...
> for(j in 1:(d-1)) { dC1<-D(expr=dC1,name=u[j]); dC2<-D(expr=dC2,name=u[j]) }
> # I wish to create a new expression equal to dC1 / dC2 and I proceed as
> follows:
> F <- paste("(",as.expression(dC1),")/(",as.expression(dC2),")",sep="")
> # To see that it works print dC1, dC2 and F and compare.
>
> # It also works for d=3, but for d=4 something strange happens:
> d<-4; u<-rep("u1",d); th<-rep("th1",d)
> for(i in 1:d) { u[i] <- paste("u",i,sep=""); th[i] <- paste("th",i,sep="") }
> dC1 <- expression((((1 - 2 + u1^(-th1)) + (((1 - 2 + u2^(-th2)) + (((1 -
> 2 + u3^(-th3)) + u4^(-th3))^(-1/th3))^(-th2))^(-1/th2))^(-th1))^(-1/th1)))
> dC2 <- expression((((1 - 2 + u1^(-th1)) + (((1 - 2 + u2^(-th2)) +
> u3^(-th2))^(-1/th2))^(-th1))^(-1/th1)))
> for(j in 1:(d-1)) { dC1<-D(expr=dC1,name=u[j]); dC2<-D(expr=dC2,name=u[j]) }
> F <- paste("(",as.expression(dC1),")/(",as.expression(dC2),")",sep="")
> # Now paste(dC1) throws away parts of the expression. Try
> paste(dC1)
> # and see the end of second element.
> # You will get a 3-dim char vector where there is something missing at
> the end of the second element.
> #-----------------------------------
>
> I am running R 2.2.1 on windows.
> Any help, suggestions or comments in general will be highly appreciated.
>
> Thank you.
>
> Best wishes,
> Daniel Berg
>
> ---------------------
> danielberg.no
>
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