R help - Nov 2006 - Symbolic derivation using D in package stats - how do I properly convert the returned call into a character string?

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

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
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

On 11/21/2006 8:48 AM, Daniel Berg 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.
That's an old version, but I think the problem is still present in the 
current one.  A simple example is

paste(as.expression(as.list(1:1000)))

which on my system truncates the output just after the 121st element of 
the list.

The problem is that deparse() has a maximum line width of around 500 
characters, and you need more to fit your entire expression on one line, 
the way paste is trying to do.

I suppose we could increase the buffer size in deparse (or make it 
dynamic), but I'm not sure that it's worthwhile.  Why would you want to 
use paste on a long expression like that?  Why not leave it as an 
expression?  deparse() works fine to print it on multiple lines; I'm 
just not sure I see the value in allowing arbitrarily long lines here.

(At a minimum, I think paste() should report a "line too long" error 
rather than silently truncating; I'm not saying the current behaviour is 
fine.)

Duncan Murdoch