R help - Mar 2007 - Replacement in an expression - can't use parse()

Dear all,

Suppose I have a very long expression e. Lets assume, for simplicity, that it is

e = expression(u1+u2+u3)

Now I wish to replace u2 with x and u3 with 1. I.e. the 'new'
expression, after replacement, should be:
> e
expression(u1+x+1)

My question is how to do the replacement?

I have tried using:
> e = parse(text=gsub("u2","x",e))
> e = parse(text=gsub("u3",1,e))
Even though this works fine in this simple example, the use of parse
when e is very long will fail since parse has a maximum line length
and will cut my expressions. I need to keep mode(e)=expression since I
will use e further in symbolic derivation and division.

Any suggestions are most welcome.

Thank you.

Regards,
Daniel Berg

Hi,
perhaps you can use:

?chartr

-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22"
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On 27/03/07, Daniel Berg <daniel@danielberg.no>
wrote:
>
> Dear all,
>
> Suppose I have a very long expression e. Lets assume, for simplicity, that
> it is
>
> e = expression(u1+u2+u3)
>
> Now I wish to replace u2 with x and u3 with 1. I.e. the 'new'
> expression, after replacement, should be:
>
> > e
> expression(u1+x+1)
>
> My question is how to do the replacement?
>
> I have tried using:
>
> > e = parse(text=gsub("u2","x",e))
> > e = parse(text=gsub("u3",1,e))
>
> Even though this works fine in this simple example, the use of parse
> when e is very long will fail since parse has a maximum line length
> and will cut my expressions. I need to keep mode(e)=expression since I
> will use e further in symbolic derivation and division.
>
> Any suggestions are most welcome.
>
> Thank you.
>
> Regards,
> Daniel Berg
>
> ______________________________________________
> R-help@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.
>
	[[alternative HTML version deleted]]

Try substitute:

e <- expression(u1 + u2 + u3)
L <- list(u2 = as.name("x"), u3 = 1)
as.expression(do.call(substitute, list(as.call(e), L))[[1]])


On 3/27/07, Daniel Berg <daniel at danielberg.no>
wrote:
> Dear all,
>
> Suppose I have a very long expression e. Lets assume, for simplicity, that
it is
>
> e = expression(u1+u2+u3)
>
> Now I wish to replace u2 with x and u3 with 1. I.e. the 'new'
> expression, after replacement, should be:
>
> > e
> expression(u1+x+1)
>
> My question is how to do the replacement?
>
> I have tried using:
>
> > e = parse(text=gsub("u2","x",e))
> > e = parse(text=gsub("u3",1,e))
>
> Even though this works fine in this simple example, the use of parse
> when e is very long will fail since parse has a maximum line length
> and will cut my expressions. I need to keep mode(e)=expression since I
> will use e further in symbolic derivation and division.
>
> Any suggestions are most welcome.
>
> Thank you.
>
> Regards,
> Daniel Berg
>
> ______________________________________________
> 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.
>

Daniel Berg said the following on 3/27/2007 6:56 AM:
> Dear all,
> 
> Suppose I have a very long expression e. Lets assume, for simplicity, that
it is
> 
> e = expression(u1+u2+u3)
> 
> Now I wish to replace u2 with x and u3 with 1. I.e. the 'new'
> expression, after replacement, should be:
> 
>> e
> expression(u1+x+1)
> 
> My question is how to do the replacement?
> 
> I have tried using:
> 
>> e = parse(text=gsub("u2","x",e))
>> e = parse(text=gsub("u3",1,e))
> 
> Even though this works fine in this simple example, the use of parse
> when e is very long will fail since parse has a maximum line length
> and will cut my expressions. I need to keep mode(e)=expression since I
> will use e further in symbolic derivation and division.
> 
> Any suggestions are most welcome.
> 
> Thank you.
> 
> Regards,
> Daniel Berg
> 
> ______________________________________________
> 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.

Take a look at ?substitute (or ?bquote)

 > as.expression(substitute(u1+u2+u3, list(u1 = quote(x))))
expression(x + u2 + u3)
 > as.expression(bquote(u1+u2+.(u3), list(u3 = 1)))
expression(u1 + u2 + 1)

HTH,

--sundar

you could try something like the following (untested):

new.e <- eval(substitute(expression(u1+u2+u3), list(u2 = x, u3 = 1)))


I hope it helps.

Best,
Dimitris

----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm



----- Original Message ----- 
From: "Daniel Berg" <daniel at danielberg.no>
To: <r-help at stat.math.ethz.ch>
Sent: Tuesday, March 27, 2007 3:56 PM
Subject: [R] Replacement in an expression - can't use parse()

> Dear all,
>
> Suppose I have a very long expression e. Lets assume, for 
> simplicity, that it is
>
> e = expression(u1+u2+u3)
>
> Now I wish to replace u2 with x and u3 with 1. I.e. the 'new'
> expression, after replacement, should be:
>
>> e
> expression(u1+x+1)
>
> My question is how to do the replacement?
>
> I have tried using:
>
>> e = parse(text=gsub("u2","x",e))
>> e = parse(text=gsub("u3",1,e))
>
> Even though this works fine in this simple example, the use of parse
> when e is very long will fail since parse has a maximum line length
> and will cut my expressions. I need to keep mode(e)=expression since 
> I
> will use e further in symbolic derivation and division.
>
> Any suggestions are most welcome.
>
> Thank you.
>
> Regards,
> Daniel Berg
>
> ______________________________________________
> 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.
> 

Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

A way to do this is through substitute:

e1 <- substitute(expression(u1 + u2 + u3), list(u2=quote(x), u3=1))

But I am not sure whether you will run into similar limitations regarding
the length of the expression to be substituted. 

-Christos

Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
 

> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch 
> [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Daniel Berg
> Sent: Tuesday, March 27, 2007 9:57 AM
> To: r-help at stat.math.ethz.ch
> Subject: [R] Replacement in an expression - can't use parse()
> 
> Dear all,
> 
> Suppose I have a very long expression e. Lets assume, for 
> simplicity, that it is
> 
> e = expression(u1+u2+u3)
> 
> Now I wish to replace u2 with x and u3 with 1. I.e. the 'new'
> expression, after replacement, should be:
> 
> > e
> expression(u1+x+1)
> 
> My question is how to do the replacement?
> 
> I have tried using:
> 
> > e = parse(text=gsub("u2","x",e))
> > e = parse(text=gsub("u3",1,e))
> 
> Even though this works fine in this simple example, the use 
> of parse when e is very long will fail since parse has a 
> maximum line length and will cut my expressions. I need to 
> keep mode(e)=expression since I will use e further in 
> symbolic derivation and division.
> 
> Any suggestions are most welcome.
> 
> Thank you.
> 
> Regards,
> Daniel Berg
> 
> ______________________________________________
> 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.
> 
>

Daniel Berg wrote:
> Dear all,
>
> Suppose I have a very long expression e. Lets assume, for simplicity, that
it is
>
> e = expression(u1+u2+u3)
>
> Now I wish to replace u2 with x and u3 with 1. I.e. the 'new'
> expression, after replacement, should be:
>
>   
>> e
>>     
> expression(u1+x+1)
>
> My question is how to do the replacement?
>
> I have tried using:
>
>   
>> e = parse(text=gsub("u2","x",e))
>> e = parse(text=gsub("u3",1,e))
>>     
>
> Even though this works fine in this simple example, the use of parse
> when e is very long will fail since parse has a maximum line length
> and will cut my expressions. I need to keep mode(e)=expression since I
> will use e further in symbolic derivation and division.
>
> Any suggestions are most welcome.
>   
The short answer is substitute().

However, this is not entirely trivial to apply if you have your
expression already inside an expression() object.

The easy thing to do is
> substitute(u1+u2+u3, list(u2=quote(x),u3=1))
u1 + x + 1

but notice that this "autoquotes" the first argument, so
> substitute(e, list(u2=quote(x),u3=1))
e

which is pretty much useless.

(Arguably it would have been a better design to avoid this feature and
require substitute(quote(.....)....) for the former case.)

The way around this is to add a further layer of substitute() to insert
the value of e:
>
eval(substitute(substitute(call,list(u2=quote(x),u3=1)),list(call=e[[1]])))
u1 + x + 1

Notice that substitute will not go inside expression objects, so we need
to extract the mode "call" object using e[[1]]. Also, the result is
"call" not "expression". You may need an as.expression
construct around
the result to get exactly what you asked.

-- 
   O__  ---- Peter Dalgaard             ?ster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark          Ph:  (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)                  FAX: (+45) 35327907

On Tue, 2007-03-27 at 15:56 +0200, Daniel Berg wrote:
> Dear all,
> 
> Suppose I have a very long expression e. Lets assume, for simplicity, that
it is
> 
> e = expression(u1+u2+u3)
> 
> Now I wish to replace u2 with x and u3 with 1. I.e. the 'new'
> expression, after replacement, should be:
> 
> > e
> expression(u1+x+1)
> 
> My question is how to do the replacement?
> 
> I have tried using:
> 
> > e = parse(text=gsub("u2","x",e))
> > e = parse(text=gsub("u3",1,e))
> 
> Even though this works fine in this simple example, the use of parse
> when e is very long will fail since parse has a maximum line length
> and will cut my expressions. I need to keep mode(e)=expression since I
> will use e further in symbolic derivation and division.
> 
> Any suggestions are most welcome.
> 
> Thank you.
> 
> Regards,
> Daniel Berg
Here is one possibility, depending upon how complicated the
substitutions end up being:

e <- expression(u1+u2+u3)

TMP <- unlist(strsplit(as.character(e), " "))
> TMP
[1] "u1" "+"  "u2" "+"  "u3"

TMP <- gsub("u2", "x", TMP)

TMP <- gsub("u3", "i", TMP)
> TMP
[1] "u1" "+"  "x"  "+"  "i" 
> as.expression(paste(TMP, collapse = ""))
expression("u1+x+i")

HTH,

Marc Schwartz