Passing parameter to a function

I am trying to pass a couple of variable names to a xtabs formula:
tab <- function(x,y){
xtabs(time~x+y, data=D)
}

But when I run:
tab(A,B)
I get:

Error in eval(expr, envir, enclos) : object "A" not found

I am quite sure that there is some easy way out, but I have tried with different combinations of deparse(), substitute(), eval(), etc without success, can someone help?

Thanks,
Luca

Luca Meyer
www.lucameyer.com
IBM SPSS Statistics release 19.0.0
R version 2.12.1 (2010-12-16)
Mac OS X 10.6.5 (10H574) - kernel Darwin 10.5.0
I am trying to pass a couple of variable names to a xtabs formula:

 tab<- function(x,y){
     xtabs(time~x+y, data=D)
}

But when I run:

 tab(A,B)
I get:

Error in eval(expr, envir, enclos) : object "A" not found

I am quite sure that there is some easy way out, but I have tried with different combinations of deparse(), substitute(), eval(), etc without success, can someone help?
I assume that A and B are columns in D?  If so, you could use

tab(D$A, D$B)

to get what you want.  If you really want tab(A,B) to work, you'll need 
to do messy work with substitute, e.g. in the tab function, something like

fla <- substitute(time ~ x + y, list(x = substitute(x), y = substitute(y))
xtabs(fla, data=D)

Duncan Murdoch
Hi Duncan,

Yes, A and B are columns in D. Having said that I and trying to avoid 

tab(D$A,D$B)

and I would prefer:

tab(A,B)

Unfortunately the syntax you suggest is giving me the same error:

Error in eval(expr, envir, enclos) : object "A" not found

I have tried to add some deparse() but I have got the error over again. The last version I have tried:

function(x,y){
    z <- substitute(time ~ x + y, list(x = deparse(substitute(x)), y = deparse(substitute(y))))
    xtabs(z, data=D)

gives me another error:

Error in terms.formula(formula, data = data) : 
  formula models not valid in ExtractVars 

Any idea on how I should modify the function to make it work?

Thanks,
Luca

Il giorno 20/dic/2010, alle ore 19.28, Duncan Murdoch ha scritto:
On 20/12/2010 1:13 PM, Luca Meyer wrote:
I am trying to pass a couple of variable names to a xtabs formula:

 tab<- function(x,y){
    xtabs(time~x+y, data=D)
}

But when I run:

 tab(A,B)
I get:

Error in eval(expr, envir, enclos) : object "A" not found

I am quite sure that there is some easy way out, but I have tried with different combinations of deparse(), substitute(), eval(), etc without success, can someone help?
I assume that A and B are columns in D?  If so, you could use

tab(D$A, D$B)

to get what you want.  If you really want tab(A,B) to work, you'll need to do messy work with substitute, e.g. in the tab function, something like

fla <- substitute(time ~ x + y, list(x = substitute(x), y = substitute(y))
xtabs(fla, data=D)

Duncan Murdoch
Duncan's solution works for me. I strongly suspect that you have not
read the Help file for xtabs carefully enough. Does the "time" column
in D consist of a vector of counts?
A <- factor(c("A","A","B","B"))
 B <- factor(c(1,2,1,2))
 D <- data.frame(A,B)
 D$time <- c(10,12,14,16)
tab(A,B)
B
A    1  2
  A 10 12
  B 14 16

-- Bert
Hi Duncan,

Yes, A and B are columns in D. Having said that I and trying to avoid

tab(D$A,D$B)

and I would prefer:

tab(A,B)

Unfortunately the syntax you suggest is giving me the same error:

Error in eval(expr, envir, enclos) : object "A" not found

I have tried to add some deparse() but I have got the error over again. The last version I have tried:

function(x,y){
? ?z <- substitute(time ~ x + y, list(x = deparse(substitute(x)), y = deparse(substitute(y))))
? ?xtabs(z, data=D)

gives me another error:

Error in terms.formula(formula, data = data) :
?formula models not valid in ExtractVars

Any idea on how I should modify the function to make it work?

Thanks,
Luca

Il giorno 20/dic/2010, alle ore 19.28, Duncan Murdoch ha scritto:

On 20/12/2010 1:13 PM, Luca Meyer wrote:
I am trying to pass a couple of variable names to a xtabs formula:

?tab<- function(x,y){
? ? xtabs(time~x+y, data=D)
}

But when I run:

?tab(A,B)
I get:

Error in eval(expr, envir, enclos) : object "A" not found

I am quite sure that there is some easy way out, but I have tried with different combinations of deparse(), substitute(), eval(), etc without success, can someone help?
I assume that A and B are columns in D? ?If so, you could use

tab(D$A, D$B)

to get what you want. ?If you really want tab(A,B) to work, you'll need to do messy work with substitute, e.g. in the tab function, something like

fla <- substitute(time ~ x + y, list(x = substitute(x), y = substitute(y))
xtabs(fla, data=D)

Duncan Murdoch

______________________________________________
R-help at r-project.org 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.

Bert Gunter
Genentech Nonclinical Biostatistics
-----Original Message-----
From: r-help-bounces at r-project.org 
[mailto:r-help-bounces at r-project.org] On Behalf Of Luca Meyer
Sent: Monday, December 20, 2010 11:03 AM
To: Duncan Murdoch
Cc: R-help at r-project.org
Subject: Re: [R] Passing parameter to a function

Hi Duncan,

Yes, A and B are columns in D. Having said that I and trying to avoid 

tab(D$A,D$B)

and I would prefer:

tab(A,B)

Unfortunately the syntax you suggest is giving me the same error:

Error in eval(expr, envir, enclos) : object "A" not found
You didn't show what you did, but Duncan's suggestion
works for me when it is in a function:
f0 <- function (a, b) {
aExpr <- substitute(a)
    bExpr <- substitute(b)
    formula <- substitute(time ~ e1 + e2, list(e1 = aExpr, e2 = bExpr))
    xtabs(formula, data = D)
}
D <- data.frame(time=2^(0:9),A=rep(1:2,each=5),B=rep(11:13,c(3,3,4)))
z <- f0(A, B)
print(z)
B
A    11  12  13
  1   7  24   0
  2   0  32 960
xtabs(time~A+B, data=D)
B
A    11  12  13
  1   7  24   0
  2   0  32 960

The main problem with f0() is that the call attribute of
xtabs' output is always xtabs(formula=formula, data=D).
If you want the formula expanded to its value you have to
do more tricks.  E.g., use call() and eval() to pass
some arguments by value and some by name:

f1 <- function (a, b) {
    aExpr <- substitute(a)
    bExpr <- substitute(b)
    formula <- substitute(time ~ e1 + e2, list(e1 = aExpr, e2 = bExpr))
    # pass formula by value and D by name
    theCall <- call("xtabs", formula, data = quote(D))
    # may have to use enclos= argument to get D from right place
    eval(theCall)
}

or use substitute a fourth time to patch up the call
attribute:

f2 <- function (a, b) {
    aExpr <- substitute(a)
    bExpr <- substitute(b)
    formula <- substitute(time ~ e1 + e2, list(e1 = aExpr, e2 = bExpr))
    retval <- xtabs(formula, data = D)
    attr(retval, "call") <- do.call("substitute", list(attr(retval, 
        "call"), list(formula = formula)))
    retval
}

  > z <- f2(A,B) # f0 does the same
  > attr(z, "call")
  xtabs(formula = time ~ A + B, data = D)
  > z
     B
  A    11  12  13
    1   7  24   0
    2   0  32 960
I have tried to add some deparse() but I have got the error 
over again. The last version I have tried:

function(x,y){
    z <- substitute(time ~ x + y, list(x = 
deparse(substitute(x)), y = deparse(substitute(y))))
    xtabs(z, data=D)

gives me another error:

Error in terms.formula(formula, data = data) : 
  formula models not valid in ExtractVars 
Look at what 'z' is in the above:
  > g <- 
  + function(x,y){
  +     z <- substitute(time ~ x + y, list(x = 
  +         deparse(substitute(x)), y = deparse(substitute(y))))
  +     z
  + }
  > print(g(A,B))
  time ~ "A" + "B"
The strings "A" and "B" don't make sense there.  They need
to be names (a.k.a. "symbols").  Remove the calls to deparse()
and it will work.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Any idea on how I should modify the function to make it work?

Thanks,
Luca

Il giorno 20/dic/2010, alle ore 19.28, Duncan Murdoch ha scritto:

On 20/12/2010 1:13 PM, Luca Meyer wrote:
I am trying to pass a couple of variable names to a xtabs formula:

 tab<- function(x,y){
    xtabs(time~x+y, data=D)
}

But when I run:

 tab(A,B)
I get:

Error in eval(expr, envir, enclos) : object "A" not found

I am quite sure that there is some easy way out, but I 
have tried with different combinations of deparse(), 
substitute(), eval(), etc without success, can someone help?
I assume that A and B are columns in D?  If so, you could use

tab(D$A, D$B)

to get what you want.  If you really want tab(A,B) to work, 
you'll need to do messy work with substitute, e.g. in the tab 
function, something like
fla <- substitute(time ~ x + y, list(x = substitute(x), y = 
substitute(y))
xtabs(fla, data=D)

Duncan Murdoch

______________________________________________
R-help at r-project.org 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.