proper work-flow with 'formula' objects and lm()

Dear all
I have a work-flow issue with lm(). When I use
lm(y1~x1, anscombe)
Call:
lm(formula = y1 ~ x1, data = anscombe)

Coefficients:
(Intercept)           x1
     3.0001       0.5001

I get as expected the formula, "y1 ~ x1", in the print()ed results or
summary(). However, if I pass through a formula object
(form <- formula(y1~x1))
y1 ~ x1
lm(form, anscombe)
Call:
lm(formula = form, data = anscombe)

Coefficients:
(Intercept)           x1
     3.0001       0.5001

then I only get the object name in the call, 'form', instead of the
formula. When passing a 'formula' object to lm, is it possible to
retain in the resulting 'lm' object the actual formula used for the
regression: "y1 ~ x1"?

Regards
Liviu
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Dear all
I have a work-flow issue with lm(). When I use
 lm(y1~x1, anscombe)
Call:
lm(formula = y1 ~ x1, data = anscombe)

Coefficients:
(Intercept)           x1
      3.0001       0.5001

I get as expected the formula, "y1 ~ x1", in the print()ed results or
summary(). However, if I pass through a formula object
 (form<- formula(y1~x1))
y1 ~ x1
 lm(form, anscombe)
Call:
lm(formula = form, data = anscombe)

Coefficients:
(Intercept)           x1
      3.0001       0.5001

then I only get the object name in the call, 'form', instead of the
formula. When passing a 'formula' object to lm, is it possible to
retain in the resulting 'lm' object the actual formula used for the
regression: "y1 ~ x1"?
It is retained.  terms(fit) will give it to you, if fit is an lm object.

Duncan Murdoch
On Thu, Nov 24, 2011 at 9:14 PM, Duncan Murdoch
It is retained. ?terms(fit) will give it to you, if fit is an lm object.

Thank you. The following works nicely.
(form <- formula(y1~x1))
y1 ~ x1
x <- lm(form, anscombe)
formula(terms(x))
y1 ~ x1

However, I was hoping that there was a way to input the 'form' object
so that summary(x) would print the underlying formula used, not the
formula object name. I was thinking of something in the style of
x <- lm(deparse(form), anscombe)
summary(x)

Can this be done? Regards
Liviu

On Thu, Nov 24, 2011 at 9:14 PM, Duncan Murdoch
<murdoch.duncan at gmail.com> wrote:
It is retained. ?terms(fit) will give it to you, if fit is an lm object.

Thank you. The following works nicely.
(form <- formula(y1~x1))
y1 ~ x1
x <- lm(form, anscombe)
formula(terms(x))
y1 ~ x1

However, I was hoping that there was a way to input the 'form' object
so that summary(x) would print the underlying formula used, not the
formula object name. I was thinking of something in the style of
x <- lm(deparse(form), anscombe)
summary(x)

Can this be done? Regards
Yes.  That's a job for substitute (the second time today).
form <- formula(y1~x1)
x <- eval(substitute(lm(f, anscombe), list(f = form)))
summary(x)
Call:
lm(formula = y1 ~ x1, data = anscombe)
...
Liviu

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Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595
On Thu, Nov 24, 2011 at 10:25 PM, Prof Brian Ripley
Yes. ?That's a job for substitute (the second time today).

form <- formula(y1~x1)
x <- eval(substitute(lm(f, anscombe), list(f = form)))
summary(x)
Call:
lm(formula = y1 ~ x1, data = anscombe)

That's what I wanted. Thanks!

However, I do want to simplify the syntax and define a new function:
x.lm <-
  function(formula, data, ...)
{
  eval(substitute(lm(f, data, ...), list(f = formula)))
}

For the simple case it works just fine
(form <- formula(y1~x1))
y1 ~ x1
x <- x.lm(form, anscombe)
But it fails when I try to input more lm() arguments:
(x <- x.lm(form, anscombe, subset=-5))
Error in eval(expr, envir, enclos) :
  ..1 used in an incorrect context, no ... to look in

Am I doing something obviously wrong? Regards
Liviu
Dear all
I have a work-flow issue with lm(). When I use
lm(y1~x1, anscombe)
Call:
lm(formula = y1 ~ x1, data = anscombe)

Coefficients:
(Intercept) ? ? ? ? ? x1
? ? 3.0001 ? ? ? 0.5001

I get as expected the formula, "y1 ~ x1", in the print()ed results or
summary(). However, if I pass through a formula object
(form <- formula(y1~x1))
y1 ~ x1
lm(form, anscombe)
Call:
lm(formula = form, data = anscombe)

Coefficients:
(Intercept) ? ? ? ? ? x1
? ? 3.0001 ? ? ? 0.5001

then I only get the object name in the call, 'form', instead of the
formula. When passing a 'formula' object to lm, is it possible to
retain in the resulting 'lm' object the actual formula used for the
regression: "y1 ~ x1"?

Try this:
form <- formula(y1~x1)
do.call("lm", list(form, quote(anscombe)))
Call:
lm(formula = y1 ~ x1, data = anscombe)

Coefficients:
(Intercept)           x1
     3.0001       0.5001
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
You would get exactly the same problem with ...,, anway.

Here's a commonly used approach in R sources:

x.lm <- function(formula, data, ...)
{
     Call <- match.call(expand.dots = TRUE)
     Call[[1]] <- as.name("lm")
     Call$formula <- as.formula(terms(formula))
     eval(Call)
}

On Thu, Nov 24, 2011 at 10:25 PM, Prof Brian Ripley
<ripley at stats.ox.ac.uk> wrote:
Yes. ?That's a job for substitute (the second time today).

form <- formula(y1~x1)
x <- eval(substitute(lm(f, anscombe), list(f = form)))
summary(x)
Call:
lm(formula = y1 ~ x1, data = anscombe)

That's what I wanted. Thanks!

However, I do want to simplify the syntax and define a new function:
x.lm <-
 function(formula, data, ...)
{
 eval(substitute(lm(f, data, ...), list(f = formula)))
}

For the simple case it works just fine
(form <- formula(y1~x1))
y1 ~ x1
x <- x.lm(form, anscombe)
But it fails when I try to input more lm() arguments:
(x <- x.lm(form, anscombe, subset=-5))
Error in eval(expr, envir, enclos) :
 ..1 used in an incorrect context, no ... to look in

Am I doing something obviously wrong? Regards
Liviu

Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595
On Thu, Nov 24, 2011 at 11:55 PM, Prof Brian Ripley
You would get exactly the same problem with ...,, anway.

Here's a commonly used approach in R sources:

x.lm <- function(formula, data, ...)
{
? ?Call <- match.call(expand.dots = TRUE)
? ?Call[[1]] <- as.name("lm")
? ?Call$formula <- as.formula(terms(formula))
? ?eval(Call)
}

From limited testing this seems to do exactly what I wanted!
I thank all those who came up with suggestions. Regards
Liviu

On Thu, 24 Nov 2011, Liviu Andronic wrote:

On Thu, Nov 24, 2011 at 10:25 PM, Prof Brian Ripley
<ripley at stats.ox.ac.uk> wrote:
Yes. ?That's a job for substitute (the second time today).

form <- formula(y1~x1)
x <- eval(substitute(lm(f, anscombe), list(f = form)))
summary(x)
Call:
lm(formula = y1 ~ x1, data = anscombe)

That's what I wanted. Thanks!

However, I do want to simplify the syntax and define a new function:
x.lm <-
?function(formula, data, ...)
{
?eval(substitute(lm(f, data, ...), list(f = formula)))
}

For the simple case it works just fine
(form <- formula(y1~x1))
y1 ~ x1
x <- x.lm(form, anscombe)
But it fails when I try to input more lm() arguments:
(x <- x.lm(form, anscombe, subset=-5))
Error in eval(expr, envir, enclos) :
?..1 used in an incorrect context, no ... to look in

Am I doing something obviously wrong? Regards
Liviu

--
Brian D. Ripley, ? ? ? ? ? ? ? ? ?ripley at stats.ox.ac.uk
Professor of Applied Statistics, ?http://www.stats.ox.ac.uk/~ripley/
University of Oxford, ? ? ? ? ? ? Tel: ?+44 1865 272861 (self)
1 South Parks Road, ? ? ? ? ? ? ? ? ? ? +44 1865 272866 (PA)
Oxford OX1 3TG, UK ? ? ? ? ? ? ? ?Fax: ?+44 1865 272595

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