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Russell Senior

2002-Jul-09 10:48 UTC

[R] building formula objects

I want to write a function to take an argument as the response
variable of a linear model, e.g. to do anova's across a list of
variables, something like the following (except, of course, this
doesn't work):

  function(x) { anova(lm(x ~ my.factor,data=my.data)) }

The x in lm() above is getting evaluated at the wrong level.  How
can I make this work? 

-- 
Russell Senior         ``The two chiefs turned to each other.        
seniorr at aracnet.com      Bellison uncorked a flood of horrible       
                         profanity, which, translated meant, `This is
                         extremely unusual.' ''
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Jonathan Baron

2002-Jul-09 11:38 UTC

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[R] building formula objects

On 07/09/02 03:48, Russell Senior wrote:>
>I want to write a function to take an argument as the response
>variable of a linear model, e.g. to do anova's across a list of
>variables, something like the following (except, of course, this
>doesn't work):
>
>  function(x) { anova(lm(x ~ my.factor,data=my.data)) }
>
>The x in lm() above is getting evaluated at the wrong level.  How
>can I make this work? 
I don't know.  But what I do instead is this:

my.coeffs <- matrix(NA,mynumber,3) # set up empty matrix
for (i in 1:my.number) {
 mycoeffs[i] <- as.numeric(try(coef(lm(Y[i]~X1[i]+X2[i]+X3[i]))[2:4]))}

The "try()" is in case it doesn't work for a given value of i
(e.g., too many missing data), and that makes "as.numeric()"
necessary. The coef() extracts the coefficients, excluding the
constant,, which is all I was interested in, but you could store
the whole thing if you use a list instead of my.coeffs.

I would like to know how to do it with a function.

Jon
-- 
Jonathan Baron, Professor of Psychology, University of Pennsylvania
R page:               http://finzi.psych.upenn.edu/
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Bill.Venables@cmis.csiro.au

2002-Jul-09 12:33 UTC

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[R] building formula objects

In response to a query from Russel Senior, Jon Baron writes:
>  -----Original Message-----
> From: 	Jonathan Baron [mailto:baron at cattell.psych.upenn.edu] 
> Sent:	Tuesday, July 09, 2002 9:38 PM
> To:	Russell Senior
> Cc:	r-help at stat.math.ethz.ch
> Subject:	Re: [R] building formula objects
> 
> On 07/09/02 03:48, Russell Senior wrote:
> >
> >I want to write a function to take an argument as the response
> >variable of a linear model, e.g. to do anova's across a list of
> >variables, something like the following (except, of course, this
> >doesn't work):
> >
> >  function(x) { anova(lm(x ~ my.factor,data=my.data)) }
> >
> >The x in lm() above is getting evaluated at the wrong level.  How
> >can I make this work? 
> 
> I don't know.  But what I do instead is this:
> 
> my.coeffs <- matrix(NA,mynumber,3) # set up empty matrix
> for (i in 1:my.number) {
>  mycoeffs[i] <- as.numeric(try(coef(lm(Y[i]~X1[i]+X2[i]+X3[i]))[2:4]))}
> 
> The "try()" is in case it doesn't work for a given value of i
> (e.g., too many missing data), and that makes "as.numeric()"
> necessary. The coef() extracts the coefficients, excluding the
> constant,, which is all I was interested in, but you could store
> the whole thing if you use a list instead of my.coeffs.
> 
> I would like to know how to do it with a function.
[WNV]  It's not too bad, but a little trickier than most people would
expect.  See the Programmer's Niche article in the latest edition of R news
for a general strategy for handling this kind of problem (at the risk of
touting my own stuff).

Bill Venables
> Jon
> -- 
> Jonathan Baron, Professor of Psychology, University of Pennsylvania
> R page:               http://finzi.psych.upenn.edu/
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Thomas Lumley

2002-Jul-09 15:35 UTC

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[R] building formula objects

On 9 Jul 2002, Russell Senior wrote:
>
> I want to write a function to take an argument as the response
> variable of a linear model, e.g. to do anova's across a list of
> variables, something like the following (except, of course, this
> doesn't work):
>
>   function(x) { anova(lm(x ~ my.factor,data=my.data)) }
>
> The x in lm() above is getting evaluated at the wrong level.  How
> can I make this work?
>
When I tried this first I thought it did work. It provides a nice example
of why R does things this way (as well as why it's useful to give an
example in help questions)

Consider a data frame
	my.data<-data.frame(a=rep(0:1,12),b=rep(0:2,8),y=rnorm(24))
and
      f<-function(x)  { anova(lm(x ~ my.factor,data=my.data)) }

If we generate a new response vector (perhaps for simulations) and want to
regress it on a and b then the function above works
nicely> z<-rnorm(24)
> f(z)Analysis of Variance Table

Response: x
          Df  Sum Sq Mean Sq F value Pr(>F)
a          1  0.0086  0.0086  0.0078 0.9305
b          1  0.1013  0.1013  0.0921 0.7645
Residuals 21 23.0914  1.0996

Here the response variable is the value of z. Presumably the reason it
`doesn't work' is that the question was different.  If we want to
specify
a column in `my.data' as the response variable we need to pass in the name
of the variable and somehow get that name into the function.

This can be done with substitute(), as discussed in more detail in the R
Newsletter. This case is fairly simple and we can use

     g<-function(x) { ff<-eval(substitute(x~a+b))
 			anova(lm(ff,data=my.data)) }
> g(y)Analysis of Variance Table

Response: y
          Df  Sum Sq Mean Sq F value Pr(>F)
a          1  2.1294  2.1294  2.3335 0.1415
b          1  0.1537  0.1537  0.1685 0.6856
Residuals 21 19.1634  0.9125

which is the correct answer as it matches
> anova(lm(y~a+b,data=my.data))Analysis of Variance Table

Response: y
          Df  Sum Sq Mean Sq F value Pr(>F)
a          1  2.1294  2.1294  2.3335 0.1415
b          1  0.1537  0.1537  0.1685 0.6856
Residuals 21 19.1634  0.9125


	-thomas

Thomas Lumley			Asst. Professor, Biostatistics
tlumley at u.washington.edu	University of Washington, Seattle

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