# [R] defining a formula method for a weighted lm()

Michael Friendly friendly at yorku.ca
Fri Jan 7 15:39:53 CET 2011

```Thanks, Martin

Now I understand 'standard non-standard evaluation' magic, and the code in
http://developer.r-project.org/model-fitting-functions.txt
explains how this works.

Still, I can't help but think of this as evil-magic, for which some
anti-magic would be extremely useful, so that
a simple function like

my.lm <- function(formula, data, subset, weights, ...) {
lm(formula, data, subset, weights, ...)
}

would work as expected. Oh dear, I think I need some syntactic sugar in
my coffee!

-Michael

On 1/6/2011 2:16 PM, Martin Maechler wrote:
>>>>>> Michael Friendly<friendly at yorku.ca>
>>>>>>      on Thu, 06 Jan 2011 09:33:25 -0500 writes:
>      >  No one replied to this, so I'll try again, with a simple example.  I
>      >  calculate a set of log odds ratios, and turn them into a data frame as
>      >  follows:
>
>      >>  library(vcdExtra)
>      >>  (lor.CM<- loddsratio(CoalMiners))
>      >  log odds ratios for Wheeze and Breathlessness by Age
>
>      >  25-29    30-34    35-39    40-44    45-49    50-54    55-59    60-64
>      >  3.695261 3.398339 3.140658 3.014687 2.782049 2.926395 2.440571 2.637954
>      >>
>      >>  (lor.CM.df<- as.data.frame(lor.CM))
>      >  Wheeze Breathlessness   Age      LOR        ASE
>      >  1  W:NoW          B:NoB 25-29 3.695261 0.16471778
>      >  2  W:NoW          B:NoB 30-34 3.398339 0.07733658
>      >  3  W:NoW          B:NoB 35-39 3.140658 0.03341311
>      >  4  W:NoW          B:NoB 40-44 3.014687 0.02866111
>      >  5  W:NoW          B:NoB 45-49 2.782049 0.01875164
>      >  6  W:NoW          B:NoB 50-54 2.926395 0.01585918
>      >  7  W:NoW          B:NoB 55-59 2.440571 0.01452057
>      >  8  W:NoW          B:NoB 60-64 2.637954 0.02159903
>
>      >  Now I want to fit a linear model by WLS, LOR ~ Age, which can do directly as
>
>      >>  lm(LOR ~ as.numeric(Age), weights=1/ASE, data=lor.CM.df)
>
>      >  Call:
>      >  lm(formula = LOR ~ as.numeric(Age), data = lor.CM.df, weights = 1/ASE)
>
>      >  Coefficients:
>      >  (Intercept)  as.numeric(Age)
>      >  3.5850          -0.1376
>
>      >  But, I want to do the fitting in my own function, the simplest version is
>
>      >  my.lm<- function(formula, data, subset, weights) {
>      >  lm(formula, data, subset, weights)
>      >  }
>
>      >  But there is obviously some magic about formula objects and evaluation
>      >  environments, because I don't understand why this doesn't work.
>
>
>      >>  my.lm(LOR ~ as.numeric(Age), weights=1/ASE, data=lor.CM.df)
>      >  Error in model.frame.default(formula = formula, data = data, subset =
>      >  subset,  :
>      >  invalid type (closure) for variable '(weights)'
>      >>
>
> Yes, the "magic" has been called "standard non-standard evaluation"
> for a while (since August 2002, to be precise),
> and the http://developer.r-project.org/ web page has had two
> very relevant links since then, namely those mentioned in the
> following two lines there:
> ----------------------------
> # Description of the nonstandard evaluation rules in R 1.5.1 and some suggestions. (updated). Also an R function and docn for making model frames from multiple formulas.
>
> # Notes on model-fitting functions in R, and especially on how to enable all the safety features.
> ----------------------------
>
> For what you want, I think (but haven't tried) the second link, which is
>         http://developer.r-project.org/model-fitting-functions.txt
> is still very relevant.
>
> Many many people (package authors) had to use something like
> that or just directly taken the lm function as an example..
> {{ but then probably failed the more subtle points on how to
>     program residuals() , predict() , etc functions which you can
>     also learn from model-fitting-functions.txt}}
>
>
>      >  A second question: Age is a factor, and as.numeric(Age) gives me 1:8.
>
>      >  What simple expression on lor.CM.df\$Age would give me either the lower
>      >  limits (here: seq(25, 60, by = 5)) or midpoints of these Age intervals
>      >  (here: seq(27, 62, by = 5))?
>
> With
>
>    data(CoalMiners, package = "vcd")
>
> here are some variations :
>
>   >  (Astr<- dimnames(CoalMiners)[[3]])
>   [1] "25-29" "30-34" "35-39" "40-44" "45-49" "50-54" "55-59" "60-64"
>   >  sapply(lapply(strsplit(Astr, "-"), as.numeric), `[[`, 1)
>   [1] 25 30 35 40 45 50 55 60
>   >  sapply(lapply(strsplit(Astr, "-"), as.numeric), `[[`, 2)
>   [1] 29 34 39 44 49 54 59 64
>   >  sapply(lapply(strsplit(Astr, "-"), as.numeric), mean)
>   [1] 27 32 37 42 47 52 57 62
>
> Or use the 2-row matrix and apply(*, 1)  to that :
>
>   >  sapply(strsplit(Astr, "-"), as.numeric)
>        [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
>   [1,]   25   30   35   40   45   50   55   60
>   [2,]   29   34   39   44   49   54   59   64
>
> Regards,
> Martin Maechler, ETH Zurich
>

--
Michael Friendly     Email: friendly AT yorku DOT ca
Professor, Psychology Dept.
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