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

Martin Maechler maechler at stat.math.ethz.ch
Thu Jan 6 20:16:48 CET 2011

```>>>>> 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

```