Dear R-devel,
There seems to be a discrepancy in the order in which lm and rlm evaluate their
arguments. This causes rlm to sometimes produce an error where lm is just fine.
Here is a little script that illustrate the issue:
> library(MASS)
> ## create data
> n <- 100
> dat <- data.frame(x=rep(c(-1,0,1), n), y=rnorm(3*n))
>
> ## call lm, works fine
> summary(lm(y ~ as.factor(x), data=dat, subset=x!=0))
Call:
lm(formula = y ~ as.factor(x), data = dat, subset = x != 0)
Residuals:
Min 1Q Median 3Q Max
-2.60619 -0.82160 0.06307 0.65501 2.56677
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.061010 0.100027 0.610 0.543
as.factor(x)1 0.001332 0.141459 0.009 0.992
Residual standard error: 1 on 198 degrees of freedom
Multiple R-squared: 4.479e-07, Adjusted R-squared: -0.00505
F-statistic: 8.868e-05 on 1 and 198 DF, p-value: 0.9925
> ## call rlm, error
> summary(rlm(y ~ as.factor(x), data=dat, subset=x!=0))
Error in rlm.default(x, y, weights, method = method, wt.method = wt.method, :
'x' is singular: singular fits are not implemented in rlm
>
My guess is that rlm first converts x to a factor, which becomes a three-level
factor, then subsets on x!=0, which effectively eliminates a level, and then
creates a "regression" matrix, which becomes singular due to the absence of
data for a level.
Is there a simple way to work around it. The simplest I could think of is
with(subset(dat, x!=0), rlm(y ~ as.factor(x))
which is ok, but most of my scripts make use of data arg to regressions and I'd
like to stay consistent as much as practical.
Thanks,
Vadim
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