[R] Elegant way to express residual calculation in R?

[Gabor Grothendieck]

Here is one further idea:

Xm <- as.matrix(X)
Xm. <- scale(Xm, scale = FALSE)
Xm.. <- t(scale(t(tmp), scale = FALSE))
sum(Xm..^2)


On 2/27/06, Gabor Grothendieck <ggrothendieck at gmail.com> wrote:
> Try this:
>
> Xm <- as.matrix(X)
> X.lm <- lm(c(Xm) ~ factor(col(Xm)) + factor(row(Xm)))
> sum(resid(X.lm)^2)
>
> or if the idea was to do it without using lm try replacing
> your calculation of X.predicted with this:
>
> X.predicted <-
>  outer(operator.adjustment, machine.adjustment, "+") + mean(mean(X))
>
>
> On 2/27/06, John McHenry <john_d_mchenry at yahoo.com> wrote:
> >    Hi All,
> >
> >    I am illustrating a simple, two-way ANOVA using the following data and I'm
> >    having difficulty in expressing the predicted values succinctly in R.
> >
> >    X<- data.frame(read.table(textConnection("
> >        Machine.1    Machine.2    Machine.3
> >        53           61           51
> >        47           55           51
> >        46           52           49
> >        50           58           54
> >        49           54           50"
> >    ), header=TRUE))
> >    rownames(X)<- paste("Operator.", 1:nrow(X), sep="")
> >    print(X)
> >
> >    # I'd like to know if there is a more elegant way to calculate the residuals
> >    # than the following, which seems to be rather a kludge. If you care to read
> >    # the code you'll see what I mean.
> >
> >    machine.adjustment<-  colMeans(X) - mean(mean(X))    # length(machine.adjustment)==3
> >    operator.adjustment<- rowMeans(X) - mean(mean(X))    # length(operator.adjustment)==5
> >    X.predicted<- numeric(0)
> >    for (j in 1:ncol(X))
> >    {
> >        new.col<- mean(mean(X)) + operator.adjustment + machine.adjustment[j]
> >        X.predicted<- cbind(X.predicted, new.col)
> >    }
> >    print(X.predicted)
> >    X.residual<- X - X.predicted
> >    SS.E<- sum( X.residual^2 )
> >
> > It seems like there ought to be some way of doing that a little bit cleaner ...
> >
> > Thanks,
> >
> > Jack.
> >
> >
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