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

[Liaw, Andy]

Not sure if the following would qualify as elegant...  It's easier to work
with a matrix instead of data frame:

x.mat <- data.matrix(X)
xmean <- mean(x.mat)
operator.eff <- ave(x.mat, row(x.mat)) - xmean
machine.eff <- ave(x.mat, col(x.mat)) - xmean
(predicted.x <- xmean + operator.eff + machine.eff)
resid.x <- x.mat - predicted.x
sum(resid.x^2)

You can also use aov(), but you already know that...

Andy

From: John McHenry
> 
>     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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