From: John McHenry <john_d_mchenry_at_yahoo.com>

Date: Tue 28 Feb 2006 - 12:46:58 EST

), header=TRUE))

rownames(X)<- paste("Operator.", 1:nrow(X), sep="") print(X)

print(X.predicted)

X.residual<- X - X.predicted

SS.E<- sum( X.residual^2 )

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Tue Feb 28 12:56:46 2006

Date: Tue 28 Feb 2006 - 12:46:58 EST

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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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Tue Feb 28 12:56:46 2006

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