On Thursday 18. April 2013 12.18.03 Milan Bouchet-Valat wrote:
> First, completely stop looking at virtual memory: it does not mean much, if
> anything. What you care about is resident memory. See e.g.:
> http://serverfault.com/questions/138427/top-what-does-virtual-memory-size-m
> ean-linux-ubuntu
On Wednesday 6. March 2013 14.50.23 Ben Bolker wrote:
>Just a quick thought (sorry for removing context): what happens if
> you use sum-to-zero contrasts throughout, i.e.
> options(contrasts=c("contr.sum", "contr.poly")) ... ?
Ah, I've got it now, this pointed me in the right direction. Thanks
On Wednesday 6. March 2013 16.33.34 Peter Claussen wrote:
> But you don't have enough data points to estimate all of the possible
> interactions; that's why you have NA in your original results.
Yes, but it seems to me that lm is doing the right thing, or at least the
expected thing, here, the NA
On 03/06/2013 04:18 PM, Peter Claussen wrote:
I'll ignore the rest of your question, in the hope that this will answer them
sufficiently.
OK!
You probably want a simple linear model, specified in R using "+" instead of
"*".
>leaf.lm <- lm(yavg ~ B + C + D + E + Q, data=leaf)
>leaf.lm
Cal
On 03/06/2013 02:50 PM, Ben Bolker wrote:
Just a quick thought (sorry for removing context): what happens if
you use sum-to-zero contrasts throughout, i.e. options(contrasts=c("contr.sum",
"contr.poly")) ... ?
That works (except for the sign)! What would this mean?
Kjetil
oef()
differs from effects() in the case of fractional factorial experiments,
and the other is the factor 1/4 between the coefficients used by Wu &
Hamada and the values returned by effects() as I would think from theory
I've read that it should be a factor 2.
Best regards,
Kjetil
--
Kj
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