Hi David
This should do the trick:
data <- cbind(height, sex, age)
data.scaled <- data.frame(apply(data, 2, scale))
fit.scaled <- lm(height ~ age + sex - 1, data = data.scaled)
summary(fit.scaled)
HTH
Jo
==
Hello
ps://svn.r-project.org/R-packages/trunk/cluster/R/plotpart.q
I am trying to figure that out now...
FYI, as a test set, one could just delete columns until they are <= to the
number of rows...
clusplot has some nice extras, but I am also looking at just plotting
w/pca...
Thank you again,
Jo
lt(x, scores = TRUE, cor = ncol(x) != 2) :
'princomp' can only be used with more units than variables
Thank you for R and any assistance you may offer!
Jo
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lt(x, scores = TRUE, cor = ncol(x) != 2) :
'princomp' can only be used with more units than variables
Thank you for R and any assistance you may offer!
Jo
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How do I multiple comparison tests of the main plot factor, subplot factor and
interaction? I cannot see how to do this using TukeyHSD and use the correct
error term.
Thanks
Regards
Jo
Joanne Stringer
Senior Biometrician
BSES Limited
Postal Address: PO Box 86
Street Address: 50 Meiers R
using x.in. I used different stopping rules and tolerance levels, but
this gives the same result.
As I said, I think it's a numerical problem. Has anyone had similar
experiences using optim() and could you give some coding advice?
Thanks in advance,
Jo Reynaerts
Thnaks a lot!
Could you recommend a manual or book or something for string manipulation in
R,
or give me a direction where I can get a list of string manipulation
functions?
hyunjo
I
2009/10/14 Duncan Murdoch
> On 13/10/2009 6:44 PM, You Hyun Jo wrote:
>
>> Hello,
>>
Hello,
Do you have a function returns codes of given characters?
Best Regards,
hyunjo
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PLEASE do read the posting guide http:
> 0.5, V2)$V2, gcFirst=T)
user system elapsed
26.678 0.004 26.678
> system.time(df[df$V1>0.5, "V2"], gcFirst=T)
user system elapsed
0.060 0.000 0.057
> system.time(df$V2[df$V1>0.5], gcFirst=T)
user system elapsed
0 0 0
2009/9
Hello,
Suppose that you have a data frame 'df' with variables 'V1', 'V2', 'V3',
etc.
Is there any (performance) difference (except the difference of the return
types)
between the following two computations?
subset(df, V1 > 0, V2)
and
df$V2[df$V1 > 0]
Best Regards,
hyunjo
[[alternati
is then added to the data frame (and is the same for all
rows with the same m indicator).
How can I program this efficiently without running loops?
Thanks in advance
Jo Reynaerts
Ph.D. student
LICOS Centre for Institutions and Economic Performance
Katholieke Universiteit Leuven
Debériotstraat
Thanks for the post-processing ideas. But is there any way to do that
in one step?
On Thu, Sep 10, 2009 at 7:20 PM, Henrique Dallazuanna wrote:
>
> Try this:
>
> xy <- merge(x, y, by = c("a","b"),all = TRUE)
> xy$c <- ifelse(rowSums(!is.na(.x <- xy[, c('c.x', 'c.y')])) > 1, .x[,1],
> rowSums(.x,
I am trying to understand what the lda function in the MASS package calculates
when there are more dimensions than examples. It is my understanding that the
Fisher Linear Discriminant is not applicable in this case, because the inverse
of the covariance matrix cannot be calculated.
My question
graph. I have tried:
graphics.off()
plot(Aids ~ Year, data = aids)
line(glm(Aids ~ Year, data=aids, family=poisson()))
line(glm(Aids ~ Year+I(Year^2), data=aids, family=poisson()))
but this does not work.
Can anyone help me?
Thanks
Jo
BSES Limited Disc
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