Hi Johannes,
Looking at the code for cutree(...), if h is provided but not k, then
cutree(...) calculates k from h and calls a C function with k
to cut the tree (but all this only if height is sorted). So we can short
circuit the test this way:
cutree.h <- function(tree,h) {
# this line adapt
Yes, absolutely! Sorry about that:
zeta.sq <- function(data) {
z <- scale(data)
zeta.sq <- ifelse(z>0,1+z^2,1/(1+z^2))
}
-Original Message-
From: Michael Dewey [mailto:i...@aghmed.fsnet.co.uk]
Sent: Monday, July 14, 2014 6:57 AM
To: jlh.membership; 'Miles Y
Hi Miles,
If I read the paper correctly, zeta-squared is simply: (1+z^2) for z>=0, and
1/(1+z^2) for z<=0, where z is the z-score (Eqn. 11 in the paper). Z-scores can
be calculated in R using the scale(...) function. So this should produce a
zeta-squared transformation.
zeta.sq <- function(dat
I am encountering a problem using *geom_polygon(., fill=.)* with a dataframe
created using fortify(map) *if the polygon has holes*.
It appears that fill=. ignores the holes - -eg. fills them as if they are not
present. geom_path(.) on the other hand does
recognize the boundaries of the holes.
Not sure this helps but...
##
# data frame with 30,000 ID's, each with 5 "dates", plus some random data...
df <- data.frame(id=rep(1:3, each=5),
date=rep(1:5, each=3),
x=rnorm(15), y=rnorm(15,
mean=1),z=rnor
Same thing using ggplot...
# your data
x <- c(rep(1,100),rep(2,100),rep(3,100))
y <- c(rnorm(100,1),rnorm(100,2),rnorm(100,3))
df <- data.frame(x=x, y=y)
library(ggplot2)
ggp <- ggplot(df) + labs(x="X", y="Y")
ggp <- ggp + geom_boxplot(aes(x=factor(x), y=y))
ggp <- ggp + geom_smooth(formula=y~
Googling "R for psychology students" I found this:
http://health.adelaide.edu.au/psychology/ccs/docs/lsr/lsr-0.3.pdf
and this:
https://personality-project.org/r/
The latter has links to many short courses and tutorials.
If you do end up using R, I find the following sites extremely helpful:
Qui
An approach using data tables:
###
library(data.table)
# dt: some data arranged by group
dt <- data.table(group=c(rep("a",5), rep("b",10), rep("c",15)), values=1:30)
# summarize by group
smry <- dt[,list(min=min(values), max=max(values), range=diff(range(values))),
by="group"]
smry
###
-Ori
This definitely looks like a bug and should really be reported. Denes'
diagnosis is right on.
I get the bug here:
> z <- lm(rnorm(10)~I(1:10))
>
> predict(z, int="conf", scale=1)
Error in predict.lm(z, int = "conf", scale = 1) : object 'w' not found
And also here:
> z <- lm(rnorm(10)~I(1:10))
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