(little correction below)
Dear list,
This is off-topic, but perhaps there is an expert out there who can help
me in
a bootstrapping test.
I have two samples A, B from, say, a normal distribution. A third sample R
is the vector of pairwise minima of the same random variables:
A =
Thanks Thierry
It works perfect now.
Regards,
TL
On Fri, Feb 1, 2008 at 11:28 PM, ONKELINX, Thierry
[EMAIL PROTECTED] wrote:
Tribo
That function was added during the latest update. So you should update
to this version (0.5.7). You can update packages with update.packages()
Thierry
Hi all,
Can anyone here please tell me whether is it possible to produce a chart
displayed in http://www.datawolf.blogspot.com/ in R for visualizing
multivariate time series? If possible how?
Regards,
-
[[alternative HTML version deleted]]
Megh Dal [EMAIL PROTECTED] [Sun, Feb 03, 2008 at 12:17:00PM CET]:
Can anyone here please tell me whether is it possible to
produce a chart displayed in http://www.datawolf.blogspot.com/
in R for visualizing multivariate time series? If possible how?
Assuming you mean
Dear all,
I have a set of shapefiles which contents are polygons.
I would like to know if is there a way of I obtain the links between the
polygons, something like a polygon spanning tree. For point patterns I can
use de spatstat package, but I need to do it for polygons.
Below I send a sample
Hello List,
I saw the example in Writing R Extensions where a convolve function
is implemented in C, which is makes the whole business of calling C
from R easy to understand.
Is there a similar example of how to call Fortran 90 from R somewhere?
Thanks,
Max
Dear R users,
I have a linear model of the kind
outcome ~ treatment + covariate
where 'treatment' is a factor with three levels (0, 1, and 2),
and the covariate is continuous. Treatments 1 and 2 both have
regression coefficients significantly different from 0 when using
treatment contrasts with
On Feb 1, 2008 7:30 AM, Falco tinnunculus [EMAIL PROTECTED] wrote:
Dear all,
Is it possible with two random effects in lme()?
lmefit1- lme(Handling ~ Mass + factor(Prey)+ Mass*factor(Prey), random = ~
1 |Place+Age)
I assume that Place and Age are not nested. If so, it's possible to
fit
Perhaps,
letters - c(f,a, j,i,d,e)
a - c(18,15,12,12,9,6)
df - data.frame(letters, a)
library(lattice)
barchart(reorder(df$letters, df$a) ~ df$a)
-Lauri
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
On 2/3/08, K. Elo [EMAIL PROTECTED] wrote:
Hi,
I have a small problem when using barchart. I have the following data:
letters a
6f 18
1a 15
10 j 12
9i 12
4d 9
5e 6
The data is from a survey and summaries the alternatives selected in
On Fri, 01-Feb-2008 at 03:23PM +0200, Atte Tenkanen wrote:
| I had a problem:
|
| I saved a matrix:
|
| save(Tondistmatrix1, file=/Users/atte/Skriptit/Tondistmatrix1)
|
| then I tried to open it with R:
| Tondistmatrix1=load(/Users/atte/Skriptit/Tondistmatrix1)
| Tondistmatrix1[1:10,]
|
Dear all,
To draw a lowess line on a plot was a piece of cake; to draw a loess
line, however, seems not that easy. Is the loess plotting implemented
at all in relation to the loess function, or do I have to look in
add-on packages?
Thanks,
Marcin
__
Try this:
cars.lo - loess(dist ~ speed, cars)
with(cars, plot(speed, dist))
lines(predict(cars.lo), col=blue)
On 03/02/2008, Marcin Kozak [EMAIL PROTECTED] wrote:
Dear all,
To draw a lowess line on a plot was a piece of cake; to draw a loess
line, however, seems not that easy. Is the loess
Thanks a lot, it works.
Any ideas what's going on here?
y-c(1.75, 1.41, 1.96, 1.03, 2.38, 2.19, 1.81, 1.91, 1.47, 2.25, 1.53,
2.79, 2.54, 2.36, 2.65,
2.69, 2.58, 3.27, 3.52, 2.93)
x-c(0.59, 0.49, 0.65, 0.41, 0.84, 0.87, 0.69, 0.72, 0.67, 0.93, 0.76,
1.04, 0.87, 0.92, 0.92,
1.04, 0.94, 1.15,
Try this:
plot(x, y)
points(x, predict(fit), pch=16, col=blue)
On 03/02/2008, Marcin Kozak [EMAIL PROTECTED] wrote:
Thanks a lot, it works.
Any ideas what's going on here?
y-c(1.75, 1.41, 1.96, 1.03, 2.38, 2.19, 1.81, 1.91, 1.47, 2.25, 1.53,
2.79, 2.54, 2.36, 2.65,
2.69, 2.58, 3.27,
Dear Kimmo,
It doesn't appear that anyone has yet mentioned pareto.chart() in the
qcc package; it may serve your purposes. Please note that it does not
require the data frame to be ordered beforehand.
x - DATA[[2]]
names(x) - DATA[[1]]
library(qcc)
pareto.chart(x)
Best wishes,
Jay
On Feb
hits=-2.6 tests�YES_00
X-USF-Spam-Flag: NO
On Sun, 2008-02-03 at 22:26 +0100, Marcin Kozak wrote:
Thanks a lot, it works.
Any ideas what's going on here?
y-c(1.75, 1.41, 1.96, 1.03, 2.38, 2.19, 1.81, 1.91, 1.47, 2.25, 1.53,
2.79, 2.54, 2.36, 2.65,
2.69, 2.58, 3.27, 3.52, 2.93)
x-c(0.59,
lattice has a type = smooth for this:
library(lattice)
xyplot(dist ~ speed, cars, type = c(smooth, p))
Or with a colored line:
xyplot(dist + dist ~ speed, cars, type = c(p, smooth), col =
1:2, distribute.type = TRUE)
In ggplot2 its also a one linear:
library(ggplot2)
Yes, this is what I wanted and needed. Thank you all for your replies.
Marcin
On Feb 4, 2008 1:28 AM, Gabor Grothendieck [EMAIL PROTECTED] wrote:
lattice has a type = smooth for this:
library(lattice)
xyplot(dist ~ speed, cars, type = c(smooth, p))
Or with a colored line:
The motivation for the question comes from Figure 3 of this paper
http://www.ma.hw.ac.uk/~mcneil/ftp/cad.pdf, which shows that a
confidence interval for a statistic is possible. Does there exist a
function in R for such a calculation? If not, how would one go about
doing it in R?
Any pointers
Sorry, but the paper is wrong, or at least the language is very loose.
It does speak about a 'confidence interval' for a statistic, but that
makes no sense. The author apparently means a confidence interval for
the parameter for which the statistic is an estimate, and nominates the
profile
Dear all,
I need to get all points for each polygons and save these points in a
data.frame. I tryed to use the slot() function, but I can´t access the
coords.
grd - GridTopology(c(1,1), c(1,1), c(10,10))
polys - as.SpatialPolygons.GridTopology(grd)
centroids - coordinates(polys)
x -
R-helpers:
I have a question regarding the crr function of the cmprsk package for
performing competing risks regression. Specifically, I was wondering
if the standard likelihood ratio test for a categorical covariate
applies. For example:
# Make up a fake
try
tmp- slot(ex_1.7.selected, 'polygons')
sub.tmp - slot(tmp[[1]],'Polygons')
[EMAIL PROTECTED]
will get you there.
taka
Jarek Jasiewicz wrote:
Milton Cezar Ribeiro wrote:
Dear all,
I need to get all points for each polygons and save these points in a
data.frame. I tryed to use the
Dear Roland Kaiser,
I think there may different pixel-to-point scaling in different
devices. This scaling seems not only to vary with the device but even
with your physical screen. This is what happens with my current box
(MacBook):
quartz()
par(mar)/par(mai)
[1] 6 6 6 6
par(cra)
[1]
I am trying to make an array c(3,8) that contains the averages of what is in
another array c(9,8). I want to average rows 1:3, 4:6, 7:9 and have a loop
replace the generic 1:24 values in my array with the average of the three
rows.
The problem I am having is that R only replaces the last
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