On Fri, 15 Sep 2006, Kiermeier, Andreas (PIRSA - SARDI) wrote:
The figure margins come from what is set in par(mar), eg
layout(matrix(c(1:10),5,2),heights=c(1,rep(2,4)))
par(mar)
[1] 5.1 4.1 4.1 2.1
There is not enough space left to plot anything with those margins. You
will need to
I am trying to run a cross-correlation using the ccf() function. When
I select plot = TRUE in the ccf() I get a graph which has ACF on the
y-axis, which would suggest that these y-values are the auto-correlation
values.
How should I adjust the code to produce a plot that provides the
Hi,
I'm wondering if anyone is aware of any R packages that do kernel
smoothing with at least 4 predictors. The dataset I'm working with now
has 4 predictors, and I can fit a loess smooth, but loess has the
disadvantage that it can produce fitted values outside the range of the
response-something
Hello,
I am a new user of R statistical project. So, I have a matrix (4 columns and
330 rows) and I would like to have PCA (Principal Component Analysis). Besides,
I would like to draw every 30 rows from 330 with a different colour but I dont
know how.
Could you help me by showing how can
Hi
Linux SuSE 10
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major 2
minor 3.1
year 2006
month 06
day01
svn rev38247
language R
version.string Version 2.3.1
Do you mean something like this?
mat - cbind(rnorm(330),rnorm(330),rnorm(330),rnorm(330))
head(mat)
[,1][,2][,3] [,4]
[1,] -0.4668818 0.03015618 -0.72568113 0.6053179
[2,] -0.2625691 -2.20022333 -0.93728544 0.5455864
[3,] -3.4599164 0.71049089 -0.91545599
I am learning about using logistic regression with glm.
Suppose I have dataset:
duration -
c(45,15,40,83,90,25,35,65,95,35,75,45,50,75,30,25,20,60,70,30,60,61,65,15,20,45,15,25,15,30,40,15,135,20,40)
type -
c(0,0,0,1,1,1,rep(0,5),1,1,1,0,0,1,1,1,rep(0,4),1,1,0,1,0,1,0,0,rep(1,4))
sore -
On Thu, 14 Sep 2006, Werner,Arelia [PYR] wrote:
I am trying to run a cross-correlation using the ccf() function. When
I select plot = TRUE in the ccf() I get a graph which has ACF on the
y-axis, which would suggest that these y-values are the auto-correlation
values.
But cross-correlations
On Fri, 15 Sep 2006, Rainer M Krug wrote:
Hi
Linux SuSE 10
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major 2
minor 3.1
year 2006
month 06
day01
svn rev
Prof Brian Ripley wrote:
On Fri, 15 Sep 2006, Rainer M Krug wrote:
Hi
I have two questions concerning the source(test.R) command.
1) Is there any command which I can put into the test.R script file
which aborts the execution of the script? At the moment I use
CodeToBeExecutedInScript
if
On 9/15/2006 6:43 AM, Mark Pinkerton wrote:
Hi Duncan,
Thanks for having a look at this. Find attached a zip with all the
relevant files to run the simulation. I am running this on Windows XP, R
version 2.3.1.
Does the error still occur in a recent alpha build? It's downloadable
from CRAN,
I have just installed 2.4.0 alpha and the problem persists. Here is the
output of the run:
# Summary stats
summary(totals.losses1)
Min. 1st Qu.Median Mean 3rd Qu. Max.
0 0 1284 1617000685100 21920
mean(totals.losses1)
[1] 1617219
On 9/15/2006 7:51 AM, Mark Pinkerton wrote:
I have just installed 2.4.0 alpha and the problem persists. Here is the
output of the run:
Thanks. I'll try your script and see if I can track down what's going on.
Duncan Murdoch
# Summary stats
summary(totals.losses1)
Min. 1st Qu.
On 9/15/2006 7:51 AM, Mark Pinkerton wrote:
I have just installed 2.4.0 alpha and the problem persists. Here is the
output of the run:
When I run it, I get a series of warning messages from qbeta. Do you
get those?
Duncan Murdoch
# Summary stats
summary(totals.losses1)
Min. 1st
Dear all,
I am new to R. I wish to use R's multiple imputation to deal with missing
data. I have a data set with the size around 300 observations and 150
variables. I checked the help function in R and could not locate how to
write the codes for this. can anyone give a hand?
Do appreciate your
Yes, indeed I do. Is there any way I can dig into these a bit more? I
have also just tried using the OO inverse beta from the distr package
and this seems to work.
Mark Pinkerton
Risk Management Solutions
Peninsular House
30 Monument Street
London EC3R 8HB
UK
www.RMS.com
Tel: +44 20 7444
On 9/15/2006 9:06 AM, Mark Pinkerton wrote:
Yes, indeed I do. Is there any way I can dig into these a bit more? I
have also just tried using the OO inverse beta from the distr package
and this seems to work.
This is pretty irritating. You were getting warnings from R that the
calculations
Hello,
I got stuck with a graphics question: I've 3 figures that I present on a single
page (window) via 'layout'. The layout is
layout(matrix(c(1,1,2,3), 2, 2, byrow=TRUE));
so that the frst plot spans the both columns in row one. Now I'd like to
magnify the fist figure so that it takes 20%
You've not given us much information to go on! Have you tried
help(aregImpute,package=Hmisc)
On 15/09/06, Qiong Wang [EMAIL PROTECTED] wrote:
Dear all,
I am new to R. I wish to use R's multiple imputation to deal with missing
data. I have a data set with the size around 300 observations
[EMAIL PROTECTED] wrote:
Hello,
I got stuck with a graphics question: I've 3 figures that I present on a
single page (window) via 'layout'. The layout is
layout(matrix(c(1,1,2,3), 2, 2, byrow=TRUE));
so that the frst plot spans the both columns in row one. Now I'd like to
magnify
Use the heights parameter in the layout function, as shown in ?layout. For
example, to get the first figure to be twice as tall as the other two, use:
layout(matrix(c(1,1,2,3),2,2,byrow=TRUE),heights=c(2,1))
layout.show(3)
On 15/09/06, [EMAIL PROTECTED] [EMAIL PROTECTED]
wrote:
Hello,
I
See MICE
http://www.multiple-imputation.com/
-Original Message-
From: David Barron [mailto:[EMAIL PROTECTED]
Sent: sexta-feira, 15 de Setembro de 2006 14:38
To: Qiong Wang; r-help
Subject: Re: [R] missing data codes
You've not given us much information to go on! Have you tried
[EMAIL PROTECTED] wrote:
Hello,
I got stuck with a graphics question: I've 3 figures that I present
on a single page (window) via 'layout'. The layout is
layout(matrix(c(1,1,2,3), 2, 2, byrow=TRUE));
so that the frst plot spans the both columns in row one. Now I'd like
to magnify the
(re)-Hello
I actually thought about another possibility with a 1 column, a sum
(instead of a mean), and a division of the columns for which I want
the mean:
DF = data.frame( V1=c(A,A,A,B,B,C) , V2=c(1,3,2,0.5,0.9,5.0),
V3=c(200,800,200,20,50,70), V4=c(ID1,ID1,ID1,ID2,ID2,ID3))
DF2 =
Hello,
I have the following data:
Km Sex
250 1
300 2
290 2
600 1
450 2
650 1
.
I would like to obtain one histogram where the data (or the part) of each
sex is visible, it is like cumulative histogram or spinogram.
To be more comprehensible, i would like to know if the following graph is
Good idea. You could write it compactly like this:
transform(aggregate(cbind(DF[2:3], o = 1), DF[c(1,4)], sum, na.rm = TRUE),
+ V2 = V2/o, V3 = V3/o)
V1 V4 V2 V3 o
1 A ID1 2.0 400 3
2 B ID2 0.7 35 2
3 C ID3 5.0 70 1
On 9/15/06, Emmanuel Levy [EMAIL PROTECTED] wrote:
(re)-Hello
I
David Barron wrote:
You've not given us much information to go on! Have you tried
help(aregImpute,package=Hmisc)
And for that sample size you'll have to tell aregImpute to force all
continuous variables to act linearly
Frank
On 15/09/06, Qiong Wang [EMAIL PROTECTED] wrote:
Dear all,
Anupam Tyagi wrote:
justin bem justin_bem at yahoo.fr writes:
Mr Harrell,
After reading discussion about R output and SAS output , I will like to use
rreport package. I a windows XP
user
Sincerly
See:
http://biostat.mc.vanderbilt.edu/twiki/bin/view/Main/Rreport
Anupam.
Alexandre Depire wrote:
Hello,
I have the following data:
Km Sex
250 1
300 2
290 2
600 1
450 2
650 1
.
I would like to obtain one histogram where the data (or the part) of each
sex is visible, it is like cumulative histogram or spinogram.
To be more comprehensible, i would
Hi all:
Newbies (and others!) may find useful the R Reference Card made available by
Tom Short of EPRI Solutions at http://www.rpad.org/Rpad/Rpad-refcard.pdf or
through
the Contributed link on CRAN (where some other reference cards are also
linked). It categorizes and organizes a bunch
On Fri, 15 Sep 2006 16:45:31 +0200 Alexandre Depire wrote:
Hello,
I have the following data:
Km Sex
250 1
300 2
290 2
600 1
450 2
650 1
.
I would like to obtain one histogram where the data (or the part) of
each sex is visible, it is like cumulative histogram or spinogram.
I could use some help understanding how nls parses the formula argument
to a model.frame and estimates the model. I am trying to utilize the
functionality of the nls formula argument to modify garchFit() to handle
other variables in the mean equation besides just an arma(u,v)
specification.
My
Dear R users,
This is a trivial question, there might even be an R function for it, but I have
to do it many times and wonder if there is an efficient for it.
Suppose we have a data frame like this:
d - data.frame(x=sample(seq(0.1:1, by=0.01), size=100, replace=TRUE),
y=rnorm(100, 0.2, 0.6))
Perhaps using 'ave' and 'cut':
df - data.frame(x=runif(100, 0.1, 1), y=rnorm(100, 0.2, 0.6))
df$xcut-cut(df$x, seq(0, 1, 0.1))
df$z-ave(df$y, df$xcut)
df[order(df$x),]
Stefano
-Messaggio originale-
Da: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] conto di
[EMAIL PROTECTED]
Inviato:
Hi,
I would like to paired ID_ with Cod for analysis in spdep.
Any ideas?
head(bai$att.data)
ID_ NAME1_ NAME2_ PARTS_ POINTS_ LENGTH_ AREA_
1 410690205001 410690205001 NA 1 158 5.243338 1.2668820
2 410690205009 410690205009 NA 1 159 6.071286
Hi,
1. How do I construct 95% prediction interval for new x values, for example -
x = 3?
2. How do I construct 95% confidence interval?
my dataframe is as follows :
dt
structure(list(y = c(2610,
6050, 1620, 3070, 7010, 5770, 4670,
In dotplot, what's the best way to suppress the unused levels of 'y' on
a per-panel basis? This is useful for the case that 'y' is a factor
taking perhaps thousands of levels, but for a given panel, only a
handfull of these levels ever present.
Thanks,
Ben
predict(s.lm,data.frame(x=3),interval=prediction)
fit lwr upr
[1,] 16073985 -9981352 42129323
predict(s.lm,data.frame(x=3),interval=confidence)
fit lwr upr
[1,] 16073985 5978125 26169846
On 15/09/06, Sachin J [EMAIL PROTECTED] wrote:
Hi,
1.
Benjamin Tyner said the following on 9/15/2006 2:36 PM:
In dotplot, what's the best way to suppress the unused levels of 'y' on
a per-panel basis? This is useful for the case that 'y' is a factor
taking perhaps thousands of levels, but for a given panel, only a
handfull of these levels
Hi,
I want to do 2-dimentional loess smoothing in gam through the command
gam(Y~lo(X1,X2,span=span,degree=1) )
in library gam.
The span is the percentage of data points to define the neighborhood and used
for the smoothing.
Does this command do one 2-dimentional smoothing with
David,
Thanks for the quick reply.
Just confirming, does predict(s.lm,data.frame(x=3),interval=prediction)
gives prediction interval or tolerance interval?
Thanks
Sachin
David Barron [EMAIL PROTECTED] wrote:
predict(s.lm,data.frame(x=3),interval=prediction)
On 9/15/06, Benjamin Tyner [EMAIL PROTECTED] wrote:
In dotplot, what's the best way to suppress the unused levels of 'y' on
a per-panel basis? This is useful for the case that 'y' is a factor
taking perhaps thousands of levels, but for a given panel, only a
handfull of these levels ever
I resend this to expect more responses. Thanks!
Inspired by the responses, I tried to do this analytically.
The idea is that truncated mean and standard deviation could be expressed as
integral forms. So if given truncated mean, sd and truncated point (mut, sdt,
thre), an optim( ) function
I am updating the CRAN package LMGene, and I'm having trouble installing the
new version.
I changed the working package name to WLMG so I could use the new version
without removing the old version from the R directory.
I set up R_LIBS to allow installation in a different directory.
When I did
R
On Sun, 2006-09-10 at 20:36 +0100, Prof Brian Ripley wrote:
I am however interested in areas where the probability of success is
noticeably higher than 5%, for example 20%. I've tried rpart and the
weights option, increasing the weights of the success-observations.
You are 'misleading'
Dear All,
Is there a function in R which can do a periodogram of Schuster ?
Thanks in advance !
Guillaume Blanchet
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Hi,
Is there an R implementation of least angle regression for binary response
modeling? I know that this question has been asked before, and I am also
aware of the lasso2 package, but that only implements an L1 penalty, i.e.
the Lasso approach.
Madigan and Ridgeway in their discussion
On Fri, 2006-09-15 at 18:49 -0400, Ravi Varadhan wrote:
Hi,
Is there an R implementation of least angle regression for binary response
modeling? I know that this question has been asked before, and I am also
aware of the lasso2 package, but that only implements an L1 penalty, i.e.
the Lasso
hi all,
I have a simple question that does power spectral analysis related to
capacity dimension, information dimension, lyapunov exponent, hurst
exponent.
If yes then please show me the way. I am newbie in the world of chaos.
Sayonara With Smile With Warm Regards :-)
G a u r a v Y
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