Hi folks,
can anybody give me a hint how to solve the following problem?
I have several plots in one window like this:
layout(matrix(c(1,2,3,4), nrow = 2, byrow = TRUE))
plot(rnorm(100))
plot(rnorm(200))
plot(rnorm(300))
plot(rnorm(400))
Now, I'd like to create a legend below each plot and
Hello,
I have a problem with data reshaping. Here's my data
DF -
structure(list(idvar1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(patient1, patient2
), class = factor), idvar2 = structure(c(1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label =
Hey Marc,
sorry I forgot to mention it. I am the owner of the path, which is in my home
directory. I can change the read/write/execute privileges. All users have - at
least - read privileges for all files and the directory:
hpc36 40: ls -la *.tped
-rw-r--r-- 1 cremer C020 205487231 Dec 3
I need some help in comparing t values avalaible from two data frames. I
have two data frames with each containing a column for t valuse
For example one data frame looks like as follows
IDlogFCt P.Value adj.P.Val
B
39 a39 -1.737037118
gauravbhatti wrote:
...
I have to plot the t values obtained for each id against each other. Since
the order of the IDs in the two data frames is different , I am finding it
difficult to plot the t values. Can anyone help in arranging the t values in
the increasing order of the ID values like
Hi,
I?m looking for a package which includes a test for seasonality in time
series.
Any help and input would be greatly appreciated.
Thanks,
Matthias
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
On Thu, 4 Dec 2008, Andrew Hooker wrote:
Hi,
I am wondering if there is a way to change the value of the record option
in a graphics device that is already open (and accepts this option). I
don't want to open a new device with, for example dev.new(record=T), but
just want to change the
Hi all
Here is a rather naive implementation of the SEPA algorithm for generating
lexical permutations:
lexperm3 - function(x, n=length(x)) {
perms - list()
k - 1
perms[[k]] - x
k - k + 1
for (y in 1:(factorial(n)-1)) {
i - n-1
while (x[i] x[i+1] i 0) {
i - i - 1
}
# i is
Here's something that may help you get started:
library(ts)
?decompose
?stl
Thanks,
-Girish
On Dec 5, 1:55 pm, Matthias [EMAIL PROTECTED] wrote:
Hi,
I?m looking for a package which includes a test for seasonality in time
series.
Any help and input would be greatly appreciated.
Thanks,
Dear Wacek,
I've thought a bit more about this problem, and recall that I originally
wrote Strsplit() [and replacements for sub() and gsub(), which were not then
in S-PLUS] for the version of the car package that I released for S-PLUS,
because other functions in the package used these. The
Rory
there are several packages that perform this.
I would use permn() of the combinat library, then, if
lexicographical order is important, sort it explicitly.
HTH
rksh
[EMAIL PROTECTED] wrote:
Hi all
Here is a rather naive implementation of the SEPA algorithm for generating
lexical
John Fox wrote:
Dear Wacek,
I've thought a bit more about this problem, and recall that I originally
wrote Strsplit() [and replacements for sub() and gsub(), which were not then
in S-PLUS] for the version of the car package that I released for S-PLUS,
because other functions in the package
Dear R-users,
I want to construct a relative path in R, but I am not able to do it or to
find a function that does it.
My solution is somewhat awkward as it uses string manipulation.
The problem: I have the following folder structure.
project_folder
|
|- data
|-
On Fri, 5 Dec 2008, Mark Heckmann wrote:
Dear R-users,
I want to construct a relative path in R, but I am not able to do it or to
find a function that does it.
My solution is somewhat awkward as it uses string manipulation.
The problem: I have the following folder structure.
project_folder
Dear All,
I ma having a trouble in generating a figure containing 3 insets with
the gridBase package.
I always get an error message of the kind:
Error in gridPLT() : Figure region too small and/or viewport too large
No matter which parameters I choose. The plots works nicely with two
insets
Dear colleagues
Please could someone kindly explain the following inconsistencies I've
discovered when performing logical calculations in R:
8.8 - 7.8 1
TRUE
8.3 - 7.3 1
TRUE
Thank you,
Emma Jane
[[alternative HTML version deleted]]
On 12/5/2008 7:23 AM, emma jane wrote:
Dear colleagues
Please could someone kindly explain the following inconsistencies I've
discovered when performing logical calculations in R:
8.8 - 7.8 1
TRUE
8.3 - 7.3 1
TRUE
See R FAQ 7.31
Dear Emma,
On Fri, 5 Dec 2008 04:23:53 -0800 (PST)
emma jane [EMAIL PROTECTED] wrote:
Please could someone kindly explain the following inconsistencies
I've discovered__when performing logical calculations in R:
8.8 - 7.8 1
TRUE
8.3 - 7.3 1
TRUE
Gladly: FAQ 7.31
Hi there,
I am interested in the inner workings of wilcox.test:
wilcox.test
function (x, ...)
UseMethod(wilcox.test)
environment: namespace:stats
how can I get at the code, if it is R-code? For Methods one should be able to
learn what extension to use, but here default or such do not help.
Hello,
I hope this question is not too stupid. I would like to know how to update
levels after subsetting data from a data.frame.
df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4))
names(df) - c(X1,X2,X3)
my.sub - subset(df, X1 == a | X1 == b)
levels(my.sub$X1)
# still gives
Hi all,
does anybody know about R implementations for quantile regression for
longitudinal data? I am just aware of a very basic version of R.
Koenker's approach using fixed effects.
Thanks in advance
Armin
__
R-help@r-project.org mailing list
ExpeRts,
I'm trying to compare three survival curves using the function survdiff in
the survival package. Following is my code and corresponding error message.
survdiff(Surv(st_months, status) ~ strata(BOR), data=mydata)
Error in survdiff(Surv(st_months, status) ~ strata(BOR), data = mydata) :
try this:
df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4))
names(df) - c(X1,X2,X3)
my.sub - subset(df, X1 == a | X1 == b)
levels(my.sub$X1)
[1] a b c
my.sub$X1 - factor(my.sub$X1)
levels(my.sub$X1)
[1] a b
On Fri, Dec 5, 2008 at 7:50 AM, Antje [EMAIL PROTECTED] wrote:
I'm trying to omit NA:s in this DF and produce a reduced DF. The
problem is that I cannot completely omit NA rows.
I tried
library(reshape)
g - melt(DF, id=c(idvar1, idvar2))
g - na.omit(g)
You're missing an id variable:
DF$idvar3 - 1:2
g - melt(DF, id=c(idvar1, idvar2, idvar3), na.rm =
I do the following for a subsetted dataframe:
cleanfactors - function(mydf){
outdf-mydf
for (i in 1:dim(mydf)[2]){
if (is.factor(mydf[,i]))
outdf[,i]-factor(mydf[,i])
}
outdf
}
Antje wrote:
Hello,
I hope this question is not too stupid. I would like to know how to
update levels
On Fri, 5 Dec 2008, Christian Hoffmann wrote:
Hi there,
I am interested in the inner workings of wilcox.test:
wilcox.test
function (x, ...)
UseMethod(wilcox.test)
environment: namespace:stats
how can I get at the code, if it is R-code? For Methods one should be able to
learn what
On Fri, Dec 5, 2008 at 6:50 AM, Antje [EMAIL PROTECTED] wrote:
Hello,
I hope this question is not too stupid. I would like to know how to update
levels after subsetting data from a data.frame.
df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4))
names(df) - c(X1,X2,X3)
my.sub
On Fri, 5 Dec 2008, jim holtman wrote:
try this:
df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4))
names(df) - c(X1,X2,X3)
my.sub - subset(df, X1 == a | X1 == b)
levels(my.sub$X1)
[1] a b c
my.sub$X1 - factor(my.sub$X1)
I find
my.sub$X1 - my.sub$X1[drop=TRUE]
a lot more
Berwin A Turlach wrote:
Dear Emma,
On Fri, 5 Dec 2008 04:23:53 -0800 (PST)
emma jane [EMAIL PROTECTED] wrote:
Please could someone kindly explain the following inconsistencies
I've discovered__when performing logical calculations in R:
8.8 - 7.8 1
TRUE
8.3 - 7.3 1
Hi all
(I'm sure this question has been asked before, but I cant find it).
If I have two character vectors:
x - c(aaa,bbb,ccc)
y - c(1,2,3)
How can I get the cartesian product of the string values?
expand.grid(x,y)
Gives me a data frame with separate columns...however, I cant seem to get
Hi All,
I'm trying to split my dataset, into multiple datasets that i'll analyse
later, i wanted to do this dynamically as i might need to rerun the code
later.
I was looking at doing this via a loop, (Are other methods more appropriate?
Would a function be better?)
However i'm not sure in R how
Hi all,
I would like to ask one simple question. I wanna know how
you can ask R to change the order of x factor when you draw
interaction plot. I find that R puts factors in an alphabetic
order (e.g., dog, lamb, and monkey etc). But I simply want to
control the order. How can I do this?
Aloha,
Check out the 'levels' option on 'factor'
On Fri, Dec 5, 2008 at 8:32 AM, Kota Hattori [EMAIL PROTECTED] wrote:
Hi all,
I would like to ask one simple question. I wanna know how
you can ask R to change the order of x factor when you draw
interaction plot. I find that R puts factors in an
Hi the list
I would like to add a legend under a graph but at a fixed distance from
the graphe. Is it possible ?
More precisely, here is my code :
--- 8
symboles - c(3,4,5,6)
dn - rbind(matrix(rnorm(20),,5),matrix(rnorm(20,2),,5))
listSymboles - rep(symboles,each=2)
Dear Mr. Daalgard.
thank you very much for your reply, it helped me to progress a bit.
The following works fine:
dd - expand.grid(C = 1:7, B= c(r, l), A= c(c, f))
myma - as.matrix(myma) #myma is a 12 by 28 list
mlmfit - lm(myma~1)
mlmfit0 - update(mlmfit, ~0)
anova(mlmfit, mlmfit0, X= ~C+B, M =
I am sorry. I am not sure if the mail a send before to this list was rejected
because of header (subject).
I've changed it. The first maybe was not appropriate.
I did a function (sec_conop) whose arguments are syndic,
well and wellconop.
layout(matrix(c(1,2,3,4), nrow = 2, byrow = TRUE))
plot(rnorm(100))
plot(rnorm(200))
plot(rnorm(300))
plot(rnorm(400))
Now, I'd like to create a legend below each plot and generate a common
title.
How can I do that?
If you are laying plots out in grids like this then lattice graphics
This query of why do SAS and S give different answers for Cox models comes
up every so often. The two most common reasons are that
a. they are using different options for the ties
b. the SAS and S data sets are slightly different.
You have both errors.
First, make sure I have
Rodrigo,
This is an old and quite basic Krig, my data was continuous
measurements in lat, long so binned first.
library(geoR)
counts-bins2d(long, lat,bin=c(0.1,0.1),plot=FALSE, nlevels=15,
color.palette=heat.colors, xaxs='i', yaxs='i', las=1, main='')
countsgeo-as.geodata(counts)
Dear all,
I'm looking for a method to quantify assortative mixing in undirected graphs,
preferably something that is compatible with the igraph-library for network
analysis. Is anybody aware of an already existing function/library for this or
an easy way to implement this functionality in R?
Does this satisfy?
levels(interaction(x,y))
[1] aaa.1 bbb.1 ccc.1 aaa.2 bbb.2 ccc.2 aaa.3 bbb.3
ccc.3
--
David Winsemius
On Dec 5, 2008, at 8:12 AM, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
Hi all
(I'm sure this question has been asked before, but I cant find it).
If I have two
Hi David
Perfect - if I specify sep=, it gives me exactly what I need.
Cheers
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: David Winsemius [mailto:[EMAIL PROTECTED]
Sent: 05 December 2008 14:16
To: WINSTON, Rory, GBM
Cc:
Sorry, I spoke too soon...
interaction() only works for sequences of equal length. Anyone know a method
that works with unequal-length vectors?
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Hello everybody,
I'm trying to perform a ESDA on some data about per capita GDP among
European regions and I'm trying to learn how to make R and GeoDa interact.
I'm reading that in order to import a GAL file from GeoDa to R it's best to
source into R the read.gal3.R file. But where can I get it
Thanks a lot!!!
the drop thing was exactly what I was looking for (I already used it some
time ago but forgot about it).
Thanks to everybody else too.
Antje
Prof Brian Ripley schrieb:
On Fri, 5 Dec 2008, jim holtman wrote:
try this:
df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8),
I hope this question is not too stupid. I would like to know how to
update
levels after subsetting data from a data.frame.
df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8),
c(9,1,2,3,4))
names(df) - c(X1,X2,X3)
my.sub - subset(df, X1 == a | X1 == b)
levels(my.sub$X1)
# still
Hi Lars,
I am having trouble replicating this problem, using the default
file/folder permissions on my Fedora 10 system.
When creating a new folder, I get:
mkdir Test
drwxrwxr-x 2 marcs marcs 4096 2008-12-05 08:17 Test
When creating a new file, I get:
touch Test.txt
-rw-rw-r-- 1 marcs
Got it...I completely overlooked the collapse argument to paste():
apply(expand.grid(x,y),1,function(x) paste(x,collapse=))
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: WINSTON, Rory, GBM
Sent: 05 December 2008 14:30
To: WINSTON, Rory,
Wiindows XP
R 2.7
I am using sink() to send the results of my analyses to a text file.
Unfortunately my graphs do not become part of the file. Is there anyway that I
can have both the text and graphic output of my analyses appear in a file?
Thanks,
John
John David Sorkin M.D., Ph.D.
Chief,
Dear Nils,
You might also take a look at the Anova() function in the car package, which
though less flexible than anova() should get you the tests that you want
more simply. ?Anova has an example of a repeated-measures ANOVA with two
within-subject and two between-subject factors.
I hope that
Try this also:
with(expand.grid(x,y), paste(Var1, Var2, sep = ))
On Fri, Dec 5, 2008 at 1:08 PM, [EMAIL PROTECTED] wrote:
Got it...I completely overlooked the collapse argument to paste():
apply(expand.grid(x,y),1,function(x) paste(x,collapse=))
Rory Winston
RBS Global Banking Markets
I am using sink() to send the results of my analyses to a text file.
Unfortunately my graphs do not become part of the file. Is there
anyway that I can have both the text and graphic output of my
analyses appear in a file?
You can create a latex document with text, graphs and R-code using
I am having trouble writing a code for matching two pairs of sequences with
differing lengths:
for example sequence1= 1,2,3,4,5,6,7 sequence2=1,2,3,4,5,6,7,8,9,10
I want to create several new pairs of sequences in several dataframes such
that:
1st dataframe is 1,10 (start of sequence1, end of
I am using sink() to send the results of my analyses to a text
file. Unfortunately my graphs do not become part of the file.
Is there anyway that I can have both the text and graphic
output of my analyses appear in a file?
Well, how would you expect the graphs to be represented in a text
In case anyone other than me was interested, a pretty efficient
circular tophat can be made using the fields function fields.rdist.near:
CircHat=function (x, y, h=1, gridres = c((max(x)-min(x))/25,(max(y)-
min(y))/25), lims = c(range(x), range(y)),density=FALSE)
{
require(fields)
Try this:
s1 - 1:7
s2 - 1:10
lapply(seq(length(s1)), function(i)cbind(head(s1, i), tail(s2, i)))
On Fri, Dec 5, 2008 at 1:44 PM, emj83 [EMAIL PROTECTED] wrote:
I am having trouble writing a code for matching two pairs of sequences with
differing lengths:
for example sequence1=
Dear all,
I wrote the following code to calculate the density functions for two data
sets, respectively.
den_str -density(str_data$Similarity);
den_non_str -density(nonstr_data$Similarity);
However, I would like to knowing the difference between den_str and
den_non_str, that is, the
I am having trouble writing a code for matching two pairs of sequences with
differing lengths:
for example sequence1= 1,2,3,4,5,6,7 sequence2=1,2,3,4,5,6,7,8,9,10
I want to create several new pairs of sequences in several dataframes such
that:
1st dataframe is 1,10 (start of sequence1, end of
Dear R-users,
I have a question concerning the use of lattice plots within for-loops.
I want to create a png file containing a lattice histogram which works out
fine (part 1).
When I loop the whole code, the graphic file does not contain anything (part
2).
I can fix it by wrapping the histogram
[R] Strplit code
pomchip at free.fr pomchip at free.fr
Wed Dec 3 20:52:21 CET 2008
Dear R-users,
The strsplit function does not exist in S-plus and I would like to
use it. How
could I reproduce the function in Splus or access to its source code?
Thank you in advance,
I have a question concerning the use of lattice plots within for-loops.
I want to create a png file containing a lattice histogram which works
out
fine (part 1).
When I loop the whole code, the graphic file does not contain anything
(part
2).
I can fix it by wrapping the histogram
Thank you so much, this was very helpful.
Svetlana
Terry Therneau wrote:
This query of why do SAS and S give different answers for Cox models comes
up every so often. The two most common reasons are that
a. they are using different options for the ties
b. the SAS and S
It is best to create the graphics device at the final size desired, then do the
plotting and add the legend. For getting a fixed distance, look at the
function grconvertY for one possibility.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
A similar question was posed and answered:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/119793.html
Two aspects needed to be addressed ... specifying the same domain, and
getting the x-values to line up prior to the subtraction (or
whatever function is desired).
What are you going to
I cannot understand the following error printed out when I try to get the
extrema of my time series.
I would appreciate some suggestion as I really cannot interpret the error. I
might not be using a proper
set of parameters in calling such functions. I am learning by doing ...
aa.peak -
2008/12/5 Philipp Pagel [EMAIL PROTECTED]:
I am using sink() to send the results of my analyses to a text
file. Unfortunately my graphs do not become part of the file.
Is there anyway that I can have both the text and graphic
output of my analyses appear in a file?
Well, how would you expect
I think Sweave and/or odfWeave are the real answer, though. Obviously, a
bigger and more elaborate kettle of fish.
-- Bert Gunter
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Barry Rowlingson
Sent: Friday, December 05, 2008 9:21 AM
To: Philipp Pagel
Others mentioned Sweave, which is the best way to go if you have a planned set
of routines to do. If you are more just playing and want a transcript of what
you are doing (unpreplanned) that includes graphics in the final version, then
look at etxtStart from the TeachingDemos package. You
Hi,
I have about 900 files that I need to run the same R script on. I
looked over the R Data Import/Export Manual and couldn't come up with
a way to read in a sequence of files.
The files all have unique names and are in the same directory. What I
want to do is:
1) Create a list of
Dear R users,
RMySQL 0.7-2 is now available on CRAN:
http://cran.r-project.org/web/packages/RMySQL/index.html
From the NEWS file:
* New maintainer is Jeffrey Horner [EMAIL PROTECTED].
* We no longer distribute libmysql.dll. This library is now found
either by reading the MYSQL_HOME
Have you thought about using one of the Python/R interface modules?
http://www.omegahat.org/RSPython/
http://rpy.sourceforge.net/
Admittedly, I have not had much success in getting these to work on my
machine, but I know others who have.
Kyle H. Ambert
Graduate Student
Department of Medical
This is almost a macro problem. It could be done in SAS language using the
WPS product (660 USD) I think.
It is a familiar problem and I would be quite interested in the result.
Is there any concept of Macros in R or a package to do the same.
Regards,
Ajay
On Fri, Dec 5, 2008 at 11:31 PM,
2008/12/5 Chris Poliquin [EMAIL PROTECTED]:
Hi,
I have about 900 files that I need to run the same R script on. I looked
over the R Data Import/Export Manual and couldn't come up with a way to
read in a sequence of files.
The files all have unique names and are in the same directory.
R has quite a few functions to get and manipulate filenames to facilitate
exactly what you want to do. See ?files and especially the links at the end
to the file name manipulation functions.
e.g. dir(pathname) lists all file names in the directory pathname.
?list.files gives details.
-- Bert
#my suggestion would be to use the morlet wavelet as opposed to the
mexican hat wavelet (default).
aa - (structure(list(X.0.85 = c(-1.02, -1.17, -1.29, -1.39, -1.46,
-1.5, -1.52, -1.5, -1.46, -1.39, -1.3, -1.19, -1.07, -0.93, -0.79,
-0.65, -0.5, -0.36, -0.22, -0.08, 0.05, 0.18, 0.3, 0.41, 0.52,
Thanks, Barry. I'll use that in the future.
---Kyle.
On Fri, Dec 5, 2008 at 11:01 AM, Barry Rowlingson
[EMAIL PROTECTED] wrote:
2008/12/5 Chris Poliquin [EMAIL PROTECTED]:
Hi,
I have about 900 files that I need to run the same R script on. I looked
over the R Data Import/Export
Use dir to get the names and then lapply over them with a
custom anonymous function where L is a list of the returned
values:
# assumes file names are those in
# current directory that end in .dat
filenames - dir(pattern = \\.dat$)
L - lapply(filenames, function(x) {
DF - read.table(x,
I can't believe the two 'solutions' already posted. It's easy:
Me neither.
?list.files
That's what I would use, too. If the OP is on a UNIX platform,
run the R-script in a loop in the shell is an alternative.
Something like this (bourne shell syntax):
for datafile in *.csv ; do
Is there a way to list only the files in a given directory without
passing pattern=... to list.files()?
On Fri, Dec 5, 2008 at 5:10 PM, Kyle. [EMAIL PROTECTED] wrote:
Thanks, Barry. I'll use that in the future.
---Kyle.
On Fri, Dec 5, 2008 at 11:01 AM, Barry Rowlingson
[EMAIL PROTECTED]
Try this:
dir()[!file.info(dir())$isdir]
On Fri, Dec 5, 2008 at 2:30 PM, Gustavo Carvalho
[EMAIL PROTECTED] wrote:
Is there a way to list only the files in a given directory without
passing pattern=... to list.files()?
On Fri, Dec 5, 2008 at 5:10 PM, Kyle. [EMAIL PROTECTED] wrote:
Thanks,
This is almost a macro problem. It could be done in SAS language using
the WPS product (660 USD) I think. ...
OUCH! Why do it the complicated way??? Check out ?dir, ?list.files, and
then ?lapply for a simple start.
Don't give up so soon! When it comes to R there is no need to punt - you
can
Thanks for the solution . I especially liked the analogy along with the code
of course.
Regards,
Ajay
www.decisionstats.com
On Sat, Dec 6, 2008 at 1:23 AM, [EMAIL PROTECTED] wrote:
This is almost a macro problem. It could be done in SAS language using
the WPS product (660 USD) I think. ...
It seems that you have 900 files with the same parameters in each file (I
might be reading more between the lines here than you inferred). However if
this is the case, why not import each of the files into a common database
and then link the database using ODBC connectivity options. If that is
Dear R users,
Suppose I have the following data.frame:
myID myType myNum1 myNum2 myNum3
a Single 10 11 12
b Single 15 25 35
c Double 22 33 44
d Double4 6 8
and I want
In general, comparing two continuous densities is difficult because they
can differ on a set of measure 0 (i.e., at a single point) and yet have
the same distribution function.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Chang Jia-Ming
Sent: Friday,
Andy,
Thanks for your email.
I understand that by default, the sampsize variable will use the behavior
variable that we are classifying as the strata variable.
Then, I could set sampsize=c(no=89, yes=11). I implemented that but I got
99% classification error rate on the yes value. When I
Am 03.12.2008 um 09:06 schrieb Liviu Andronic:
Dear all,
I'm looking for an alternative way to replicate the 2, string for an
x number of times, and end up with one string containing 2, x times.
I can partly achieve this using replicate().
y - rep(2,, times=3)
y
JFTR: replicate() is a
Using lmer() on my data results in an error. The problem,
I think, is my model specification. However, lm() works
ok.
I recreated this error with a more simple dataset. (See
code below.)
# word and letter recognition data
# two within factors:
# word length: 4, 5, 6 letters
# letter position:
This should get you close:
x - read.table(textConnection(myID myType myNum1 myNum2 myNum3
+ a Single 10 11 12
+ b Single 15 25 35
+ c Double 22 33 44
+ d Double4 6 8),
On Fri, Dec 5, 2008 at 3:44 PM, B. Meijering [EMAIL PROTECTED] wrote:
Using lmer() on my data results in an error. The problem, I think, is my
model specification. However, lm() works ok.
I recreated this error with a more simple dataset. (See code below.)
# word and letter recognition data
Thanks much Jim.
On Fri, Dec 5, 2008 at 2:05 PM, jim holtman [EMAIL PROTECTED] wrote:
This should get you close:
x - read.table(textConnection(myID myType myNum1 myNum2 myNum3
+ a Single 10 11 12
+ b Single 15 25 35
+ c
Rory.WINSTON at rbs.com writes:
Sorry, I spoke too soon...
interaction() only works for sequences of equal length. Anyone know a method
that works with
unequal-length vectors?
Something like
c(outer(vec1,vec2,paste,sep=)) ?
(totally untested)
Ben Bolker
On Thu, Dec 4, 2008 at 6:21 PM, Bert Gunter [EMAIL PROTECTED] wrote:
Folks:
Suppose I have 3 random effects, A,B, and C. Using the older lme() function
(in nlme) it was possible (using the pdMat classes) to specify that they are
uncorrelated with identical variances. Is it possible to do
Here is a solution using sqldf
library(sqldf)
DF2 -
structure(list(myID = structure(1:4, .Label = c(a, b, c,
d), class = factor), myType = structure(c(2L, 2L, 1L, 1L), .Label
= c(Double,
Single), class = factor), myNum1 = c(10, 15, 22, 4), myNum2 = c(11,
25, 33, 6), myNum3 = c(12, 35, 44, 8)),
Skotara wrote:
Dear Mr. Daalgard.
thank you very much for your reply, it helped me to progress a bit.
The following works fine:
dd - expand.grid(C = 1:7, B= c(r, l), A= c(c, f))
myma - as.matrix(myma) #myma is a 12 by 28 list
mlmfit - lm(myma~1)
mlmfit0 - update(mlmfit, ~0)
anova(mlmfit,
Here is a solution using doBy and Gabor's DF2 created below:
library( doBy )
newrows - summaryBy ( myNum1 + myNum2 + myNum3 ~ myType , DF2, keep.names =
TRUE )
newrows[,myID] - +
rbind ( DF2, newrows)
- Original message -
From: Gabor Grothendieck [EMAIL PROTECTED]
To: Ferry
thanks Gabor, and rmailbox.
On Fri, Dec 5, 2008 at 3:32 PM, [EMAIL PROTECTED] wrote:
Here is a solution using doBy and Gabor's DF2 created below:
library( doBy )
newrows - summaryBy ( myNum1 + myNum2 + myNum3 ~ myType , DF2, keep.names
= TRUE )
newrows[,myID] - +
rbind ( DF2, newrows)
Hi All,
Please pardon me if I am missing something obvious here. How do I get
the Kaplan-Meier estimate function that is created by survfit and
plotted by the code.
fit - survfit(Surv(time, status) , data=aml)
plot(fit)
That is, I need a function that will give me the survival estimate at
a
on 12/05/2008 09:10 PM Ritwik Sinha wrote:
Hi All,
Please pardon me if I am missing something obvious here. How do I get
the Kaplan-Meier estimate function that is created by survfit and
plotted by the code.
fit - survfit(Surv(time, status) , data=aml)
plot(fit)
That is, I need a
You can do (1) and (2) [with some additional coding] using mvspec.R, which
you can download from http://www.stat.pitt.edu/stoffer/tsa2/chap7.htm ...
scroll down to the Spectral Envelope section and you'll find it there. You
can look at the top part of the examples to get an idea of how to use
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