Dear all,
I have a stat question that may not be related to R, but I would like to
have your advice.
I have just read a medical paper in which the authors report the 1-p (where
p is the cumulative survival probability from the Kaplan Meier curve) as
incidence of disease.
Specifically,
Dear S-users.
This should be an easy one: How do I change pages on an X11 graphics
device under linux?
I thought that the page-up/page-down keys were supposed to do the trick,
but the frame (window) seems to be kind of immune to any kind of keyboard
input. The only reaction I ever see is that
Dear all,
I know that ggplot2 documentation is coming along,
but at the moment I can't find how to do the following:
a) change the title of the legend
b) get rid of the closing line at the bottom of the
density line.
I also observed that the density lines (after limiting the
x-scale) extend a
Hi Ido,
On 7/5/07, Ido M. Tamir [EMAIL PROTECTED] wrote:
Dear all,
I know that ggplot2 documentation is coming along,
but at the moment I can't find how to do the following:
a) change the title of the legend
There's lot of examples in the documentation - and you seem to have
figured how to
the problem I have is that userid's are not just sequential from
1:n_users. if they were, of course I'd have made a big matrix that was
n_users x n_fields and that would be that. but, I think what I cando is
just use the hash to store the index into the result matrix, nothing
more. then the rest
Hi All, I am trying to make a loop for a function and I am using the
following codes. p and var are some matrix obtained before. I would like
to apply the function gpdlow for i in 1:12 and get the returnlow for i
in 1:12. But when I ask for returnlow there are warnings and it turns out
some
Dear R-Help,
I have an array 1260x1260, upper triangle consisting of numbers between 0 and
100, and lower triangle all NA. I can extract the index of those values say
above 99 using the following code:
which(myArray=99 , ind.arr=T)
which returns:
row col
5475 252 253
45423 764 765
ndx - which(myArray=99 , ind.arr=T)
cbind(ndx, myArray[ndx])
Best regards
Frede Aakmann Tøgersen
Scientist
UNIVERSITY OF AARHUS
Faculty of Agricultural Sciences
Dept. of Genetics and Biotechnology
Blichers Allé 20, P.O. BOX 50
DK-8830 Tjele
Phone: +45 8999 1900
Direct: +45 8999 1878
I have a model with 3 fixed factors (type, stress, MorD) and two
significant two-way interactions (type*stress, stress*MorD).
x$summary
# Estimate Std.Error DF t.value pvals ci950 ci990 ci999
#(Intercept)241.738 8.757 994 27.606 0e+00 TRUE TRUE TRUE
#typePsPr
Hello
Ive got a large CSV file (500M) with statistical data. Its devided in
12 columns and I dont know how many lines.
The second column is the date and the second is a unique code for the
location, the rest is (lets say different whether data. See example
below.
070704, 25,
Exactly how are you accessing it and what warnings are you getting. Your
loop is just returning a single value; the last time i=12. If you want a
vector of values back, the you have to do:
returnlow - numeric(12)
for (i in 1:12){
gpdlow - function(u){
p[,i]$beta -u*p[,i][[2]]
}
You are getting two very different results in what you are comparing.
system.time(lapply(1:10^4, mean))
user system elapsed
1.310.001.31
is returning a list with 10,000 values in it. It is taking time to allocate
the space and such.
system.time(for(i in 1:10^4) mean(i))
user
Hubertus wrote:
Dear list,
if I do
smooth.spline(tmpSec, tmpT, all.knots=T)
with the attached data, I get this error-message:
Note that you cannot attach data that way. You migth want to upload it
to some web space and send us the link.
Uwe Ligges
Error in smooth.spline(tmpSec,
Smith, Phil (CDC/CCID/NCIRD) wrote:
Hi R-ers:
I'm drawing a plot and have used different line types (lty) for
different race/ethnicity groups. I want a legend that explains what line
types correspond to the different race/ethnicity groups. I used the
following code:
legend( 1992 , 42
You do not have matching parentheses in this line
returnlow - gpdlow(var[,i][var[,i](p[,i][[2]])
most likely there is a syntax error that halts the execution of the
assignment statement?
--- livia [EMAIL PROTECTED] wrote:
Hi All, I am trying to make a loop for a function and I am using
Hi,
I have a list myList (see below) that consists of id's $'7',,$'10'
and each id has an individual array, the length of which can vary from
id to id.
How can I convert this list into an array, ie. stacking the individual
arrays into one large array?
Thanks
Zava
myList:
$`7`
I'm trying to plot a variogram of a time series, but I can't find anything
that is suitable for a time series. Can anyone help me.
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
Better still IMHO - try NX or FreeNX. That runs an xclient (also of course an
xserver underneath) on your WIndows desktop. Windows pop up as required.
NX compression means that you get very impressive speed, mouse response etc.
A server also needs installing on the Linux box so you may have
Thanks a lot. I have corrected this. But it still does not work. Any thought?
Stephen Tucker wrote:
You do not have matching parentheses in this line
returnlow - gpdlow(var[,i][var[,i](p[,i][[2]])
most likely there is a syntax error that halts the execution of the
assignment statement?
Nguyen Dinh Nguyen wrote:
Dear all,
I have a stat question that may not be related to R, but I would like to
have your advice.
I have just read a medical paper in which the authors report the 1-p (where
p is the cumulative survival probability from the Kaplan Meier curve) as
[[diverted from R-bugs to R-help by the list maintainer]]
Dear Friend and distinguished R gurus,
first of all really thank you very much for the marvellous tool that is R.
I am using R 2.5.0, windows XP - italian language.
I was perfoming some calculation on fractional
What does
gpdlow(var[,i][var[,i](p[,i][[2]])
return? Is it a vector; if so, how long? Your declaration of
returnlow- matrix(,12)
str(returnlow)
logi [1:12, 1] NA NA NA NA NA NA ...
is a matrix of 12 rows and one column. You may be getting the error message
is gpdlow is returning a
I tried str(gpdlow(var[,i][var[,i](p[,i][[2]])) and it returns num [1:49]
-1.92 -1.69 -2.20 -1.65 -2.13 ...
It is the number when i=1, I guess it does not loop. In fact, the number
should be different when loop between i.
jim holtman wrote:
What does
gpdlow(var[,i][var[,i](p[,i][[2]])
I don't get your point, because
exp(-(-3)^2.2)
[1] NaN
is correct. A negative value to the power of a non-integer is undefined
in IR. Of course it is defined as a complex number:
exp(-(-3+0i)^2.2)
[1] 1.096538e-04-3.47404e-05i
Uwe Ligges
Giuseppe PEDRAZZI wrote:
The 1-Pr(disease free survival) estimate from KM is not appropriate if
competing risk of mortality (from causes other than the disease of interest)
are present. In that case, 1-Pr(disease free survival) over-estimates the
cumulative incidence of disease. The larger the hazard of mortality, the
Thanks. Yes, gpdlow is indeed return a vector longer than one. The length of
the vector is different for i in 1:12, each equals to
length(var[,i][var[,i](p[,i][[2]]).
jim holtman wrote:
What does
gpdlow(var[,i][var[,i](p[,i][[2]])
return? Is it a vector; if so, how long? Your
It is going to be easy question to you. I've started to interest in
model-based clustering.
Adrian E. Raftery Recent Advances in Model-Based Clustering: Image
Segmentation and Variable Selection (www.stat.washington.edu/Raftery)showed
that we can compare different classification methods using
On Thu, 5 Jul 2007, Giuseppe PEDRAZZI wrote:
I am using R 2.5.0, windows XP - italian language.
I was perfoming some calculation on fractional exponential and
I found a strange behaviour. I do not know if it is really a bug, but I would
expect
a different answer from R.
I was trying the
Hi All,
is there Levene' test in R ? If not ,Could you give me some
advice about Levene test with R?
Thanks a lot! I am waiting for yours.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
adschai : this isn't particularly helpful but when I am using a
function from a package called xxx that
I have little knowledge about, I take the source as is and create my own
function out of
It called my.xxx and then put print statements
Inside it to see what's going on. This is probably an
Hi,
RSiteSearch(Levene Test, restrict=functions)
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 05/07/07, along zeng [EMAIL PROTECTED] wrote:
Hi All,
is there Levene' test in R ? If not ,Could you give me some
advice about Levene test with R?
Thanks
I got it!
Thank all of you,Sorry I am a freshman of R.
2007/7/5, Chuck Cleland [EMAIL PROTECTED]:
along zeng wrote:
Hi All,
is there Levene' test in R ? If not ,Could you give me some
advice about Levene test with R?
Thanks a lot! I am waiting for yours.
Did you try
along zeng wrote:
Hi All,
is there Levene' test in R ? If not ,Could you give me some
advice about Levene test with R?
Thanks a lot! I am waiting for yours.
Did you try RSiteSearch(levene, restrict=function), which points
to a funtion in the car package?
if by array, you just mean vector, then the following would work .
stack(myList)
If you want to take off the names of myList that get put in the second
column
stack(myList)[,-2,drop=FALSE]
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Aydemir, Zava
I got it!
Thank all of you,Sorry I am a freshman of R.
2007/7/5, Tomas Goicoa [EMAIL PROTECTED]:
Hi,
library(car)
?levene.test
At 16:36 5/7/2007, along zeng wrote:
Hi All,
is there Levene' test in R ? If not ,Could you give me some
advice about Levene test with R?
I would like to add points to a wireframe but with a conditioning variable. I
found a solution for this without a conditioning variable here,
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/65321.html. Does anyone know
how to plot a wireframe conditioned on a variable and add the points
hello,
I have a problem running a R script actually I'm using source() and sink() and
it doesn't work
source(T:/agents/melyakhlifi/R/essai_rep.r)
to execute a file and the file contain
sink(T:/agents/melyakhlifi/R/sortie.html)
cat(htmlbodypre\n)
matrix.merge2
cat(/pre/body/html\n)
sink()
I
Hi,
library(car)
?levene.test
At 16:36 5/7/2007, along zeng wrote:
Hi All,
is there Levene' test in R ? If not ,Could you give me some
advice about Levene test with R?
Thanks a lot! I am waiting for yours.
__
Hi Lars,
I haven't tried this, but I believe there were a couple of messages on
the list recently on reading large files that basically used scan with
connections, and reading in by blocks.
see ?scan, ?connections
HTH
steve
Lars Modig wrote:
Hello
I’ve got a large CSV file (500M) with
All,
Is there an efficient way to apply say mean or median to a dataframe
according to say all combinations of two variables in the dataframe?
Below is a simple example and the outline of a manual solution that
will work but is not very efficient
(could also generalize this to a function).
I don't know what is causing your problem, But if you goal is to
produce html then you may want to look at the R2HTML package. It may do
what you want without using sink.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801)
Adschai:
here is an example for class.weights (isn't it on the help page?):
data(iris)
i2 - iris
levels(i2$Species)[3] - versicolor
summary(i2$Species)
wts - 100 / table(i2$Species)
wts
m - svm(Species ~ ., data = i2, class.weights = wts)
Cheers,
David
Try this:
aggregate(dat.ex[2:3], dat.ex[4:5], mean)
On 7/5/07, Afshartous, David [EMAIL PROTECTED] wrote:
All,
Is there an efficient way to apply say mean or median to a dataframe
according to say all combinations of two variables in the dataframe?
Below is a simple example and the
On 7/5/07, jim holtman [EMAIL PROTECTED] wrote:
You are getting two very different results in what you are comparing.
system.time(lapply(1:10^4, mean))
user system elapsed
1.310.001.31
is returning a list with 10,000 values in it. It is taking time to allocate
the space and
Afshartous, David afshart at exchange.sba.miami.edu writes:
All,
Is there an efficient way to apply say mean or median to a dataframe
according to say all combinations of two variables in the dataframe?
..[snip]..
See function summaryBy in package doBy
my.data - data.frame(
trts - rep(c('Drug 1','Drug2'), each = 10),
doses - rep(c('Low dose','High dose'), 10),
resp - rnorm(20)
)
tapply(my.data$resp, list(my.data$trts, my.data$doses), mean)
Jim
Afshartous, David wrote:
All,
Is there an efficient way to
Please show use the statements that you used. Did you put the 'str' inside
the loop? It is hard to tell what is happening without reproducible code.
On 7/5/07, livia [EMAIL PROTECTED] wrote:
I tried str(gpdlow(var[,i][var[,i](p[,i][[2]])) and it returns num [1:49]
-1.92 -1.69 -2.20 -1.65
Hi all,
Does anyone know where to find more information on the by class?
OR
Does anyone know how to coerce an object of the by class into a
data.frame containing the results of FUN and the values of the
grouping variables?
Pietrzykowski, Matthew (GE, Research)
[EMAIL PROTECTED] wrote:
Hello all-
I would appreciate any guidance that can be provided.
I am new to R and am using it exclusively in a statistics
program I am undertaking that mainly references
Minitab. My focus is on data modeling and further
Dear all,
I have a very small script to plot a function. Here it is:
##
sinca - function(N,th)
{
return(sin((N+0.5)*th)/sin(0.5*th))
}
plot_sinca - function(N)
{
x - seq(-5*pi,5*pi,by=pi/100)
y - rep(0,length=length(x))
for (i in 1:length(x))y[i] -
Hi all,
I want to make a vector with the third column of a matrix, but only for the
2+3n rows of the matrix, with n being an entire number from 0 to a million.
How can I do that in an easy way?
Thanks in advance,
Juan Pablo
[[alternative HTML version deleted]]
Hi Zeng,
I just glanced at the link, but I think this is what you are after:
x=rnorm(1000)#1000 random samples from N(0,1)
y=rlnorm(1000)#1000 random samples from Lognormal(0,1)
fx=ecdf(x)#Empirical cumulative density function of x
fy=ecdf(y)#Empirical cumulative density function of y
Dear All,
I am not a statistician, and was wondering if anyone could help me
with the following.
Greenacre, in his Correspondence Analysis in Practice (1993, p.173)
gives a method for testing the significance of an axis in CA where:
$\chi^2 = \lambda \times n$ where \lambda is the the
Hi,
Example:
n - 1:10
mat[2+3*n,3] #Where mat is the matrix
Is what you want?
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 05/07/07, Juan Pablo Fededa [EMAIL PROTECTED] wrote:
Hi all,
I want to make a vector with the third column of a matrix, but only for
Does anyone know how to print out the current function name and line
number (similar to how warning(call.=TRUE) operates). I'd like this for
a logging package I'm working on, but I can't seem to find a way.
Thanks for your help,
John Scillieri
This e-mail and any attachments are confidential,
On 7/5/07, Mark Lyman [EMAIL PROTECTED] wrote:
I would like to add points to a wireframe but with a conditioning variable. I
found a solution for this without a conditioning variable here,
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/65321.html. Does anyone know
how to plot a wireframe
Or an alternative to Henrique's if you want to select
all the rows from row 2 up to the 3*n row this may
work.
n - 2
myvector - data1[2:(2*n), 3]
--- Juan Pablo Fededa [EMAIL PROTECTED] wrote:
Hi all,
I want to make a vector with the third column of a
matrix, but only for the
2+3n
I'm trying to hunt down an appropriate kriging package for my specific
application, and I was hoping someone on the R list might have some pointers
-- I'm interested in performing kriging and related spatial interpolations
with one of the R packages, but I need to be able to provide my own
I'm trying to hunt down an appropriate kriging package for my specific
application, and I was hoping someone on the R list might have some pointers
-- I'm interested in performing kriging and related spatial interpolations
with one of the R packages, but I need to be able to provide my own
On 7/5/2007 11:35 AM, elyakhlifi mustapha wrote:
hello,
I have a problem running a R script actually I'm using source() and
sink() and it doesn't work
source(T:/agents/melyakhlifi/R/essai_rep.r)
to execute a file and the file contain
sink(T:/agents/melyakhlifi/R/sortie.html)
Deepayan Sarkar deepayan.sarkar at gmail.com writes:
On 7/5/07, Mark Lyman mark.lyman at gmail.com wrote:
I would like to add points to a wireframe but with a conditioning
variable. I
found a solution for this without a conditioning variable here,
Very simple; it is your function. You need to step through and see that you
are evaluating close to zero:
x[701]
[1] 6.283185
sin((4.5*x[701]))
[1] -1.666142e-14
sin(.5*x[701])
[1] -1.653896e-15
sin((4.5*x[701]))/sin(.5*x[701])
[1] 10.07404
With numbers that small you might be losing
What value should your formula give when x is a multiple of 2*pi?
You seem to believe 9 is correct but in fact NaN is.
Element 701 of x is approximately but not exactly 2*pi: on my system
it is about 7*.Machine$double.eps different. You cannot expect sin(N*pi)
to be exactly zero for N != 0.
On Thu, 5 Jul 2007, Paul Matthias Diderichsen wrote:
Dear S-users.
This is the help forum for R users
This should be an easy one: How do I change pages on an X11 graphics
device under linux?
I thought that the page-up/page-down keys were supposed to do the trick,
It is baffling,
True.
I've corrected my function in this way:
###
sinca - function(N,th)
{
mod_th - th%%(2*pi)
if (mod_th == 0)
{
ff - 2*N+1
}
else
{
ff - sin((N+0.5)*th)/sin(0.5*th)
}
return(ff)
}
###
The function is equal to
How on Earth can I know what are the arguments of any of the functions of
the tcl/tk package? I tried hard to find, using all search engines
available, looking deep into keywords of R, python's tkinter and tcl/tk, but
nowhere I found anything remotely similar to a help.
For example, what are
One approach is to use the fact that vectors are automatically
replicated to the correct length when subscripting, so you can do
something like:
my.matrix[ c(FALSE,TRUE,FALSE), 3 ]
To get every 3rd element starting at the 2nd element, and the 3rd
column.
Hope this helps,
--
Gregory (Greg)
see FAQ 7.31 and read What Scientists should know about floating point
numbers
On 7/5/07, James Foadi [EMAIL PROTECTED] wrote:
True.
I've corrected my function in this way:
###
sinca - function(N,th)
{
mod_th - th%%(2*pi)
if (mod_th == 0)
{
ff -
Alberto Monteiro [EMAIL PROTECTED] wrote:
How on Earth can I know what are the arguments of any of the functions of
the tcl/tk package? [...]
My impression is that you as supposed to look in tck/tk manuals.
For example, Googling on
tk tck getopenfile
pointed to this Web page:
The problem is that you are dividing two numbers that are both very small.
Any small imprecision in the denominator creates a big error.
Here you can re-write your function using a trig identity to get accurate
results:
sinca - function(N,th) {
#return(sin((N+0.5)* th)/sin(0.5*th))
return(
Alberto Monteiro wrote:
How on Earth can I know what are the arguments of any of the functions of
the tcl/tk package? I tried hard to find, using all search engines
available, looking deep into keywords of R, python's tkinter and tcl/tk, but
nowhere I found anything remotely similar to a
Hi Alberto,
It took me approximately 20 seconds to find all the arguments for this
function. Here were the steps I took.
1.) Look at R help page ?tkgetOpenFile
2.) Hmmm. Lots of functions, but little info. But wait, what's this?
Details:
[snip]
There are far too many of these
Dear R list,
In the course of learning to work with Rmpi, we are confused about a few
points. The following simple program is based on some examples we retrieved
from the web. Each slave is writing the same output line multiple times (a
multiple equal to the number of slaves). In other words,
James MacDonald wrote:
3.) Type tkgetOpenFile at R prompt.
tkgetOpenFile
function (...)
tcl(tk_getOpenFile, ...)
environment: namespace:tcltk
4.) Google tk_getOpenFile.
5.) http://www.tcl.tk/man/tcl8.5/TkCmd/getOpenFile.htm
Thanks, you (all who helped) are so nice. I even
H == Hubertus [EMAIL PROTECTED]
on Wed, 4 Jul 2007 14:58:44 +0200 writes:
H Dear list,
H if I do
H smooth.spline(tmpSec, tmpT, all.knots=T)
H with the attached data,
Thanks for providing the data via URL (see below)
H with the attached data, I get this error-message:
Hi,
I would like to generate below triangular matrix for some a.
0
0
0
0
0
0
1
0
0
0
0
0
a
1
0
0
0
0
a^2
a
1
0
0
0
a^3
a^2
a
1
0
0
a^4
a^3
a^2
a
1
0
What's an efficient way to do this? (this matrix being KxK, K very
large)
Thanks
Zava
On 05/07/2007 4:38 PM, Peter Dalgaard wrote:
Alberto Monteiro wrote:
How on Earth can I know what are the arguments of any of the functions of
the tcl/tk package? I tried hard to find, using all search engines
available, looking deep into keywords of R, python's tkinter and tcl/tk, but
On 7/5/07, Paul Matthias Diderichsen
[EMAIL PROTECTED] wrote:
Dear S-users.
This should be an easy one: How do I change pages on an X11 graphics
device under linux?
I thought that the page-up/page-down keys were supposed to do the trick,
but the frame (window) seems to be kind of immune to
Hi All,
Sorry if I ask an obvious thing, I am still new to R ...
I created a data frame of given dimensions to which I gave strings as
column names. I want to write to elements of the data frame by indexing
them with the row number and column name (string). The problem is that I
can read
Dear Friends,
Suppose I have a vector as follows
RI Value
1 10
2 11
3 8
4 4
6 12
I would like a function which returns therow index number for the minimum
value of VALUE. Can somebody please give me some direction on how I can do
this. In effect I'm trying to find a
?which.min
x
RI Value
1 110
2 211
3 3 8
4 4 4
5 612
which.min(x$Value)
[1] 4
On 7/5/07, Anup Nandialath [EMAIL PROTECTED] wrote:
Dear Friends,
Suppose I have a vector as follows
RI Value
1 10
2 11
3 8
4 4
6 12
I would like a
Hi R Gurus.
I'm trying to install the Rmpi package.
Here are the errors:
checking for stdint.h... yes
checking for unistd.h... yes
checking mpi.h usability... no
checking mpi.h presence... no
checking for mpi.h... no
Try to find libmpi or libmpich ...
checking for main in -lmpi... no
libmpi not
Have you seen The Basics of S Plus, by Krause and Olson? It's really good too
.
On 7/5/07, Mike Prager [EMAIL PROTECTED] wrote:
Pietrzykowski, Matthew (GE, Research)
[EMAIL PROTECTED] wrote:
Hello all-
I would appreciate any guidance that can be provided.
I am new to R and am using it
Hi List,
I want replace characters within a vector. Outside R I could use sed,
but I'd like to automate it in R. For example
vectorx
xxxyyz
xxxyyza
xxxyyzzb
I want to change to:
vectorx
aaayyz
aaayyza
aaayyzzb
The obvious replace command only deals with whole data entries?
Any hints would
On Fri, 2007-07-06 at 12:40 +1000, [EMAIL PROTECTED] wrote:
Hi List,
I want replace characters within a vector. Outside R I could use sed,
but I'd like to automate it in R. For example
vectorx
xxxyyz
xxxyyza
xxxyyzzb
I want to change to:
vectorx
aaayyz
aaayyza
aaayyzzb
The
On 5 July 2007 at 20:19, Edna Bell wrote:
| Hi R Gurus.
|
| I'm trying to install the Rmpi package.
|
| Here are the errors:
| checking for stdint.h... yes
| checking for unistd.h... yes
| checking mpi.h usability... no
| checking mpi.h presence... no
| checking for mpi.h... no
| Try to find
Check out ?chartr
On 7/5/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Hi List,
I want replace characters within a vector. Outside R I could use sed,
but I'd like to automate it in R. For example
vectorx
xxxyyz
xxxyyza
xxxyyzzb
I want to change to:
vectorx
aaayyz
aaayyza
sub(xxx, aaa, vectorx)
or maybe gsub, depending on your application.
Cheers,
Simon.
On Fri, 2007-07-06 at 12:40 +1000, [EMAIL PROTECTED] wrote:
Hi List,
I want replace characters within a vector. Outside R I could use sed,
but I'd like to automate it in R. For example
vectorx
xxxyyz
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