I had all these headaches with Tinn-R on Vista - tried reading all the
message boards, reconfiguring .Rprofile, etc... ...no luck.
I finally gave up and started using Eclipse with StatET. Now, it's actually
easier to run RCMDs to check and release a package, and with SVN integration
in Eclipse
I have tried to compile the working Fortran-90 routines (see attachment) to
interface them with R.
The compiler automatically caled by R CMD command line prints a bunch of error
messages (as follows).
I think I'm better off translating the Fortran implementation into R ... which
is what I hoped
Here is a suggestion. Let your data frame be 'dat':
> dat
X YZ
123 31
234 31
345 42
456 32
Try this:
bigData <- data.frame(with(dat,
rbind(cbind(X = rep(X, Z), Y = rep(Y,Y), Z = 1),
cbind(X = rep(X, Y-Z), Y = rep(Y,
Dear Dan,
Try this:
with(dummy,tapply(as.numeric(len),as.factor(V1),summary))
See ?with, ?tapply and ?summary for more information.
HTH,
Jorge
On Fri, Mar 27, 2009 at 10:15 PM, Dan Kortschak <
dan.kortsc...@adelaide.edu.au> wrote:
> Hello R users,
>
> I have a data set which is a set of len
Dear Colin,
Try this:
# Data
DF <- structure(c(0, 0, 0, 4, 27, 0, 1, 0, 7, 16, 1, 0, 0, 16, 44,
0, 0, 0, 2, 39), .Dim = c(5L, 4L), .Dimnames = list(c("L01.1",
"L01.2", "L01.3", "L01.4", "L01.5"), c("P01", "P02", "P03", "P04"
)))
# Distances
Names<-t(combn(colnames(DF),2))
apply(Names,1,function(x
Hello R users,
I have a data set which is a set of lengths and types of objects. I want
to calculate the mean length for each type of object as opposed to the
mean of all the objects in the set.
This is in order to make a comparison between the lengths of each type
of objects and the number of th
Perhaps BA[, 2] is a factor? What you might need is something like
BA[order(BA[, 1], -as.numeric(BA[, 2]), ]
?
Bill Venables.
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Jun Shen [jun.shen...@gmail.com]
Sent: 28 March
All:
Jan de Leeuw and I have taken the initiative to organize a R users group
in Los Angeles with regular meetings.
Our aim is to bring together practitioners (from industry and academia
alike) in order to facilitate a free and open exchange of knowledge and
experience in solving data analysis, s
On 28/03/2009, at 1:56 PM, T.D.Rudolph wrote:
Is there any package or operation in R designed to conduct or
facilitate
Quadrat Variance analysis of spatial data? Any leads would be much
appreciated as I have found very little in my searches thus far.
For spatial ***point pattern*** data,
Is there any package or operation in R designed to conduct or facilitate
Quadrat Variance analysis of spatial data? Any leads would be much
appreciated as I have found very little in my searches thus far.
Tyler
--
View this message in context:
http://www.nabble.com/Quadrat-Variance-analysis-tp
Is there any way to identify or infer the inflection points in a smooth
spline object? I am doing a comparison of various methods of time-series
analysis (polynomial regression, spline smoothing, recursive partitioning)
and I am specifically interested in obtaining the julian dates associated
wit
Since you want to generate your own variable names instead of using a
list, do the operation in three steps:
x <- get(paste("arunoff_",table_year, sep=''))
colnames(x)[4] <- "COUNT"
assign(paste("arunoff_",table_year, sep=''), x)
On Fri, Mar 27, 2009 at 4:01 PM, Steve Murray wrote:
>
> Dear all
Its possible to have otherwise illegal names by surrounding them
in backticks:
> `...@b` <- 3
> `...@b`
[1] 3
> `if` <- 55
> `if`
[1] 55
Also note use of check.names=FALSE:
> Lines <- "a...@b if
+ 1 2"
> DF <- read.delim(textConnection(Lines), header = TRUE, check.names = FALSE)
> DF
a...@b i
Adam Wilson wrote:
Greetings all,
First of all, thanks to all of you for creating such a useful, powerful
program.
I regularly work with very large datasets (several GB) in R in 64-bit Fedora
8 (details below). I'm lucky to have 16GB RAM available. However if I am
not careful and load too muc
Hi, everyone,
I was trying to sort a data frame by two columns, one increasing, the other
decreasing and got an error.
"Error in FUN(left) : invalid argument to unary operator",
The command is "BA[order(BA[1],-BA[2]),]". BA is the data frame. It was
working if I used increasing on both columns.
Arafat Ud Zaman [Tue, Mar 24, 2009 at 06:26:59PM CET]:
> Dear sir,
> I want to know about R command for parameter estimation for generalized
> Poisson regression.
>
Do you mean the method of Consul and Shoukri?
I once tried Generalized Poisson parameter estimation by their suggestion
side by si
Bert Gunter a écrit :
Based on a simple scatterplot of pourcma vs transat, a 4 parameter logistic
looks like wild overfitting, and that may be the source of your problems.
Given the huge scatter, a straight line is about as much as would seem
sensible. I think this falls into the "Why ever would
Hi listers,
Does anybody knows if the function qda for quadratic discriminant analysis
provides the coefficients of quadratic discriminants...
Well, I find out that for the linear discriminant analysis lda, the fonction
provides the coefficients...
Thanks in advance,
Marcio
--
View this message
Florin,
How do you obtain x from (Y, b), i.e. x = g(Y,b)?
I don't follow how a "discontinuity" is introduced, when you plug in x(Y, b)
into f. If f(.) is smooth and all the g(.) are smooth, then the composition
f(g(.)) will also be smooth. If this is not the case, what type of
discontinuity
2009/3/27 huiming song :
> hi, everybody, please help me with this question:
>
> If I want to do iteration for 1000 times, however, for the 500th iteration,
> there is NaN appears. Then the iteration will stop. If I don't want the stop
> and want the all the 1000 iterations be done. What shall I do
Dear everyone,
I don't know much about Integer Programming but am afraid I am facing
a problem that can only be solved via Integer Programming. I was
wondering if those of you who have experience with it could recommend
an R package.
I found the following R packages:
Rglpk
glpk
lpSolve
lpSolveAPI
redirected to r-devel, because there are implementational details of
[.data.frame discussed here. spoiler: at the bottom there is a fairly
interesting performance result.
Romain Francois wrote:
>
> Hi,
>
> This is a bug I think. [.data.frame treats its arguments differently
> depending on the num
Hi all,
I wanted to let you know about our training seminar on predictive analytics
- coming April, May, Oct, and Nov in San Jose, NYC, Stockholm, Toronto and
other cities. This is intensive training for marketers, managers and
business professionals to make actionable sense of customer data by
p
William, The function keepWarnings that you wrote did the trick. Thanks for the
help!
Aaron
> Subject: Re: [R] deleting/removing previous warning message in loop
> Date: Fri, 27 Mar 2009 13:33:51 -0700
> From: wdun...@tibco.com
> To: awell...@hotmail.com
>
> You try a usi
Brigid Mooney wrote:
My apologies, the files version 4-3 files I downloaded came from the link:
http://cran.r-project.org/contrib/extra/batchfiles
Ah, those files, then please ask the author of those files. I'd never
use them, since running
R CMD BATCH TestBatch.R testoutput.txt
without
Dear all,
I've been trying to implement the advice given to me, but without much success
so far. I thought I'd provide the code in full in the hope that it might make
more sense. Just to reiterate, I'm attempting to change the header of the 4th
column of every table to "COUNT".
year<- 1951:2
Based on a simple scatterplot of pourcma vs transat, a 4 parameter logistic
looks like wild overfitting, and that may be the source of your problems.
Given the huge scatter, a straight line is about as much as would seem
sensible. I think this falls into the "Why ever would you want to do such a
t
The number of variables is larger that the number of functions constraints.
You are right I can rewrite my problem like this
max f =h1(x11;x12;..;x1n;Y,b)+ h2(x21,x22, ... x2m;Y,b)
x,b
I know Y and for given values of b I can compute {x11,x1n} as
one system of equations
and {x21,x22
My apologies, the files version 4-3 files I downloaded came from the link:
http://cran.r-project.org/contrib/extra/batchfiles
2009/3/27 Uwe Ligges :
>
>
> Brigid Mooney wrote:
>>
>> Hello,
>>
>> I got a new computer, and am trying to reinstall R and have run into a
>> bit of a problem when runni
Thank you so much again!
Laura
2009/3/27 jim holtman :
> I meant type in "?sample" to see what the 'sample' function does. For
> example, to select 50 random rows,
>
>> sample(2000,50)
> [1] 1004 857 1219 1850 480 543 1466 1498 1854 922 1713 621 414
> 1830 105 817 906 825 1329 1500 18
"in non linear modelling finding appropriate starting values is
something like an art"... (maybe from somewhere in Crawley , 2007) Here
a colleague and I just want to compare different response models to a
null model. This has worked OK for almost all the other data sets except
that one (dumped b
On Fri, 2009-03-27 at 15:11 -0400, Laura Rodriguez Murillo wrote:
> Hi dear list,
>
> I have a list of around 2000 identifiers aranged in a dataframe in one
> column and I would like to choose a random subset of these. I wonder
> if somebody can tell me if I could do this with R...
Not sure what
I meant type in "?sample" to see what the 'sample' function does. For
example, to select 50 random rows,
> sample(2000,50)
[1] 1004 857 1219 1850 480 543 1466 1498 1854 922 1713 621 414
1830 105 817 906 825 1329 1500 1806 1623 186 352 1270 1727 1170
518 1743 370 964 20 710 870
Hi,
Looking at your problem, it seems like you can simply transform it to an
unconstrained problem:
Maximize h(x1, x2, ..., xn)
where h(x1, x2, ..., xn) = f(g1(x), g2(x), ..., gn(x)).
Am I missing something or haven't you provided all the information?
Ravi.
__
Sorry, I'm attaching part of the file I have. They go up to 2000 and I
would like to take a subset of around 300.
5001 1
5001 2
5001 3
5001 5
5001 6
5002 1
5002 2
5002 5
5002 6
5002 11
5002 15
5002 16
5003 1
5003 2
5003 5
5003 6
5004 1
5004 2
5004 5
5004 6
5004 10
5004 11
5004 12
5004 15
5004 16
See, e.g., Paul Murrell's slides from a keynote at useR! 2006, "Can R
Draw Graphs?": http://www.stat.auckland.ac.nz/~paul/Talks/rgraphs.pdf
or even better buy his book on R Graphics and pay attention to the
pixmap package with the addlogo() function.
Uwe Ligges
augusto.sanab...@ga.gov.au wro
?sample
On Fri, Mar 27, 2009 at 3:11 PM, Laura Rodriguez Murillo
wrote:
> Hi dear list,
>
> I have a list of around 2000 identifiers aranged in a dataframe in one
> column and I would like to choose a random subset of these. I wonder
> if somebody can tell me if I could do this with R...
>
> Than
Brigid Mooney wrote:
Hello,
I got a new computer, and am trying to reinstall R and have run into a
bit of a problem when running the BATCH command.
For reference, the OS is Windows Vista, 64 bit.
I installed R 2.8.1 and have the 4-3 files from the following link
extracted with the containing
Hi dear list,
I have a list of around 2000 identifiers aranged in a dataframe in one
column and I would like to choose a random subset of these. I wonder
if somebody can tell me if I could do this with R...
Thank you so much!
Laura RM
__
R-help@r-proj
huiming song wrote:
hi, everybody, please help me with this question:
If I want to do iteration for 1000 times, however, for the 500th iteration,
there is NaN appears. Then the iteration will stop. If I don't want the stop
and want the all the 1000 iterations be done. What shall I do?
suppos
Hello R Users,
I am having difficulty deleting the last warning message in a loop so that the
only warning that is produced is that from the most recent line of code. I
have tried options(warn=1), rm(last.warning), and resetting the last.warning
using something like:
> warning("Resett
Can you tell us more about your obj function, f, and the equality constraints
g_k?
Do you really have as many equality constraints as the number of variables?
Are these all non-linear? Can't you find the roots of this system of
equations? If yes, you could find all the roots (with multiple
Camel case will cause database issues, too, at least if you ever
need to move from one db to another--every one seems to have its
own idea whether to preserve case, up-case, or down-case object
names. You may not care if you only use one db and only from R.
SomeOfUsFindCamelCaseUnreadableToo.
--
Look at what
paste("Fekete_",index$year, index$month, sep='')
is creating in the 'assign'; it is a character vector that have the
same value each time through the loop. What you may want to do is
something like this:
index <- expand.grid(year = sprintf("%04d", seq(1986, 1995)), month =
sprintf(
Hi,
I'm pretty new to R and VERY new to RExcel. I've just finished setting
it up and I'm having some problems. My first Issue is that if I attempt
to use a function like RApply("sum", A1:A4) from within excel it gives
me an error saying "There seems to be no R process connected to the
server"
The variable name in your call to assign should vary within the for
loop, otherwise you're always assigning the value to the same variable.
Consider the following example,
listOfNames <- c("a", "b", "c")
listOfVariables <- c("vara", "varb", "varc")
for(index in seq_along(listOfNames)){
Read ?importance, especially the "scale" argument.
Andy
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Li GUO
> Sent: Friday, March 27, 2009 1:24 PM
> To: r-help@r-project.org
> Subject: [R] Random Forest Variable Importance
Dear all,
I think I'm nearly there in writing R code which will read in files with two
variable parts to the file name and then assigning these file names to objects,
which also have two variable parts. I have got the code running without
encountering errors, however, I receive 50+ of the same
Duncan Murdoch wrote:
On 3/27/2009 11:46 AM, 93354504 wrote:
Hi there,
Given a positive definite symmetric matrix, I can use chol(x) to
obtain U where U is upper triangular
and x=U'U. For example,
x=matrix(c(5,1,2,1,3,1,2,1,4),3,3)
U=chol(x)
U
# [,1] [,2] [,3]
#[1,] 2.236068
Hello everyone!
I have a piece of code that works and does what I need but...:
# I have 3 slots:
nr.of.slots<-3
# My data frame is "new.a":
new.a<-data.frame(x=c("john",
"mary"),y=c("pete","john"),z=c("mary","pete"),stringsAsFactors=FALSE)
print(new.a)
# Creating all possible combinations of the
Hi
I need advice regarding constraint optimization with large number of
variables.
I need to solve the following problem
max f(x1,...,xn)
x1,..xn
x1=g1(x1,...,xn)
.
.
xn=gn(x1,...,xn)
I am using Rdonlp2 package which works well until 40 variables in my
case. I ne
Recent versons of snow signal an error if the value returned from a
worker indicates an error. The error handling facilities in snow are
still evolving; for now id you don't want an error on a worker to
become an error an the master you need to catch the error in the
worker yourself and produce a
Very nice, Duncan.
Here is a little function called loch() that implements your idea for the
Lochesky factorization:
loch <- function(mat) {
n <- ncol(mat)
rev <- diag(1, n)[, n: 1]
rev %*% chol(rev %*% mat %*% rev) %*% rev
}
x=matrix(c(5,1,2,1,3,1,2,1,4),3,3)
L <- loch(x)
all.equal(x, t(L) %*
Very nice, Duncan.
Here is a little function called loch() that implements your idea for the
Lochesky factorization:
loch <- function(mat) {
n <- ncol(mat)
rev <- diag(1, n)[, n: 1]
rev %*% chol(rev %*% mat %*% rev) %*% rev
}
x=matrix(c(5,1,2,1,3,1,2,1,4),3,3)
L <- loch(x)
all.equal(x, t(L) %*
You want a factorizzation of the form: A = L' L. Am I right (we may name this
a "Lochesky factorization)?
By convention, Cholesky factorization is of the form A = L L', where L is a
lower triangular matrix, and L', its transpose, is upper traingular. So, all
numerical routines compute L accord
hi, everybody, please help me with this question:
If I want to do iteration for 1000 times, however, for the 500th iteration,
there is NaN appears. Then the iteration will stop. If I don't want the stop
and want the all the 1000 iterations be done. What shall I do?
suppose I have x[1:1000] and z
Hello,
I have written a small function ('JostD' based upon a recent molecular
ecology paper) to calculate genetic distance between populations (columns in
my data set). As I have it now I have to tell it which 2 columns to use (X,
Y). I would like it to automatically calculate 'JostD' for all co
hi, so, please bear with me as I am new to the wonderful world of
computers...
i am trying to answer the following question, and having no luck:
Focus your analysis on a comparison between respondents labeled “Low” (coded
1) on attend4 and respondents labeled “High” (coded 4). Then, examine the
v
On 3/27/2009 11:46 AM, 93354504 wrote:
Hi there,
Given a positive definite symmetric matrix, I can use chol(x) to obtain U where
U is upper triangular
and x=U'U. For example,
x=matrix(c(5,1,2,1,3,1,2,1,4),3,3)
U=chol(x)
U
# [,1] [,2] [,3]
#[1,] 2.236068 0.4472136 0.8944272
#
Hello,
I have an object of Random Forest : iris.rf (importance = TRUE).
What is the difference between "iris.rf$importance" and "importance(iris.rf)"?
Thank you in advance,
Best,
Li GUO
[[alternative HTML version deleted]]
__
R-help@
Greetings all,
First of all, thanks to all of you for creating such a useful, powerful
program.
I regularly work with very large datasets (several GB) in R in 64-bit Fedora
8 (details below). I'm lucky to have 16GB RAM available. However if I am
not careful and load too much into R's memory, I
I have a very large file with many rows and columns. I want to create a plot
with lowess.
If I try the following it works fine:
data(PrecipGL)
plot(PrecipGL)
lines(lowess(time(PrecipGL),PrecipGL),lwd=3, col=2)
In my file, 2 columns are "nox" and "sdate", and are both typeof() = double.
On 3/27/2009 11:18 AM, Christos Hatzis wrote:
Hi,
Is there a way to find which functions are flagged for debugging in a given
session?
The isdebugged() function (which is new in 2.9.0) will tell you this.
Duncan Murdoch
__
R-help@r-project.org m
On 3/27/2009 10:12 AM, Gerrit Voigt wrote:
Dear list,
Latex/Sweave has trouble processing Sveave-output coming from the
summary-command of a linear Model.
>summary(lmRub)
The output line causing the trouble looks in R like this
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
In
See either textplot in the gplots package or addtable2plot in the plotrix
package, or for even more flexibility learn Sweave or its variants.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-
We will try to quickly get out a new version of Design that checks for
the version of survival that is installed and uses a different .C call
accordingly. This will involve ignoring (for now) the new weights
option Terry has implemented.
Frank
Terry Therneau wrote:
A couple additions to Th
Dear Sebastian,
Consider matplot() for this. Here is an example (taken from Baptiste
Auguie's post):
date <- factor(letters[1:9])
d <- data.frame(x1=seq(1, 9), x2=seq(2, 10), date=date)
matplot(d[,-3],pch=16,xaxt='n',las=1,ylab='Some label here',xlab='Date')
axis(1,d[,3],d[,3])
legend('toplef
I think the problem is is that 'diff' does not have a 'units'
parameter; 'difftime' does. Here is a way of doing it:
> x
[1] "2009-03-27 13:00:00 EDT" "2009-03-27 13:00:35 EDT" "2009-03-27
13:01:10 EDT" "2009-03-27 13:01:45 EDT" "2009-03-27 13:02:20 EDT"
"2009-03-27 13:02:55 EDT" "2009-03-27 13:
How can I print a data.frame to a PDF with pdf()...dev.off()
Paulo E. Cardoso
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PLEASE do read the posting gu
Hi there,
Given a positive definite symmetric matrix, I can use chol(x) to obtain U where
U is upper triangular
and x=U'U. For example,
x=matrix(c(5,1,2,1,3,1,2,1,4),3,3)
U=chol(x)
U
# [,1] [,2] [,3]
#[1,] 2.236068 0.4472136 0.8944272
#[2,] 0.00 1.6733201 0.3585686
#[3,] 0
You are misusing source.
Write a function to do what you want. An "Introduction to R" documents how.
Have you read it?
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
650-467-7374
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Be
Here is some code that may get you started (if I am understanding your question
correctly):
library(TeachingDemos)
myfunc <- function(zmin=90, zmax=195, ncol=100, pal='heat') {
cols <- switch(pal,
heat=heat.colors(ncol),
terrain=terrain.colors(ncol),
Petr,
I'd like to be able to change the ramp to other than grey shades.
Please see my previous message with some data.
Paulo E. Cardoso
-Mensagem original-
De: Petr PIKAL [mailto:petr.pi...@precheza.cz]
Enviada: sexta-feira, 27 de Março de 2009 15:12
Para: Paulo E. Cardos
Thanks, that's great - just what I was looking for.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-conta
How do I pass parameters to R script in Rgui ? Currently, I am using
source("foo.R").
Thanks
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
a
Here's my suggestion using the ggplot2 package (but you may prefer to
stick with base functions),
date = factor(letters[1:9])
d <- data.frame(x1=seq(1, 9), x2=seq(2, 10), date=date)
head(d) # dummy data that resembles yours
str(d)
library(reshape)
md <- melt(d, id="date") # creates a data.fr
OK
I got it working partially.
The plot results in a ramp varying from white -> red -> black, doing this:
plot(grelha,ol="grey80", #! Gráfico com grelha de amostragem e gradiente de
abundância
fg=color.scale(((1-(quad_N_sp$x)/max(quad_N_sp$x))),c(0,1,1),c(0,1),c(0,1)),
cex.lab=0.7,
cex.axi
Hello,
I got a new computer, and am trying to reinstall R and have run into a
bit of a problem when running the BATCH command.
For reference, the OS is Windows Vista, 64 bit.
I installed R 2.8.1 and have the 4-3 files from the following link
extracted with the containing folder in my system PATH
Alice Lin gmail.com> writes:
>
>
> I have a vector of binary data – a string of 0’s and 1’s.
> I want to weight these inputs with a normal kernel centered around entry x
> so it is transformed into a new vector of data that takes into account the
> values of the entries around it (weighting th
Sorry for the mistake. As you probably already guesed, I am just
starting using R. I could not name the difference between a matrix and a
data.frame.
> str(b)
'data.frame': 9 obs. of 7 variables:
$ Datum: Factor w/ 9 levels "06.03.","07.03.",..: 1 2 3 4 5 6 7 8 9
$ X1 : int 408 335 2123 4
Hi,
Is there a way to find which functions are flagged for debugging in a given
session?
Thank you.
-Christos
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PLEASE d
the result of read.table is a data.frame, not a matrix as you first
suggested. Can you copy the result of str(b) so we know what your data
is made of?
I'm guessing the most elegant solution will be to use the reshape
package, followed by ggplot2 or lattice.
baptiste
On 27 Mar 2009, at 14
Hi
r-help-boun...@r-project.org napsal dne 27.03.2009 15:36:23:
> I'm certainly missing something.
>
> In fact the ramp I need must be scaled according to a vector of values
(in
> this case species abundance in each grid cell), as in the example vector
> below:
>
> > length(quad_N_sp$x) # wher
Hello,
I'm trying to calculate an integration and x-axis is a time (format :
%Y-%m-%d %H:%M:%S").
I use diff(date, units="mins") in a loop for but sometimes the results stay
in seconds (95% is ok).
Examples for 2 sets of data are given below (first result stays in seconds
whereas the second in m
dear giovanni---
thanks for answering on r-help to me as well as privately. I very much
appreciate your responding. I read the plm vignette. I don't have the book,
so I can't consult it. :-(. I am going to post this message now (rather
than just email it privately), because other amateurs ma
Unfortunately, I could not solve the problem of plotting all columns of
a matrix against the first column
I used:
b=read.table("d:\\programme\\R\\übungen\\Block 1b.txt", header=T)
"b" is a table with the first column using Dates and the following
columns with vectors.
apply(b[,-1], 2, plot
Can you provide a minimal example that we can run directly after copy
and paste (using a standard data set or dummy data)?
It's always helpful to try and nail down the core of your question
(often you'll find the answer while formulating your question in
minimal terms).
baptiste
On 27
Hi Terry,
My "this" was your (a), i.e. the smoothed hazard rate function.
I apologize if I came across as being rude. I was only curious to see if you
had any scientific/statistical rationale for not including the smoothed hazard
option in your "survival" package, which is, by far, the most w
Dear Prof. Therneau,
Thank you for your views on this subject. I think all R users who play
with survival analysis are most grateful for the functions you have
already supplied us with.
I'm guessing Ravi is wondering why you have not implemented the
smoothing of the baseline hazard from the Cox
I'm certainly missing something.
In fact the ramp I need must be scaled according to a vector of values (in
this case species abundance in each grid cell), as in the example vector
below:
> length(quad_N_sp$x) # where x is the abundance value
[1] 433
quad_N_sp$x
[1] 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0
Dear R users,
I want to get the D1 details reconstructed
to the level of my time series. My original time series is NH$annual[,]
and it has 118 elements. This is the code I use and the results:
library(wavelets)
NHj <- extend.series(X=NH$annual[,], method="reflection",
length="powerof2", j=7);
deta
At 18:22 25/03/2009, Jonathan Baron wrote:
On 03/25/09 19:06, soeren.vo...@eawag.ch wrote:
> Can't make sense of calculated results and hope I'll find help here.
>
> I've collected answers from about 600 persons concerning three
> variables. I hypothesise those three variables to be components (o
That's because do.call wants a list:
what about this one:
> do.call( sprintf, append( list("C:\\Documents and
Settings\\Data\\comp_runoff_hd_%04d%02d.asc"), expand.grid(
seq(1986,1995), 1:12) ) )
Romain
Steve Murray wrote:
Dear all,
Thanks for the help in the previous posts. I've conside
Does this give you what you want (just did it in two steps);
> x <- expand.grid(year = sprintf("%04d", seq(1986, 1995)), month =
> sprintf("%02d", 1:12))
> filelist <- paste("C:\\Documents and Settings\\Data\\comp_runoff_hd_",
> paste(x$year, x$month, sep=''), '.asc', sep='')
>
>
>
>
> filelist
On Fri, 27 Mar 2009, Tsjerk Wassenaar wrote:
Hi,
Have a look at:
?rainbow
?rgb
?heatmap
Furthermore, the packages colorspace, ggplot, plotrix, and RColorBrewer
have useful tools for this.
For the ides behind the palettes in colorspace, see
Achim Zeileis, Kurt Hornik, and Paul Murrell (2
A couple additions to Thomas's message.
The 'survest' function in design directly called C routines in the survival
package. The argument list to the routines changed due to the addition of
weights; calling a C routine with the wrong arguments is one of the more
reliable ways to crash a prog
Dear list,
Latex/Sweave has trouble processing Sveave-output coming from the
summary-command of a linear Model.
>summary(lmRub)
The output line causing the trouble looks in R like this
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
In my Sweaved Tex-file that line looks like this
On 3/27/2009 9:57 AM, bastian250...@freenet.de wrote:
Hello,
I would like to create a 3D plot with the following data formats:
a <- 1:100
b <- 1:100
c <- matrix(, 100, 100)
i.e.
c(i,j) = f ( a(i) , b(j) )
each of the 1 elements i,j in matrix c is a function of a(i) and
b(j). I would li
Hello,
I have a program that used to run well in October, it uses library snow.
Since then, one change has ocurred (snow library has been updated) and
another could have ocurred (I've unadvertently modified something).
Anyway, now when I make the call:
parallel.model.results <- clusterApply(cl,p
Dear all,
Thanks for the help in the previous posts. I've considered each one and have
nearly managed to get it working. The structure of the filelist being produced
is correct, except for a single space which I can't seem to eradicate! This is
my amended code, followed by the first twelve row
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