P.Dalgaard wrote:
>
>
>> IF TYPE='TRUCK' and count=12 THEN VEHICLES=TRUCK+((CAR+BIKE)/2.2);
>
> vehicles <- ifelse(TYPE=='TRUCK' & count=12, TRUCK+((CAR+BIKE)/2.2), NA)
>
>
Read both versions to an audience, and you will have to admit that this is
one of the cases where SAS is superior.
D
Tony Breyal wrote:
>
> Is there an alternative website which uses a similar structure to
> google groups? I had a quick browse on the R Wiki
> (http://wiki.r-project.org/rwiki/doku.php?id=links:links) but didn't
> see a page with this sort of info.
>
>
For reading and searching, nothing beat
Manuel Morales wrote:
>
> I'm trying to fit a model like beta[trt]/(1+alpha*x) where the data
> include some grouping factor. The problem is that the estimate for alpha
> is undefined for some of the treatments - any value greater than 20 is
> equally good and a step function would suffice. Ign
Hi there,
I just wonder how to draw this kind of picture...
http://www.nabble.com/file/p24158796/b.jpg
http://www.nabble.com/file/p24158796/a.jpg
and this is what i have done
%
library(shape)
library(diagram)
curve(sin(x),bty="n",-8,8,yaxt="n",ylab="",xaxt="n",type="n",xlab="")
axis(1,la
I think you should explain (to yourself primarily) what it means to
have a non-significant intercept. If you can justify on a theoretic
basis the exclusion of an intercept, then you may get more assistance.
However, if you are just naively questing after some mythical concept
of "significan
I posted this question way down at teh end of another thread realted to an
error in step, but that was stupid since it really is another matter
altogether. I should have posted it separately, as I have now done.
The code below creates a data.frame comprising three marginally noisy
surfaces. The c
try:
'x' %in% colnames(df)
hth,
Kingsford
On Mon, Jun 22, 2009 at 9:17 PM, R_help Help wrote:
> Hi,
>
> I have a matrix or data.frame (df). When I would like to check if a column
> named "x" exists, it becomes my habit to do the following:
>
> if (length(which(colnames(df)=="x"))>0)
>
> Is there
I think that you want the the @nord tag which is in development. I am eager
for the release of the Roxygen version with this tag too!
See their mailing list on R-Forge for more details on this.
cheers,
Ian
Ken-JP wrote:
>
> I don't want any zoo.Rd to be generated - I am a user of the library
Hi,
I have a matrix or data.frame (df). When I would like to check if a column
named "x" exists, it becomes my habit to do the following:
if (length(which(colnames(df)=="x"))>0)
Is there any smarter way to do this? Thank you.
adschai
[[alternative HTML version deleted]]
__
David Winsemius wrote:
>
>
> ...
>
> Perhaps:
>
> by(data2, data2$grp, function(x) step(lm(model, data=x)))
>
>
> David Winsemius, MD
> Heritage Laboratories
> West Hartford, CT
>
>
Perhaps ;) You know of course that this worked. It took a moment to
understand the screen output bec
Dieter Menne wrote:
>
> ...
>
> Looks like an environment problem. I could not find a workaround quickly,
> but you might have a look at
>
> http://finzi.psych.upenn.edu/R/Rhelp02a/archive/16599.html
>
> We call it "Ripley's Game" here, because variants of it can help you quite
> often.
>
On 23/06/2009, at 11:38 AM, m.gha...@yahoo.fr wrote:
Hi,
I'm a beginner using R and I'm modeling a time series with ARIMA.
I'm looking for a way to determine the p-values of the coefficients
of my model.
Does ARIMA function return these values? or is there a way to
determine them easily?
Try
txt[- grepl(pattern, txt)]
txt[! grepl(pattern, txt)]
On Mon, Jun 22, 2009 at 5:45 PM, orzack wrote:
> Does anybody know how to negate a string in a grep command, i.e., what I
> need is to return only strings that do NOT contain a second string anywhere
> in the entire string.
>
>
> for
Hi Myriam,
I'll take a stab at it, but can't offer elegance in the solution such
as the more experienced R folks might deliver.
I believe that the ARIMA function provides both point estimates and
their standard errors for the coefficients. You can use these as you
might a mean and standard error
You can try somthing about like this:
setdiff(txt, grep("(oob|boo|\\w[^oo]\\w)", txt, value = TRUE))
On Mon, Jun 22, 2009 at 6:45 PM, orzack wrote:
> Does anybody know how to negate a string in a grep command, i.e., what I
> need is to return only strings that do NOT contain a second string an
David,
Once again, many thanks for your very useful and timely feedback, and
for your patience with my learning curve.
Sincerely,
Cliff
On Mon, Jun 22, 2009 at 7:11 PM, David Winsemius wrote:
>
> On Jun 22, 2009, at 7:55 PM, David Winsemius wrote:
>
>>
>> On Jun 22, 2009, at 6:16 PM, Clifford
Check for unbalanced quotes or comments. try:
count.fields("RWM Shopper Tracker - RAW DATA - 22JUN09 - Copy.csv",
quote='', comment.char='',
sep=",")
On Mon, Jun 22, 2009 at 9:38 PM, Chris Howden wrote:
> Morning all,
>
> I'm trying to read in a csv file and R is having some problems. For some
>
Morning all,
I'm trying to read in a csv file and R is having some problems. For some
reason its not 'seeing' all the columns for each row, and as such is not
reading in the file.
I've opened the file in EXCEL and I can't see any problems with it. All rows
have the correct number of columns.
The
Hi all,
Raymond Wan wrote:
I am creating dendrograms using agnes and was wondering if it is
possible to add color to the leaves (and just the leaves).
I think I can answer my own question :-). Just found out about the A2R
package, which seems to do what I am looking for... Seems to work f
Thanks a lot! There is no MSVC project for R on Windows?
On Mon, Jun 22, 2009 at 5:32 PM, Duncan Murdoch wrote:
> On 22/06/2009 6:52 PM, Michael wrote:
>>
>> Hi all,
>>
>> I am thinking of extending a package by directly adding stuff to its
>> C++ code. And then I have to recompile the package.
>>
If you want to use regular expressions, try:
> x
V1 V2 V3 V4
1 1 2009-06-20 00:53:00 73
2 2 2009-06-20 01:08:00 73
3 3 2009-06-20 01:44:00 72
4 4 2009-06-20 01:53:00 71
5 5 2009-06-20 02:07:00 72
> x[grep(":53:", x$V3),]
V1 V2 V3 V4
1 1 2009-06-20 00:53:00 73
4
Try this:
> Lines <- "dt,tf
+ 2009-06-20 00:53:00,73
+ 2009-06-20 01:08:00,73
+ 2009-06-20 01:44:00,72
+ 2009-06-20 01:53:00,71
+ 2009-06-20 02:07:00,72"
> DF <- read.csv(textConnection(Lines), colClasses = c("POSIXct", "numeric"))
> DF[format(DF$dt, "%M") == "53",]
dt tf
1 2009
Does anybody know how to negate a string in a grep command, i.e.,
what I need is to return only strings that do NOT contain a second
string anywhere in the entire string.
for
txt <- c("boo","goo","doob","foo","boofoo")
I need a grep command that returns strings with "oo" except when "b"
is
Convert dt to POSIXct class:
wtd$dt <- as.POSIXct(wtd$dt)
subset(wtd, format(dt, '%M') == 53)
On Mon, Jun 22, 2009 at 9:58 PM, Keith Jones wrote:
> Hi,
>
> I have a data frame with two columns: dt and tf. The dt column is datetime
> and the tf column is a temperature.
>
>
Hi,
I'm a beginner using R and I'm modeling a time series with ARIMA.
I'm looking for a way to determine the p-values of the coefficients of my model.
Does ARIMA function return these values? or is there a way to determine them
easily?
Thanks for your answer
Myriam
Hi,
I have a data frame with two columns: dt and tf. The dt column is
datetime and the tf column is a temperature.
dt tf
1 2009-06-20 00:53:00 73
2 2009-06-20 01:08:00 73
3 2009-06-20 01:44:00 72
4 2009-06-20 01:53:00 71
5 2009-06-20 02:07:00 72
...
I need a
On 22/06/2009 6:52 PM, Michael wrote:
Hi all,
I am thinking of extending a package by directly adding stuff to its
C++ code. And then I have to recompile the package.
Do I have to download the whole R source repository, in order to do
the recompilation? What is the minimal setup requirement for
On Jun 22, 2009, at 7:56 PM, Jeffrey Edgington wrote:
Greetings,
I have two files which contain responses to a series of multiple
choice questions. One
file contains responses before an "intervention" and the other
contains the responses afterward.
There were three possible responses to
Try this:
table(factor(pre, levels = c("D", "F", "T")),
factor(post, levels = c("D", "F", "T")))
On Mon, Jun 22, 2009 at 8:56 PM, Jeffrey Edgington wrote:
> Greetings,
>
> I have two files which contain responses to a series of multiple choice
> questions. One
> file contains responses
On Jun 22, 2009, at 7:55 PM, David Winsemius wrote:
On Jun 22, 2009, at 6:16 PM, Clifford Long wrote:
Hi David,
I appreciate the advice. I had coerced 'list4' to as.list, but
forgot
to specify "list=()" in the call to aggregate. I made the
correction,
and now get the following:
slo
Greetings,
I have two files which contain responses to a series of multiple
choice questions. One
file contains responses before an "intervention" and the other
contains the responses afterward.
There were three possible responses to each question: D, F, T (for
Don't Know, False, and Tru
On Jun 22, 2009, at 6:19 PM, Mark Na wrote:
Hi R-helpers,
I have been struggling with calculating row and column statistics,
e.g. standard deviation.
I know that
datac$Mean<-rowMeans(datac,na.rm=TRUE)
will give me row means.
I have tried to replicate those row means with the apply function
On Jun 22, 2009, at 6:16 PM, Clifford Long wrote:
Hi David,
I appreciate the advice. I had coerced 'list4' to as.list, but forgot
to specify "list=()" in the call to aggregate. I made the correction,
and now get the following:
slope.mult = simarray[,1]
adj.slope.value = simarray[,2]
adj.sl
Try this:
> unname(t(do.call(cbind, lapply(a, ts
[,1] [,2] [,3] [,4]
[1,] "a" "b" "c" NA
[2,] "d" "e" NA NA
[3,] "f" "g" "h" "i"
On Mon, Jun 22, 2009 at 7:15 PM, Kenneth Takagi wrote:
> Hi,
>
>
>
> I have a list made up of character strings with each item a different
> length
Thanks for all the feedback; I found a previous post covering this
question:
https://stat.ethz.ch/pipermail/r-help/2009-February/189232.html
Problem solved!
Kenneth Takagi
kat...@psu.edu
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Monda
Try this:
matrix(unlist(lapply(a, '[', 1:max(sapply(a, length, ncol = 4, byrow =
TRUE)
or
do.call(rbind, lapply(a, '[', 1:max(sapply(a, length
On Mon, Jun 22, 2009 at 8:15 PM, Kenneth Takagi wrote:
> Hi,
>
>
>
> I have a list made up of character strings with each item a different
>
try this:
> a
[[1]]
[1] "a" "b" "c"
[[2]]
[1] "d" "e"
[[3]]
[1] "f" "g" "h" "i"
> # find max row length
> rowMax <- max(sapply(a, length))
> # now create output matrix by lengthening rows
> do.call(rbind, lapply(a, function(x){
+ length(x) <- rowMax
+ x
+ }))
[,1] [,2] [,3] [,4]
[1,]
Hi,
I have a list made up of character strings with each item a different
length (each item is between 1and 6 strings long). Is there a way to
convert a "ragged" list to a matrix such that each item is its own row?
Here is a simple example:
a=list();
a[[1]] = c("a", "b", "c");
a[[2]] = c
I am trying to run jags.model and need the data read in a suitable
format. The package (rjags) documentation describes "read.data", while
the "classic-bugs" examples use "read.jagsdata". I am running R (R
2.8.1) on a PowerBook G4 (Mac OS X 10.5.7), and my installation
recognizes neither read
I have used all 3 packages for decision trees (SAS/EM, CART and R). As
another user on the list commented, the algorithms CART uses are
proprietary. I also know that since the algorithms are proprietary, the
decision tree that you get from SAS is based on a "slightly different"
algorithm so as to n
Hi all,
I am thinking of extending a package by directly adding stuff to its
C++ code. And then I have to recompile the package.
Do I have to download the whole R source repository, in order to do
the recompilation? What is the minimal setup requirement for such a
recompilation?
I am using MSVC
Mark Na wrote:
Hi R-helpers,
I have been struggling with calculating row and column statistics,
e.g. standard deviation.
I know that
datac$Mean<-rowMeans(datac,na.rm=TRUE)
will give me row means.
I have tried to replicate those row means with the apply function:
datac$Mean2<-apply(datac,2,m
Tena koe Mark
I think you might want
apply(datac, 1, mean)
i.e., apply the function to the first dimension (rows) rather than the
second (columns).
HTH ...
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ma
thanks Deepayan, that works great!
2009/6/19 Deepayan Sarkar :
> On 6/18/09, Katharina May wrote:
>> Hi there,
>>
>> sorry for troubling everybody once again, I've got a problem rewriting
>> Sarkar's function for
>> rewriting the tick locations in a logaritmic way (s.
>> http://lmdvr.r-forge
Hi R-helpers,
I have been struggling with calculating row and column statistics,
e.g. standard deviation.
I know that
> datac$Mean<-rowMeans(datac,na.rm=TRUE)
will give me row means.
I have tried to replicate those row means with the apply function:
> datac$Mean2<-apply(datac,2,mean)
so that I
Hi David,
I appreciate the advice. I had coerced 'list4' to as.list, but forgot
to specify "list=()" in the call to aggregate. I made the correction,
and now get the following:
> slope.mult = simarray[,1]
> adj.slope.value = simarray[,2]
> adj.slope.level = simarray[,2]
> qc.run.violation = sim
Hi R-list,
I'll apologize in advance for (1) the wordiness of my note (not sure
how to avoid it) and (2) any deficiencies on my part that lead to my
difficulties.
I have an application with several stages that is meant to simulate
and explore different scenarios with respect to product sales (in
Hi John,
The SIDS examples in the spatial graphics gallery may be helpful:
http://r-spatial.sourceforge.net/gallery/
For a general reference, see Applied Spatial Data Analysis with R.
Note that the book's webpage includes figures with code
http://www.asdar-book.org/
There's also the spatial Ta
Hello list,
I'm trying to fit a model like beta[trt]/(1+alpha*x) where the data
include some grouping factor. The problem is that the estimate for alpha
is undefined for some of the treatments - any value greater than 20 is
equally good and a step function would suffice. Ignoring the grouping
stru
On Mon, 22 Jun 2009, charles78 wrote:
It is not the case as you described. In any case, the total area should be 1
and labeled fraction on y axis should be far less than 1, since I have more
than 1 data points. I also test differerent bin size by change the
break.
Try reading Greg's res
On Monday 22 June 2009, Gonzalo Quiroga wrote:
> Hi, I need help to perform a Shapiro.test on a data frame, I know that
> this test works only with vector but I guess there most be a way to
> permor it on a data frame instead of vactor by vector (i.e. I've got 40
> variables to analyze and its kind
(a) Your code is unnecessarily convoluted.
(b) The example of things *not* working is not reproducible. (Read
the posting guide!!!)
(c) Nonetheless the phenomenon you describe is weird/interesting.
On my system, the following runs without error:
fit.list <- NULL
a <- factor(rep(1:10,each=
Try this:
x <- data.frame(A = runif(10), B = rnorm(10))
lapply(x, shapiro.test)
On Mon, Jun 22, 2009 at 3:15 PM, Gonzalo Quiroga
wrote:
> Hi, I need help to perform a Shapiro.test on a data frame, I know that
> this test works only with vector but I guess there most be a way to
> permor it on a
It is not the case as you described. In any case, the total area should be 1
and labeled fraction on y axis should be far less than 1, since I have more
than 1 data points. I also test differerent bin size by change the
break.
I draw the graph using only 1 group, the same result was obtaine
Emmanuel Charpentier wrote:
>
> I do not understand the problem as stated. if x[i] and n[i] are known,
> and unless sum(n)=0, your dataset reduces to a set of nrow(dataset)
> independent linear equations with nrow(dataset) unknowns (the beta[i]),
> whose solution is trivially beta[i]=log(x[i]/n[
Hi, I need help to perform a Shapiro.test on a data frame, I know that
this test works only with vector but I guess there most be a way to
permor it on a data frame instead of vactor by vector (i.e. I've got 40
variables to analyze and its kinda annoying to do it one by one)
Thanks to anyone that
Hey,
A basic question. Is there anyone that knows a good text on using the
R packages map and maptools for R? I have read the references for both
but having trouble getting started.
What I want to do is to load a map (.shp file) and display statistics
on this maps using colours.
Thanks in advanc
Gmane has already been mentioned, and you might also want to consider
Nabble -- judging from my referrer logs many people use it to read
r-help. If you use Gmail, you might also want to consider subscribing
to the list and using a simple filter. Details and links at
blog.revolution-computing.com, h
Le lundi 22 juin 2009 à 09:35 -0700, francogrex a écrit :
> Hello, I have this generalized linear formula:
> log(x[i]/n[i])=log(sum(x)/sum(n)) + beta[i]
I do not understand the problem as stated. if x[i] and n[i] are known,
and unless sum(n)=0, your dataset reduces to a set of nrow(dataset)
indepe
Mark Na wrote:
Dear R-helpers,
I am helping a SAS user run some analyses in R that she cannot do in
SAS and she is complaining about R's peculiar (to her!) way of
recoding variables. In particular, she is wondering if there is an R
package that allows this kind of SAS recoding:
IF TYPE='TRUCK'
On 6/22/2009 2:27 PM, Mark Na wrote:
> Dear R-helpers,
>
> I am helping a SAS user run some analyses in R that she cannot do in
> SAS and she is complaining about R's peculiar (to her!) way of
> recoding variables. In particular, she is wondering if there is an R
> package that allows this kind of
On Jun 22, 2009, at 2:27 PM, Mark Na wrote:
Dear R-helpers,
I am helping a SAS user run some analyses in R that she cannot do in
SAS and she is complaining about R's peculiar (to her!) way of
recoding variables. In particular, she is wondering if there is an R
package that allows this kind of
Dear R-helpers:
May I ask a question related to storing a number of lmer model fit into a
list.
Basically, I have a for-loop (see towards the bottom of this email)
in the loop, I am very sure that the i-th model fit (i.e.,fit_i) is
successfully generated and the character string (i.e., tmp_i) is c
Dear R-helpers,
I am helping a SAS user run some analyses in R that she cannot do in
SAS and she is complaining about R's peculiar (to her!) way of
recoding variables. In particular, she is wondering if there is an R
package that allows this kind of SAS recoding:
IF TYPE='TRUCK' and count=12 THEN
On Jun 22, 2009, at 12:35 PM, francogrex wrote:
Hello, I have this generalized linear formula:
log(x[i]/n[i])=log(sum(x)/sum(n)) + beta[i]
where the the x[i] and the n[i] are known.
Is there a way to program the GLM procedure to input the formula
above and
get the beta[i] estimates? If not t
On Jun 22, 2009, at 1:21 PM, Jorge Ivan Velez wrote:
Hi David,
It works for me when handling data frames mixing characters and
numeric by
using either print() or a function called "foo":
# Some data
x <- c("A", "B", "C", 3.5, 1, 0, NA, "a character",NA, "another
character")
mydata <- dat
Hi David,
It works for me when handling data frames mixing characters and numeric by
using either print() or a function called "foo":
# Some data
x <- c("A", "B", "C", 3.5, 1, 0, NA, "a character",NA, "another character")
mydata <- data.frame(x = x, y = sample(x))
mydata
# Option 1: print()ing
pr
When freq=FALSE then the y axis is not the proportion in each group (what I am
assuming you mean by fraction), but rather is scaled so that the total area of
the histogram is 1 (making comparing to theoretical densities easier). If all
the data values are between 0 and 1, then the height of at
Hello, I have this generalized linear formula:
log(x[i]/n[i])=log(sum(x)/sum(n)) + beta[i]
where the the x[i] and the n[i] are known.
Is there a way to program the GLM procedure to input the formula above and
get the beta[i] estimates? If not the GLM is there another procedure to do
that? The aim
Dear R users,
I am having problems in trying to run R2jags on a Linux Cluster. When I
tried to run the model given in the R2jags manual I get the following error
error message:
Error in FUN(X[[3L]], ...) : object 'J' not found
Can any one help me with this problem?
Looking to hear from you all.
Karl,
You may want to look at the paper by Dwyer on "Some applications of matrix
derivatives in multivariate analysis" (JASA 1967), especially the Table 2 on
p. 617.
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Profe
> "DM" == Duncan Murdoch
> on Mon, 22 Jun 2009 09:41:12 -0400 writes:
DM> On 6/22/2009 9:23 AM, Tobias Verbeke wrote:
>> On Mon, Jun 22, 2009 at 2:18 PM, Douglas Bates
wrote:
>>> On Mon, Jun 22, 2009 at 7:12 AM, Martin
>>> Maechler wrote:
> "TobiasV" == To
I have been trying to draw histogram for my manscript and found some strange
things that I could not figure out why.
Using the same code listed below I have successfully draw histograms for a
few figures with fraction labeled on Y axis less than 1 (acturally between 0
to 0.1). But one dataset gi
Ravi Varadhan skreiv:
I am not aware of any. May I ask what your purpose is? You don't really
need this if you are going to use it in optimization, since most optimizers
use a simple finite-difference approximation if you don't provide the
gradient. Using the numerical approximation from "numD
Jorge,
Thanks a lot for your reply. It's indeed working in this case. However, with
data frames mixing numeric and character is not working anymore. Morever I
am working with many variables and I don't want to modify them. What I would
really appreciate is a global option (in the Rprofile?) that
On 6/22/2009 9:23 AM, Tobias Verbeke wrote:
On Mon, Jun 22, 2009 at 2:18 PM, Douglas Bates wrote:
On Mon, Jun 22, 2009 at 7:12 AM, Martin
Maechler wrote:
"TobiasV" == Tobias Verbeke
on Sun, 21 Jun 2009 08:25:07 +0200 writes:
TobiasV> Hi Ken,
>> I have been using R for a while. Rec
On Jun 22, 2009, at 12:04 PM, Clifford Long wrote:
Hi R-list,
I'll apologize in advance for (1) the wordiness of my note (not sure
how to avoid it) and (2) any deficiencies on my part that lead to my
difficulties.
I have an application with several stages that is meant to simulate
and explore
Resending, as am not sure about the original "To:" address. Sorry for
any redundancy.
- Cliff
-- Forwarded message --
From: Clifford Long
Date: Mon, Jun 22, 2009 at 11:04 AM
Subject: question about using _apply and/or aggregate functions
To: r-h...@lists.r-project.org
Hi R-lis
There are many ways to do this in R. See R News 4/1 for info on dates.
Here is one method:
> library(chron)
> mdy <- month.day.year("05/16/2008")
> mdy
$month
[1] 5
$day
[1] 16
$year
[1] 2008
> mdy$month
[1] 5
On Mon, Jun 22, 2009 at 11:36 AM, Jack Luo wrote:
> Hi,
>
> I have a simple questi
Dear All,
I used an AR(1) model to explain the process of the stationary residual and
have used an 'ar' command in R. From the results, i tried to extract
the standard error and p-value for the estimated parameter, but unfortunately,
i never find any way to extract it from the output.
What
Couple of choices depending on what you want to do with the data:
> x <- as.POSIXct("05/16/2008", format="%m/%d/%Y") # if you want to use the
date
> format(x, "%d") # day
[1] "16"
> format(x, "%m") # month
[1] "05"
> format(x, "%Y") # year
[1] "2008"
> # or
> y <- strsplit("05/16/2008", '/') # t
Hi,
I have a simple question, suppose I have the date "05/16/2008", what would
be the command to get the month, day and year?
Thanks,
-Jack
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman
Hello R Users,
I have a question regarding fitting a model with GAM{mgcv}. I have data
from several predictor (X) variables I wish to use to develop a model to
predict one Y variable. I am working with ecological data, so have data
collected many times (about 20) over the course of two years. P
On 6/22/2009 10:30 AM, Steve Murray wrote:
Thanks again for a very useful comment. That seems to have separated the text
and put it onto separate lines.
However, whilst this results in the text being centralised in relation to the
axis, it means that the lower line is left-justified in relatio
On Jun 22, 2009, at 4:08 AM, Dieter Menne wrote:
Chris Friedl gmail.com> writes:
I have two questions about the built-in function step. Ultimately I
want to
apply a lm fitting and subsequent step procedure to thousands of
data sets
groups by a factor defined as a unique ID.
Q1. The code
Hi Karl,
I am not aware of any. May I ask what your purpose is? You don't really
need this if you are going to use it in optimization, since most optimizers
use a simple finite-difference approximation if you don't provide the
gradient. Using the numerical approximation from "numDeriv" will be
Thanks again for a very useful comment. That seems to have separated the text
and put it onto separate lines.
However, whilst this results in the text being centralised in relation to the
axis, it means that the lower line is left-justified in relation to the upper
line, rather than being cent
What error is it giving? Please include the exact error.
What happens if you do this:
if (file.exists(findings)) cat('File',findings,'exists\n') else
cat('File',findings,not found\n')
Your description suggests that you are using 'if' expression inside a
loop. If that is the case, try
i
Hi all, Eigen vectors obtained from the function eigen() are ortho-normal? I
see the documentation however there is no formal mention on that. If no,
then is there any direct function to do the same?
--
View this message in context:
http://www.nabble.com/Eigen-value-calculation-tp24147807p241478
Does anybody know of a function that implements the derivative (gradient) of
the multivariate normal density with respect to the *parameters*?
It’s easy enough to implement myself, but I’d like to avoid reinventing the
wheel (with some bugs) if possible. Here’s a simple example of the result
I’
Look at show.settings() and str(trellis.par.get()). This will show you
what the default settings are. The group colors are set by the
superpose.* elements (e.g. superpose.line is for group lines). To set
them, I usually create a list and pass it to par.settings. For
example,
my.theme <- list(super
Dear list,
I have been struggling to find how I would go about changing the
bakground colors of groups in a lattice barchart in a way so that the
auto.key generated also does the right thing and pick it up for the
key.
I have used the "col" argument (which I guess is sent to par()) in a
way so tha
Fredrik Nilsson-5 wrote:
>
>>From an earlier post I got the impression that one could promote
> warnings from a glm to errors (presumably by putting
> options(warn=1)?), then try() would flag them as errors. I’ve spent
> half the day trying to do this, but no luck. Do you have an explicit
> sol
On Mon, Jun 22, 2009 at 2:18 PM, Douglas Bates wrote:
> On Mon, Jun 22, 2009 at 7:12 AM, Martin
> Maechler wrote:
>>> "TobiasV" == Tobias Verbeke
>>> on Sun, 21 Jun 2009 08:25:07 +0200 writes:
>>
>> TobiasV> Hi Ken,
>> >> I have been using R for a while. Recently, I have begun c
Auty, Dave forestry.gsi.gov.uk> writes:
>
> I'm running the following code to produce lattice plots of microfibril
> angle versus ring number in Scots pine. There are 12 trees and 5 sample
> positions ("Position") in each tree:
>
> xyplot(MFA ~ RN | Tree, data = MFA.data,
>
>groups =
Dear list,
>From an earlier post I got the impression that one could promote
warnings from a glm to errors (presumably by putting
options(warn=1)?), then try() would flag them as errors. I’ve spent
half the day trying to do this, but no luck. Do you have an explicit
solution?
My problems is that
Dear David,
Try this:
x <- c("A", "B", "C", NA)
x[is.na(x)] <- "."
x
HTH,
Jorge
On Mon, Jun 22, 2009 at 2:11 AM, David_D wrote:
>
> Dear R-users,
>
> For reporting purpose (using Sweave and LaTeX), I am creating complex
> tables
> with the cat function such as
>
> > x<-c("A", "B", "C", NA)
>
On 6/22/2009 7:23 AM, Tony Breyal wrote:
Greetings,
I usually read this mailing list through google groups
(http://groups.google.com/group/r-help-archive), but when I opened the
webpage this morning it said: "The group named r-help-archive has been
removed because it violated Google's Terms Of S
On 6/22/2009 7:58 AM, Steve Murray wrote:
Thanks for the response, however, whilst this eliminates the 'new line'
character from appearing, it doesn't actually cause a new line to be printed!
Instead, I have the last few characters of the first character string
overlapping with the first few c
Greetings,
I usually read this mailing list through google groups
(http://groups.google.com/group/r-help-archive), but when I opened the
webpage this morning it said: "The group named r-help-archive has been
removed because it violated Google's Terms Of Service."
Is there an alternative website w
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