Hi r-users,
I try to fit the t copula using the gamma marginals. But I got error message
which I don't really understand.
Thank you for any help given.
myCop.t <- ellipCopula(family = "t", dim = 2, dispstr = "toep", param = 0.5, df
= 8)
myCop.t
myMvd <- mvdc(copula = myCop.t, margins = c("ga
R error messages go to stderr, not stdout, so you need to redirect the
latter. In most shells, including cmd.exe (which as you don't say is
likely what you are using) you can use
Rscript test.R xxx.txt > error.log 2>&1
to redirect both stdout and stderr to the same file (see the rw-FAQ
Q2.12
edit: I found out how to declare empty variables in R, but the code still
does not work. I get the index out of bounds error since my data is
irregular (some have more dates than others, and the matrix will not allow
for different sized rows)
Dear R Gurus,
Thanks for any help in advance!
Date.f
Hi.
first things first ... thanks for ggplot2.
Now my question. I'm using qplot to generate a plot as follows where X,Y,Z,
A are columns in a dataframe.
qplot(X, Y, data=XYDATA, color=Z, geom=c("point"), size=A)
This works as expected. Factor A has three levels so there are three sizes
of the
Thanks for the excellent help on my recent question on this topic in which
the data frame had been reshaped by cast.
Now, I would like to access and change erroneous data in a data frame that
has not been reshaped.
The file is lupepn1, with identifier variables bushno & bout and
dependent variabl
Dear R Gurus,
Thanks for any help in advance!
Date.frame: Returns.names
X id ticker date_ adjClose totret RankStk
258060 258060 13645T10 CP 2001-06-29 18.125 1877.758
My data frame is in the above format. I would like to filter by period, per
id (every 125 days)
Seb,
Thanks. That doesn't solve the problem of combining two expressions.
My aa and bb are expressions constructed somewhere else and passed to the
current function which wants to use them together. Your solution moves the
construction
of aa and bb into the function and is equivalent to my aabb.
On Wed, 9 Jun 2010 00:15:19 -0400,
"RICHARD M. HEIBERGER" wrote:
> text(5,1, parse(text=paste(deparse(aa[[1]]), deparse(bb[[1]]),
> sep="~"))) text(5,2, parse(text=paste(deparse(aa[[1]]),
> deparse(bb[[1]]), sep="~', '~")))
> Is there a cleaner way of combining the expressions aa and bb to get
>
Is there a cleaner way of combining two expressions.
This example works and gives what I want
plot(1:10)
aa <- expression(alpha==.05)
bb <- expression(beta ==.80)
aabb <- expression(alpha==.05 ~ ", " ~ beta ==.80)
text(5, 10, aa)
text(5, 9, bb)
text(5, 8, aabb)
text(5,1, parse(text=paste(depa
I'm having trouble with the ordisurf function in the vegan package.
I have created an ordination plot (cmdscale) of 60 samples based on
Bray-Curtis dissimilarities, and would like to overlay various soil edaphic
characteristics as possible clues to the clustering I observe in my plot.
However, I f
hi ,R user folks .
Nowadays I read a paper which draw a probability ellipse circle figure
shown in the appendix.
I wonder how to draw this figure by R ?
the x-axis and y-axis both express the error but in different direction .
--
TANG Jie
Email: totang...@gmail.com
Tel: 0086-2154896104
Shanghai
Shifting this to r-devel ...
On 6/6/2010 11:20 PM, baptiste auguie wrote:
Hi,
I've just added width/heightDetails methods following Paul's
suggestion. I kept a duplicate of all on-the-fly grob size
calculations; it is necessary to ensure that the table cells adjust to
the content which can be
Hi there,
I use Rscript.exe for batch run (actually it is used in ASP.net code)
c:>"C:\Program Files\R\R-2.10.1\bin\Rscript.exe" test.r xxx.txt
Where test.r is the r program and xxx.txt is a file name test.r will read in,
it comes from
a web form. This works fine, when the file is in required form
At 10:56 AM +0800 6/8/10, elaine kuo wrote:
Hello,
I searched the archives but found no answers.
How to modify the hisgram color of function pairs.panels of Psych package ?
I tried col() but it was the line color modified.
Thanks.
Elaine Elaine,
Good question.
Right now, without going int
Ross,
My apologies, I just discovered your email (from April) to the
R-help list serve asking about ICC in psych.
ICC does not remove missing data but rather lets the ANOVA handle it.
It is probably more appropriate work on complete cases (as does the
icc in the irr package).
that is
my.d
Hi
grid.polygon() can do multiple polygons in a single call, but rather
than using NA's to separate sub-polygons, it uses an 'id' argument (or
an 'id.lengths' argument) to identify sub-polygons within the vectors of
x- and y-values (see the examples in ?grid.polygon). So a ggplot2 patch
that
Hi there,
I use Rscript.exe for batch run (actually it is used in ASP.net code)
c:>"C:\Program Files\R\R-2.10.1\bin\Rscript.exe" test.r xxx.txt
Where test.r is the r program and xxx.txt is a file name test.r will read in,
it comes from
a web form. This works fine, when the file is in required f
Try
v1<.3 | v1 > .7
HTH,
Jorge
On Tue, Jun 8, 2010 at 6:38 PM, Worik R <> wrote:
> If I create a vector thusly
>
> > v1 <- runif(20, min=0, max=1)
> > v1
> [1] 0.9754443 0.6306228 0.3238158 0.3175769 0.6791534 0.6956507 0.3840803
> [8] 0.1421328 0.8592398 0.4388306 0.9472040 0.4727435 0.5645
Hello,
This should work
v1 <- runif(20, min=0, max=1)
v2 <- v1 <.3 | v1 >.7 #you just need one | not two
HTH,
Josh
On Tue, Jun 8, 2010 at 3:38 PM, Worik R wrote:
> If I create a vector thusly
>
>> v1 <- runif(20, min=0, max=1)
>> v1
> [1] 0.9754443 0.6306228 0.3238158 0.3175769 0.6791534 0.6
Here is one way
...
DF4 <- cast(formula=Date~V2,data=DF3,value="X1",fill=0)
d <- with(DF4, seq(min(Date), max(Date), by = 1)) ### full set
m <- as.Date(setdiff(d, DF4$Date)) ### missing dates
if(length(m) > 0) {
extras <- cbind(data.frame(Date = m), cat = 0, dog = 0, tre
If I create a vector thusly
> v1 <- runif(20, min=0, max=1)
> v1
[1] 0.9754443 0.6306228 0.3238158 0.3175769 0.6791534 0.6956507 0.3840803
[8] 0.1421328 0.8592398 0.4388306 0.9472040 0.4727435 0.5645302 0.7391616
[15] 0.6116199 0.2727754 0.2657867 0.5261744 0.8764804 0.2032126
And I want to cre
In the code fragment, I used 'by' to actually compute the min value (part of
the statement with the eval) - and I agree that an apply would work there
wonderfully.
However, my hope was to use an apply for the subsetting of the data.frame's
columns, so that I could then use an apply to compute
Hi,
I am relatively new to R; when creating functions, I run into problems with
missing values. I would like my functions to ignore rows with missing values
for arguments of my function) in the analysis (as for example is the case in
STATA). Note that I don't want my function to drop rows if there
Given the following snippet
m.nf.xts <- xts(rep(0, length(index(m.xts))), order.by=index(m.xts))
Does R know to cache the index(m.xts) or is it more efficient to say...
m.i <- index(m.xts)
m.nf.xts <- xts(rep(0, length(m.i)), order.by=index(m.i))
?
cheers
Worik
[[alternative HTML
don't forget to make the "by" option a list :
vegMeans = aggregate(SoilVegHydro[3:37],list(SoilVegHydro['Physiogomy']),mean)
and
vegSd = aggregate(SoilVegHydro[3:37],list(SoilVegHydro['Physiogomy']),sd)
see also ?aggregate
on a side note, it would be handy if that transformation to a list
would
Not much to go on, but you might find
vegMeans = aggregate(SoilVegHydro[3:37],SoilVegHydro['Physiogomy'],mean)
and
vegSd = aggregate(SoilVegHydro[3:37],SoilVegHydro['Physiogomy'],sd)
more suitable for your needs. (Not run because I don't know what
SoilVegHydro is.)
I did not go too deep into your zoology problem ;-) but as far as I
understood you, you want to omit all rows where
ID and TO_ID are A1 and A1.1, (or A2) correct?
If the data you send us is all the data and if there do not occour any
different situations the following should be sufficient:
Guys, many thanks. lapply works. Did not occur to me as I thought lapply
returns a list and the receiving entity is a data frame.
H
-Original Message-
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: Tuesday, June 08, 2010 2:22 PM
To: Horace Tso; r-help@r-project.org
Subject:
You need lapply here:
df[2:3] <- lapply(df[2:3], as.Date, '%m/%d/%Y')
On Tue, Jun 8, 2010 at 6:19 PM, Horace Tso wrote:
> Folks, i thought it should be straightforward but after a few hours poking
> around, I decided it's best to post my question on this list.
>
> I have a data frame consis
Hello,
vegMeans <- by(SoilVegHydro[3:37] , SoilVegHydro$Physiogomy, mean)
vegSD <- by(SoilVegHydro[3:37] , SoilVegHydro$Physiogomy, sd)
write.table(vegMeans,
file="A:\\Work_Area\\Steve\\Hydrology_Data\\data\\vegMeans.txt")
Error in as.data.frame.default(x[[i]], optional = TRUE, stringsAsFacto
The short answer is, don't use apply() on data.frame's.
Use lapply to loop over the columns of a data.frame.
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Horace Tso
> Sent: Tuesday, June 08, 2010 2:19 PM
> To: r-help@r-pr
Folks, i thought it should be straightforward but after a few hours poking
around, I decided it's best to post my question on this list.
I have a data frame consisting of a (large) number of date columns, which are
read in from a csv file as character string. I want to convert them to Date
type
Is this what you are looking for:
> # assume females start with "A"
> # extract first part if female from ID
> x.id <- sub("(A[[:digit:]]+).*", "\\1", x$ID)
> # now see if this pattern matches first part of TO_ID
> x.match <- x.id == substring(x$TO_ID, 1, nchar(x.id))
> # here are the ones that wo
Once again my message got held up for moderator approval so I
am deleting it and trying again. Hopefully this one goes through.
In general, we will get the simplest usage if we match the problem to
the appropriate OO class. In this case we are using time series so it
is advantageous to use a time
you should have found a solution for that in the help page of apply.
just run
min.values = apply(the.data,1,min)
the '1' marks the direction (e.g. whether apply is applied to rows or
columns), it could be a 2 as well. Check that yourself in the apply
documentation.
Then run rbind(the.data,m
Ekaterina Pek wrote:
Hi, all.
Yet another beginner to R : )
I wonder, how it's possible to get the value of a coefficient from the
object produced by cor.test() ?
cor.test(a, b, method="spearman")
You can always assign the value of a function to a variable, and then
use ?str to see the s
result <- cor.test(a,b,method="spearman")
result$estimate
Cheers
Joris
On Tue, Jun 8, 2010 at 10:40 PM, Ekaterina Pek wrote:
> Hi, all.
>
> Yet another beginner to R : )
>
> I wonder, how it's possible to get the value of a coefficient from the
> object produced by cor.test() ?
>
>> cor.test(a,
Hi, all.
Yet another beginner to R : )
I wonder, how it's possible to get the value of a coefficient from the
object produced by cor.test() ?
> cor.test(a, b, method="spearman")
Spearman's rank correlation rho
data: a and b
S = 21554.28, p-value = 2.496e-11
alternative hypothesis: tru
Try this:
xtabs( ~ V1 + V2, transform(dog3.df, V1 = factor(V1, levels =
as.character(seq(min(dog3.df$V1), max(dog3.df$V1), by = "days")
On Tue, Jun 8, 2010 at 4:52 PM, Erin Hodgess wrote:
> Dear R People:
>
> So thanks to your help, I have the following:
>
>
> > dog3.df <-
> read.delim("c:/
Hi R users,
I am trying to omit rows of data based on partial matches an example of my
data (seal_dist) is below:
A quick break down of my coding and why I need to answer this - I am dealing
with a colony of seals where for example A1 is a female with pup and A1.1 is
that female's pup, the impor
I am mainly a Java/C++ programmer, so my mind is used to iterating over data
with for loops. After a long break, I am trying to get back into the "R
mindset", but I could not find a solution in the documentation for the applys,
aggregate, or by.
I have a data.frame where each row is an entry wi
On Tue, 8 Jun 2010, Sachi Ito wrote:
Hi,
I'm analyzing my data using GEE, which looks like below:
interact <- geeglm(L ~ O + A + O:A,
+ data = data1, id = id,
+ family = binomial, corstr = "ar1")
summary(interact)
Call:
geeglm(formula = lateral ~ ontask + attachment + ontask:attachment,
Hi,
I'm analyzing my data using GEE, which looks like below:
> interact <- geeglm(L ~ O + A + O:A,
+ data = data1, id = id,
+ family = binomial, corstr = "ar1")
> summary(interact)
Call:
geeglm(formula = lateral ~ ontask + attachment + ontask:attachment,
family = binomial, data = firstgroup
Dear R People:
So thanks to your help, I have the following:
> dog3.df <-
> read.delim("c:/Users/erin/Documents/dog1.txt",header=FALSE,sep="\t")
> dog3.df
V1 V2
1 1/1/2000 dog
2 1/1/2000 cat
3 1/1/2000 tree
4 1/1/2000 dog
5 1/2/2000 cat
6 1/2/2000 cat
7 1/2/2000 cat
8 1
Arnau Mir uib.es> writes:
>
> Hello.
>
> Somebody knows how to compute generalized hypergeometric series in R?
> (see
> http://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/02/
> to understand what I mean)
library(sos)
findFn("generalized hypergeometric function")
-> see
While it is possible to set your own dash patterns as you show below, it is
unlikely that the resulting graph will be very meaningful. Most people cannot
keep the detailed dash patterns separate, and if they need to refer to a legend
then it makes it even harder (See Bert Gunter's rant on the "
> On 06/08/2010 05:29 AM, Mark Seeto wrote:
>>
>>> On 06/06/2010 10:49 PM, Mark Seeto wrote:
Hello,
I have a couple of questions about the ols function in Frank Harrell's
rms
package.
Is there any way to specify variables by their column number in the
data
>
Hi list,
I have two question relating to the Rsymphony package:
a) is there a way to use 'warm starts' ?
i.e. if i successively call Rsymphony_solve_LP()
with only one constraints change, does it perform the optimization
from the simplex-origin or does it uses the previously found solution
[i.e.
Hi Joris,
thanks for your help. I just had to alter it slightly (basically just
transposing):
tmp <- array(rbind(t(test),t(test2)),
dim=c(9,2,2),
dimnames=list(colnames(test),rownames(test),c("Test","Test2")))
ftable(tmp)
Thanks again!
Best,
Stefan
On Tue, Jun 8, 2010 at
On 08-Jun-10 18:00:18, Bert Gunter wrote:
> Mount soapbox; begin rant {
>
> ... However I think it should be added that rarely does this work with
> more than about a half dozen different symbols: a viewer of a graphic
> simply cannot keep the distinctions straight -- or often even decode
> them.
Luis Felipe Parra wrote:
Hello I am using POSIXlt date format and I am having the following problem,
I've got two dates called FechaIni and FechaFin, one in 2008 and the other
in 2009 but when I do FechaIni$year and FechaFin$year to call the year I am
getting the smae year for both.
FechaIni
Hello I am using POSIXlt date format and I am having the following problem,
I've got two dates called FechaIni and FechaFin, one in 2008 and the other
in 2009 but when I do FechaIni$year and FechaFin$year to call the year I am
getting the smae year for both.
> FechaIni
[1] "2008-11-13 UTC"
> Fecha
Mount soapbox; begin rant {
... However I think it should be added that rarely does this work with more
than about a half dozen different symbols: a viewer of a graphic simply
cannot keep the distinctions straight -- or often even decode them. Using
color to distinguish groups is typically more ef
Hello.
Somebody knows how to compute generalized hypergeometric series in R?
(see
http://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/02/
to understand what I mean)
Thanks in advance,
Arnau.
__
R-help@r-project.org mailing list
Thank you. It works like a charm.
A. Huang
From: Joshua Wiley
Cc: r-help@r-project.org
Sent: Tue, June 8, 2010 9:19:19 AM
Subject: Re: [R] Convert a data frame to a 2D array?
Hi,
Does this work?
array(data=unlist(yourdataframe), dim=c(n,m))
Josh
> Hi t
Hi All ,
For an academic project I am trying to do the following
Step 1 ) Draw and cluster a N ( lets say 3 ) column dataset by dbscan
algorithm using R-projectâs fpc package ( let say they are "training
clusters" ) ,
Using dbscan as number of clusters are not kno
The my.symbols function (TeachingDemos package) allows for defining your own
symbols to use in plots using base graphics (see ms.filled.polygon for an
example), there is also panel.my.symbols which works with lattice (possibly
with general grid, but I have not tested it that way).
Those may giv
You really need to do some studying on mixed effects models. Some resources
are at: http://lme4.r-forge.r-project.org/
Your formula below is wrong, you fit only the intercept as a fixed effect and
you are fitting a random slope on animal by day for the random effect, which
does not make much s
Because contourplot comes from the lattice package, I think you'll want to
look at these help pages:
+ help(trellis.focus)
+ help(lpoints)
Below, I've used the example from help(contourplot) to demonstrate how one
might add points and text to a lattice plot.
-tgs
#
#
test <- data.frame(x=1:10,y=1:10)
as.array(as.matrix(test))
does the job too
Cheers
On Tue, Jun 8, 2010 at 6:07 PM, A Huang wrote:
> Hi there,
>
> I've read a file into a data frame. The data is n rows by m columns, all
> values are numbers.
> Is there a way to convert the data frame to a 2D ar
Hi,
Does this work?
array(data=unlist(yourdataframe), dim=c(n,m))
Josh
On Tue, Jun 8, 2010 at 9:07 AM, A Huang wrote:
> Hi there,
>
> I've read a file into a data frame. The data is n rows by m columns, all
> values are numbers.
> Is there a way to convert the data frame to a 2D array? I tri
Hi there,
I've read a file into a data frame. The data is n rows by m columns, all values
are numbers.
Is there a way to convert the data frame to a 2D array? I tried as.array(), but
got some
error messages.
Thanks
A. Huang
[[alternative HTML version deleted]]
_
I could get something close to what you asked using a little hack,
emulating a table using an array based on your two matrices :
test <- matrix(rpois(18,10),ncol=9,nrow=2)
colnames(test) <- paste("Dis",1:9,sep="")
rownames(test) <- c("2010","2020")
test2 <- matrix(rpois(18,10),ncol=9,nrow=2)
coln
Hi All,
I'm trying to estimate some parameters in my model via GMM using the
function gmm(), but I keep getting the message "The covariance matrix of
the coefficients is singular". I've changed the moment conditions and
the initial value of the parameters, and I still get this message. Are
th
On Tue, Jun 8, 2010 at 7:10 AM, Enrico Colosimo wrote:
> Hello,
> I am having some trouble running a very simple
> example. I am running a logistic regression entering the SAME data set
> in two different forms and getting different values for the deviance residual.
>
> Just look with this naive
The first time I posted this it got held up for approval so I am
trying it again. Hopefully this time it gets through right away.
As with your prior post we can use read.zoo(..., split=...).
Alternatives are reshape and reshape::cast.
# read data into DF
Lines <- "V1 V2
1 1/1/2000 dog
2 1/1
We wish to announce the new package:
fdth - Frequency Distribution Table and Associated Histogram.
The package contains a a set of high level function which easily allows the
user
to make a frequency distribution table (fdt) and its associated plots.
The fdt can be formatted in many ways which may
I am looking at a new project involving time series analysis. I know I can
complete the tasks involving VARMA using either dse or mAr (and I think
there are a couple others that might serve).
However, there is one task that I am not sure of the best way to proceed.
A simple example illustrates w
Am 08.06.2010 17:04, schrieb Erin Hodgess:
> Here is a particular way to solve the problem:
>
If you solve your own problem then please reply to your own message
otherwise things get confused. How should one know what your problem was
without knowing your first e-mail - if you reply your own e-
Hi R Users,
I want to distinguish different condition by different symbols by pch in
function grid.points, but the symbols needed should be with solid or hollow,
in this way only 21 to 25 in pch worked, is there any other symbols could be
used like this? or does it exist any other way to draw
Am 08.06.2010 16:52, schrieb Erin Hodgess:
>
> I would like to set up 3 time series; one for dog, one for cat, one
> for tree, such that each runs from 1/1/2000 to 1/3/2000, with 0 if
> there is no entry for the day.
>
>
>
Before using zoo or zooreg you should transform your data.frame as such
Here is a particular way to solve the problem:
> test3 <- seq(from=as.Date("1/1/2000","%m/%d/%Y"),to=as.Date("1/3/2000",
+ "%m/%d/%Y"),length=3)
> test3
[1] "2000-01-01" "2000-01-02" "2000-01-03"
> zoo(table(dog.df$V1,dog.df$V2)[,1],order=test3)
2000-01-01 2000-01-02 2000-01-03
1
Dear R People:
I have the following data frame:
> str(dog.df)
'data.frame': 7 obs. of 2 variables:
$ V1: Factor w/ 3 levels "1/1/2000","1/2/2000",..: 1 1 1 2 2 3 3
$ V2: Factor w/ 3 levels "cat","dog","tree": 2 1 3 1 3 2 3
> dog.df
V1 V2
1 1/1/2000 dog
2 1/1/2000 cat
3 1/1/2000 t
On 08.06.2010 15:17, Paco Pastor wrote:
Hi everyone
I want to install Rmpi to use R in parallel mode in a Linux cluster
(Ubuntu, Hardy Heron). It seems to be properly installed but a problem
appears when loading Rmpi library.
R version 2.11.1 (2010-05-31)
> library("Rmpi")
Error: package 'R
Try this:
m <- matrix(Corr, ncol = 3, dimnames = list(unique(NodesRow),
unique(NodesCol)))
m[col(m) == row(m) | upper.tri(m)] <- NA
subset(as.data.frame.table(m), !is.na(Freq))
On Tue, Jun 8, 2010 at 10:29 AM, Matthew DiLeo wrote:
> I have a large correlation matrix that I'm trying to convert
The author's site is here:
http://www.bio.ic.ac.uk/research/mjcraw/therbook/
and you can read the file right off his site like this:
URL <- "http://www.bio.ic.ac.uk/research/mjcraw/therbook/data/Gain.txt";
gg <- read.table(URL, header = TRUE)
On Tue, Jun 8, 2010 at 9:37 AM, Andrew Kelly wrote:
Dear all,
an hopefully quick table question.
I have the following data:
Two objects that are 2*9 matrix with nine column names (Dis1, ...,
Dis9) and the row names (2010,2020). The content are frequencies
(numeric).
In want to create a table that is along the lines of
ftable(UCBAdmissions) and sho
Hi,)
I am just getting started with R but have hit an early snag. I am working
through Crawley (2008) The R Book and on page 6, 'Significance Stars', I am
trying to enter the commands given. However, 'Gain.txt' does not seem to have
been downloaded when I downloaded the R programme.
I have sear
I have a large correlation matrix that I'm trying to convert to a list of
every connection (edge) between every two nodes with its accompanying
correlation value (for Cytoscape). I figured out how to do this and to
remove the connections that nodes have to themselves but I can't figure out
how to g
Hi everyone
I want to install Rmpi to use R in parallel mode in a Linux cluster
(Ubuntu, Hardy Heron). It seems to be properly installed but a problem
appears when loading Rmpi library.
R version 2.11.1 (2010-05-31)
> library("Rmpi")
Error: package 'Rmpi' was built before R 2.10.0: please re
Thanks for the advice. I found the function : summary(mcmc(x)), from the
coda package.
2010/6/4 Gavin Simpson
> On Thu, 2010-06-03 at 23:44 +0200, Jimmy Söderly wrote:
> > Thanks for your help.
> >
> > Does it have something to do with the mcmc package, the coda package, or
> the
> > lattice pa
Hello,
I am having some trouble running a very simple
example. I am running a logistic regression entering the SAME data set
in two different forms and getting different values for the deviance residual.
Just look with this naive data set:
The only goals I remember are that the "Hand of God" was at 6 min second period
and the
"Goal of the Century" at 11 min second period
The others dont count.
HG
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Paul [p...@paulhu
Hi
r-help-boun...@r-project.org napsal dne 08.06.2010 14:21:10:
> Thank you to all for your help!
>
> I received
> two equal alternative solutions that bypassed elegantly the problem
> from peter.l.e.koni...@gmail.com
> and
> rafael.bj...@gmail.com
>
> rr.dia2.corr <- rr.dia.2m
> rr.dia2.corr[w
On 06/08/2010 05:29 AM, Mark Seeto wrote:
On 06/06/2010 10:49 PM, Mark Seeto wrote:
Hello,
I have a couple of questions about the ols function in Frank Harrell's
rms
package.
Is there any way to specify variables by their column number in the data
frame rather than by the variable name?
For
Thank you to all for your help!
I received
two equal alternative solutions that bypassed elegantly the problem
from peter.l.e.koni...@gmail.com
and
rafael.bj...@gmail.com
rr.dia2.corr <- rr.dia.2m
rr.dia2.corr[which(med.hyper == 1)] <- rr.dia.2m[which(med.hyper == 1)] - 5
From pbu...@pburns.sea
yahoo.com> writes:
>
> Hi,
>
> I have used DeSolve package for my ODE problem regarding
> infectious disease transmission and currently am
> trying to pass lots (roughly a thousand) of model parameters
> to the C compiled model (I have to use C
> compiled code instead of R code purely because
On Tuesday, June 8, 2010, christiaan pauw wrote:
> Hi everybody
>
> I have found something (for me at least) strange with duplicated(). I will
> first provide a replicable example of a certain kind of behaviour that I
> find odd and then give a sample of unexpected results from my own data. I
> ho
Hi All,
I want to add one point to contourplot(). I used contourplot() in my code like
contourplot(z ~ a + b |c, data)
I understand there is plot.axes argument for filled.contour(), but it did not
work for my code. I also tried plot() and text() for contourplot(), but got
this error: "plot.new
Thanks for your help Petr
I think I understand better now.
> > > Masechaba$unique[which(is.na(unique(Masechaba$PROPDESC))==FALSE)]=TRUE
>^^^
> This seems to be strange. At first sight I am puzzlet what result I shall
> expect from s
Hello everyone,
This is just a quick double check. It concerns the 'scatterplot function' in
R.
I have 6 curves and I wish to represent each of them by a different kind of
line (their colour must be black).
The curves are derived from the cuminc function...the coordinates of which are
in '
Hi
Hm, maybe you can first make a sequence of all required dates and ids,
construct empty data frame with all possible dates, merge your existing
data frame with empty one just to fill in all dates, get rid of duplicated
dates and ids if necessary and finally use na.locf from zoo library to
fi
First, read the posting guides.
Then, supply us with a bit more information, like the package you
used, example code that reproduces the error, information about the
data, the complete error message, the traceback (use the function
traceback() right after you got the error).
Otherwise we ain't goi
Install the caret package and see ?train. There is also:
http://cran.r-project.org/web/packages/caret/vignettes/caretTrain.pdf
http://www.jstatsoft.org/v28/i05/paper
Max
On Tue, Jun 8, 2010 at 5:34 AM, azam jaafari wrote:
> Hi
>
> I want to do leave-one-out cross-validation for multinom
That will be R 2.10.1 if I'm correct.
For reading in csv files, there's a function read.csv who does just that:
los <- read.csv("file.csv",header=T)
But that is just a detail. You have problems with your memory, but
that's not caused by the size of your dataframe. On my system, a
matrix with 100,
As far as my knowledge goes, nnet doesn't have a built-in function for
crossvalidation. Coding it yourself is not hard though. Nnet is used
in this book : http://www.stats.ox.ac.uk/pub/MASS4/ , which contains
enough examples on how to do so.
See also the crossval function in the bootstrap package.
Hi
r-help-boun...@r-project.org napsal dne 08.06.2010 11:46:17:
> Hi,
>
> You are using if/then/else which is a logical control statement and so
doesn't
> return a value, see
> ?if
> for details.
>
> You are probably looking for the ifelse function.
Or use possibility of easy conversion logi
Hi
r-help-boun...@r-project.org napsal dne 08.06.2010 08:44:39:
> Hi everybody
>
> I have found something (for me at least) strange with duplicated(). I
will
> first provide a replicable example of a certain kind of behaviour that I
> find odd and then give a sample of unexpected results from m
On 06/07/2010 10:05 PM, ogbos okike wrote:
Greetings to you all.
I have two datasets - Time and magnitude. For a particular location, the
magnitude of the parameter varies with time. I wish to obtain a polar
coordinate distribution of time (0-24h) and magnitudes so as to visualize
how magnitude
> On 06/06/2010 10:49 PM, Mark Seeto wrote:
>> Hello,
>>
>> I have a couple of questions about the ols function in Frank Harrell's
>> rms
>> package.
>>
>> Is there any way to specify variables by their column number in the data
>> frame rather than by the variable name?
>>
>> For example,
>>
>> l
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