Hi all,
I am fairly new to R and I am trying to run mvpart and create a MRT using
explanatory variables and covariables. I've been following the procedures in
Numerical Ecoogy with R.
The command (no covariables) which works fine -
ABUNDTMRT <- mvpart(abundance ~
.,factors,margin=0.08,cp=0,xv="1
Hi,
Does anyone know where I can find a package which implements this network?
http://en.wikipedia.org/wiki/Radial_basis_function_network
The package "neural" was good for that I think, but it doesn't exist anymore
(I don't know why).
Thanks a lot
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Friends
I am looking at Rcpp and I am a bit stuck on a simple matter.
(I am calling R from c++, if there is a better way...)
Given this simple example using the TTR package and the SMA function which
returns a simple moving average
Rcpp::NumericVector rv;
for(int i = 0; i < 100; i++){
'f$V1' is a factor.
try
as.POSIXct(as.character(f$V1), format="%m/%d/%Y %H:%M:%S")
You need to convert to a character first.
On Mon, Sep 12, 2011 at 7:24 PM, bradford wrote:
> I don't know R, so maybe I've done something wrong, but I'm working off an
> example I saw on the web and wondering wh
eeadie unm.edu> writes:
> Now I have a new problem with the same model that I've been working on. Here
> is the model and the error message:
>
> >
modelnbbb<-glmmadmb(total_bites_rounded~age_class_back+
(1|focal_individual)+(1|food.dif.id)+
offset(log(forage_time)),data=data,family="nbi
I have uploaded a datafile that contains the following two variables: time
(X value) and response (Y value). This is a fairly extensive file (with >
16000 entries). I have two questions:
1. I want to use the following equation to regress Y on X: Y-hat = min +
(max-min)/(1 + (X/EC50)^Hillslope).
I don't know R, so maybe I've done something wrong, but I'm working off an
example I saw on the web and wondering why as.POXIXct isn't returning the
same result on f$V1 as it is on z. Did I do something wrong? Or is it a
problem with my build?
> f$V1
[1] 09/11/2011 13:46:39 09/11/2011 13:45:18
Hi All,
I have a quick question on random forests. Simply, I am not sure how to read
the values of independent variables related to the highest value of a response
variable from all trees generated from random forests. It is easy to do this in
a single regression tree. But I am not clear how to
joerg stephan rhrk.uni-kl.de> writes:
>
> Hi,
>
> I tried to do a nested Anova with the attached Data. My response
> variable is "survivors" and I would like to know the effect of
> (insect-egg clutch) "size", "position" (of clutch on twig) and "clone"
> (/plant genotype) on the survival of
On 13/09/11 11:27, Carl Witthoft wrote:
Love the page.
Just out of interest, is this an updated version or the same ol'
Inferno document?
And why do I keep thinking you (Patrick Burns) are the Hab's coach? :-)
Now *that's* a blast from the past! Burns coached the Canadiens
way back whe
YAddo gmail.com> writes:
>
> Dear All:
>
> I am calculating the relative importance of a regressor in a linear model.
> Does anyone know how I can obtain/install the 'pmvd' computation type? I am
> a US user.
>
I didn't know what the heck you were talking about, but having looked
at http:
Thank you a lot Morgan. Your suggestion helped me to speed up my code. But
I still believe that the inefficience is an S4 issue.
Best regards,
André Rossi
2011/9/12 Martin Morgan
> Hi André...
>
>
> On 09/12/2011 07:20 AM, André Rossi wrote:
>
>> Dear Martin Morgan and Martin Maechler...
>>
Love the page.
Just out of interest, is this an updated version or the same ol'
Inferno document?
And why do I keep thinking you (Patrick Burns) are the Hab's coach? :-)
--
-
Sent from my Cray XK6
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https:/
I already provided the link to the task view, which provides a list of
the more popular machine learning algorithms for R.
Do you have a particular algorithm or technique in mind? Does it have a
name?
How does sequential classification differ form running a one-off
classifier for each run?
Dear All:
I am calculating the relative importance of a regressor in a linear model.
Does anyone know how I can obtain/install the 'pmvd' computation type? I am
a US user.
Regards,
Y
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Justin,
Thanks for your help.
On Mon, Sep 12, 2011 at 2:19 PM, Justin Haynes wrote:
>
> the data you've given is all character vectors!
Yes, I'm sorry about that. I should not have used cbind when forming
my data.frame. It changed my numeric data to character. This command
would have been be
On Mon, Sep 12, 2011 at 1:32 PM, Claudia Stocker
wrote:
> Dear all,
>
> I have a problem in writing a variable to a NetCDF-File.
> My code works pretty well until the step put.var.ncdf():
>
>
[...code omitted...]
> R prints the following error:
> #-
>
> Error in put.var.ncdf(sp
This suggests that this is a dangerous office to be in because this is a
basic question. I am sure somebody in your office knows this. Anyway, the
baseline gives you the average value of the group that constitutes the
baseline when all other covariates are zero. Let's say you measure whether
men or
jonas garcia googlemail.com> writes:
> I am trying to fit some mixed models using packages lme4 and nlme.
>
> I did the model selection using lmer but I suspect that I may have some
> autocorrelation going on in my data so I would like to have a look using the
> handy correlation structures avai
You can do this using ifelse(). See example below.
x<-rpois(100,100)
NA.x<-sample(1:100,40)
x[NA.x]=NA
y<-rpois(100,100)
NA.y<-sample(1:100,40)
y[NA.y]=NA
z<-ifelse(!is.na(y),y,ifelse(!is.na(x),x,NA))
HTH,
Daniel
holly shakya wrote:
>
> I have 2 columns for weight. There are NAs in each col
Dear all,
I have a problem in writing a variable to a NetCDF-File.
My code works pretty well until the step put.var.ncdf():
# Get variables
#-
data1 <- open.ncdf("PREC_me_03-1500.nc")
prec1 <- get.var.ncdf(data1,"PRECT")
dim.time <- get.var.ncdf(data1,"time2")
close.ncdf(
You are asking a question about a package that seldom appears on
rhelp, leading me to infer that there is not a large user community
that reads this mailing list. You are also not providing the data
needed to reproduced the problem. You would be better off taking the
time to contact the pac
I have read it three times and still no concrete idea what you are actually
trying to do, mainly because there is no information as to which
level/variable you are aggregating on. It'd help if you provided the
aggregated data (or sample rows thereof) so that we know what you want the
result to be.
I have 2 columns for weight. There are NAs in each column but not for the
same observation. Some observations have values for both. I would want to
prioritize the WT2 values so I would like to do the following:
>From this:
ID WT1WT2
1 134 NA
2 145 155
1
Hi,
I tried to do a nested Anova with the attached Data. My response
variable is "survivors" and I would like to know the effect of
(insect-egg clutch) "size", "position" (of clutch on twig) and "clone"
(/plant genotype) on the survival of eggs (due to predation). Each plant
was provided with
Thanks, Bettina.
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Sent from the R help mailing list archive at Nabble.com.
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Dear list,
I am trying to fit some mixed models using packages lme4 and nlme.
I did the model selection using lmer but I suspect that I may have some
autocorrelation going on in my data so I would like to have a look using the
handy correlation structures available in nlme.
The problem is th
I am using the package SPACECAP which provides an interface for R. I used
this interface to import the csv file into R as well as the other two csv
files that are required.
This is the output I get when querying the error message
Error in NN[i, 1:length(od)] <- od : subscript out of bounds
> str(
Hi I am having difficulty interpretive the multiple regression output. I
would like to know what it means when one of the factors is assigned as the
intercept?
In my data I am looking at the relationship between environmental parameters
and biological production.
One of my variables in the analys
Setting par(usr=something) does not survive the creating of a new high level
plot. Using par(new=TRUE) is to be avoided if at all possible (it just leads
to problems like yours), it would be better for you to use matlines instead of
matplot which adds lines to the current plot.
If you want to
Hello,
I am trying to port one of my plotting S+ functions to R and I am having
difficulties!!! I am including here only the troublesome code!
I first produce a barplot, saving the positions of the bar's centers.
par(mar = c(6.1, 5.1, 4.1, 4.1), mgp = c(3, 3.0, 0))
ticks.loc <- barplot(su
Hi all,
I have a dataframe that includes data on individuals that are distributed
across multiple rows. I have aggregated the data using ddply, but I have
columns in the original data frame that are factors ( such as sites "A",
"B", and "C") that I would like to include in the new data frame. I
Hi,
I am trying to learn to use ggplot2 for what I had hoped would be a
fairly simple task. I have a relatively small data.frame (100 by 4).
The first column contains symbols. The 2nd, 3rd and 4th columns
represent percentage weightings for each symbol using 3 different
methodologies. For examp
I may be totally off base with this, but I'm wondering what exactly this
would suggest or why you want to do it. Specifically "multiple regression
with only intercept" -- how is it multiple if you don't have any regressors?
Furthermore, you want to run a "regression" on a single data point --
reall
Luis Felipe Parra wrote:
>
>
> and as you can see in the results some coefficients (for example ar2 and
> ar8) are different in the different R versions. does anybody know what
> might
> be going on. Was there any change in the arima function between the two
> versions?
>
You asked the sam
Marc's links lists many packages. Of those, I would recommend
XLConnect.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Marc Schwartz
Sent: Monday, September 12, 2011 12:53 PM
To: Damian Abalo
Cc: r-help
Subject: Re: [R] Writting
Em 9/9/2011 14:32, array chip escreveu:
Thanks all again for the suggestions. I agree with Wolfgang that
mcnemar.test() is what I am looking for. The accuracy is the
proportion of correct diagnosis compared to a gold standard, and I am
interested in which diagnosis test is better, not particular
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Benjamin Høyer
> Sent: Monday, September 12, 2011 6:19 AM
> To: r-help@r-project.org
> Cc: t...@novozymes.com
> Subject: [R] 1 not equal to 1, and rep command
>
> Hi
>
> I need
Here is another approach. A linear regression with a single binomial predictor
will give the same results as a pooled t-test (if you insist on non-pooled then
use sapply as previously suggested). The lm function will do multiple
regressions if given a matrix as the y-variable, so you can do a
Go to your R session.
First off, tell the R-help list how you got your csv file into R as NN.
read.csv() ? read.table()?
Then tell us exactly what code you're using. Where did i come from?
Also type in exactly the commands I gave you (four lines), and share the
output with the R-help list.
str(
I updated the code as follows:
dev.new(width=2.5, height=3,mar=c(0,0,0,0))
par(mfrow=c(5,1),mar=c(.5, .5, 1.5, .5), oma=c(.4, 0,.5, 0))
barplot(c(1,1,1,1,1),col=c("white","white","red","white","white"), axes =
FALSE,border=NA)
barplot(c(1,1,1,1,1),col=c("orange","white","white","white","yellow"), a
worked beautifully.
Thanks.
On Mon, Sep 12, 2011 at 9:45 AM, Gabor Grothendieck
wrote:
> On Mon, Sep 12, 2011 at 11:57 AM, steven mosher
> wrote:
>> Gabor.. thanks.
>>
>> zr <- zooreg(rnorm(24), as.chron("2011-01-01"), frequency = 24)
>>
>> a couple issues: my date data has missing days and
Not so strange, in fact this is FAQ 7.31, and has to do (as you guess)
with the way that computers store numbers.
You need to do as you did, and use round() or floor() or similar to
ensure that you get the results you expect.
Sarah
2011/9/12 Benjamin Høyer :
> Hi
>
> I need to use rep() to get a
Here is the new code. It works just like I wanted.
dev.new(width=6, height=6.5,mar=c(0,0,0,0))
par(mfrow=c(5,1),mar=c(.5, .5, 1.5, .5), oma=c(.4, 0,.5, 0))
barplot(c(1,1,1,1,1),col=c("white","white","red","white","white"), axes =
FALSE,border=NA)
barplot(c(1,1,1,1,1),col=c("orange","white","white"
I will try stacking 5 barplots (with 5 bars per plot) and somehow only
showing the middle bar for the top and bottom plots and the two end bars for
the two middle plots.
-
In theory, practice and theory are the same. In practice, they are not - Albert
Einstein
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Hi André...
On 09/12/2011 07:20 AM, André Rossi wrote:
Dear Martin Morgan and Martin Maechler...
Here is an example of the computational time when a slot of a S4 class
is of another S4 class and when it is just one object. I'm sending you
the data file.
Thank you!
Best regards,
André Rossi
On Sep 12, 2011, at 10:28 AM, Damian Abalo wrote:
> Hello.
> I need to generate, using R code, an excel file with multiple sheets,
> I wonder if any of you know how to do so.
>
> Thanks for the help
See the following:
R Data Import/Export Manual: http://cran.r-project.org/doc/manuals/R-data.ht
On Mon, 2011-09-12 at 03:24 -0700, Briony wrote:
> Thank you very much for the suggestion. And while I'm here, thank you for
> vegan and the documentation that goes with it.
>
> >Function ordiplot3d uses scatterplot3d, and it returns also all
> >scatterplot3d items, like functions xyz.converter an
Hello.
I need to generate, using R code, an excel file with multiple sheets,
I wonder if any of you know how to do so.
Thanks for the help
__
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PLEASE do read the posting guide
Dear Martin Morgan and Martin Maechler...
Here is an example of the computational time when a slot of a S4 class is of
another S4 class and when it is just one object. I'm sending you the data
file.
Thank you!
Best regards,
André Rossi
##
On Mon, Sep 12, 2011 at 11:57 AM, steven mosher wrote:
> Gabor.. thanks.
>
> zr <- zooreg(rnorm(24), as.chron("2011-01-01"), frequency = 24)
>
> a couple issues: my date data has missing days and missing hours..
> Sorry if I was not clear on
> that.. I input it to a data frame and dates are of t
Hi all,
I have a time series a column vector with the ordered data so that the first
column is the first observation and so on.
The fact is that I want to run a multiple regression with only intercept.
My first task is to run the regression on the first observation (1 from 276)
and at the same t
Hi,
On Mon, Sep 12, 2011 at 11:07 AM, vkent wrote:
> I would be grateful if anyone could tell me what the error message:
>
> Error in NN[i, 1:length(od)] <- od : subscript out of bounds
It means that either i ends up being larger than the number of rows in
NN, or that length(od) ends up being la
Hi
I need to use rep() to get a vector out, but I have spotted something very
strange. See the reproducible example below.
N <- 79
seg <- 5
segN <- N / seg # = 15.8
d1 <- seg - ( segN - floor(segN) ) * seg
d1# = 1
rep(2, d1) # = numeric(0), strange - why doesn't
Hello dear members,
I need to calculate "by hand" a local lineal regression so I need to compute a
kernel weight.
Does somebody knows how to get a kernel to use as weighted? I can calculate a
density kernel function and after pre-multiply it by the sample size. However I
know this is not w
I would be grateful if anyone could tell me what the error message:
Error in NN[i, 1:length(od)] <- od : subscript out of bounds
means for a large .csv file containing gps coordinates. I am using package
SPACECAP and have successfully run it with other .csv files but now keep
getting this error
Gabor.. thanks.
zr <- zooreg(rnorm(24), as.chron("2011-01-01"), frequency = 24)
a couple issues: my date data has missing days and missing hours..
Sorry if I was not clear on
that.. I input it to a data frame and dates are of the form 20110101
and hours are in the format
0,100,200
The end goa
Hello , I have estimated the following model, a sarima:
p=9
d=1
q=2
P=0
D=1
Q=1
S=12
In R 2.12.2
Call:
arima(x = xdata, order = c(p, d, q), seasonal = list(order = c(P, D, Q),
period = S),
optim.control = list(reltol = tol))
Coefficients:
ar1 ar2 ar3 ar4 ar5 ar
I've created a chart with times that employees have entered data on named tasks
as in the following example:
Employee <- c(rep("Tom", 127),
rep("Dick", 121),
rep("Sally", 130)
)
Time <- c(seq(as.POSIXct("2011-09-12 07:00:00"), as.POSIXct("2011-09-12
14:00:00"), 200),
seq(as.POSIXct("2011-09-12
I'm not sure this is the right location (maybe R-devel would be better).
-
In theory, practice and theory are the same. In practice, they are not - Albert
Einstein
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Sent from the
dev.new(width=6, height=1.5,mar=c(0,0,0,0))
par(mfrow=c(1,1),mar=c(.5, .5, 1.5, .5), oma=c(.4, 0,.5, 0))
barplot(c(1,1,1,1,1,1),col=c("blue","purple","red","green","orange","yellow"),
axes = FALSE)
I have a barplot that returns six colors in a line. I would like to get the
same six color blocks in
On 09/11/2011 03:42 PM, Jay wrote:
> What R packages are available for performing classification tasks?
> That is, when the predictor has done its job on the dataset (based on
> the training set and a range of variables), feedback about the true
> label will be available and this information shoul
Try this
> ltr<-LETTERS[1:3]
> unique(apply(expand.grid(ltr,ltr,ltr),1,function(x)
> paste("Var",unique(sort(x)),collapse="+",sep="")))
[1] "VarA" "VarA+VarB" "VarA+VarC" "VarA+VarB+VarC" "VarB"
"VarB+VarC" "VarC"
>
Andrej
--
Andrej Blejec
National Institute
On 09/12/2011 04:28 PM, vioravis wrote:
I am using 'tm' package for text mining and facing an issue with finding the
frequently occuring terms. From the definition it appears that findFreqTerms
and minDocFreq are equivalent commands and both tries to identify the
documents with terms appearing mo
On 12/09/2011 9:03 AM, amir wrote:
Hi,
Is there any function or command in R that show that how many times a
number is repeated in an array?
?table
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PLEASE do read the
Hi,
Is there any function or command in R that show that how many times a
number is repeated in an array?
Regards,
Amir
--
___
Amir Darehshoorzadeh |Comp. Architecture Dept.
PhD Student |UPC-Campus Nord, C6-221
Em
Thank you for your help
Yes i wanted to do the t test for all columns except for the grouping
column.
2011/9/12 Uwe Ligges
>
>
> On 12.09.2011 13:16, Raphael Saldanha wrote:
>
>> Hi!
>>
>> Try something like this:
>>
>> subset(example, disease==TRUE)
>> subset(example, disease==FALSE)
On 12.09.2011 12:30, Nevil Amos wrote:
A very basic query
This code plots OK the axis values are in bold but the axis labels are
not. how do I get them in bold too?
Add
font.lab=2
Uwe Ligges
thanks
Nevil Amos
plot(c(1,1),xlim=c(0,450),ylim=c(0.7,1.4),xlab="Distance (cells) from
edge
On Mon, Sep 12, 2011 at 4:59 AM, Laurent Fernandez Soldevila
wrote:
> Good afternoon,
>
>
> After cuting a hierarchical tree using cutree(), how to check correspondances
> between classes and branches?
> This is what we do:
>
> srndpchc <- hclust(dist(srndpc$x[1:1000,1:3]),method="ward") #creatio
one option is the following:
varNames <- c("varA", "varB", "varC", "varD")
f <- function (i) {
combn(length(varNames), i,
function (x) paste(varNames[x], collapse = " + "))
}
lapply(seq_along(varNames), f)
However, in case you're interested in performing a linear regression
with t
Good afternoon,
After cuting a hierarchical tree using cutree(), how to check correspondances
between classes and branches?
This is what we do:
srndpchc <- hclust(dist(srndpc$x[1:1000,1:3]),method="ward") #creation of
hierarchical tree
plclust(srndpchc,hmin=2) #visualisation
srndpchc2
Hi!
How can I make a PerMANOVA in R comparing treatments in a matrix that looks
something like this:
Treatment 1 Treatment 2 Treatment 3
Species 1 0.6
0.2 0
Species 2 0
0.7 0.3
Species 3
A very basic query
This code plots OK the axis values are in bold but the axis labels are
not. how do I get them in bold too?
thanks
Nevil Amos
plot(c(1,1),xlim=c(0,450),ylim=c(0.7,1.4),xlab="Distance (cells) from
edge of grid",ylab="Resistance distance",
type="l",col="white",lwd=2,font=2
How can I display a heatmap.2 with a column dendrogram without reordering
neither column or row?
library(vegan)
dissimilaritymatrix<-data.matrix(vegdist(step3,method="bray"))
library(gplots)
heatmap<-heatmap.2(dissimilaritymatrix,dendrogram="column",Colv=T,
Rowv=F,key=TRUE, symkey=FALSE, density.i
How can I display a heatmap.2 with a column dendrogram without reordering
neither column or row?
library(vegan)
dissimilaritymatrix<-data.matrix(vegdist(step3,method="bray"))
library(gplots)
heatmap<-heatmap.2(dissimilaritymatrix,dendrogram="column",Colv=T,
Rowv=F,key=TRUE, symkey=FALSE, density.i
Hello,
I'd like to generate automatically all the possible combinations of a set of 8
variables (there are 535, too many to do it by hand). For example:
input: varA, varB, varC
output: varA+varB+varC
varA+varB
varA+varC
varB+varC
varA
Szeptember 12-től 26-ig irodán kívül vagyok, és az emailjeimet nem érem el.
Sürgős esetben kérem forduljon Kárpáti Edithez (karpati.e...@gyemszi.hu).
Üdvözlettel,
Mihalicza Péter
I will be out of the office from 12 till 26 September with no access to my
emails.
In urgent cases please contact
In my mind this sequential classification task with feedback is
somewhat different from an completely offline, once-off,
classification. Am I wrong?
However, it looks like the mentality on this topic is to refer me to
cran/google in order to look for solutions myself. Oblivious I know
about these s
Thank you very much for the suggestion. And while I'm here, thank you for
vegan and the documentation that goes with it.
>Function ordiplot3d uses scatterplot3d, and it returns also all
>scatterplot3d items, like functions xyz.converter and points3d that
>can be used for tuning labels.
I tried or
Dear Ehsan,
the cluster option is not implemented in 'eha', although you obviously get
no error if trying
I'll fix this. Thanks for the report. (So, use 'coxph' with cluster).
Göran
On Mon, Sep 12, 2011 at 4:43 AM, Ehsan Karim wrote:
> Sorry: there was an error in the weight calculation,
On Mon, Sep 12, 2011 at 3:42 AM, marcel wrote:
> I have data of the form
>
> tC <- textConnection("
> Subject Date parameter1
> bob 3/2/99 10
> bob 4/2/99 10
> bob 5/5/99 10
> bob 6/27/99 NA
> bob 8/35/01 10
> bob 3/2/02 10
> steve 1/2/99 4
> steve 2/2/00 7
> s
On 12.09.2011 13:16, Raphael Saldanha wrote:
Hi!
Try something like this:
subset(example, disease==TRUE)
subset(example, disease==FALSE)
Hmmm, I think the actual answer to the question is something along this
line:
sapply(example[names(example)!="disease"],
function(x) t.test(x ~
Hi!
Try something like this:
subset(example, disease==TRUE)
subset(example, disease==FALSE)
On Mon, Sep 12, 2011 at 4:54 AM, C.H. wrote:
> Dear R experts,
>
> Suppose I have an data frame likes this:
>
> > example <- data.frame(age=c(1,2,3, 4,5,6),
> height=c(100,110,120,130,140,150), disease
On Mon, Sep 12, 2011 at 1:58 AM, steven mosher wrote:
> I have date data as a numeric and hourly data in 0 to 2300 hours in a
> dataframe.
>
> d <- rep(20110101,24)
> h <- seq(from = 0, to = 2300, by = 100)
>
> df <- data.frame(LST_DATE = d, LST_TIME = h, data = rnorm(24, 0, 1))
Please quote the prior thread, otherwise readers of this mailing list
will not get the context.
Uwe Ligges
On 12.09.2011 01:41, Briony wrote:
Thank you very much for the suggestion. And while I'm here, thank you for
vegan and the documentation that goes with it.
I tried ordilabel(pl$arrows
Far be it from me to misquote someone.
On 12/09/2011 09:17, peter dalgaard wrote:
On Sep 12, 2011, at 09:41 , Patrick Burns wrote:
R-help is all about solving R problems.
So here ya go:
http://www.portfolioprobe.com/2011/09/12/solve-your-r-problems/
Grin.
Incidentally, I don't think it is
Hi:
Here's one approach:
# date typo fixed in record 5 - changed 35 to 5
tC <- textConnection("
Subject Dateparameter1
bob 3/2/99 10
bob 4/2/99 10
bob 5/5/99 10
bob 6/27/99 NA
bob 8/5/01 10
bob 3/2/02 10
steve 1/2/99 4
steve 2/2/00 7
steve 3/2/01 10
steve
Hello,
I have a time-series that has some missing samples.
I was thinking on completing them using either zero-order hold or linear
interpolation.
I am looking for an efiicient way (other than a loop...) of identifiying the
missing time slots and filling them.
Can you think of any methods that mi
On Mon, Sep 12, 2011 at 9:21 AM, Twaha Mlwilo wrote:
>
> Hello all,
> Good day,
> I have problem on how to remove the source code from the pdf output.Here I
> mean this.
> code in sweave Rnw files
>
> <<>>=
> x<-c(1,2,3,4,5,6)
> x
> mean(x)
> sd(x)
> @
> then would like it appear as
> mean =
Dear Tom,
I think you failed to generate simulated outcome from the correct model. Hence
the zero variance of your random effects. Here is a better working example.
library(lme4)
fake2 <- expand.grid(Bleach = c("Control","Med","High"), Temp =
c("Cold","Hot"), Rep = factor(seq_len(3)), ID = seq
http://cran.r-project.org/web/views/
Look for 'machine learning'.
Dennis
On Sun, Sep 11, 2011 at 11:33 PM, Jay wrote:
> If the answer is so obvious, could somebody please spell it out?
>
>
> On Sep 11, 10:59 pm, Jason Edgecombe wrote:
>> Try this:
>>
>> http://cran.r-project.org/web/views/Mach
On Mon, Sep 12, 2011 at 9:41 AM, Patrick Burns wrote:
> R-help is all about solving R problems.
> So here ya go:
> http://www.portfolioprobe.com/2011/09/12/solve-your-r-problems/
>
Sweet. :)
May I suggest a font change: anything but the default CM should do the
trick. For one I prefer Palatino &
Hi Steven:
How about this?
d <- rep(20110101,24)
h <- sprintf('%04d', seq(0, 2300, by = 100))
df <- data.frame(LST_DATE = d, LST_TIME = h, data = rnorm(24, 0, 1))
df <- transform(df, datetime = as.POSIXct(paste(LST_DATE, LST_TIME),
format = '%Y%m%d %H%M'))
library(zoo
On Sep 12, 2011, at 09:41 , Patrick Burns wrote:
> R-help is all about solving R problems.
> So here ya go:
> http://www.portfolioprobe.com/2011/09/12/solve-your-r-problems/
Grin.
Incidentally, I don't think it is quite true that I called you "that infernal
guy" at useR. I might have done so (
I have data of the form
tC <- textConnection("
Subject Dateparameter1
bob 3/2/99 10
bob 4/2/99 10
bob 5/5/99 10
bob 6/27/99 NA
bob 8/35/01 10
bob 3/2/02 10
steve 1/2/99 4
steve 2/2/00 7
steve 3/2/01 10
steve 4/2/02 NA
steve 5/2/03 16
kevin 6/5/04 2
Hello all,
Good day,
I have problem on how to remove the source code from the pdf output.Here I mean
this.
code in sweave Rnw files
<<>>=
x<-c(1,2,3,4,5,6)
x
mean(x)
sd(x)
@
then would like it appear as
mean = 3.5
sd=1.3
x=1,2,3,4,5,6
thank you in advance
Dear R experts,
Suppose I have an data frame likes this:
> example <- data.frame(age=c(1,2,3, 4,5,6), height=c(100,110,120,130,140,150),
> disease=c(TRUE, TRUE, TRUE, FALSE, FALSE, FALSE))
> example
age height disease
1 1100TRUE
2 2110TRUE
3 3120TRUE
4 4130
R-help is all about solving R problems.
So here ya go:
http://www.portfolioprobe.com/2011/09/12/solve-your-r-problems/
--
Patrick Burns
pbu...@pburns.seanet.com
twitter: @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of 'Some hints for the R beginner'
and 'The
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