Hi eric,
Try
colnames(x)
colnames(x)[1] <- 'newname'
colnames(x)
HTH,
Jorge
On Sun, Jan 16, 2011 at 11:28 PM, eric <> wrote:
>
> How do I change the name of one column in a data frame ? Suppose I have a
> data frame x with 5 columns. If the names were date, col1, col2, col3, col4
> and I want
Hello,
Back again,
I thought the problem was solved but I realised that the only reason I was
getting the correct answer was because my data set happened to only have two
"rfts" to choose from, so it looked correct.
I have been using:
onlyfirstresponseafterrft<-which(!diff(as.numeric(factor(
How do I change the name of one column in a data frame ? Suppose I have a
data frame x with 5 columns. If the names were date, col1, col2, col3, col4
and I wanted to simply change the name of date, what would the command be ?
I tried the following and it didn't seem to work :
names(x[1]) <- "newn
Dear R community,
I have been getting this warning message after running a function sourced
from an R script, and can't seem to work out why R is looking for a folder
that wasn't even specified (it attaches a \NA to the specified directory,
where assess_rev has not asked to do so at all. R code h
What are the R equivalents to the Stata command egen tag
and
egen count?
egen station_week_tag = tag(station week)
Thank you
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Dear Solomon,
On Sun, Jan 16, 2011 at 10:27 PM, Solomon Messing
wrote:
> Dear Soren and R users:
>
> I am trying to use the summaryBy function with weights. Is this possible?
> An example that illustrates what I am trying to do follows:
>
> library(doBy)
> ## make up some data
> response = rno
Dear Soren and R users:
I am trying to use the summaryBy function with weights. Is this possible? An
example that illustrates what I am trying to do follows:
library(doBy)
## make up some data
response = rnorm(100)
group = c(rep(1,20), rep(2,20), rep(3,20), rep(4,20), rep(5,20))
Hi, I am wondering if there is a similar effects package for mixed models, just
like what effects package does for linear, generalized linear models?
Specifically I am looking for a way to calculate the SAS-co-called least
squared
means (LS means) in mixed models (I understand there is a substa
On Sun, Jan 16, 2011 at 11:58 AM, Hugo Mildenberger
wrote:
> Thank you very much for your qualified answers, and also for the
> link to the Tukey paper. I appreciate Tukey's writings very much.
Yes, thanks to Hadley for the nice reference, I hadn't seen it before.
> Looking at the lattice code (
That's possible, but without knowing exactly what you did I
really can't say for certain what went wrong.
I'm glad you've got it working.
Sarah
On Sun, Jan 16, 2011 at 8:06 PM, Freddy Gamma wrote:
> Well, how about some information:
> What version of Windows are you running?
> XP PRO 2000
> Wha
Hello,
I've been waiting to see if anyone else would answer this.
I've previously used random reallocation of objects to groups
(clusters) as a monte-carlo test of the informativeness of groups, as
described here:
http://lastresortsoftware.blogspot.com/2010/09/monte-carlo-testing-of-classificati
Hi Emmanuel,
Try the following:
1) removing unnecessary programs from memory, this might give u a larger
contiguous memory block for R
2) remove unnecessary data from R's memory, so many of the preceding data
sets U no longer need can be removed. use the rm() command. U might need
to run gc() aft
or
rm(list=ls(pattern="^NY")) if you only want those objects that begin with
"NY" ...
HTH
Pete
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On 11-01-16 5:32 PM, barbara.r...@uniroma1.it wrote:
I have lost the work of 2 days for problems to my pc. Can I get back it?
Probably not unless you explicitly saved it to disk. R doesn't do
automatic backups for you.
Duncan Murdoch
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I have lost the work of 2 days for problems to my pc. Can I get back it?
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PLEASE do read the posting guide http://www.R-project.
Hi dear all, i am triying to do jackknife-after bootstrap for detection of
influential observation.
my data and resamples are following ;
e <- rnorm(n=50, mean=0, sd=sqrt(0.5625))
x0 <- c(rep(1,50))
x1 <- rnorm(n=50,mean=2,sd=1)
x2 <- rnorm(n=50,mean=2,sd=1)
x3 <- rnorm(n=50,mean=2,sd=1)
x4 <- rn
How about
ls(pattern="NY")
#rm(list=ls(pattern="NY"))
On Sunday 16 January 2011 22:08:16 Erin Hodgess wrote:
> Dear R People:
>
> I have the following:
>
> > ls(pattern="NY*")
> [1] "CRAN_df" "CRAN_df0" "CRAN_df1" "CRAN_mat" "CRAN_sp"
> [6] "CRAN_spdf1" "CRAN_spdf2
On Jan 16, 2011, at 4:17 PM, David Winsemius wrote:
On Jan 16, 2011, at 4:08 PM, Erin Hodgess wrote:
Dear R People:
I have the following:
ls(pattern="NY*")
[1] "CRAN_df" "CRAN_df0" "CRAN_df1" "CRAN_mat"
"CRAN_sp"
[6] "CRAN_spdf1" "CRAN_spdf2" "CRAN_spdf4" "delaune
On Sun, Jan 16, 2011 at 1:08 PM, Erin Hodgess wrote:
> Dear R People:
>
> I have the following:
>
>> ls(pattern="NY*")
> [1] "CRAN_df" "CRAN_df0" "CRAN_df1" "CRAN_mat" "CRAN_sp"
> [6] "CRAN_spdf1" "CRAN_spdf2" "CRAN_spdf4" "delauney_NY" "dist2_NY"
> [11] "dist3_NY" "G
On Jan 16, 2011, at 4:08 PM, Erin Hodgess wrote:
Dear R People:
I have the following:
ls(pattern="NY*")
[1] "CRAN_df" "CRAN_df0" "CRAN_df1" "CRAN_mat"
"CRAN_sp"
[6] "CRAN_spdf1" "CRAN_spdf2" "CRAN_spdf4" "delauney_NY"
"dist2_NY"
[11] "dist3_NY" "Gabriel_NY"
Dear R People:
I have the following:
> ls(pattern="NY*")
[1] "CRAN_df" "CRAN_df0" "CRAN_df1" "CRAN_mat" "CRAN_sp"
[6] "CRAN_spdf1" "CRAN_spdf2" "CRAN_spdf4" "delauney_NY" "dist2_NY"
[11] "dist3_NY" "Gabriel_NY" "NY8a_nb" "rel_neigh_NY" "scot_BNG"
[16] "SOI_NY"
Thank you very much for your qualified answers, and also for the
link to the Tukey paper. I appreciate Tukey's writings very much.
Looking at the lattice code (below), a possible implementation might
involve binning, not so?
I see a problematic part here:
xx <- sort(unique(x))
Unique cer
Thank you for the informations, especially for the cumhaz tip.
I'll explain a bit more of my experimental design:
I have 4 different replicate populations of drosophila, from which I
take samples to infect, in 5 independent replicates (tubes) of 10
individuals each. That makes 200 individuals per
I am not sure if this type of posting is allowed, but I'm sure many fellow R
enthusiasts will agree with me on this:
I want to encourage everyone with any type of R-related question, to post it
on StackOverflow.com, tagged with the keyword "r", for the following
reasons:
1. There are many R ex
Using lattice and the rainfall$Time series as proposed below by Dennis
gives also a nice result:
rainfall$Time <- seq(from = as.Date('1993-01-01'),
to = as.Date('2007-12-01'), by = 'month')
xyplot(rainfall~Time,data=rainfall,type=c("g","p","l","smooth"))
On
Hi,
I have read several threads about memory issues in R and I can't seem to
find a solution to my problem.
I am running a sort of LASSO regression on several subsets of a big dataset.
For some subsets it works well, and for some bigger subsets it does not
work, with errors of type "cannot alloca
Hi:
Arggh, it's too early in my morning...apologies.
With the same adjustment to the data frame as given previously, this
'works':
xyplot(rainfall ~ Time, data = rainfall, type = 'l')
Gabor's solution is nice because it gives you more control over the date
format and the zoo plot method works d
On Sun, Jan 16, 2011 at 5:48 AM, wrote:
> Here is one way
>
> Here is one way:
>
>> con <- textConnection("
> + ID TIME OBS
> + 001 2200 23
> + 001 2400 11
> + 001 3200 10
> + 001 4500 22
> + 003 3900 45
>
On Sun, Jan 16, 2011 at 8:01 AM, Kang Min wrote:
> Hi,
>
> I would like to plot time against rainfall data (data is at the end)
> using xyplot.
>
> The basic code looks like this: xyplot(rainfall~time, type="a")
> When I do this, the graph looks ok except that the x-axis has too many
> values. I w
Hi all,
I try to print a table with the grid.draw(tableGrob(...)) function applied
to an ftable object. The output only give me a table without
colnames either rownames, but with edit.grid() I can add only one of them.
And this is the problem, I have a table with a two rows head, like this:
thanks so much - that did it.
I am new to this - so the help is greatly appreciated.
Matthew
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Type 'search()' to see what is on the search path to see if anything is
missing. Also try 'apropos('read')' to see if there are other functions
containing 'read'
Sent from my iPhone
What is the problem you are trying to solve?
On Jan 16, 2011, at 8:49, Freddy Gamma wrote:
> Egregious,
>
>
If b is the coefficient from the meta-regression model that indicates the
(average) change in the outcome measure for a one-unit increase in the
corresponding explanatory variable, then 5*b is the (average) change in the
outcome measure for a 5-unit increase in the explanatory variable.
Or equi
Hi:
Try this, since your data have no missing months:
rainfall$Time <- seq(from = as.Date('1993-01-01'), to =
as.Date('2007-12-01'), by = 'month')
g <- ggplot(rainfall, aes(x = Time, y = rainfall))
g + geom_path()
HTH,
Dennis
On Sun, Jan 16, 2011 at 5:01 AM, Kang Min wrote:
> Hi,
>
> I would
Well, how about some information:
What version of Windows are you running?
What version of R are you running?
How did you start R? Did you load an existing Rdata file? Are you
using one of the GUIs, or piping through an editor?
What happens if you start a clean session of R without any add-ons?
I
Hi,
I would like to plot time against rainfall data (data is at the end)
using xyplot.
The basic code looks like this: xyplot(rainfall~time, type="a")
When I do this, the graph looks ok except that the x-axis has too many
values. I would just like to display the years and not the months on
the x-
Hi All,
I am trying to use the synonyms function of wordnet. If I try for example
synonyms("happy"), it returns character(0).
But if i start up the program wordnet, "happy" is part of the database.
And if I try for example:
synonyms("company") it returns:
c(\"caller\", \"company\")"
Egregious,
look what just happened few minutes ago while opening up my program and
running the scripts
> A<-read.table("C:\\Documents and Settings\\
+ me\\Desktop\\TESI\\generale.txt",head=T)
Errore: non trovo la funzione "read.table" #CAN'T FIND read.table
FUNTION!!!??? ARE WE ok???
>
>
Thank you both very much !
This helped me a lot.
Best,
Holger
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Hi Achim!
On 16 January 2011 16:37, Achim Zeileis wrote:
> Arne: Unless I'm missing something, hausman.systemfit() essentially does the
> right thing and computes the right statistic and p-value (see my other mail
> to Holger). Maybe some preliminary check on the input objects could be used
> for
I haven't seen an answer so far, hence I try:
As far as I can see, everything is correct here, since you have an
element polygons in you data.frame that is a valid vector of mode list.
Nevertheless, since printing /plotting etc. on the dataframe won't give
desired results now, it would be mor
On Sun, 16 Jan 2011, Arne Henningsen wrote:
Hi Holger!
On 16 January 2011 15:53, Holger Steinmetz wrote:
One follow up question. The Hausman-test always gives me a p-value of 1 - no
matter how small the statistic is.
I now generated orthogonal regressors (X1-X3) and the test gives me
On Sun, 16 Jan 2011, Holger Steinmetz wrote:
Dear Achim,
thank you very much.
One follow up question. The Hausman-test always gives me a p-value of 1
- no matter how small the statistic is.
I now generated orthogonal regressors (X1-X3) and the test gives me
Hausman specification t
Hi Holger!
On 16 January 2011 15:53, Holger Steinmetz wrote:
> One follow up question. The Hausman-test always gives me a p-value of 1 - no
> matter how small the statistic is.
>
> I now generated orthogonal regressors (X1-X3) and the test gives me
>
>
> Hausman specification test for cons
On Sat, Jan 15, 2011 at 4:26 PM, Matthew Strother wrote:
> I have a data set with several thousand observations across time, grouped by
> subject (example format below)
>
> ID TIME OBS
> 001 2200 23
> 001 2400 11
> 001 3200 10
> 001
On Sun, 16 Jan 2011, Hadley Wickham wrote:
The normal distribution is a continuous distribution, i.e., the frequency
for each observed value will essentially be 1/n and not converge to the
density function. Hence, you would need to look at histogram or smoothed
densities. Rootograms, on the othe
On 16/01/2011 9:13 AM, Patrick Hausmann wrote:
> Dear all,
>
> for each A == 3 in 'df' I would like to change the variables B and K.
> My result should be the whole df and not the subset (A==3)...
>
> df<- data.frame(A = c(1,1,3,2,2,3,3),
>B = c(2,1,1,2,7,8,7),
>
Dear Achim,
thank you very much.
One follow up question. The Hausman-test always gives me a p-value of 1 - no
matter
how small the statistic is.
I now generated orthogonal regressors (X1-X3) and the test gives me
Hausman specification test for consistency of the 3SLS estimation
data:
Try this:
df[df$A == 3, c('B', 'K')] <- with(df[df$A == 3, c('B', 'K')], cbind(5,
gsub("f", "m", K)))
On Sun, Jan 16, 2011 at 12:19 PM, Patrick Hausmann <
patrick.hausm...@uni-bremen.de> wrote:
> Arrg, sorry - of course I don't want *new* variables. So this is my correct
> example:
>
>
> df <- d
> The normal distribution is a continuous distribution, i.e., the frequency
> for each observed value will essentially be 1/n and not converge to the
> density function. Hence, you would need to look at histogram or smoothed
> densities. Rootograms, on the other hand, are intended for discrete
> di
On Sun, 16 Jan 2011, Holger Steinmetz wrote:
Hi,
can anybody tell me how the Hausman test for endogenty works?
I have a simulated model with three correlated predictors (X1-X3). I also
have an instrument W for X1
Now I want to test for endogeneity of X1 (i.e., when I omit X2 and X3 from
the
On Sun, 16 Jan 2011, Hugo Mildenberger wrote:
Using R-2.12.1 and latticeExtra-0.6-14, I would like to understand
why a rootogram displaying samples from the Poisson distribution looks like I
expected it, whereas a rootogram using the normal distribution does not:
library(latticeExtra)
rootogram
Arrg, sorry - of course I don't want *new* variables. So this is my
correct example:
df <- data.frame(A = c(1,1,3,2,2,3,3),
B = c(2,1,1,2,7,8,7),
K = c("a.1", "d.2", "f.3",
"a.1", "k.4", "f.9", "f.5"))
x1 <- within(df[df$A ==3, ], {
Dear all,
for each A == 3 in 'df' I would like to change the variables B and K.
My result should be the whole df and not the subset (A==3)...
df <- data.frame(A = c(1,1,3,2,2,3,3),
B = c(2,1,1,2,7,8,7),
K = c("a.1", "d.2", "f.3",
"a.1", "k
See ?try in order to get the automated process going and for ignoring or
separately post-processing stuff where the method does not work.
Uwe Ligges
On 15.01.2011 21:48, Jinrui Xu wrote:
Hello everyone,
I am using fgev from evd package to fitting my data with general extreme
value distribut
Hi,
can anybody tell me how the Hausman test for endogenty works?
I have a simulated model with three correlated predictors (X1-X3). I also
have an instrument W for X1
Now I want to test for endogeneity of X1 (i.e., when I omit X2 and X3 from
the equation).
My current approach:
library(system
On Sun, 16 Jan 2011, alessandro.sarre...@inwind.it wrote:
Dear R experts,
I'm trying to install rgdal in my R 2.11.1 (Ubuntu 10.10).
I have as an ouput the following messages. It seems that there are problems
with libgdal1.7.0 or sqlite3...
Could someboby help me?
You would do better to ask on
On Sat, Jan 15, 2011 at 11:20 PM, rnick wrote:
>
> Hi all,
> I have run into a problem and some help would be highly appreciated.
> I have a .csv with the following columns:
> Date Time Open High Low Close
> 1/2/2005 17:05 1.3546 1.3553 1.3546 1.35495
> 1/2/2005
On Sun, 16 Jan 2011, Peter Ehlers wrote:
Adelchi,
Since % is the LaTeX comment character, you may want to
try escaping it.
As package 'base' does at
https://svn.r-project.org/R/trunk/src/library/base/man/matmult.Rd
Peter Ehlers
On 2011-01-16 02:35, Adelchi Azzalini wrote:
[Hope this is th
I am a relative newbie to survival analysis and R in general, but
would like to use the coxme package to analyse some data I currently
have.
The data is relative to survival times of drosophila melanogaster
populations to infection with pathogens, and has the variables:
Time,
Status,
Treatment (4 t
On Sun, Jan 16, 2011 at 03:01:49PM +0800, r-help wrote:
> I have a data frame with 10 columns: A:J and I want to have the output as a
> data frame with 11 columns, the value of 11th column is??
>
>
> for each row, if any column can be divided by 13, then the 11th column has a
> values of 1, oth
Using R-2.12.1 and latticeExtra-0.6-14, I would like to understand
why a rootogram displaying samples from the Poisson distribution looks like I
expected it, whereas a rootogram using the normal distribution does not:
library(latticeExtra)
rootogram(~rpois(1000, lambda = 50), dfun = function(x)
Dear R experts,
I'm trying to install rgdal in my R 2.11.1 (Ubuntu 10.10).
I have as an ouput the following messages. It seems that there are problems
with libgdal1.7.0 or sqlite3...
Could someboby help me?
Thanks a lot!
Ale
> install.packages()
Warning in install.packages() :
argument 'lib' is
Here is one way
Here is one way:
> con <- textConnection("
+ ID TIMEOBS
+ 001 220023
+ 001 240011
+ 001 320010
+ 001 450022
+ 003 3900 45
+ 003 5605
Adelchi,
Since % is the LaTeX comment character, you may want to
try escaping it.
Peter Ehlers
On 2011-01-16 02:35, Adelchi Azzalini wrote:
[Hope this is the right list where to send...]
An attempt to update package 'mnormt' involves the addition of a
small new function called 'pd.solve'. Wh
On 2011-01-15 18:54, Kumar Mainali wrote:
Dear R users,
I cannot display ylab with the following code while trying to stack 7 rows
of 3 histograms each. Thanks in advance!
par(mfrow=c(7,3))
par(mar=c(2,2,2,2))
par(oma=c(5,5,0,0))
with (newdata<- subset(table, Month1 == "1"), hist(newdata$Carap
[Hope this is the right list where to send...]
An attempt to update package 'mnormt' involves the addition of a
small new function called 'pd.solve'. When I come to the package
checking stage, an error occurs in parsing pd.solve.Rd.
The full transcript of the outcome is copied below (it include
Assuming your csv file looks like:
Date,Time,Open,High,Low,Close
1/2/2005,17:05,1.3546,1.3553,1.3546,1.35495
1/2/2005,17:10,1.3553,1.3556,1.3549,1.35525
1/2/2005,17:15,1.3556,1.3556,1.35515,1.3553
1/2/2005,17:25,1.355,1.3556,1.355,1.3555
You could do something like:
aa <- read.csv("tmp.csv",head
On Sun, Jan 16, 2011 at 03:01:49PM +0800, r-help wrote:
> I have a data frame with 10 columns: A:J and I want to have the output as a
> data frame with 11 columns, the value of 11th column is??
>
>
> for each row, if any column can be divided by 13, then the 11th column has a
> values of 1, oth
Hi all,
I have run into a problem and some help would be highly appreciated.
I have a .csv with the following columns:
DateTimeOpenHighLow Close
1/2/200517:05 1.3546 1.3553 1.3546 1.35495
1/2/200517:10 1.3553 1.3556 1.3549 1.35525
1/2/200517:15
I have a data set with several thousand observations across time, grouped by
subject (example format below)
ID TIMEOBS
001 220023
001 240011
001 320010
001 450022
003 390
Thank you Mike - that has worked brilliantly.
I have one further question that I was hoping someone on the forum may be
able to help with.
For most of my meta regressions I am interested in the impact of a one unit
increase in the explanatory variable, however for one I want to estimate the
effe
I have a data frame with 10 columns: A:J and I want to have the output as a
data frame with 11 columns, the value of 11th column is£º
for each row, if any column can be divided by 13, then the 11th column has a
values of 1, otherwise, it has a value of 0. How to do that?
input is
a=matrix(1:
Dear friends:
Any of you know how to do nonparametric bootstrapping on the smoothed
frequency (2nd column in the table) ? For example, after smoothing is done,
I have the following data in the table. "Smoothed"=smoothed frequency. I
found some options with no clear examples. Options include *
Is it possible to simultaneously bootstrap the mean of multiple vectors
(columns) in a dataframe producing a matrix where each row provides the mean
for each vector for a given bootstrap sample? Thank you for any assistance
possible.
KAM
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Hello everyone,
I am using fgev from evd package to fitting my data with general
extreme value distribution. The command I used is simple: y <- fgev(x).
The error is "observed information matrix is singular;" How can I
solve these problem. I tried all methods in optim and it did not work.
Dear R users,
I cannot display ylab with the following code while trying to stack 7 rows
of 3 histograms each. Thanks in advance!
par(mfrow=c(7,3))
par(mar=c(2,2,2,2))
par(oma=c(5,5,0,0))
with (newdata <- subset(table, Month1 == "1"), hist(newdata$CarapWid,
breaks=5*(0:80), col="gray80", main="A
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