Hi,
Here i have a data frame like this,
frame name is "dat"
*NAME AGE ELIGIBLE*
A 20
B 14
C 35
D 10
E 50
In this, i want to change the column '*ELIGIBLE*' status to '*YES
He folks,
I want to use quantile regression for doing a test of symmetrie of a
distribution. Following Buchinsky I want to test, whether the square of \tau
= \beta(p)+\beta(1-p)-2*\beta(0.5) (\beta(\tau) is the estimated slope
parameter for quantile \tau).Unfortunately I do not know how to implemen
On 29.05.2012 09:17, Rantony wrote:
Hi,
Here i have a data frame like this,
frame name is "dat"
*NAME AGE ELIGIBLE*
A 20
B 14
C 35
D 10
E 50
In this, i want to change the column
Hi everybody.
I'm trying to do a correlation matrix in a list of files. Each file contains
2 columns: "capt1" and "capt2". For the example, I merged all in one
data.frame. My data also contains many missing data. The aim is to do a
correlation matrix for the same data for course (one correlation m
Dear all,
I'm using the mgcv library by Simon Wood to fit gam models with interactions
and I have been reading (and running) the "factor 'by' variable example"
given on the gam.models help page (see below, output from the two first models
b, and b1).
The example explains that both b and b1 fit
Dear all,
I would like to ask how my data set contains problems.
Here is what I have done.
SandP = read.csv("Price.csv")
library("forecast")
auto.arima(SandP)
and
R shewed this to me
Error in model.frame.default(formula = x ~ 1, drop.unused.levels = TRUE) :
invalid type (list) f
Dear R-users,
I keep getting an ERROR saying " *Error in model.matrix.default(formula,
mframe) : model frame and formula mismatch in model.matrix()* " when i fit
poLCA with more than 63 variables. Below are the details.
I am trying to do a Latent Class Analysis using poLCA. My data set contains
Dear all,
I am implementing a stochastic utility model that will eventually
make use of multinomial logit. I found that there is a package in R
called mlogit. I am not sure whether I have already found the correct
package or software. May I ask am I correct?
Basically, let's say
I have obse
Hi Michel,
In this function, I want to apply âORâ operator instead of â&â
Condition.
dat[dat$AGE<=40 & da$AGE>=30,"TRUE/FALSE"]<-TRUE
How it is possible ?
- Thanks
Antony
From: Michael Weylandt [via R]
[mailto:ml-node+s789695n4631610...@n4.nabbl
Dear fellow R users,
I'm using the package tm for text mining, and have a problem with
reading in a corpus from XML files.
When I copy the example from "Introduction to the tm package" of the
small reuters subset "crude", everything goes well, and I get a corpus
with the required meta data.
Wh
Hi Uwe,
How can we give "OR" Operator instead of "AND" here.
dat[, tmpCol] <- ifelse(dat$AGE<=35 & da$AGE>18, "YES", " ")
I want change this to "OR"
dat[, tmpCol] <- ifelse(dat$AGE<=35 OR da$AGE>18, "YES", " ")
"OR" - How it is possible ?
From: Uwe Ligges-3 [via R]
[mailto:m
On Tue, May 29, 2012 at 8:30 AM, stefan23 wrote:
> He folks,
> I want to use quantile regression for doing a test of symmetrie of a
> distribution. Following Buchinsky I want to test, whether the square of \tau
> = \beta(p)+\beta(1-p)-2*\beta(0.5) (\beta(\tau) is the estimated slope
> parameter fo
Hello,
Without trying it, I very much doubt that your code runs: First you use
'asim' and create it after.
So I've moved asim <- 1000 up some lines.
And it still doesn't work.
1. Run your own code before posting it.
2. Learn to use indentation.
3. Learn to use spaces to make your code easier
On 29/05/2012 09:44, Actuarial_Labor wrote:
Dear all,
I would like to ask how my data set contains problems.
Here is what I have done.
SandP = read.csv("Price.csv")
library("forecast")
auto.arima(SandP)
That's a data frame, not (from the help):
x: a univariate time serie
Hello r-help members,
thank you very much for your reply, Rui Barradas.
> Your data file has more than one line.
Yes, each line is a new record and I read several such data files into one
data.frame.
> I've called it "sabrina.txt" and then processed with:
>
> x <- readLines("sabrina.txt")
>
>
Hello,
Your model is equivalent of
y = b1(X1 + X3) + b2X2
(plus error)
So, use I() to add X1 and X3. You don't need to create an extra variable
X13 <- X1 + X3. See the help page for it. The point on function formula.
?I
linMod2 = lm(Y ~ -1 + I(X1 + X3) + X2, data=data.set)
summary(linMod2
Just a note, mainly to people who maintain larger R installations, that we
intend to have a patch release version on June 22. The nickname will be
"Roasted Marshmallows".
This is somewhat later than previous .1 releases, but we are also now aiming
for an annual rather than biannual release sche
Hello,
The error message means that 'x' is not a character vector. Can't you
try it only with the text in the link you've posted,
http://pastebin.com/mYZNDXg6 ?
I'm asking this because I've just checked it and it doesn't give any eror.
Em 29-05-2012 12:39, Sabina Arndt escreveu:
Hello r-hel
Hi,
You can try:
colnames(mymatrix)<-NULL
Best
Ozgur
-
Ozgur ASAR
Research Assistant
Middle East Technical University
Department of Statistics
06531, Ankara Turkey
Ph: 90-312-2105309
http://www.stat.metu.edu.tr/people/assistants/ozgur/
--
View this mess
Sorry, I misunderstood your question.
colnames(mymatrix)
would select column names of your matrix.
Best
ozgur
-
Ozgur ASAR
Research Assistant
Middle East Technical University
Department of Statistics
06531, Ankara Turkey
Ph: 90-312-2105309
http://www.s
That did it. Thanks! One more follow-up questions. How do I set a parameter
to a particular value? I tried I(.5*X2), but that didn't do what I
expected.
On Tue, May 29, 2012 at 6:39 AM, Rui Barradas wrote:
> Hello,
>
> Your model is equivalent of
>
> y = b1(X1 + X3) + b2X2
>
> (plus error)
>
> S
Simple, just make y2 <- y - 0.5*X2
Rui Barradas
Em 29-05-2012 13:55, Dustin Fife escreveu:
That did it. Thanks! One more follow-up questions. How do I set a parameter
to a particular value? I tried I(.5*X2), but that didn't do what I
expected.
On Tue, May 29, 2012 at 6:39 AM, Rui Barradas wro
offset() fixes the parameter to 1. So offset(I(.5*X2)) should do the trick.
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2
Hello,
Instead of 'sum' use 'mean'
ok <- apply(tbl, 2, function(x) mean(!is.na(x)) >= 0.5)
cor(tbl[, ok], use="pairwise.complete.obs")
Hope this helps,
Rui Barradas
Em 29-05-2012 10:03, jeff6868 escreveu:
Hi everybody.
I'm trying to do a correlation matrix in a list of files. Each file cont
Great. Thank you!
On Tue, May 29, 2012 at 8:11 AM, ONKELINX, Thierry wrote:
> offset() fixes the parameter to 1. So offset(I(.5*X2)) should do the trick.
>
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
> Forest
> team Biometrie & Kwaliteitszo
It would behoove you to read some introductory materials -- use
help.start() to get be provided with the standard "An Introduction to
R."
More generally, the help system can be accessed by typing help("name")
or, as a shortcut, ?name. In your case, help("&") would lead to the
correct answer.
The
I need to determine whether a variable y2e increases with a covariate xt within
individuals, where individuals a-j are measured several times.
Is it possible to test whether within-group coefficients are significantly
different from zero?
The coefficients from an lmList fitted to my data give:
>
Hello,
thank you very much for your reply, Rui Barradas.
OK, I did what you said:
> x <- readLines("sabina.txt")
> s <- strsplit(x, ";[[:space:]]\\[")
> r <- lapply(s, function(x) sapply(strsplit(x, "[[:blank:]]"), tail, 1))
> length(r)
[1] 20
I don't know why your result here was 21 since the
Greetings
I want to take a fitted regression and replace all uses of a variable
in a formula. For example, I'd like to take
m1 <- lm(y ~ x1, data=dat)
and replace x1 with something else, say x1c, so the formula would become
m1 <- lm(y ~ x1c, data=dat)
I have working code to finish that part of
Hello everyone,
I would like to create a new R mirror en Chile and I would be very placed if
any of you who are familiar doing it can help us.
We are looking for manuals and materials need it.
Thank you in advance,
José Bustos
Data Scientist at www.aespro.cl
[[alternative HTML version de
Dear all,
I'm trying to understand how the strucchange package is working and I have been
looking at the examples given for the Fstats() function.
The first example (Nile), shows one peak in the F-stats and one breakpoint is
estimated, that can be plotted using the following code
## Nile data wi
Hello
I am currently working on forecasting hourly electricity prices where I am
required to do wavelet transformation. First I need to calculate
decomposition coefficients of the wavelet transform, then I need to select
wavelet function using multiresolution technique. I would be really
grateful
Hi,
i have a matrix with headers. How can i get header name in that marix ?
i tried in R, with
names(MyMatrix[1]) - its getting exactly.
But when i try in eclips, its not receiving
ColHdr <- names(MyMatrix[1])
- it getting NULL value
- could you please help me ?
- Thanks
Antony.
--
View this
Hi,
i have a function like this,
*GetCsv[GetCsv[,ColHdr]ArrMinMax[iCol,2],ColHdr_QF]<-"RC"*
Here,
*GetCsv* -- It contain a matrix
*ColHdr* -- Is a column Header
*ArrMinMax* - It is a array of 2 columns
-
Here, this is the function to change
Dear R-users,
I keep getting an ERROR saying " Error in model.matrix.default(formula,
mframe) : model frame and formula mismatch in model.matrix() " when i fit
poLCA with more than 63 variables. Below are the details.
I am trying to do a Latent Class Analysis using poLCA. My data set contains
bi
ITS OK. And thanks. Its working !
From: Ãzgür Asar [via R] [mailto:ml-node+s789695n4631684...@n4.nabble.com]
Sent: Tuesday, May 29, 2012 6:04 PM
To: Akkara, Antony (GE Energy, Non-GE)
Subject: Re: Matrix Header Name Select
Sorry, I misunderstood your question.
colnames(mymatrix)
woul
Dear all,
is anyone aware of an R package for handling posets, drawing Hasse diagrams
etc.? I browsed through CRAN but was not successful...
Thanks in advance,
Christian
--
Christian Kampichler
Roghorst 91
6708 KD Wageningen, Niederlande
Tel +31 (0)317 417527
Mob +
Hello
I am currently working on forecasting hourly electricity prices where I am
required to do wavelet transformation. First I need to calculate
decomposition coefficients of the wavelet transform, then I need to select
wavelet function using multiresolution technique. I would be really
grateful
Hello,
I am trying to run the random Forest function on a data.frame using the
following code..
myrf <- randomForest (y=sample_data_metal, x=Train, importance=TRUE,
proximity=TRUE)
However, an error occurs saying, "can not handle categorical predictors with
more than 32 categories".
My
Dear all,
I am experiencing problems using the glmmPQL function in the MASS package
(Venables & Ripley 2002) to model binomial data with spatial
autocorrelation.
My question - is the presence of birds affected by various hydrological
parameters?
Presence/absence data were collected from 83 sit
Dear list,
I need to include in a superscript number (-1) in a y-axis label of a
boxplot graphic. I tried the following line:
>boxplot(five$gr13~five$Code, xlab="Species code",
ylab=substitute(paste("DGL growth (cm.yr"^{-1},")")))
It returns the following error message:
Error in "DGL growth (cm
Hi,
i have a matrix like this
*DWATT *
10
20
30
30
32
Here i need to create a new column named "*DWATT_QF*" with filling value as
per the following condition,
if DWATT_QF = RC When DWATT value less than 30
Hi,
I am hoping to use the RODBC package to write a dataframe to an Excel .xlsx
file. The dataframe contains at least one field with character elements
that exceed 255 bytes, which appears to be the cell width limit in Excel.
Below is example code and the warning message received:
library(RODBC)
Hi!!
My question is related to Rcmdr. I have Mac OS and I use R since more or less
1 year, not as an expert but I really like it. Now I want to install the Rcmdr
package, I already downloaded it, but when I want to "charged" R tell me that
it needed the tcltk package, so I downloaded it too
Dear R-users!
I am using the poLCA package to find latent classes in a dataset with binary
outcomes. Several hundreds of persons can engage hundreds of items. Assume
100 items being engaged by 500 persons, the resulting dataframe is as shown
below (5 persons).
item1 item2 item3 item4 item5 .
Hi Antony,
Try this,
> Obj_Name<-da$AGE
> Obj_Name
[1] 20 30 40 30 25
da[Obj_Name>=30 & Obj_Name<=40, "TRUE/FALSE"]<-TRUE
colnames(da)<-c("NAME","Obj_Name","PLACE","ELIGIBLE")
> da
NAME Obj_Name PLACE ELIGIBLE
1 ABC 20 INDIA NA
2 XYZ 30 FRANCE TRUE
3 PQR
R version 2.15.0 (2012-03-30)
Copyright (C) 2012 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: i386-pc-mingw32/i386 (32-bit)
package ‘survey’ successfully unpacked and MD5 sums checked
package ‘odfWeave.survey’ successfully unpacked and MD5 sums checked
> library(odfWe
I'm trying to get a minimally intuitive way of plotting zoo objects
extracted for given periods of time
I do:
> plot(alyL32007z$NEE_st,col=cod,type="b",pch=18,xlim=time(alyL32007z)[c(3000,15000)])
and get a correct plot but the expression for xlim is hard to read by humans.
I try
> time(alyL3200
Please See John Fox's notes
http://socserv.socsci.mcmaster.ca/jfox/Misc/Rcmdr/installation-notes.html
and perhaps other parts of his website.
On Tue, May 29, 2012 at 10:06 AM, mabel alsina harris
wrote:
>
> Hi!!
> My question is related to Rcmdr. I have Mac OS and I use R since more or
> less 1
Hello, again.
See comments inline
Em 29-05-2012 16:28, Sabina Arndt escreveu:
Hello,
thank you very much for your reply, Rui Barradas.
OK, I did what you said:
x<- readLines("sabina.txt")
s<- strsplit(x, ";[[:space:]]\\[")
r<- lapply(s, function(x) sapply(strsplit(x, "[[:blank:]]"), tail, 1)
Hi Paul,
I haven't quite thought through this yet, but might it not be easier
to convert your formula to a character and then use gsub et al on it
directly?
Something like this
# Using m2 as you set up below
m2 <- lm(y ~ log(x1) + x2*x3, data=dat)
f2 <- formula(m2)
as.formula(paste(f2[2], f2[1
RSiteSearch("wavelet") at the R command line prompt, or
http://rseek.org
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Perhaps ?str is the command you seek?
-Original Message-
From: sagarnikam123
Sent: Saturday, May 26, 2012 10:24 PM
To: r-help@r-project.org
Subject: Re: [R] how to check given number seq. is time series or not?
Yes,sir index of above/below numbers is time (both dataset are same)
Michael:
m2 is a model fit, not a formula. So I don't think what you suggested will work.
However, I think your idea is a good one. The trick is to protect the
model specification from evaluation via quote(). e.g.
> z <- deparse(quote(lm(y~x1)))
> z
[1] "lm(y ~ x1)"
Then you can apply your sug
See
http://cran.r-project.org/mirror-howto.html
Uwe Ligges
On 29.05.2012 18:08, Jose Bustos Melo wrote:
Hello everyone,
I would like to create a new R mirror en Chile and I would be very placed if
any of you who are familiar doing it can help us.
We are looking for manuals and materials need i
On Tue, May 29, 2012 at 11:47 AM, Agustin Lobo
wrote:
> I'm trying to get a minimally intuitive way of plotting zoo objects
> extracted for given periods of time
>
> I do:
>> plot(alyL32007z$NEE_st,col=cod,type="b",pch=18,xlim=time(alyL32007z)[c(3000,15000)])
>
> and get a correct plot but the ex
I should have added:
If the formula is just assigned to a name, quote() and
eval(parse(...)) are not needed:
fm1 <- y ~ x1 ## a formula
w <- gsub( "x1","log(x1)", deparse(fm1))
fm2 <- formula(w)
This is probably the btter way to do it.
-- Bert
On Tue, May 29, 2012 at 10:01 AM, Bert Gunter
On Tue, May 29, 2012 at 11:43 AM, Paul Johnson wrote:
> Greetings
>
> I want to take a fitted regression and replace all uses of a variable
> in a formula. For example, I'd like to take
>
> m1 <- lm(y ~ x1, data=dat)
>
> and replace x1 with something else, say x1c, so the formula would become
>
>
>>
>
> Try substitute:
>
>> do.call("substitute", list(newFmla, setNames(list(as.name("x1c")), "x1")))
> y ~ log(x1c) + x2 * x3
>
Damn. That's pretty. I'd say "setNames" a magic bullet too.
Thanks very much.
The approach suggested by Michael and Bert has the little shortcoming
that grepping for "
On Tue, May 29, 2012 at 1:39 PM, Agustin Lobo
wrote:
> Thanks!
> Using as.POSIXct() instead of as.Date() solves the problem:
>> plot(alyL32007z$NEE_st,col=cod,type="b",pch=18,xlim=as.POSIXct(c("2007-03-04
>> 11:30:00 UTC" ,"2007-11-09 11:30:00 UTC"))) #works fine
>> delme <- window(alyL32007z, st
Deparse... that's it -- was disappointed with having to turn
as.character.formula inside out once and again. Merci!
But, as always, we all loose to Gabor ;-)
Michael
On Tue, May 29, 2012 at 1:16 PM, Bert Gunter wrote:
> I should have added:
>
> If the formula is just assigned to a name, quote()
Hi Sarah,
I can not for the life of me understand what was wrong with my post?!? Can
you please explain, so that i will not make the same mistake again?
Regards, m
--
View this message in context:
http://r.789695.n4.nabble.com/importing-multiple-file-form-folder-tp4631637p4631734.html
Sent fr
Hi Rui,
Thanks a lot, it works like I wanted. I am sorry about my late response
though...
Is it possible to get the name of the file (Channe 1. ) as a name of the
column in the resulting table (dataframe)?
Many thanks,
m
--
View this message in context:
http://r.789695.n4.nabble.com/im
Lattice experts:
Can you think of a way to produce a levelplot as below and then add a histogram
of the z variable to the top margin of the plot that would sit on top of the
color key?
x <- seq(pi/4, 5 * pi, length.out = 100)
y <- seq(pi/4, 5 * pi, length.out = 100)
r <- as.vector(sqrt(outer(
Hi again Rui,
i got the column names settled like this :
>colnames(Temp) <- paste(flist, seq_len(ncol(Temp)), sep=".")
Thanks again, m
--
View this message in context:
http://r.789695.n4.nabble.com/importing-multiple-file-form-folder-tp4631637p4631737.html
Sent from the R help mailing list a
Paul et. al:
I think Gabor's incantation qualifies as my desired alternative to
eval(parse())). It is, unfortunately, rather tricky, imo.
However, the objection you raise to the gsub(deparse()) solution is
easily overcome through the use of an appopriate regex: e.g.:
gsub("\\","log(x)","x+xc+cx
Le mardi 29 mai 2012 à 10:03 +0200, Ad Feelders a écrit :
> Dear fellow R users,
>
> I'm using the package tm for text mining, and have a problem with
> reading in a corpus from XML files.
> When I copy the example from "Introduction to the tm package" of the
> small reuters subset "crude", ever
Hello,
I am having a problem making use of some data outputted from an
instrument in a somewhat weird format. The instrument outputs two
columns - one called JulianDay.Hour and one called Minutes.Seconds. I
would like to convert these columns into a single column with a time.
So I was using substr
Thanks!
Using as.POSIXct() instead of as.Date() solves the problem:
> plot(alyL32007z$NEE_st,col=cod,type="b",pch=18,xlim=as.POSIXct(c("2007-03-04
> 11:30:00 UTC" ,"2007-11-09 11:30:00 UTC"))) #works fine
> delme <- window(alyL32007z, start =
> as.POSIXct("2007-03-04",tz="UTC"),end=as.POSIXct("20
This is my current version of R:
R version 2.15.0 (2012-03-30)
Copyright (C) 2012 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-pc-mingw32/x64 (64-bit)
OS: Windows
I get the following:
> memory.size()
[1] 23.04
> memory.limit()
[1] 190600
>
I also have the foll
Given this example
#start code
a<-c(0,70,50,100,70,650,1300,6900,1780,4930,1120,700,190,940,
760,100,300,36270,5610,249680,1760,4040,164890,17230,75140,1870,22380,5890,2430)
b<-c(0,0,10,30,50,440,1000,140,70,90,60,60,20,90,180,30,90,
3220,490,20790,290,740,5350,940,3910,0,640,850,260)
Dear Prof. Eduardo,
Try this:
rot<-expression(paste("DGL growth (cm.yr"^"-1",")"))
boxplot(five$gr13~five$Code, xlab="Species code", ylab=rot)
In this page there are some useful mathematical annotations in R:
http://stat.ethz.ch/R-manual/R-patched/library/grDevices/html/plotmath.html
Best wishe
Paul Johnson writes:
> Greetings
>
> I want to take a fitted regression and replace all uses of a variable
> in a formula. For example, I'd like to take
>
> m1 <- lm(y ~ x1, data=dat)
>
> and replace x1 with something else, say x1c, so the formula would become
>
> m1 <- lm(y ~ x1c, data=dat)
So
Apologies. I was searching using the wrong search terms. This is
clearly a string issue. I've added the solution below.
Sam
On Tue, May 29, 2012 at 11:39 AM, Sam Albers wrote:
> Hello,
>
> I am having a problem making use of some data outputted from an
> instrument in a somewhat weird format. Th
?plot.trellis.
In general something like
mlp<- levelplot(...
mhist<- histogram(...
plot(mlp,split=c(1,2,1,2),more=T)
plot(mhist,split=c(1,1,1,2),more=F)
You will need to do some work on the padding and layout widths to get the
distances right (I assume the key is to be the "x-axis" of the histogr
Hello,
I noticed something odd when working with data frames and xts objects.
If I read in a CSV file, R creates a nice data.frame. This works well.
If I then convert to an XTS object, I see that all the values in the data are
now quoted. My data is a mix of numeric and character. This is us
On Tue, May 29, 2012 at 2:11 PM, Noah Silverman wrote:
> Hello,
>
> I noticed something odd when working with data frames and xts objects.
>
> If I read in a CSV file, R creates a nice data.frame. This works well.
>
> If I then convert to an XTS object, I see that all the values in the data are
Hi Kelly,
The function has a limitation that it cannot handle any column in your "x" that
is a categorical variable with more than 32 categories. One possibility is to
see if you can "bin" some of the categories into one to get below 32 categories.
Andy
-Original Message-
From: r-hel
Only convert numeric columns to time series. This is a fundamental limitation
of ts, xts, and zoo.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go.
Just one more thing ...
my colnames are as follows :
c("channel 1 20120509 153744 1.ddf", "channel 1 20120509 154744
1.ddf",
"channel 1 20120509 155744 1.ddf", "channel 1 20120509 160744
1.ddf",
"channel 1 20120509 161744 1.ddf", "channel 1 20120509 162744
1.ddf",...)
H
This is a well-known limitation. You have to group categorical attributes
together to work around.
--
Weifeng (aaron) liu | retail systems pricing | sr research scientist
-Original Message-
From: r-h
- Forwarded Message -
From: arun
To: tarun trivedi
Cc:
Sent: Tuesday, May 29, 2012 12:24 PM
Subject: Re: [R] Help needed in wavelet transforming
Hi Tarun,
Have you looked into LS2w?
http://rgm2.lab.nig.ac.jp/RGM2/func.php?rd_id=LS2W:imwd
Hope this helps.
A.K.
- Original M
Hi Antony,
Try this,
I changed matrix to data frame to make it work and change it back again to
matrix. There might be a easy way to do this.
GetCsv<-as.matrix(data.frame("DWATT"=c(10,20,30,30,32)))
GetCsv<-data.frame(GetCsv)
GetCsv[GetCsv$DWATT<30 | GetCsv$DWATT >31, "DWAT_QF"] <-"RC"
Get
Hi,
I am using the glm.nb" model type to fit some count data with TWO offset
variables.
I have successfully used this approach to build scores of models for
several datasets but I am having problems with two in particular.
Depending on the model I try to fit for a given dataset, I obtain th
HI!
I am working with a table with quantitative and qualitative data; one of
the columns is "Level of education", which is an ordered factor. I want to
get rid of the ordered factor, and set my own scale. How to do that? All
the time I get the following error:
In `[<-.factor`(`*tmp*`, educa = "Le
HI!!!
I have a table containing qualitative and quantitative data; one of the
columns contains "Level of education", and the possibilities are "none",
"High school", "college"; I want to give the value 0 to "none", the value 1
to "High school", and 2 to "college", but I got the following error:
Hello R lovers,
For my GUI, I am using gslider of Gwidgets. I need to add the "to=" value
dynamically ie., maximum value from a data frame column(something like
to=max(df[,3])). But I am getting error message like
Error in checkPtrType(object, "GtkWidget") :
object of class NULL isn't a GtkW
or use quote()...
-- Bert
On Tue, May 29, 2012 at 10:48 AM, wrote:
> Paul Johnson writes:
>
>> do.call( substitute, list( frm, list( x = as.name("z") ) ) )
## or using quote()
>>do.call( substitute, list( frm, list( x = quote(z
> y ~ log(z) * (w + u)
>
>
> HTH,
>
> Chuck
>
>
>>
>> I h
Thanks Joshua,
Appreciate it!
--
Noah Silverman
UCLA Department of Statistics
8208 Math Sciences Building
Los Angeles, CA 90095
On May 29, 2012, at 12:17 PM, Joshua Ulrich wrote:
> On Tue, May 29, 2012 at 2:11 PM, Noah Silverman
> wrote:
>> Hello,
>>
>> I noticed something odd when working w
Read this:
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
Read the posting guide: http://www.R-project.org/posting-guide.html
and try again.
It's entirely impossible to help you as is.
Michael
On Tue, May 29, 2012 at 2:01 PM, RBB wrote:
> HI!!!
>
> I h
Hello,
thank you very much for your reply, Rui Barradas.
> In my workspace everything was converted either to non-empty strings or
> NULLs.
> This is how to do it.
>
>
> r1 <- lapply(r, function(x) x[nchar(x) > 0])
> r1 <- lapply(r1, function(x) if(length(x)) x else NULL) # second pass
> cou
Please don't double post. (And see my reply to your other thread)
Michael
On Tue, May 29, 2012 at 1:50 PM, Raúl Bajo wrote:
> HI!
>
> I am working with a table with quantitative and qualitative data; one of
> the columns is "Level of education", which is an ordered factor. I want to
> get rid of
karthicklakshman gmail.com> writes:
>
> Hello R lovers,
>
> For my GUI, I am using gslider of Gwidgets. I need to add the "to=" value
> dynamically ie., maximum value from a data frame column(something like
> to=max(df[,3])). But I am getting error message like
>
You update the items to sel
As long as you can remember that the summaries such as variable importance, OOB
predictions, and OOB error rates are not applicable, I think that should be
fine.
Andy
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Nikita Desai
...and
On Tue, May 29, 2012 at 1:04 PM, R. Michael Weylandt
wrote:
> Read this:
> http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
>
> Read the posting guide: http://www.R-project.org/posting-guide.html
>
> and try again.
>
> It's entirely impossible to help
Here is how to get the number:
> x <- c("channel 1 20120509 153744 1.ddf", "channel 1 20120509 154744
> 1.ddf",
+ "channel 1 20120509 155744 1.ddf", "channel 1 20120509 160744 1.ddf",
+ "channel 1 20120509 161744 1.ddf", "channel 1 20120509 162744 1.ddf")
> # get the numbe
Dear R users,
I am trying to use xyplot to draw group mean and CI. The following is the
sample code. But I want:
1. Use different colors and symbols to draw individual points, CI and
the lines connect group means from different time points;
2. Add jitters to x axis to allow CIs not be ov
What is the result after each step? Could you use dput to post them?
dput(head(r))
dput(head(r1)) # after the first lapply
Copy the output of those instructions and paste them here.
I'm asking this because I've tried with your dataset and it worked.
Rui Barradas
Em 29-05-2012 21:06, Sabina A
I am showing two alternatives, both from the HH package.
install.packages("HH")
The first one does essentially what you request,
smithdata <- src[c("x", "trt")]
smithdata$timefactor <- factor(src$time)
position(smithdata$timefactor) <- as.numeric(levels(smithdata$timefactor))
position(smithdata$t
is there a statement of work for the summer of code work.
On May 29, 2012 12:19 PM, "Joshua Ulrich" wrote:
> On Tue, May 29, 2012 at 2:11 PM, Noah Silverman
> wrote:
> > Hello,
> >
> > I noticed something odd when working with data frames and xts objects.
> >
> > If I read in a CSV file, R creat
1 - 100 of 121 matches
Mail list logo
