You need to parse it before you evaluate it.
eval(parse(text = avar))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jeremiah H. Savage
Sent: Wednesday, 30 June 2010 3:45 PM
To: r-help@r-project.org
Subject: [R] evaluate a string
Hello,
I was wondering how to evaluate a string variable in R.
eg.
> avar <- "getwd()"
> eval(avar)
gives:
[1] "getwd()"
but I would like to see:
[1] "/home/myhomedir"
Thanks,
Jeremiah
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mitools is useful too, and I can vouch for mice. mice is easy to use,
and easy to write new imputation methods too. So it is also very flexible.
Simon.
On 30/06/10 15:31, Jeremy Miles wrote:
Hi Daniel
First, newer versions of SPSS have dramatically improved their ability
to do stuff with miss
Hi Daniel
First, newer versions of SPSS have dramatically improved their ability
to do stuff with missing data - I believe it's an additional module,
and in SPSS-world, each additional module = $$$.
Analyzing missing data is a 3 step process. First, you impute,
creating multiple datasets, then y
Hi,
I am a long time SPSS user but new to R, so please bear with me if my
questions seem to be too basic for you guys.
I am trying to figure out how to analyze survey data using logistic
regression with multiple imputation.
I have a survey data of about 200,000 cases and I am trying to predict t
On Tue, Jun 29, 2010 at 5:58 PM, Jonathan Greenberg
wrote:
> Folks:
>
> I have two sets of dates, and one set of data:
>
> ***
>
> require("chron")
> require("zoo")
> reference_dates=seq.dates("01/01/92", "12/31/92", by = "months")
> data_dates=seq.dates("01/15/91", "12/15/93", by = "months")
> da
On Tue, 29 Jun 2010, Jonathan Greenberg wrote:
Folks:
I have two sets of dates, and one set of data:
***
require("chron")
require("zoo")
reference_dates=seq.dates("01/01/92", "12/31/92", by = "months")
data_dates=seq.dates("01/15/91", "12/15/93", by = "months")
data=1:length(data_dates)
refe
On Jun 29, 2010, at 7:17 PM, Jonathan Greenberg wrote:
> Pardon the barrage of time series related questions, but another issue
> I'm trying to solve is how to determine a sequence of dates a la
> seq.dates() except going BACKWARDS in time, e.g. if seq.dates()
> allowed for the "to" variables to b
Dear All:
I apologize for asking such an elementary question, but I could not find an
adequate response on line. I am hoping to receive some help with the
interpretation of the Intercept coefficient in the gam model below.
I1 through I3 are dummy coded "Item difficulty" parameters in a data s
Owo! Many thanks to you all. A lot of answers in just a few minutes!! I'd
like to thank you Dennis, Joshua, David, Erik and Bill.
I'll try these solutions and read something about xyplot.
Best regards to you.
On Tue, Jun 29, 2010 at 10:54 PM, Dennis Murphy wrote:
> Hi:
>
> This is to add a co
You can do this in reshape package as mentioned earlier.
However, if you need a solution with aggregate here it is
a <- with(data, aggregate(cbind(v1,v2), by=list(x,y,z),sum))
names (a) <- c("x","y","z","v1","v2")
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Caro
Hi:
This is to add a couple of bells and whistles to the excellent replies of
David and Bill. You may not want the 0.5 increments in the x-axis ticks, and
if you want all groups in a single plot, then you may want a legend. Below
are some basic ways to specify these (to head off the obvious follow
Pardon the barrage of time series related questions, but another issue
I'm trying to solve is how to determine a sequence of dates a la
seq.dates() except going BACKWARDS in time, e.g. if seq.dates()
allowed for the "to" variables to be set alone, rather than the from=.
Ultimately, I'd like to hav
Thank you very much for response. Finally I took David's way. Others' work
well for this specific case. But I find problem is there are more than one
column of character variables.
z=c('ab','ah','bc','ah','dv')
x=substr(z,start=1,stop=1)
y=substr(z,start=2,stop=2)
v1=5:9
v2=7:11
data=data.frame(x,
Great help. It works when the first and the second columns are ordered the
same way. But aggregate does not work for the following case:
z=c('ab','ah','bc','ah','dv')
x=substr(z,start=1,stop=1)
y=substr(z,start=2,stop=2)
v1=5:9
v2=7:11
data=data.frame(x,y,z,v1,v2)
> data
x y z v1 v2
1 a b ab 5
Hello,
There are many ways to do what you want in R. I am showing one way
using one of my favorite graphics packages, ggplot2. If you do not
have it installed yet, uncomment the first line.
#install.packages("ggplot2")
library(ggplot2) #load the package
#Read in data
samp.dat <- structure(list
require(reshape)
cast(data, first+second~ ., sum)
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
nikhil.l...@gmail.com
On Jun 29, 2010, at 3:05 PM, Yi wrote:
first=c('u','b','e','k','j','c','u','f','c','e')
second
=
c
('usa
','Brazil
','England','Korea',
Hello Steve
Thanks for quick responses its really helping me out .Ya I made the
necessary changes you had mentioned. I was not sure of that 'type' argument
where u had told me to set it to SVM . Do you mean I have to give that
argument in this line "cl <- c(c(rep("ALL",10), rep("AML",10)));" a
On Jun 29, 2010, at 3:05 PM, Yi wrote:
Hi, folks,
I am sorry that I did not state the problem correctly yesterday.
Please let me address the problem by the following codes:
first=c('u','b','e','k','j','c','u','f','c','e')
second
=
c
('usa
','Brazil
','England','Korea','Japan','China','usa','
Monday, June 28, 2010, 4:40:11 PM, you wrote:
> On Mon, Jun 28, 2010 at 7:30 PM, wrote:
>> Hi everybody,
>>
>> I'm working on the very
>> messy data, I have tried to clean it up in SAS and
>> SAS/IML but there is not enough info on how to handle certain things
>> in SAS so I have turned to R. Th
Folks:
I have two sets of dates, and one set of data:
***
require("chron")
require("zoo")
reference_dates=seq.dates("01/01/92", "12/31/92", by = "months")
data_dates=seq.dates("01/15/91", "12/15/93", by = "months")
data=1:length(data_dates)
reference_zoo=zoo(order.by=reference_dates)
data_zoo=z
Hi:
If you can deal with alphabetic order, the following seems to work:
v <- aggregate(third ~ first, data = data, FUN = sum)
v$second <- levels(data$second)
v[, c(1, 3, 2)]
first second third
1 b Brazil 2
2 c China15
3 e England13
4 f France 8
5 j Jap
> Hello Steve
> Thanks for quick responses its really helping me out .Ya I made the
> necessary changes you had mentioned. I was not sure of that 'type' argument
> where u had told me to set it to SVM . Do you mean I have to give that
> argument in this line "cl <- c(c(rep("ALL",10), rep("AML",10
> require(lattice)
Loading required package: lattice
> xyplot(CASES ~ YEAR | AREA, data = t, type = "b")
> xyplot(CASES ~ YEAR, data = t, type = "b", groups = AREA)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Pablo Cerdeira
S
Pablo Cerdeira wrote:
Hi All,
this is my first mail here.
I'm trying to plot a multiline chart grouping values with no success. I have
read a lot in the official Wiki and also searched via Google, but I did not
find anything.
I know how to do it in Excel, using a Pivot Table. But I'm trying
On Jun 29, 2010, at 8:42 PM, Pablo Cerdeira wrote:
Hi All,
this is my first mail here.
I'm trying to plot a multiline chart grouping values with no
success. I have
read a lot in the official Wiki and also searched via Google, but I
did not
find anything.
I'm importing some data from a c
Hi All,
this is my first mail here.
I'm trying to plot a multiline chart grouping values with no success. I have
read a lot in the official Wiki and also searched via Google, but I did not
find anything.
I'm importing some data from a cvs file. Here is a sample:
YEAR,AREA,CASES
1988,CONTRACTS,2
Thank you for the fast replies.
I've set the PATH env var to include the PostgreSQL bin diectory and
it's working fine.
On 29-06-2010 23:44, Joe Conway wrote:
On 06/29/2010 03:35 PM, João Gonçalves wrote:
Error: package 'RPostgreSQL' could not be loaded
exists which makes RPost
Hi
On 6/30/2010 2:17 AM, Jinsong Zhao wrote:
Hi there,
I am a Chinese R user. I hope to display Chinese character in a plot,
and than save it in PostScript format. I have read the article titled
"Non-Standard Fonts in PostScript and PDF Graphics", especially the
section about CJK fonts. I also
Dear list,
I wanna select the significant variables relative to bird distribution,
using stepwise method.
However, the result is always the best-fit model.
Please kindly suggest if it is possible to show the selection process.
Thank you
Elaine
[[alternative HTML version deleted]]
_
On Tue, Jun 29, 2010 at 11:35:44PM +0100, João Gonçalves wrote:
> Dear list users,
>
> The problem occurs when library(RPostgreSQL) is issued on R. This issue
> has previously appeared on R mailing list without any robust solution.
> The error message issued by R:
>
> Loading required package:
On 06/29/2010 03:35 PM, João Gonçalves wrote:
> Error: package 'RPostgreSQL' could not be loaded
> exists which makes RPostgreSQL loading to fail. The message appears for
> any of the following DLLs (that actually exist on
> X:/PostgreSQL_installation_directory/bin):
> To "solve" this problem the
Dear list users,
The problem occurs when library(RPostgreSQL) is issued on R. This issue
has previously appeared on R mailing list without any robust solution.
The error message issued by R:
Loading required package: RPostgreSQL
Loading required package: DBI
Error in inDL(x, as.logical(local)
Use the merge function, look at the by.x and by.y arguments, also look at the
all.x and all.y arguments as well as the suffixes argument. You may need to
delete some columns after the merge (or replace missing values in one column
with those in the same location from the next column, see the if
use 'merge'
> a.df
DATE GENDER PATIENT_ID AGE SYNDROME
1 4/16/2009 F 23686 45 RASH ON BODY
2 4/16/2009 F 13840 35 CANT URINATE
3 4/16/2009 M 12895 30 BLURRED VISION
4 4/16/2009 M 18375 33 UNABLE TO VOID
On Fri, 25-Jun-2010 at 10:09AM +0530, suman dhara wrote:
|> Sir,
|> I want to apply mclapply() function for my analysis. So, I have to install
|> multicore package. But I can not install the package.
|>
|> >install.packages("multicore")
|> It gives that package multicore is not available.
|>
|>
Hi Steve
I had done one more thing I had taken the transpose of both my test and
train files as given below:
model<- svm(t(train),cl);
pred <- predict(model,t(test));
And the result I had got is :
Result:
pred ALL AML
ALL 10 0
AML 0 10
why is there a difference in th
Hello Steve
Thanks for quick responses its really helping me out .Ya I made the
necessary changes you had mentioned. I was not sure of that 'type' argument
where u had told me to set it to SVM . Do you mean I have to give that
argument in this line "cl <- c(c(rep("ALL",10), rep("AML",10)));" a
Hi, folks,
I am sorry that I did not state the problem correctly yesterday.
Please let me address the problem by the following codes:
first=c('u','b','e','k','j','c','u','f','c','e')
second=c('usa','Brazil','England','Korea','Japan','China','usa','France','China','England')
third=1:10
data=data.
Erin,
?merge
Try
c.df=merge(a.df,b.df,by.x="PATIENT_ID",by.y="ID")
hope it helps
Weidong
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
a
On 29.06.2010 19:19, Bert Gunter wrote:
Uwe:
Did you forget to add the "fixed = TRUE" parameter to your gsub call in your
reply?
Yes, thanks.
Uwe
gsub("N\\A", "NA", "N\\A")
[1] "N\\A"
gsub("N\\A","NA","N\\A",fixed=TRUE)
[1] "NA"
I only mention it because there is already sufficie
Thank you very much for a reference to "reshape".
I found that melt(MyData) - and then resorting - gives me exactly what I want!
Dimitri
On Tue, Jun 29, 2010 at 1:46 PM, Henrique Dallazuanna wrote:
> Try this:
>
> reshape(MyData, direction = 'long', varying = list(c('jan', 'feb')), idvar =
> 2:3
On Jun 29, 2010, at 2:54 PM, Stuart Luppescu wrote:
Hello, I have to construct 5 new columns in a data frame depending on
the value of another of the columns in the data frame. The only way I
could figure out to do this was to subset the data frame five times,
do
the variable construction, a
On Tue, Jun 29, 2010 at 6:22 AM, Simon Kiss wrote:
> Dear colleagues, particularly academic ones,
> So I'm creating a Microsoft Word template for myself so that every time I
> teach a new course, I don't have to enter in the dates manually for each
> class session.
> I'd like to use an R script th
On Tue, 2010-06-29 at 13:59 -0500, Erik Iverson wrote:
> I'm sure there's an easier way, but it's going to be easiest to get a
> useful response if we have a reproducible, minimal example, as the
> posting guide requests. ?tapply is probably involved.
A minimal example of data? How about this:
Dear R People:
I have two data frames, a.df and b.df as seen here:
> a.df[1:10,]
DATE GENDER PATIENT_ID AGE SYNDROME
1 4/16/2009 F 23686 45 RASH ON BODY
2 4/16/2009 F 13840 35 CANT URINATE
3 4/16/2009 M 12895 30 BLURRED
It is hard to know exactly what you want without a description of your data or
what you want the final plot to look like. But you can do the equivalent of
plot followed by multiple calls to points by using a loop, apply functions, the
lattice package or the ggplots2 package (I'm guessing the la
A fairly simple way is to generate one series with all the Tuesdays, then
another with all the Thursdays, combine and sort.
> sort( c( seq.Date( as.Date('2010-6-29'), by='week', length.out=10),
+ seq.Date( as.Date('2010-7-1'), by='week', length.out=10) )
+ )
[1] "2010-06-29" "2010-07-01" "2010-
Also the paste solution can be abbreviated to just:
paste(as.Date(x), x - floor(x))
On Tue, Jun 29, 2010 at 2:57 PM, Gabor Grothendieck
wrote:
> The time zone independent solution, i.e. paste(as.Date(x), format(x -
> floor(x))),
> is the safer although format(as.POSIXlt(x, tz = "GMT")) seems
I'm sure there's an easier way, but it's going to be easiest to get a
useful response if we have a reproducible, minimal example, as the
posting guide requests. ?tapply is probably involved.
Stuart Luppescu wrote:
Hello, I have to construct 5 new columns in a data frame depending on
the value
The time zone independent solution, i.e. paste(as.Date(x), format(x -
floor(x))),
is the safer although format(as.POSIXlt(x, tz = "GMT")) seems to work too.
On Tue, Jun 29, 2010 at 2:44 PM, stephen sefick wrote:
> Thank you both! If I don't want to deal with a Time Zone potentailly
> converting
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Bert Gunter
> Sent: Tuesday, June 29, 2010 11:08 AM
> To: 'Jason Rupert'; 'Duncan Murdoch'
> Cc: r-help@r-project.org
> Subject: Re: [R] gsub issue in R 2.11.1, but not present in
Hello, I have to construct 5 new columns in a data frame depending on
the value of another of the columns in the data frame. The only way I
could figure out to do this was to subset the data frame five times, do
the variable construction, and then rbind the subsets back together.
Here's part of the
Thank you both! If I don't want to deal with a Time Zone potentailly
converting some of the dates, which would be your suggestions. Or,
are they all the same way to skin a cat. Again thank you for your
wonderful help.
kindest regards,
Stephen Sefick
On Tue, Jun 29, 2010 at 1:39 PM, Gabor Grot
On Tue, Jun 29, 2010 at 2:22 PM, Gabor Grothendieck
wrote:
> On Tue, Jun 29, 2010 at 2:01 PM, stephen sefick wrote:
>> the date were created with chron with this argument
>>
>> format=c(dates="Y/m/d", times="H:M:S"))
>>
>> so I have the dates being displayed as
>>
>> (10/06/22 12:00:00)
>>
>> I w
HI, Dear community,
How to distinguish bi-mode distribution from mono-mode distribution? I have
only the histograms of 3500 data set.
Thanks!
--
Sincerely,
Changbin
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On Tue, Jun 29, 2010 at 2:01 PM, stephen sefick wrote:
> the date were created with chron with this argument
>
> format=c(dates="Y/m/d", times="H:M:S"))
>
> so I have the dates being displayed as
>
> (10/06/22 12:00:00)
>
> I would like to have them displayed as
>
> "2010-06-22 12:00:00" or "%Y-%m
Hi Phani,
something like this looks promising:
#
library(forecast)
library(Mcomp)
MAPE.for.Holt <- function (x,series,bignum=10e6) {
foo <-
try(ets(series$x,model="AAN",damped=FALSE,alpha=x[1],beta=x[2],restrict=FALSE),silent=TRUE)
if ( class(f
Phil,
Thank you very much for your swift reply. I have not used POSIXct
much, but I do remember having issues with time zones in POSIXx.
Should I be aware of this, or is this not a problem. I have set the
data recorders to stay one time and one time zone and never change, so
that I don't have to
Stephen -
format(as.POSIXct(x),"%Y-%m-%d %H:%M:%S")
[1] "2009-08-07 17:00:00" "2009-08-07 17:15:00" "2009-08-07 17:29:59"
[4] "2009-08-07 17:45:00" "2009-08-07 18:00:00"
- Phil Spector
Statistical Computing Facil
Jason:
I think it's actually even a bit worse than what Duncan said, which was:
---
"You need to double the backslashes to enter them in an R string. So
gsub("N\\A", "NA", original, fixed=TRUE)
should work if original contains a single backslash, and
gsub("NA", "NA", original, fix
the date were created with chron with this argument
format=c(dates="Y/m/d", times="H:M:S"))
so I have the dates being displayed as
(10/06/22 12:00:00)
I would like to have them displayed as
"2010-06-22 12:00:00" or "%Y-%m-%d %H:%M:%S"
and then I can convert these for mergeing with another dat
dbs_influential_obs <- which(apply(fit.influential.observations$is.inf, 1,
any))
dbs_sans_influential_obs <- dbs1[-dbs_influential_obs,]
--
View this message in context:
http://r.789695.n4.nabble.com/Remove-observations-deemed-influential-by-influential-measure-tp2272474p2272524.html
Sent from
Try this:
reshape(MyData, direction = 'long', varying = list(c('jan', 'feb')), idvar =
2:3)
On Tue, Jun 29, 2010 at 2:22 PM, Dimitri Liakhovitski <
dimitri.liakhovit...@gmail.com> wrote:
> Hello, everyone!
> I have a very simple task - I have a data frame (see MyData below) and
> I need to stack
Hi,
I have a correlation matrix of 1000 x 1000 genes. I have grouped (20 groups)
genes based on function and each gene has a index. I would like to reorder the
correlation matrix based on the group index. Could you please suggest me how to
do that?
Thanks in advance.
Kind regards,
Ezhil
Dear R-users,
I have created a map with plot location using longitude/latitude coordinates
with the PlotOnStaticMap() function of the RgoogleMaps package. Everything
works fine until I try to put a polygon on the map. The polygon() function
doesnt work and I need to use the special function Plo
On Jun 29, 2010, at 1:10 PM, GL wrote:
dbs is an existing dataframe. I fit a lm and looked at influential
observations. I want now to delete the influential observations from
dbs,
fit another lm, and see how different the results are. What is the
syntax to
remove the influential observati
On Tue, Jun 29, 2010 at 12:22 PM, Dimitri Liakhovitski
wrote:
> Hello, everyone!
> I have a very simple task - I have a data frame (see MyData below) and
> I need to stack the data (see result below).
> I wrote the syntax below - it's very basic and it does what I need.
> But I am sure what I am t
Hello, everyone!
I have a very simple task - I have a data frame (see MyData below) and
I need to stack the data (see result below).
I wrote the syntax below - it's very basic and it does what I need.
But I am sure what I am trying to do is a very typical task and there
must be a much shorter/more
Uwe:
Did you forget to add the "fixed = TRUE" parameter to your gsub call in your
reply?
> gsub("N\\A", "NA", "N\\A")
[1] "N\\A"
> gsub("N\\A","NA","N\\A",fixed=TRUE)
[1] "NA"
I only mention it because there is already sufficient confusion that the
typo may totally bewilder people.
-- Bert
B
dbs is an existing dataframe. I fit a lm and looked at influential
observations. I want now to delete the influential observations from dbs,
fit another lm, and see how different the results are. What is the syntax to
remove the influential observations from dbs?
fit <- lm(NI ~ PG + log(TG), data
I am not entirely sure, but the actual amount of RAM is less than the
GB specified by the specs. Are you directing R to a max memory
size... If so it is too large.
On Tue, Jun 29, 2010 at 11:32 AM, Bogaso wrote:
>
> When I use this I am getting following warning at the time of opening R from
>
When I use this I am getting following warning at the time of opening R from
that shortcut :
"-max-mem-size=2048MB:too large and taken as 2047M"
Why I am getting this? I have 3GB ram installed and using within Vista.
Thanks
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On 29/06/2010 17:36, Marc Schwartz wrote:
On Jun 29, 2010, at 10:08 AM, Ottorino-Luca Pantani wrote:
paste("$\\pm$", 1.34, sep="")
[1] "$\\pm$1.34"
I believe you then need to tweak the sanitize.text.function argument in
print.xtable() to properly handle the backsl
> Date: Wed, 16 Jun 2010 17:53:56 +0200
> From: "Adolf STIPS"
>
> Hi,
>
> I'm using the "ncdf" library for creating ncdf files.
> But I need to create about 100 variables per file (e.g. single rivers),
> So I do not like to create each variable separately.
>
> Unfortunately I found no way to make
On Jun 29, 2010, at 10:08 AM, Ottorino-Luca Pantani wrote:
> Dear R-users,
> please consider the following minimal example:
>
> \documentclass[a4paper,titlepage,onecolumn,12pt]{article}
> \usepackage[italian]{babel}
> \usepackage{amssymb}
> \usepackage[utf8x]{inputenc}
> \usepackage[pdftex]{graph
Thanks
>
>
>
> On 29/06/2010 6:55 AM, cgenolin wrote:
>> Thanks, Duncan.
>> And where am I suppose to put the file toto.pdf ?
>> In "/myPack/doc/" or in "myPack/inst/doc/"
>>
>
>
> The latter. When your package is installed, it will be installed to
> myPack/doc.
>
> Duncan Murdoch
>
> ___
Dear R-users,
please consider the following minimal example:
\documentclass[a4paper,titlepage,onecolumn,12pt]{article}
\usepackage[italian]{babel}
\usepackage{amssymb}
\usepackage[utf8x]{inputenc}
\usepackage[pdftex]{graphicx}
\begin{document}
<>=
df.data1 <-
cbind.data.frame(A = rnorm(18),
Hi,
On Tue, Jun 29, 2010 at 7:16 AM, Aadhithya wrote:
>
> Following is the error I am getting:
> Error in svm.default(train, cl) :
> Need numeric dependent variable for regression.
Here's a problem that you may not even know you had yet.
By the looks of your code, it seems as if you want to do
If the vector elements are (still) strings, you could simply try
x<- c("2000-01-01", "2000-01-23", "2001-03-12", "2009-12-31")
substring(x, 1, 7)
# [1] "2000-01" "2000-01" "2001-03" "2009-12"
Hope this helps a little.
Allan
On 29/06/10 14:36, Thomas Jensen wrote:
Dear R Experts,
I have a
On Jun 29, 2010, at 8:36 AM, Thomas Jensen wrote:
> Dear R Experts,
>
> I have a vector of dates in character format like this:
>
> date
> "2000-01-01"
> "2000-01-23"
> "2001-03-12"
> ...
> ...
> ...
> "2009-12-31"
>
> I would like to delete the last part of the character string (i.e. the "day"
Dear R Experts,
I have a vector of dates in character format like this:
date
"2000-01-01"
"2000-01-23"
"2001-03-12"
...
...
...
"2009-12-31"
I would like to delete the last part of the character string (i.e. the
"day" part), so the vector looks like this:
date
"2000-01"
"2000-01"
"2001-03"
Dear List,
I am a little unsure how to structure my model and was after some advice. I am
a little unsure if this question is appropriate for this list, if it is not
please just delete and accept my apologise.
I have 10 factors that are categorical variables and 5 levels of response
variables
Dear List
I am a little unsure how to structure my model and was after some advice. I am
a little unsure if this question is appropriate for this list, if it is not
please just delete and accept my apologise.
I have 10 factors that are categorical variables and 5 levels of response
variables
Hi there,
I am a Chinese R user. I hope to display Chinese character in a plot,
and than save it in PostScript format. I have read the article titled
"Non-Standard Fonts in PostScript and PDF Graphics", especially the
section about CJK fonts. I also tried the code:
> pdf("chinese.pdf", width=3, h
I suspect that it was Intel's marketing department, after a few beers at the
local bar...
;-)
Regards,
Marc
On Jun 29, 2010, at 9:09 AM, Joris Meys wrote:
> *slaps forehead*
> Thanks. So out it goes, that hyperthreading. Who invented
> hyperthreading on a quad-core anyway?
>
>
> Cheers
> Jo
*slaps forehead*
Thanks. So out it goes, that hyperthreading. Who invented
hyperthreading on a quad-core anyway?
Cheers
Joris
2010/6/29 Uwe Ligges :
>
>
> On 29.06.2010 15:30, Joris Meys wrote:
>>
>> Dear all,
>>
>> I've recently purchased a new 64bit system with an intel i7 quadcore
>> processo
On Tue, 2010-06-29 at 09:09 +, Ilona Leyer wrote:
> Dear all,
> In a greenhouse experiment we tested performance of 4 different species
> (B,H,P,R) under 3 different water levels in 10 replications. As response
> variable e.g. the number of emerging sprouts were measured on three dates. A
>
On 29.06.2010 15:30, Joris Meys wrote:
Dear all,
I've recently purchased a new 64bit system with an intel i7 quadcore
processor. As I understood (maybe wrongly) that to date the 32bit
version of R is more stable than the 64bit, I installed the 32bit
version and am happily using it ever since.
Dear all,
I've recently purchased a new 64bit system with an intel i7 quadcore
processor. As I understood (maybe wrongly) that to date the 32bit
version of R is more stable than the 64bit, I installed the 32bit
version and am happily using it ever since. Now I'm running a whole
lot of models, whic
IIona-
I think you may be misinterpreting the t-test.
In model 1, consider the speciesH coefficient. A test that speciesH = 0,
essentially asks: Is speciesH the same as speciesB? The test statistic for
this hypothesis is the t-value reported in the table. (t-value= -4.2,
p-value=0.0001)
In mo
On Tue, Jun 29, 2010 at 8:02 AM, Matthew Dowle wrote:
>
>> dt = data.table(d,key="grp1,grp2")
>> system.time(ans1 <- dt[ , list(mean(x),mean(y)) , by=list(grp1,grp2)])
> user system elapsed
> 3.89 0.00 3.91 # your 7.064 is 12.23 for me though, so this
> 3.9 should be faster for y
> dt = data.table(d,key="grp1,grp2")
> system.time(ans1 <- dt[ , list(mean(x),mean(y)) , by=list(grp1,grp2)])
user system elapsed
3.890.003.91# your 7.064 is 12.23 for me though, so this
3.9 should be faster for you
However, Rprof() shows that 3.9 is mostly dispatch of mea
[I replied to Gabor only, I think i may be interesting to cc the list, for
the record.]
-- Forwarded message --
From: bruno Piguet
Date: 2010/6/29
Subject: Re: [R] Fast and simple tool for re-sampling of asynchronous time
series ?
To: Gabor Grothendieck
2010/6/25 Gabor Grothend
On Jun 28, 2010, at 10:58 AM, Martin Spindler wrote:
Hello,
currently I am estimating an ordered probit model with the function
polr
(MASS package).
Is there a simple way to obtain values for the prediction of the index
function ($X*\hat{\beta}$)?
(E..g. in the GLM function there is th
thanks everyone.
I think the motto should be "always specify the levels of a factor when
you create it
if you possibly can".
best wishes
Robin
On 06/29/2010 12:39 PM, Felix Andrews wrote:
Just use factor(), not levels(); you can pass a factor to factor() too.
x<- factor(c(rep("a",3
When I replied to this message I just hit the reply button. I am
resending it using reply to all, in case it did not go to the list.
Dear Ilona,
Looking at the estimation results you have, I think your regression
equations are correctly specified. Just thinking aloud, I do not think
the results a
Just use factor(), not levels(); you can pass a factor to factor() too.
> x <- factor(c(rep("a",3),"b","d"), levels = letters[1:5])
> table(x)
x
a b c d e
3 1 0 1 0
Cheers,
-Felix
On 29 June 2010 20:59, Robin Hankin wrote:
> Hi
>
> suppose I have a factor 'x':
>
>> x <- as.factor(c(rep("a",3),
On 29.06.2010 12:47, Jason Rupert wrote:
Previously in R 2.9.2 I used the following to convert from an improperly
formatted NA string into one that is a bit more consistent.
gsub("N\A", "NA", "N\A", fixed=TRUE)
This worked in R 2.9.2, but now in R 2.11.1 it doesn't seem to work an throws
t
Following is the error I am getting:
Error in svm.default(train, cl) :
Need numeric dependent variable for regression.
My dataset looks like this in both training and testing:
ALL ALL ALL ALL ALL ALL ALL ALL ALL ALL
AML AML AML AML AML
1 - 100 of 116 matches
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