Hi
Your question is rather cryptic. Why the output shall be 3? What has
unique count to do with match function?
Maybe you want something what is described in switch help.
See
?switch
Regards
Petr
>
> Hi
>
> My code looks like this
>
> I have two parameters x and par1. X contains values an
Dear R Users
I am maximizing a likelihood function it has two two correlated random
effects which follows bivariate normal distribution. To get the marginal
distribution I want to integrate out with respect to these two correlated
random effects. Does any body know how can I implement gaussian q
I figured as much.
Thank you greatly.
plangfelder wrote
>
> On Wed, Apr 18, 2012 at 7:25 PM, chuck.01
> wrote:
>> Let say I have an object (I hope my terminology is correct) a
>> a <- 12
>>> a
>> [1] 12
>>
>> And "a" has been assigned the number 12, or whatever
>> And let
Hi
My code looks like this
I have two parameters x and par1. X contains values and par1 contains the
function which i required to use
if par1 is max then output should be max(x).
FUN <- match.fun(par1)
result=FUN(x)
Is it possible to incorporate the unique coun
No like I said, lapply gets columnwise results (if you stopped
there, it would tell you whether each country is in the EU separately)
and the Reduce() combines them. Re-read my explanation and play around
with it on your own data set.
To convert 1 -> True and 2 -> False, you could use Boolean
On Wed, Apr 18, 2012 at 9:52 PM, Michael wrote:
> Thanks! Speed is not a concern...
>
> Accuracy and stability is the concern.
>
> I am actually using the method to find all the roots of a big polynomial...
>
> Want to see whether it's accurate or not...
Well the authors may have written it so th
Thanks! Speed is not a concern...
Accuracy and stability is the concern.
I am actually using the method to find all the roots of a big polynomial...
Want to see whether it's accurate or not...
On Wed, Apr 18, 2012 at 11:46 PM, Joshua Wiley wrote:
> Hi Michael,
>
> There are lots of options, bu
Hi Michael,
There are lots of options, but here is a simple one:
x <- matrix(rnorm(3000*3000), 3000)
e <- eigen(x)
only takes a couple minutes on my machine for a 3000 x 3000 matrix.
Cheers,
Josh
On Wed, Apr 18, 2012 at 8:55 PM, Michael wrote:
> Say a matrix of size of thousands?
>
> I am lo
I am actually looking for a "generalized" eigen-value decomposition algo in
R which works on non-symmetrical matrix...
I plan to use it for solving high order polynomial equations...
Any thoughts? Thank you!
On Wed, Apr 18, 2012 at 10:55 PM, Michael wrote:
> Say a matrix of size of thousands?
Say a matrix of size of thousands?
I am looking for an eigen-value decomposition algo in R to give good
eigenvalues...
Is that a hopeful thing?
Thank you!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.
Thanks. That is clear.
I need to seperate the concept of Array and Vector in my head. As
soon as I hit sendon that last post I realised it is a bit silly to
say vectors only contain atomic things.
W
On Thu, Apr 19, 2012 at 12:35 PM, William Dunlap wrote:
>> -Original Message-
>> From:
Appears to be a definite bug, probably caused by having more than one
correlation parameter. I hope to have this fixed within 3 days.
Frank
Mark Seeto wrote
>
> Dear R-help,
>
> I don't understand why Gls gives me an error when trying to fit a
> model with AR(2) errors, while gls (from nlme) do
Perhaps the following?
get(call_A)
See ?get for more details.
HTH,
Jorge.-
On Wed, Apr 18, 2012 at 10:25 PM, chuck.01 <> wrote:
> Let say I have an object (I hope my terminology is correct) a
> a <- 12
> > a
> [1] 12
>
> And "a" has been assigned the number 12, or whatever
> And lets say I hav
On Wed, Apr 18, 2012 at 7:25 PM, chuck.01 wrote:
> Let say I have an object (I hope my terminology is correct) a
> a <- 12
>> a
> [1] 12
>
> And "a" has been assigned the number 12, or whatever
> And lets say I have a character "call_A"
> call_A <- "a"
>>call_A
> [1] "a"
>
> What is the function "
Let say I have an object (I hope my terminology is correct) a
a <- 12
> a
[1] 12
And "a" has been assigned the number 12, or whatever
And lets say I have a character "call_A"
call_A <- "a"
>call_A
[1] "a"
What is the function "F" that allows this to happen:
> F( call_A )
[1] 12
--
View this mes
not much to speak of.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Re
how much accuracy does it have when I have a polynomial of thousands of
orders?
Thank you!
On Wed, Apr 18, 2012 at 3:49 PM, Rolf Turner wrote:
> On 19/04/12 08:14, Michael wrote:
>
>> Is there a way to find all roots of a polynomial equation?
>>
>> Lets say
>>
>> x^5+a*x^4+b*x^3+c*x^2+d*x+e=0
Dear R-help,
I don't understand why Gls gives me an error when trying to fit a
model with AR(2) errors, while gls (from nlme) does not. For example:
library(nlme)
library(rms)
set.seed(1)
d <- data.frame(x = rnorm(50), y = rnorm(50))
gls(y ~ x, data=d, correlation = corARMA(p=2)) #This works
Gl
Dear All
Does any one know if the following survival analysis exists, and if so, is it
available in R?
I am trying to parametrically model survival where some events happen at time =
0. I am particularly interested in generating the 95% confidence interval
around the survival function over its
> -Original Message-
> From: Worik R [mailto:wor...@gmail.com]
> Sent: Wednesday, April 18, 2012 5:05 PM
> To: William Dunlap
> Cc: r-help
> Subject: Re: [R] Can a matrix have 'list' as rows/columns?
>
> [snip]
> >
> > sapply. In this case I would expect M to be a list. I am gobsmacked t
On Apr 18, 2012, at 7:52 PM, Dmitriy Lyubimov wrote:
Sorry. i forgot the lines with parameters for my multivariate normal
simulation. here's the code for data simulation i used:
===
sigm1 <- diag(c(30,50))
sigm2 <- diag(c(5,3))
mu1 <- c(10,15)
mu2 <- c(80,60)
sample <- rou
[snip]
>
> sapply. In this case I would expect M to be a list. I am gobsmacked that
> > a list can be considered a vector. Is that a bug? It must be bad design?
> >
> > I have been using R for a number of years (5?) and heavilly for two years.
> > I am still getting bitten by these "features"
Hello R Help,
I have a data set with 30 variables, some of these variables are factors
while some are numeric. My response variable is binary and I have made sure
that this is set as a factor with two levels and given each level a name
"Yes" and "No". I would like to run the randomPackage to disco
Sorry. i forgot the lines with parameters for my multivariate normal
simulation. here's the code for data simulation i used:
===
sigm1 <- diag(c(30,50))
sigm2 <- diag(c(5,3))
mu1 <- c(10,15)
mu2 <- c(80,60)
sample <- round(rbind(rmvnorm(100,mu1,sigm1),rmvnorm(100,mu2,sigm2)))
Thanks David.
let's say sample data (two-modal multivariate normal simulation, needs
mvtnorm package)
=
library(mvtnorm)
sample <- round(rbind(rmvnorm(100,mu1,sigm1),rmvnorm(100,mu2,sigm2)))
sample <- sample[! (sample[,1] <1 | sample[,2] <1 | sample[,1]> 100 |
sample[,2]>
>Is there a way to find all roots of a polynomial equation?
>
>Lets say
>
>x^5+a*x^4+b*x^3+c*x^2+d*x+e=0
>
>
>how to find its all roots?
>
>
The package rootSolve might proove interesting
__
R-help@ mailing list
https://stat.ethz.ch/mailman/listin
On Apr 18, 2012, at 6:55 PM, Dmitriy Lyubimov wrote:
Hello,
I'd be very grateful for help with some ggplot2's stat_density2d
issues.
First issue is with data limits. xlim() and ylim() doesn't seem to
work; instead, estimates (and plotting) seems to be constrained to
range(x), range(y) no m
Hello,
I'd be very grateful for help with some ggplot2's stat_density2d issues.
First issue is with data limits. xlim() and ylim() doesn't seem to
work; instead, estimates (and plotting) seems to be constrained to
range(x), range(y) no matter what i do. The documentation says i can
pass in kde2d'
Hello all,
I would like to rigorously test whether observations in my dataset are
outliers. I guess all the main tests in R (Grubbs) impose the assumption
of normality. My data is surely not normal, so I would like to use
something else. As far as I can tell from wikipedia, Peirce's criterion i
Hi all,
Here is an integration function
require(pracma) # for 'quadinf'
myint=function(j) {
quadinf(function(x)
(1/(1+exp(-x)))^j*(1-1/(1+exp(-x)))^(k-j)*dnorm(x,mu,casigma),-Inf,Inf)
}
in any optimization routine. It works fine most of the time but failed with
some particular sets of va
On Wed, Apr 18, 2012 at 12:58 PM, Ben quant wrote:
> Hello,
>
> I have two date strings, say "1972-06-30" and "2012-01-31", and I'd like to
> get every quarter period end date between those dates? Does anyone know how
> to do this? Speed is important...
>
> Here is a small sample:
>
> Two dates:
>
Try this (NOT tested) or something similar:
all.comps <- ls(pattern="^res")
for(i in all.comps) {
obj <- all.comops[[i]]
gene.ids <- rownames(obj$counts)
x <- data.frame(gene.ids = gene.ids, obj$counts,
obj$e1, obj$e2,
obj$log.fc, obj$p.value,
obj$q.value)
x <-
Thanks for your help, Milan, in showing me a way to exhibit the structure of my
data.
# Again, my data sets are:
> [1] "res.Callus.Explant" "res.Callus.Regen" "res.Explant.Regen"
# Structure is:
> str(res.Callus.Explant)
List of 18
$ name : chr "two group comparison"
$ group1
x <- rep(1:10,4)
y <- runif(40)
z <- sample(letters[1:4],40,rep=TRUE)
grp <- sample(LETTERS[1:2],40,rep=TRUE)
xyplot(y~x|z, group=grp, panel = panel.superpose,
panel.groups = function(...){
panel.xyplot(...)
panel.lmline(...)
})
Keys (legends), colors, line styles,etc, can all be cu
On Wed, Apr 18, 2012 at 2:07 PM, Louis Plough wrote:
> If you could lead me to an example with code, that would help me figure out
> how to do it for my function
The states example in ?xyplot uses groups and subscripts in a panel function
>> I read it, but I guess I don't quite understand which
On Wed, Apr 18, 2012 at 07:15:45AM -0700, uday wrote:
> hi Petr ,
> Thanks for replay and sorry for typo mistake
> approx(pres, sci.pre) its nothing but approx(pre2, pre1).
>
> so for more simplicity
> x <- c(10.34615 , 52.02116, 146.17357, 243.28644, 347.41504, 431.67105,
> 521.4
Le mercredi 18 avril 2012 à 13:13 -0700, Vining, Kelly a écrit :
> Dear useRs,
>
> A colleague has sent me several batches of output I need to process, and I'm
> struggling with the format to the point that I don't even know how to extract
> a test set to upload here. My apologies, but I think t
please look at the ancova function in the HH package.
On Wed, Apr 18, 2012 at 3:34 PM, Louis Plough wrote:
> Hi,
> I am trying to use xyplot to plot the relationship between size and day
> (y~x) by a food factor that has two levels, low and high. I have 3 reps per
> factor/day. I want the plot
On 19/04/12 08:14, Michael wrote:
Is there a way to find all roots of a polynomial equation?
Lets say
x^5+a*x^4+b*x^3+c*x^2+d*x+e=0
how to find its all roots?
I believe that Galois (or maybe it was Abel) proved that this
is *NOT* possible, a couple of centuries ago.
However if you are thin
On 12-04-18 4:14 PM, Michael wrote:
Is there a way to find all roots of a polynomial equation?
Lets say
x^5+a*x^4+b*x^3+c*x^2+d*x+e=0
how to find its all roots?
polyroot() should be able to do it, assuming that a,b,c,d and e are
numeric or complex values, not just symbols.
Duncan Murdoch
Is there a way to find all roots of a polynomial equation?
Lets say
x^5+a*x^4+b*x^3+c*x^2+d*x+e=0
how to find its all roots?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-he
Dear useRs,
A colleague has sent me several batches of output I need to process, and I'm
struggling with the format to the point that I don't even know how to extract a
test set to upload here. My apologies, but I think that my issue is
straightforward enough (for some of you, not for me!) that
Oh... I see!!! I very appreciate for your clear explanation! Thank you, Uwe
Ligges.
--
View this message in context:
http://r.789695.n4.nabble.com/General-question-on-the-message-with-non-zero-exit-status-tp4551438p4568839.html
Sent from the R help mailing list archive at Nabble.com.
_
I wanted to do an F-test to compare model L2 with L3 assuming
heteroskedasticity. The variable KONK2 only has three levels and are for
that reason replaced by dummyvariables KONKD1 and KONKD2 in L3. Hence making
L4 equivalent to L2.
L2<-lm(PRMRES ~ factor(NYPR) + KONK2 + OMS + factor(NYPR)*OMS )
Cecile De Cat leeds.ac.uk> writes:
> I'm analysing reaction time data from a linguistic experiment (a variant of
> a lexical decision task). To ascertain that the data was normally
> distributed, I used *shapiro.test *for each participant (see commands
> below), but only one out of 21 returns a
If you could lead me to an example with code, that would help me figure out
how to do it for my function
On Wed, Apr 18, 2012 at 3:57 PM, Louis Plough wrote:
> I read it, but I guess I don't quite understand which arguments to pass
> panel.groups to get different lm objects based on the two gro
So the lapply function checks every country pair row for row (in my country1
and country2 column) and send back a true if both countries are EU?
How do I get a 1 instead of true and 2 instead of false?
You have been a BIG help.
I have a WHOLE assignment to get done, so maybe I'll post some more
I ues R on a Linux server running under Screen or Emacs server from a Mac
desktop.
I use Screen and/or Emacs server so that I can reattach to long running
sessions, and this works well except...
If X11 server on my desktop is shut down, or my ssh connection is broken, I get
this error on the n
dear list members
I tried to install qvalue package from biocLite, and I am confronted with
the following error:
Tcl/Tk support is not available on this system.
I installed the newest version of Tcl/Tk in my system, and then uninstall
and install R 2.15.0 again.
I did the command line as
./co
Dear R experts:
could someone please point me to a page that explains how to set up
more than 1 machine for library parallel (which is quickly becoming my
favorite!)
my dream setup would be a design where I just pass a list of
hostnames:user:password to my parallel master, and then start R
listen
Please read ?panel.superpose again and pay attention to the
panel.groups argument, where this can be specified.
-- Bert
On Wed, Apr 18, 2012 at 12:34 PM, Louis Plough wrote:
> Hi,
> I am trying to use xyplot to plot the relationship between size and day
> (y~x) by a food factor that has two lev
> If the students are in a "science research" class, does that mean they
> have data from their own research that they would want to understand
> better? I think that would be much more motivating than anything else.
It might depends on the class - most high school science experiments
aren't that
Hi,
I am trying to use xyplot to plot the relationship between size and day
(y~x) by a food factor that has two levels, low and high. I have 3 reps per
factor/day. I want the plots from each food treatment on the same axiss,
so I used this code:
xyplot(Size ~ Day, groups = Food, data = louis.data
On Apr 18, 2012, at 1:49 PM, David Winsemius wrote:
>
> On Apr 18, 2012, at 2:05 PM, Marc Schwartz wrote:
>
>> On Apr 18, 2012, at 11:58 AM, Ben quant wrote:
>>
>>> Hello,
>>>
>>> I have two date strings, say "1972-06-30" and "2012-01-31", and I'd like to
>>> get every quarter period end date
Cecile:
On Wed, Apr 18, 2012 at 8:21 AM, Cecile De Cat wrote:
> Hello,
>
> I'm analysing reaction time data from a linguistic experiment (a variant of
> a lexical decision task). To ascertain that the data was normally
> distributed, I used *shapiro.test *for each participant (see commands
> be
On Apr 18, 2012, at 2:05 PM, Marc Schwartz wrote:
On Apr 18, 2012, at 11:58 AM, Ben quant wrote:
Hello,
I have two date strings, say "1972-06-30" and "2012-01-31", and I'd
like to
get every quarter period end date between those dates? Does anyone
know how
to do this? Speed is important.
On 12-04-17 11:50 PM, David Bapst wrote:
Hi all-
Henrik, packageDescription informs me that I have Matrix version 1.0-6,
which is the newest version on CRAN.
The --as-cran check in R 2.15.0 is a little bit too aggressive: this
was a spurious error. If your package depends on another package
Easily fixed, Peter -
str(ad04)
'data.frame': 1195 obs. of 15 variables:
$ V040001 : num 1 2 3 4 5 6 7 8 9 10 ...
$ insurance: num 7 2 4 1 5 2 7 1 5 5 ...
$ jobs : num 7 3 4 1 6 5 7 5 4 7 ...
$ services : num 5 2 5 1 5 2 7 1 4 7 ...
$ ss : num 2 2 2 1 2 1 2 1 1 3 ...
$ wom
Sarah -
I have all the vars.. they all look sensible. I have been able to
trace the problem to NAs in the nominal variable, which is a bit
strange. The error that the program spits out is a class logical
equating error in presence of NAs etc. and hence was my hunch that it
was a bug.
On Wed, A
On 2012-04-18 11:18, Gaurav Sood wrote:
Using: Amelia::amelia
R version: 2.15
OS: Windows 7 Enterprise
data = National Election Studies (cross-sectional); mostly ordinal
variables, some nominal
summary(ad04)
insurance jobs services sswomen
Min.
Hi,
On Wed, Apr 18, 2012 at 2:18 PM, Gaurav Sood wrote:
> Using: Amelia::amelia
>
> R version: 2.15
> OS: Windows 7 Enterprise
>
> data = National Election Studies (cross-sectional); mostly ordinal
> variables, some nominal
>
> summary(ad04)
>
> insurance jobs services
I've had a little experience using the package, Amelia. Are you sure that
your nominal variables - race, south, etc - are in your ad04 data frame ?
David Freedman
--
View this message in context:
http://r.789695.n4.nabble.com/Amelia-error-tp4568455p4568600.html
Sent from the R help mailing lis
Using: Amelia::amelia
R version: 2.15
OS: Windows 7 Enterprise
data = National Election Studies (cross-sectional); mostly ordinal
variables, some nominal
summary(ad04)
insurance jobs services sswomen
Min. :1.000 Min. :1.000 Min. :1.000 Min
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Hadley Wickham
> Sent: Wednesday, April 18, 2012 10:37 AM
> To: Christopher W Ryan
> Cc: R-help
> Subject: Re: [R] introducing R to high school students
>
> > Now I have to pu
On Apr 18, 2012, at 11:58 AM, Ben quant wrote:
> Hello,
>
> I have two date strings, say "1972-06-30" and "2012-01-31", and I'd like to
> get every quarter period end date between those dates? Does anyone know how
> to do this? Speed is important...
>
> Here is a small sample:
>
> Two dates:
>
Hi Rainier,
Thanks for your suggestions. I should have been more specific, I am using
multiple cores on a Mac Pro running Snow Leopard. I can see where that makes a
difference.
--David
On 4/18/12 12:13 AM, "Rainer M Krug" wrote:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 17/04/12 20
Do you mean the Amelia package? Or a function named Amelia()?
What is your OS and version of R and package?
What are the commands you are using?
What do your data look like?
Can you provide a reproducible example?
Did you read the posting guide?
Your problem doesn't "seem like a bug" rather t
> Now I have to put my money where my mouth is. I've offered to visit a
> high school and introduce R to some fairly advanced students
> participating in a longitudinal 3-year science research class.
>
> I anticipate keeping things very simple:
> --objects and the fact that there is stuff inside th
Hi,
Encountering the following error using Amelia -
Error in if (sum(non.vary == 0)) { :
argument is not interpretable as logical
In addition: Warning message:
In FUN(X[[34L]], ...) : NAs introduced by coercion
Seems like a bug.
__
R-help@r-project.
On Apr 18, 2012, at 12:31 PM, Jeff Newmiller wrote:
I think that mostly avoiding the statistics and matrix capabilities
is wise. You might want to (re-)read Burns' article on Spreadsheet
Addiction for help in justifying the effort required to learn R.
In that vein, there is a classic exper
Hi Ben,
A couple of quick comments:
the semicolons at the end of each line are unnecessary (they're
harmless, but considered bad style -- most common in code of C and
MATLAB coders
yes, I know, just forgot to clean it up after translating :)
You don't need to source("getband.R") inside
Hello,
I have two date strings, say "1972-06-30" and "2012-01-31", and I'd like to
get every quarter period end date between those dates? Does anyone know how
to do this? Speed is important...
Here is a small sample:
Two dates:
"2007-01-31"
"2012-01-31"
And I'd like to get this:
[1] "2007-03-
I think that mostly avoiding the statistics and matrix capabilities is wise.
You might want to (re-)read Burns' article on Spreadsheet Addiction for help in
justifying the effort required to learn R.
In that vein, there is a classic experiment where a small ball is rolled down
an inclined pane
Try recording the data as a rectangle with 3 entries in each row
timestamp, sensor_id, sensor_reading
You can read this into an R data.frame, sensorData, with a call to read.csv
then use
z <- split(sensorData, sensorData$sensor_id)
to make a list of data.frames, one for each sensor, that you
On Wed, Apr 18, 2012 at 8:45 AM, slavrenz wrote:
I would like to display with the xyplot() function
> for several states. I will have a total of 6 plots, I need to plot the
> points of one of the states in a different color than all the rest, such
> that they can be more easily referenced in a p
Hi Michael,
that's exactly the problem. The format for time series you suggest requires
to have a reading from every timestamp of every sensor. But that is not the
case at all, unfortunately :-(
It might well be in my case, that one sensor reports 1000 readings during
one day while the other one
Hi Ivan,
We are in the process of creating an R mode for highlight.js -- will send
you a pull request as soon as its ready.
Cheers,
J.J. Allaire
j...@rstudio.org
--
View this message in context:
http://r.789695.n4.nabble.com/Support-for-R-in-highlight-js-tp4551547p4568141.html
Sent from the R
Hello Everyone,
Trying to learn a little bit about data mining. I'm working on a text mining
project that will attempt to predict whether cancer patients got a particular
type of genetic testing. A subsequent stage then will be aimed at predicting
what the results of that testing were.
I've
By the way, as I've been reading through some other posts on here, let me
clarify that this isn't for a homework assignment or anything. This is
something I am working on as part of my dissertation.
--
View this message in context:
http://r.789695.n4.nabble.com/Displaying-data-in-Trellis-tp456792
Hello,
I'm analysing reaction time data from a linguistic experiment (a variant of
a lexical decision task). To ascertain that the data was normally
distributed, I used *shapiro.test *for each participant (see commands
below), but only one out of 21 returns a p value above p.0 05.
> f = functio
Hi,
I want to access and modify a multi dimensional array but without
knowing the exact number of dimensions.
Basically. I want something like that for every i (except boundaries of
course):
field[i, ...] <- (field[i, ...] + field[i+1, ...] + field[i-1, ...])/3
In principle, ... should just rep
Your problem is that columns A & B contain something that can't be
ordered. (Likely "factor" (=categorical) data like Male / Female
rather than numeric like 10 and 5)
Use str(data_2) to see what classes your data are -- they sometimes
get converted in unexpected ways if you aren't careful in setti
Your problem is that length(x) != length(y)
approx uses linear interpolation but there's no way to make sense of
that if you can't match up the x and y coordinates -- and you can't
match up the x and y coordinates if there aren't the same number of
them.
Michael
On Wed, Apr 18, 2012 at 10:15 AM,
The preferred format for time series will be something like this:
Sensor1Sensor2 Sensor3
T1 Read1_1 Read2_1 Read3_1
T2 Read1_2 Read2_2 Read3_2
T3 Read1_3 Read2_3 Read3_3
if you can get that. CSV separations are nice but not as essential as
the columnar organization. (It's pos
hi Petr ,
Thanks for replay and sorry for typo mistake
approx(pres, sci.pre) its nothing but approx(pre2, pre1).
so for more simplicity
x <- c(10.34615 , 52.02116, 146.17357, 243.28644, 347.41504, 431.67105,
521.42712, 629.00446 ,729.95941, 827.86279, 921.55078,
956.6)
y <- c(
Thanks for your reply!
As I write the script to export the data from the database myself, I'm free
to sort the data (almost) at will.
I can either export to CSV in the form of
sensor1
timestamp1, reading1, timestamp2, reading2, timestamp2, reading2...
sensor2
timestamp1, reading1, timestamp2, r
Hi,
I am stuck with creating a conditional categorical variable in R
If my dataframe data_2 has 3 variables A,B,CI want to create variable D
which would be something like :
data_2$D <- ifelse(data_2$A < data_2$B & promotion_ind =="N",1,0),
this throws up an error "In Ops.factor(data_2$A,dat
Dear list,
I can't use the choose.files for list the 2,000 files in directory.
# list files
> names <- (tk_choose.files(default = "", caption = "Select the files
",
+multi = TRUE, filters = Filt,
+index = 1))
>
>
> names
character(0)
# names is empty
Hi Uwe
It works modifying a little bit your command instead of "file:=/home/..." I
used "file:///home/..."
install.packages("marginalmodelplots", dependencies=TRUE,
contriburl="file:///home/bioadm/install/R_packages/test/")
Thanks for your time and your patience
Regards
Joël
-Message d
This is a question relating to the Trellis function in R. I have a set of
socioeconomic data that I would like to display with the xyplot() function
for several states. I will have a total of 6 plots, with the y-variable as
"Age Group", and the x-variable as "Median Income".
Suitability of the plo
Indrajit,
As a former math teacher I understand your concerns wholly. My perspective is
that this must be approached with caution so you don't miss out on the
important learning but I think with proper guidance and scaffolding this could
be an amazing tool. We already using the graphing capab
Christopher,
I originally thought about writing off list to avoid a plethora of babble.
However here goes.
I don't see any reason why good students can 't learn the fundamentals of
R. It has lots of advance methods that perhaps are too complex to handle
for younger - less experienced people. On
Thanks for the clarification Dr. Therneau. Until I learn more about this I can
at least remember that "plain" is bad.
Thanks,
Paul
--- On Mon, 4/16/12, Terry Therneau wrote:
> From: Terry Therneau
> Subject: Re: Kaplan Meier analysis: 95% CI wider in R than in SAS
> To: r-help@r-project.org,
Thanks all for the excellent thought-provoking comments.
I want to clarify that these students are, for good or for ill, already
doing all these analytical and graphical things for their projects. They
are doing them with Excel and SPSS. One of my goals would be to teach
them how they can be done
Thanks very much! However, I tried with that
(ln -s /usr/local/texlive/2011 /usr/share/texmf)
without luck. yum is still asking for a list of dependencies, including
several texlive files.
All the best,
Petar
On Wed, 2012-04-18 at 08:58 -0400, Randy Johnson wrote:
> I can't make any guaranties,
On Wed, Apr 18, 2012 at 03:55:07AM -0700, uday wrote:
> This moment I got stuck with one interpolation issue
> the sample data which I have is as follows
>
> pre1 <- c(10.34615 , 52.02116, 146.17357, 243.28644, 347.41504, 431.67105,
> 521.42712, 629.00446 ,729.95941,82
On Apr 18, 2012, at 6:31 AM, Petar Milin wrote:
> Hello ALL!
> I am running Fedora 16 x86_64. Due to some dubious problems with the
> TeXLive (2007, which is a default), I removed it. That removed R as
> well, and some other dependent packages. Then, I installed TeXLive 2011
> from CTAN. However,
On 18.04.2012 14:01, mail.bioi...@fr.netgrs.com wrote:
Hi Uwe
I tried
install.packages("marginalmodelplots", dependencies=TRUE,
contriburl="/home/bioadm/install/R_packages/test")
Whoops, this was untested an it turns out you need:
contriburl="file:/home/bioadm/install/R_packages/test"
re
HughSt hotmail.com> writes:
> I am trying to run a logistic regression to look at the risk of malaria
> infection in individuals. I want to account for intra household correlation
> and so want to include a household level random effect. I have been using
> the lmer command in lme4 package but am
I can't make any guaranties, but placing a symbolic link at /usr/share/texmf,
pointing at /usr/local/texlive/2011 might do the trick. It's a bit of a hack,
but that often works for me.
Best,
Randy
On Apr 18, 2012, at 7:31 AM, Petar Milin wrote:
> Hello ALL!
> I am running Fedora 16 x86_64. Due
1 - 100 of 123 matches
Mail list logo