As another followup, given that you are doing numerous regression
models and (I presume) working with finance/stock data that is
strictly numeric (no need for special contrast coding, etc.), you can
substantially reduce the time spent estimating the coefficients. A
simple way is to use lm.fit dire
I am having a problem with extra digits being added to my data which I think
is a result of how I am converting my data.frame data to xts.
I see the same issue in R v2.13.1 and RStudio version 0.94.106.
I am loading historical foreign exchange data in via csv files or from a sql
server database.
Hi
>
> Dear R-users,
> I'm using lattice package and function xyplot for the first time so
> you will excuse me for my inexperience. I'm facing quite a simple
> problem but I'm having troubles on how to solve it, I've read tons of
> old mails in the archives and looked at some slides from Deepayan
Hi
>
> OK. So my original advice and warnings are correct.
>
> However, now there is an additional wrinkle because your response is a
> count, which is not a continuous measurement. For this, you'll need
glm(...,
> family = "poisson") instead of lm(...), where the ... is the stuff I
gave
> you
Hi
I would try either some tree method (mvpart) or you can expand lm model
also with users.
fit<-lm(value~variable+users, data=test.m)
Anyway I am not an ultimate expert in statistics. so you shall also
consult some appropriate literature which can be found in CRAN web. Did
you try to look i
thank you, everyone. this was very helpful to my specific task and
understanding. for the benefit of future googlers, I thought I would
post some experiments and results here.
ultimately, I need to do a by() on an irregular matrix, and I now know
how to speed up by() on a single-core, and then a
The following avoids the overhead of data.frame methods
(and assumes the data.frame doesn't include matrices
or other data.frames) and relies on split(vector,factor)
quickly splitting a vector into a list of vectors.
For a 10^6 row by 10 column data.frame split in 10^5
groups this took 14.1 seconds
Bogaso wrote:
>
> Dear all, I have a system of simultaneous equations with 2 unknowns as
> follows:
>
> x*y + (1-x) = 0.05
>
> x*(y - .5)^2 + (1-x)*0.6 = 0.56^2
>
> Ofcourse I can do it manually however wondering whether there is any
> direct
> way in R available to get the solution of this sy
collegegurl69 hotmail.com> writes:
>
> I have this error and I can't figure out whats wrong:
> "invalid or not-yet-implemented 'Matrix' subsetting"
> it pops up when I try to run this line of code:
> S <- B[indices.mod,union(mir.e.nc,mir.negatives.nc)]
Could we have a reproducible example
I tried this:
library(data.table)
N <- 1000
T <- N*10
d <- data.table(gp= rep(1:T, rep(N,T)), val=rnorm(N*T), key = 'gp')
dim(d)
[1] 10002
# On my humble 8Gb system,
> system.time(l <- d[, split(val, gp)])
user system elapsed
4.150.094.27
I wouldn't be surprise
instead of spliting the entire dataframe, split the indices and then use these
to access your data: try
system.time(s <- split(seq(nrow(d)), d$key))
this should be faster and less memory intensive. you can then use the indices
to access the subset:
result <- lapply(s, function(.indx){
do
Not having played with portfolio.opim() much, I can't guarantee this
will fix it, but if it requires a matrix rather than a vector and you
are sure about the rest of the syntax, this might do the trick:
asset_forecast[i, , drop = FALSE]
This is because:
R> x = matrix(1:9, 3)
R> is.matrix(x[,1])
Hello everyone,
I would like assistance with a snippet I have written to do a recursive
portfolio optimization given time-varying return forecasts.
In my case, I have forecast the monthly returns for nearly 55 years out on 8
asset classes.
I need to calculate the weights for the optimal (
dear R experts: apologies for all my speed and memory questions. I
have a bet with my coauthors that I can make R reasonably efficient
through R-appropriate programming techniques. this is not just for
kicks, but for work. for benchmarking, my [3 year old] Mac Pro has
2.8GHz Xeons, 16GB of RAM,
On Oct 10, 2011, at 4:48 PM, Gustaf Rydevik wrote:
Hi all,
some years ago, I sent a question to the mailing list regarding the
WHO
anthro macros. Since I've now received three mails asking how I
solved it, I
thought I'd cc R-help in for future reference. Attaching a zip file
with the rele
Hello,
I am trying to use XML package to download Facebook reviews in the following
way:
require(XML)
mydata.vectors <- character(0)
Qword <- URLencode('#IBM')
QUERY <- paste('SELECT review_id, message, rating from review where message
LIKE %',Qword,'%',sep='')
Facebook_url = paste('https://api.
I have been trying to use this test recently, following the text from this
link:
http://books.google.com/books?id=1eTyuMDND94C&pg=PA145&lpg=PA145&dq=nonparam#v=onepage&q&f=false
I ordered my data based on ranks, and ran a type III ANOVA from the car
package - something like
Anova(lm(var1~var2*var
Hi all,
some years ago, I sent a question to the mailing list regarding the WHO
anthro macros. Since I've now received three mails asking how I solved it, I
thought I'd cc R-help in for future reference. Attaching a zip file
with the relevant code parts that
I used that I'm not sure gets through (
On 11/10/11 08:21, Christoph Molnar wrote:
Hi Nellie,
hope I got you right. I guess you want something like that.
for (i in 1:7) {
oneOfNelliesArray<- eval(parse(text=paste("grp",i, sep="")))
anyFunction(oneOfNelliesArray)
}
Paste() just returns you a string. But you want R to evaluate
On Mon, Oct 10, 2011 at 4:14 PM, Joshua Wiley wrote:
> I could be waay off base here, but my concern about presplitting the data is
> that you will have your data, and a second copy of our data that is something
> like a list where each element contains the portion of the data for that
> split.
Liu Evans, Gareth liverpool.ac.uk> writes:
> In my post at https://stat.ethz.ch/pipermail/r-help/2011-October/292019.html
> I included an undefined term "ej". The problem code should be as follows.
> It seems like a simple linear programming problem, but for some reason my
> code is not finding
I have this error and I can't figure out whats wrong:
"invalid or not-yet-implemented 'Matrix' subsetting"
it pops up when I try to run this line of code:
S <- B[indices.mod,union(mir.e.nc,mir.negatives.nc)]
--
View this message in context:
http://r.789695.n4.nabble.com/invalid-or-not-yet-
Hi Nellie,
hope I got you right. I guess you want something like that.
for (i in 1:7) {
oneOfNelliesArray <- eval(parse(text=paste("grp",i, sep="")))
anyFunction(oneOfNelliesArray)
}
Paste() just returns you a string. But you want R to evaluate the
expression. So you have to parse it and t
okay so fixed what i need to do this way
finit=0
for(ri in 1:dim(xa)[1])
{
finit=finit+1
xc[ri,1:finit]<-xa[ri,1:finit]
xc[1:finit,ri]<-xb[1:finit,ri]
}
but getting error in heatmap.2
> mycol <- colorpanel(n=40,low="red",mid="white",high="blue")
> heatmap.2(xc, breaks=pairs.breaks, col=mycol, Ro
Hey all,
In order to test if a marker is a risk factor, I built two models (using cox
proportional hazard model). One model included this marker, and the other is
not.
Then, I use R package risksetROC to test how much predictive value did the
marker add to this model. I get two C statistics by an
And you might also consider packages like corrplot, corrgram etc. for
other plotting options of a correlation matrix.
They can be more informative than simply invoking image(heat)
> What a pleasant post to respond to - with self-contained code. :)
>
> heat<-matrix(0,nrow=dim(xa)[1],ncol=dim(xa)
This is the sort of thing that should be measured, rather than
speculated about, but if you're using multicore all those subsets can
be made at the same time, not sequentially, so they add up to a copy
of the whole data. Using data.table rather than a data.frame would
help, of course.
I would gu
Tena koe Sharad
If I understand you correctly, you want the lower triangle of your combined
matrix to be the lower triangle of one of the correlation matrices, and the
upper triangle to be the upper triangle from the other. If so, check
lower.tri() and upper.tri().
HTH
Peter Alspach
>
What a pleasant post to respond to - with self-contained code. :)
heat<-matrix(0,nrow=dim(xa)[1],ncol=dim(xa)[2])
heat[lower.tri(heat)]<-xa[lower.tri(xa)]
heat[upper.tri(heat)]<-xb[upper.tri(xb)]
diag(heat)<-1
heat
HTH,
Daniel
1Rnwb wrote:
>
> Hello Gurus
> I have two correlation matrices 'x
I could be waay off base here, but my concern about presplitting the data is
that you will have your data, and a second copy of our data that is something
like a list where each element contains the portion of the data for that split.
Good speed wise, bad memory wise. My hope with the techniqu
On Mon, Oct 10, 2011 at 2:55 PM, ecoc wrote:
> Hope someone can help me here.
>
> I have a daily time series, say
>
> 2003-02-01 2003-02-03 2003-02-07 2003-02-09 2003-02-14 .. 2004-02-01
> 2004-02-04
> 0.4914798 -1.1857653 -1.6982844 -0.3559572 -0.2333087 ...
> 0.44553 -0.4522
On Tue, Oct 11, 2011 at 7:54 AM, ivo welch wrote:
> hi josh---thx. I had a different version of this, and discarded it
> because I think it was very slow. the reason is that on each
> application, your version has to scan my (very long) data vector. (I
> have many thousand different cases, too.
On 10/9/2011 6:18 AM, Prof Brian Ripley wrote:
Sometimes it is better not to document things than try to give precise
details which may get changed *and* there will be useRs who misread (and
maybe even file bug reports on their misreadings). The source is the
ultimate documentation.
I can't ag
On Oct 10, 2011, at 12:52 PM, song_gpqg wrote:
Hello!
So I am handling this problem with some arrays grp1-grp7, I want to
write a
loop to avoid tedious work, but I don't know how to transform string
to
boor?
For example I used
i=1
paste("grp",i, sep="")
?get
e.g.
get( paste("grp",i, sep
Hope someone can help me here.
I have a daily time series, say
2003-02-01 2003-02-03 2003-02-07 2003-02-09 2003-02-14 .. 2004-02-01
2004-02-04
0.4914798 -1.1857653 -1.6982844 -0.3559572 -0.2333087 ...
0.44553-0.45222
I need to calculate the statistics for the overlappin
Hello Gurus
I have two correlation matrices 'xa' and 'xb'
set.seed(100)
d=cbind(x=rnorm(20)+1,
x1=rnorm(20)+1,
x2=rnorm(20)+1)
d1=cbind(x=rnorm(20)+2,
x1=rnorm(20)+2,
x2=rnorm(20)+2)
xa=cor(d,use='complete')
xb=cor(d1,use='complete')
I want to combine these two to get a third
Dear all,
I would be grateful if you could help me in some way! Well, I have one
dataset already uploaded into R with some missing data. My dataset is made
of 19 columns, one of these is the sector (naf code). There are many rows
relating to the single enterprises. What I want to do is the mean of
Hello!
So I am handling this problem with some arrays grp1-grp7, I want to write a
loop to avoid tedious work, but I don't know how to transform string to
boor?
For example I used
i=1
paste("grp",i, sep="")
I only got "grp1" instead of grp1, which can't be manipulate using mean() or
other functi
Hi Tom,
Just wanted to chime-in and let you know that the linked figures are really
cool! Keep up the good work.
On an un-related note, any talk of future GRASS training sessions?
Cheers,
Dylan
On Tuesday, October 04, 2011, thomas.ad...@noaa.gov wrote:
> Hadley,
>
> Thanks for responding. No,
Thanks bbolker, that's really helpful. I'll look out for the book too, it
could be helpful!
Yours,
Sam
--
View this message in context:
http://r.789695.n4.nabble.com/ANOVA-from-imported-data-has-only-1-degree-of-freedom-tp3887528p3891246.html
Sent from the R help mailing list archive at Nabble.c
Hi,
I am having some trouble using R 2.13.1 for generating a pmml object
of of class "c('randomForest.formula', 'randomForest')"
I see that these methods are available:
> methods(pmml)
[1] pmml.coxph*pmml.hclust* pmml.itemsets* pmml.kmeans*
pmml.ksvm* pmml.lm* pmml.multinom* pmml
hi josh---thx. I had a different version of this, and discarded it
because I think it was very slow. the reason is that on each
application, your version has to scan my (very long) data vector. (I
have many thousand different cases, too.) I presume that by() has one
scan through the vector that
As an interesting extension to David's post, try:
M4.e <- matrix(rexp(4,1), ncol=4)
Instead of the uniform and rerun the rest of the code (note the limits on the
x-axis).
With 3 dimensions and the restriction we can plot in 2 dimensions to compare:
library(TeachingDemos)
m3.unif <- matrix
On Oct 10, 2011, at 1:28 PM, Alaios wrote:
Dear all,
I have some device measurements and the time stamps I get from it
have the below format:
MyStruct$TimeStamps[1,]
[1] 2011.000 10.0006.000 16.000 23.000 30.539
I can convert them easily with ISOdate() to a number and do the
Package plyr has .parallel.
Searching datatable-help for "multicore", say on Nabble here,
http://r.789695.n4.nabble.com/datatable-help-f2315188.html
yields three relevant posts and examples.
Please check wiki do's and don'ts to make sure you didn't
fall into one of those traps, though (we don't
Hi Ivo,
My suggestion would be to only pass lapply (or mclapply) the indices.
That should be fast, subsetting with data table should also be fast,
and then you do whatever computations you will. For example:
require(data.table)
DT <- data.table(x=rep(c("a","b","c"),each=3), y=c(1,3,6), v=1:9)
se
On Oct 10, 2011, at 12:44 PM, Uwe Ligges wrote:
On 10.10.2011 18:10, Alexander Engelhardt wrote:
Hey list,
This might be a more general question and not that R-specific.
Sorry for
that.
I'm trying to draw a random vector of 4 numbers that sum up to 1.
My first approach was something lik
I haven't tried this, but I am pretty confident that using dlmFilter()
with fictitious future values of the observations set to NA should do
the job.
Hope this helps,
Giovanni Petris
On Sat, 2011-10-08 at 13:21 +, YuHong wrote:
>
> May I have a question about dlmForecast() function in the p
dear r experts---Is there a multicore equivalent of by(), just like
mclapply() is the multicore equivalent of lapply()?
if not, is there a fast way to convert a data.table into a list based
on a column that lapply and mclapply can consume?
advice appreciated...as always.
regards,
/iaw
Ivo
Dear all,
I have some device measurements and the time stamps I get from it have the
below format:
MyStruct$TimeStamps[1,]
> [1] 2011.000 10.000 6.000 16.000 23.000 30.539
I can convert them easily with ISOdate() to a number and do the calculations I
need.
One of my problems is that
Hi:
Here's one way to do it, adding the latticeExtra package:
array = rep(c("A","B","C"),each = 36) # array replicate
spot = rep(1:4,27) # miRNA replicate on each array
miRNA = rep(rep(paste("miRNA",1:9,sep="."),each=4),3) # miRNA label
exprs = rnorm(mean=2.8,n = 108) # intensity
dat = data.fram
OK. So my original advice and warnings are correct.
However, now there is an additional wrinkle because your response is a
count, which is not a continuous measurement. For this, you'll need glm(...,
family = "poisson") instead of lm(...), where the ... is the stuff I gave
you before. A backup app
On 10.10.2011 18:10, Alexander Engelhardt wrote:
Hey list,
This might be a more general question and not that R-specific. Sorry for
that.
I'm trying to draw a random vector of 4 numbers that sum up to 1.
My first approach was something like
a <- runif(1)
b <- runif(1, max=1-a)
c <- runif(1,
You probably want to generate data from a Dirichlet distribution. There are
some functions in packages that will do this and give you more background, or
you can just generate 4 numbers from an exponential (or gamma) distribution and
divide them by their sum.
--
Gregory (Greg) L. Snow Ph.D.
S
Hi Bert,
The real situation is like what you suggested, user x group interactions.
The users can be in more than one group.
In fact, the data that I am trying to analyse constitute of users, online
forums as groups and the attribute under measure is the number of posts made
by each user in a parti
Hi:
Here's one approach:
dat <- rnorm(40, 0, 2)
positions <- matrix(c(3, 4, 5, 8, 9, 10, 20, 21, 22, 30, 31, 32),
ncol = 3, byrow = TRUE)
# Subdata
t(apply(positions, 1, function(x) dat[x]))
[,1] [,2] [,3]
[1,] 0.5679765 1.429396 2.9050931
[2
Dear R-users,
I'm using lattice package and function xyplot for the first time so
you will excuse me for my inexperience. I'm facing quite a simple
problem but I'm having troubles on how to solve it, I've read tons of
old mails in the archives and looked at some slides from Deepayan
Sarkar but stil
Ivo,
Also, perhaps FAQ 2.14 helps : "Can you explain further why
data.table is inspired by A[B] syntax in base?"
http://datatable.r-forge.r-project.org/datatable-faq.pdf
And, 2.15 and 2.16.
Matthew
"Steve Lianoglou" wrote in message
news:CAHA9McPQ4P-a2imjm=szgjfxyx0faw0j79fwq2e87dqkf9j...@ma
Dear R-users,
I'm using lattice package and function xyplot for the first time so
you will excuse me for my inexperience. I'm facing quite a simple
problem but I'm having troubles on how to solve it, I've read tons of
old mails in the archives and looked at some slides from Deepayan
Sarkar but stil
Hey list,
This might be a more general question and not that R-specific. Sorry for
that.
I'm trying to draw a random vector of 4 numbers that sum up to 1.
My first approach was something like
a <- runif(1)
b <- runif(1, max=1-a)
c <- runif(1, max=1-a-b)
d <- 1-a-b-c
but this kind of
Dear R users
Function improveProb in the rms library calculate NRI and IDI for
predictions of binary outcome.
Do anyone extent its use in survival data?
Many thanks.
*Yao Zhu*
*Department of Urology
Fudan University Shanghai Cancer Center
Shanghai, China*
[[alternative HTML version del
Hi Ivo,
On Mon, Oct 10, 2011 at 10:58 AM, ivo welch wrote:
> hi steve---agreed...but is there any other computer language in which
> an expression in a [ . ] is anything except a tensor index selector?
Sure, it's a type specifier in scala generics:
http://www.scala-lang.org/node/113
Something s
Dear R-Users,
I have the following two vectors:
data <- rnorm(40, 0, 2)
positions <- c(3, 4, 5, 8, 9, 10, 20, 21, 22, 30, 31, 32)
now I would like to calculate the mean of every chunk of data-points (of the
data-vector) as defined by the positions-vector.
So I would li
An ad-hoc method is to impute missing scores of the whole data set
including subgroup1, then change imputed scores in subgroup1 into NA.
Weidong Gu
On Mon, Oct 10, 2011 at 5:35 AM, Sarah wrote:
> Dear R-users,
>
> I want to multiple impute missing scores, but only for a few subgroups in my
> dat
On Oct 10, 2011, at 7:45 AM, Spartina wrote:
Hello all,
how do I import a Fortran file (f3.1) into one column in R? I've
tried this
(I'm a total beginner as you can see):
FortranData<-read.fwf("C:\\Users\\format3_1.txt",rep(3,20))
Warning message:
In readLines(file, n = thisblock) :
inc
Groups are different treatments given to Users for your Outcome
(measurement) of interest. Take this idea forward and you will have an
answer.
Anupam.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Bert Gunter
Sent: Monday, October
I should have added...
If your design is not "nearly" balanced, main effects and interactions will
not have any "natural" interpretation because they will be (partially)
confounded. (I realize "nearly" is not a very useful characterization, but I
do not know a better one, as it probably depends on
Assuming your data are in a data frame, yourdat, as:
User Group Value
u1 1 !0
u2 2 5
u3 3 NA
...(etc)
where Group is **explicitly coerced to be a factor,**
then you want the User x Group interaction, obtained from
lm( Value ~ Group*User,data = yourdat)
Ho
Thanks a lot for your answer.
It is something that I do not understand with R how the wrapper functions work
really.
As an example, check the code below Â
TimeDiffInSeconds<-diff((ISOdate(timeMatrix[,1],timeMatrix[,2],timeMatrix[,3],timeMatrix[,4],timeMatrix[,5],timeMatrix[,6])),units="secs");Â
Hi Allan,
Please could you send the modified code where now it should appear x and y
coordinates?.
I do not fully understand the error message you get.
Regards,
Carlos Ortega
www.qualityexcellence.es
2011/10/10 Allan Sikk
> Thanks, Carlos,
>
> Tried that, but no success, still getting this err
In my post at https://stat.ethz.ch/pipermail/r-help/2011-October/292019.html I
included an undefined term "ej". The problem code should be as follows. It
seems like a simple linear programming problem, but for some reason my code is
not finding the solution.
obj <- c(rep(0,3),1)
col1 <-c(1,0
Hello all,
how do I import a Fortran file (f3.1) into one column in R? I've tried this
(I'm a total beginner as you can see):
> FortranData<-read.fwf("C:\\Users\\format3_1.txt",rep(3,20))
Warning message:
In readLines(file, n = thisblock) :
incomplete final line found on 'C:\Users\format3_1.tx
Hi,
OK.
Have you tried to run your code without the "pos" parameter?
Based on the help, "pos" should be just *one* parameter. "pos" offers a
finer adjustment of the text. But in your case, the first thing to get is
that the text label is represented at the specified coordinates. Besides
"pos" you
Dear all,
I want to use the deltavar() function from emdbook. I can use it
directly from the command terminal but within a function it behaves weird.
Working example:
--
library("emdbook")
fn <- function()
{
browser()
y <- 2
print(deltavar(y*b2, meanval=c(b2=3), Sigma=1) )
Hello,
To check the possible values of "pos" parameter you need to review "text()"
as it is indicated in the lattice help of panel.text().
In text() it says:
pos
a position specifier for the text. If specified this overrides any adj value
given. Values of 1, 2, 3 and 4, respectively indicate po
Dear R-users,
I want to multiple impute missing scores, but only for a few subgroups in my
data (variable 'subgroups': only impute for subgroups 2 and 3).
Does anyone knows how to do this in MICE?
This is my script for the multiple imputation:
imp <- mice(data, m=20, predictorMatrix=pred, post=p
This message has been passed onto the list moderators for
approval. This is because you are not a subscriber to this
list or the related squid-users list.
If the message is relevant to the squid-dev mailinglist
one of the moderators will accept the message and it gets
automatically forwarded to th
On Mon, Oct 10, 2011 at 8:49 AM, Gabor Grothendieck
wrote:
> On Mon, Oct 10, 2011 at 6:49 AM, Jurgens de Bruin wrote:
>> Hi,
>>
>> Please advice on what type of graph can be used to display the following
>> data set.
>>
>> I have the following:
>>
>> Name Class
>> a Class 1
>> a
Please ignore the venn digram as this will be to complex to read when more
than 3 categories are present
On 10 October 2011 15:36, Jurgens de Bruin wrote:
> Thanks for all the help,
>
> Their would it be possible to use a Venn Diagram for this application?
>
>
> On 10 October 2011 14:49, Gabor G
Thanks for all the help,
Their would it be possible to use a Venn Diagram for this application?
On 10 October 2011 14:49, Gabor Grothendieck wrote:
> On Mon, Oct 10, 2011 at 6:49 AM, Jurgens de Bruin
> wrote:
> > Hi,
> >
> > Please advice on what type of graph can be used to display the followi
Hello,
In package "qualityTools" you can find one way to perform this analysis
through the gageRR() function.
The effect of an operator on the mesasurement system (Reproductibility) is
to me equivalent to the effect you try to study of your users when they are
in different groups.
Regards,
Carlos
Here's the code. The problem seems to be specific for lattice as I can
easily use a vector with pos in "plot".
trellis.device(,width=600, height = 400)
xyplot(Npop~Narea,
scales=list(x=list(log=TRUE, at=my.at,labels = formatC(my.at, big.mark =
",", format="d")),
y=list(log=TRUE, at=c(1,10,100,10
On Mon, Oct 10, 2011 at 6:49 AM, Jurgens de Bruin wrote:
> Hi,
>
> Please advice on what type of graph can be used to display the following
> data set.
>
> I have the following:
>
> Name Class
> a Class 1
> a Class4
> b Class2
> b Class1
> d
try this version:
x <- read.table(textConnection("NameClass
a Class1
a Class4
b Class2
b Class1
d Class3
d Class5
e Class4
e Class2"), header = TRUE)
closeAllConnections()
# add columns of numeric v
Thanks, Carlos,
Tried that, but no success, still getting this error message:
Warning messages:
1: In if (pos == 1) { :
the condition has length > 1 and only the first element will be used
2: In if (pos == 2) { :
the condition has length > 1 and only the first element will be used
Thanks,
Jim, This should work, would it be possible to plot * and not larg rec.
On 10 October 2011 14:12, jim holtman wrote:
> See if this gives you the presentation you want:
>
> x <- read.table(textConnection("NameClass
> a Class1
> a Class4
> b Class2
> b
See if this gives you the presentation you want:
x <- read.table(textConnection("NameClass
a Class1
a Class4
b Class2
b Class1
d Class3
d Class5
e Class4
e Class2"), header = TRUE)
closeAllConnectio
Hi,
I am getting my results from the following:
*z$Date<-as.Date(as.character(z$Date),format="%d/%m/%y")*
instead of:
z$Date<-as.Date(as.character(z$Date,format="%d/%m/%y"))
Thanks again.
Regards
Vikram
On Mon, Oct 10, 2011 at 4:08 PM, Jeff Newmiller wrote:
> Convert to character first or
Hi,
I need to vary the placements of data labels but I cannot assign a
vector to "pos" option. Any vectors work fine with "cex", for example.
What could be the problem here?
xyplot(Npop~Narea, data=size,
scales=list(x=list(log=TRUE), y=list(log=TRUE),
xlab=expression(N[A]), ylab=expression(N[
Hi,
Please advice on what type of graph can be used to display the following
data set.
I have the following:
NameClass
a Class 1
a Class4
b Class2
b Class1
d Class3
d Class5
e Class4
e Class2
So
Thanks Petr. I will try it on the real data.
But that will only show that the groups are different or not.
Is there any way I can test if the users are different when they are in
different groups?
Regards
Gawesh
On Mon, Oct 10, 2011 at 11:17 AM, Petr PIKAL wrote:
> >
> > Hi Petr,
> >
> > It's
Convert to character first or use the "as.is" option to read.csv. The default
is to try to convert the underlying integer form of factors to date, which is
not what you intend.
---
Jeff Newmiller The . . Go Live...
DC
No, read ?difftime and look at as.double. There is a units parameter that you
must set if you want predictable results.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#..
Dear R users,
I have an elementary query.
I have a dataset which is taken from text file with the help of read.csv
command but when I generate the data in R file it converts the Dates into
factor.So for the above problem, I use as.Date to convert the Dates from
factor form to date format using th
>
> Hi Petr,
>
> It's not an equation. It's my mistake; the * are meant to be field
> separators for the example data. I should have just use blank spaces as
> follows:
>
> users Group1 Group2 Group3
> u110 5N/A
> u2 6 N/A 4
> u3
Hi Petr,
It's not an equation. It's my mistake; the * are meant to be field
separators for the example data. I should have just use blank spaces as
follows:
users Group1 Group2 Group3
u110 5N/A
u2 6 N/A 4
u3 5 2
Do you mean something like that?
as.double(diff(c(ISOdatetime(2011,6,1,11,59,1.09),ISOdatetime(2011,6,5,11,59,1.09))),length=20)
[1] 345600
From: Jeff Newmiller
Cc: "R-help@r-project.org"
Sent: Monday, October 10, 2011 10:42 AM
Subject: Re: [R] Handling Time
Difftime doesn't "report" things. When you print it, it automatically selects
an appropriate human-readable unit to display in, but that does not change its
internal representation. If you must convert to seconds, you can do so using
the as.double generic (as.double.difftime) with a units parame
Hi
I do not understand much about your equations. I think you shall look to
Practical Regression and Anova Using R from J.Faraway.
Having data frame DF with columns - users, groups, results you could do
fit <- lm(results~groups, data = DF)
Regards
Petr
>
> Hi,
>
> I'm a newbie to R. My
Dear All,
I want to estimate Bivariate Garch Model using MGARCHBEKK package. I am not
able to understand some part of the command this function.
mvBEKK.est(eps=lrdata, order = c(1,1), params = NULL,
fixed = NULL, method = "BFGS", verbose = F)
Here what exactly the eps refers to ? It would be rea
1 - 100 of 107 matches
Mail list logo
