I thought I'd chime in ... submitting the following:
?summary
Provides the following documentation for the default for generalized
argument (other than class="data.frame", "factor", or "matrix"):
## Default S3 method:
summary(object, ..., digits = max(3, getOption("digits")-3))
so pass
Hi Ernie,
You seem confused. sqrt(t(w) %*% V %*% w) calculates portfolio
volatility, not returns. You can calculate portfolio volatility with
PerformanceAnalytics::StdDev.
require(PerformanceAnalytics)
data(edhec)
set.seed(21)
w <- runif(ncol(edhec))
w <- w/sum(w)
sqrt(t(w) %*% cov(edhec) %*%
I resolved my program by restating RStudio . . .
Thanks you,
John
R 3.1.0, RStudio 0.98.95
Windows 7
I have written a function that uses lme:
doit<- function(TS,rho,premean,presd,RxEffect) {
.
.
.
# Prepare data frames for regression analyses.
data <- data.frame(group=c(rep("Cont",SS),rep(
R 3.1.0, RStudio 0.98.95
Windows 7
I have written a function that uses lme:
doit<- function(TS,rho,premean,presd,RxEffect) {
.
.
.
# Prepare data frames for regression analyses.
data <- data.frame(group=c(rep("Cont",SS),rep("Exp",SS)),
pre=pre,post=post)
.
.
.
previ
This is primarily about statistics, not R, and so is off topic here. I
believe you would do better posting on a stats site like
stats.stackexchange.com, where you could engage in a full (and likely
complicated) discussion of the pros and cons of Bayesian regression
vs. standard frequentist approach
Dear Xavier,
Perhaps you mean the qqPlot() function in the car package. If so, you should
install the car package. As well, if anc0 is a linear or generalized linear
model, qqPlot() has a method for plotting studentized residuals and you'd
probably prefer to use that rather than extracting the res
> ff <- reformulate(termlabels=c("time","as.factor(gvhd)"), response=yname,
intercept=TRUE)
If the right hand side of the formula were more complicated than an
additive list of terms,
say '~ time * as.factor(gvhd)' or the left side were more than a name, say
'log(yname)' you
could use bquote() ins
> On Feb 13, 2015, at 1:32 PM, CHIRIBOGA Xavier
> wrote:
>
> Hello! SORRY PROBLEMS WITH FUNCTIONS AGAIN...
>
>
>
> I NEED TO RUN A qqPlot ...I TRIED TO INSTALL IT , BUT A WARNING MESSAGE SAID
>
>
>
> qqPlot is not available for version 3.1.2.
What do you want?
Th
Let's see...
You can spell it qqplot.
You can avoid trying to install nonexistent packages with made-up names.
You can learn to use www.rseek.org
You can learn how to undo your caps lock key. ;)
But do try the first one, in the form of
?qqplot
Sarah
On Fri, Feb 13, 2015 at 4:32 PM, CHIRIBOGA Xa
Hello! SORRY PROBLEMS WITH FUNCTIONS AGAIN...
I NEED TO RUN A qqPlot ...I TRIED TO INSTALL IT , BUT A WARNING MESSAGE SAID
qqPlot is not available for version 3.1.2.
qqPlot(residuals(anc0),id.method="identify")
Error: could not find function "qqPlot"
> install.packages("qqPlot")
Instal
Hello Michael,
I have a question here. Does Bayesian paradigm deal with MSE kind of stuff?
Thanks & Regards,
Arnab
Arnab Kumar Maity
Graduate Teaching Assistant
Division of Statistics
Northern Illinois University
DeKalb, IL 60115
Email: ma...@math.niu.edu
Ph: 779-777-3428
On Fri, Feb 13,
Both files came through. The R-help list is picky. For example it will accept
cat.txt but not cat.csv.
Now I see what you are after. and I must admit I haven't a clue at the moment.
I suspect others who know more about ggplot can help. If not there is ggplot2
Google Groups that has a lot of k
You don't appear to be supplying a valid parameter set to optim.
The first argument in optim (par) must be a vector of parameters to optimise;
you're passing a vector of NAs. Thise are not finite.
Also, temp is defined as a value and optim will not be able to optimise that.
You need to define
Hi,
Try this.
> sd(unlist(sapply(1:dim(p)[1],function(i)rep(p[i,1],p[i,2]
--
View this message in context:
http://r.789695.n4.nabble.com/sd-mean-with-a-frequency-distribution-matrix-tp4703218p4703220.html
Sent from the R help mailing list archive at Nabble.com.
___
I've been working at interpreting the results of a non-metric multidimensional
scaling analysis. I have been using metaMDS in the vegan package because one of
the benefits is that it also rotates to solution to its principal components.
(Eventually I realized that there is no reason why my resul
Hello all,
I have a question concerning bayesian regression in R using the package bayesm
(version 2.2-5) by Peter Rossi. I have binary choice data and estimated
individual coefficients using the command rhierBinLogit(Data=Data,Mcmc=Mcmc).
That worked out properly, conversion plots, histograms,
I have three directories where there is inside each of them 5 files.each file
is a matrix lines 500 and columns 300. I want to perform an optimization
using values from three corresponding pixels.Finally I get a matrix of lines
500 and columns 300 for each parameter in my equation:
y=(a
I agree that the PG muddies the water a bit on this topic. However, the web
page from which you downloaded the patched version warns:
"This is not an official release of R. Please check bugs in this version
against the official release before reporting them."
When a new version is under develo
S Ellison writes:
>> When using the function boxplot() together with the argument
>> | horizontal = TRUE
>> xlim and ylim become swapped, i.e. ylim refers to the x-axis instead of the
>> y-
>> axis:
>
> It is neither a bug nor undocumented, though the documentation is a
> not in ?boxplot (becaus
Your work around is not as "easy" looking to me.
Survival times come in multiple flavors: left censored, right censored, interval censored,
left-truncated and right censored, and multi-state. Can you give me guidance on how each
of these should sort? If a sort method is added to the package i
> I want to do a boxcox transformation, but I got this:
> Error: could not find function "boxcox"
>
> What can I do?
Well, the recommended 'homework' in the posting guide would be a start.
i) ??boxcox, if you have any packages installed that include something with
that functionality.
ii) RSite
Hi Rich
As usual in R one can come to same result by different means. Using lme from
nlme
fit<- lme(value ~ day, data=test, random= ~1|item)
ranef(fit)
(Intercept)
a -4.8794330
b -3.5063635
c -1.7121614
d -0.2002288
e 0.5118846
f 1.7805122
g 3.4489819
h 4.5568079
fixef(fit)
(Interc
> When using the function boxplot() together with the argument
> | horizontal = TRUE
> xlim and ylim become swapped, i.e. ylim refers to the x-axis instead of the y-
> axis:
It is neither a bug nor undocumented, though the documentation is a not in
?boxplot (because xlim and ylim aren't defined t
Hi Jeff
Can you be more specific why question about "using" R-devel shall be directed
to R-devel help list?
Posting guide tells me:
R-help is intended to be comprehensible to people who want to use R to solve
problems but who are not necessarily interested in or knowledgeable about
programmin
This is the wrong part of my code.
>
>> idName=users[users$id %in% ext]
> idname
> 1: U03AEKWTL agreenmamba
> 2: U032FHV3S poisonivy
> 3: U03AEKYL4 vairis
>
Best is to use:
idNames <- users[pmatch(ext, users$id, duplicates.ok = T)]. This leave
me with an ordered and dupl
I have been searching for a while now, but can't put all pieces of the
puzzle together.
Goal : I want to replace all these kinds of patterns <@U032FHV3S> by
this <@agreenmamba>. In a more generic way, it is replacing 'id' by
user 'name'.
I have two df:
The first, 'history', is some message histor
Its a formatting thing, try
summary(testrow,digits=20)
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Franckx Laurent
Sent: 13 February 2015 11:00
To: r-help@r-project.org
Subject: [R] difference between max in summary table and max function
Dear all
Dear all
I have found out that the max in the summary of an integer vector is not always
equal to the actual maximum of that vector. For example:
> testrow <- c(1:131509)
> summary(testrow)
Min. 1st Qu. MedianMean 3rd Qu.Max.
1 32880 65760 65760 98630 131500
> max(tes
When using the function boxplot() together with the argument
,
| horizontal = TRUE
`
xlim and ylim become swapped, i.e. ylim refers to the x-axis instead of
the y-axis:
--8<---cut here---start->8---
x <- runif(1000)
boxplot(x)
boxplot(
x,
ylim =
Hi all,
I am trying to use R package flexmix for data clustering which I find
extremely useful. I would like to know how many clusters could be there in
the data and the assignment of trials to these clusters.
The data are hand kinematics data- 16 variables at 10 discrete time points
for 17 subje
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