Dear All,
I am facing the task to extract data from array to table. here an example
of array.
##Get A list of Matrices with unequal rows
Disease - NULL
Diseases - NULL
ListMatByGene - NULL
for(i in 1:3){
Disease[[i]] -matrix(sample(-30:30,25+(5*
i)),5+i)
rownames(Disease[[i]]) -
A~ha~!! Thank you, Prof. Peter Dalgaard, so much for your wonderful
lesson!!! Learning new things everyday from this R-help mailing list!
Chel Hee Lee
On 2/11/2015 10:37 AM, peter dalgaard wrote:
On 11 Feb 2015, at 17:11 , Chel Hee Lee chl...@mail.usask.ca wrote:
The functional form
For finance applications, I'm surprised that I am unable to find a function to
compute the portfolio return (sqrt(t(w) %*% V %*% w)) where w are portfolio
weights and V is the cov(returns). The Performance Analytics portfolio return
function seems to compute something else.
Ernie
Hi,
Here is an implementation. The image is uploaded as Rplot02.png.
gen - read.table(geno.txt,header=TRUE)
gen
Genotype E1 E2E3 E4 E5 E6 E7 E8 E9 E10 E11 E12 E13
E14 E15 E16 E17 E18
1 G1 0.79 2.11 6.21 0.56 4.06 2.13 5.61 0.20 3.32 3.01 5.12 0.77 0.78
Hello R-helpers,
It is possible donwload youtube videos with R? I made a google search and
find no options to do that.
Thanks in advanced,
Raoni
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On 11 Feb 2015, at 17:11 , Chel Hee Lee chl...@mail.usask.ca wrote:
The functional form given in the post written by Ssuhanchen captures my eyes.
It is the cumulative distribution function of Poisson when the number of
counts is less than or equal to 2 with unknown parameter mu=x/2.
The functional form given in the post written by Ssuhanchen captures my
eyes. It is the cumulative distribution function of Poisson when the
number of counts is less than or equal to 2 with unknown parameter
mu=x/2. Since it is a nonlinear function, there may be multiple
solutions but the
The set of things that R can do is not a subset of the set of things packages
may have been written for.
Have at it.
---
Jeff NewmillerThe . . Go Live...
I have created robust regression models using least trimmed squares and
MM-regression (using the R package robustbase).
I am now looking to create prediction intervals for the predicted results.
While I have seen some discussion in the literature about confidence intervals
on the estimates
Presumably you've checked out:
http://cran.r-project.org/web/views/Robust.html
If you can estimate the variance of parameter estimates, betahat, then
you can estimate the variance of a predicted value, X betahat; add the
estimated variance of individuals to this if that's what you're
looking for
On Sun, Feb 8, 2015 at 5:15 PM, Duncan Murdoch murdoch.dun...@gmail.com wrote:
On 08/02/2015 4:06 PM, Glenn Schultz wrote:
Hello All,
I am in the final stages of building my first package BondLab and the
check throughs the following warning. I think this is namespace thing. I
have not
That seems to have worked, both in the new and old version of R. I'll do more
unit testing on other files.
Thank you, Gabor.
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: Wednesday, February 11, 2015 10:22 AM
To: Doran, Harold
Cc:
On Wed, Feb 11, 2015 at 9:45 AM, Doran, Harold hdo...@air.org wrote:
I have a function written and tested using R 3.0.1 and sqldf_0.4-7.1 that
works perfectly. However, using this same code with R 3.1.2 and sqldf_0.4-10
yields the error below that I am having a difficult time deciphering.
Thanks Lee and Huang.
That is useful.
Best
On Fri, Feb 6, 2015 at 1:53 AM, Chel Hee Lee chl...@mail.usask.ca wrote:
lapply(1:2, function(x) t(A[rev(1:3),,x]))
[[1]]
[,1] [,2] [,3]
[1,] g d a
[2,] h e b
[3,] i f c
[[2]]
[,1] [,2] [,3]
[1,] p m j
[2,] q n k
[3,]
Hi,
Suppose your data frame is called data and the name of the factor column
is named tobeConverted. I have tried this and it worked. Hope this helps.
as.numeric(as.character(data$tobeConverted))
--
View this message in context:
Dear John,
Dear Dan,
Many thanks for your response. It works perfectly.
Best,
- Mail original -
De : John Fox j...@mcmaster.ca
À : 'varin sacha' varinsa...@yahoo.fr
Cc : 'r-help@R-project.org' r-help@r-project.org
Envoyé le : Mercredi 11 février 2015 3h12
Objet : RE: [R] library(Rcmdr)
I have a function written and tested using R 3.0.1 and sqldf_0.4-7.1 that works
perfectly. However, using this same code with R 3.1.2 and sqldf_0.4-10 yields
the error below that I am having a difficult time deciphering. Hence, same code
behaves differently on different versions of R and
That's exactly what I wanted, thank you very much!
My intent was to perform the SubStr operation first, and then apply the
types to the columns. I wasn't expecting the two types in the same column.
I appreciate your response!
On Tue, Feb 10, 2015 at 5:03 PM, David Winsemius
Hi Kim
fancyRpartPlot is a front-end to prp, and you can pass it all of the
prp options - it says this in the help for fancyRpartPlot, and that's
about all it says.
So you need to spend some time reading about prp options, and how to
customize your plot to get what you want. There are lots of
I suspect that your treatment levels are perfectly nested inside the
soilgroup levels. If this were the case you can either use one of the
factor in your analysis or the other, depending on what you want to
analyze.
Il giorno Wed, 11 Feb 2015 14:47:18 +
Mir Salam mir.sa...@uef.fi ha scritto:
Dear all,
I have follwing factors in my data set (G)
names(G)
relative ht, biomass, treatment, variety, soil group
relative height
biomass
treatment (4 levels)- control+lime, control+ash, contaminated soil+lime,
contaminated soil+ash
variety (4 levels)- Salix swherinii, Salix mursinifolia,
Is there any package that can replace maxent for a large number of independent
variables?
Thanks,
Ken
Sent from my iPhone
On Feb 2, 2015, at 6:07 PM, Kenneth Z kenneth...@gmail.com wrote:
Yesterday I installed the most recent R and maxent package, but it stopped
working. Even a simple
Marco Barbàra jabbba at gmail.com writes:
DeaR userRs,
I recently read this Liang-Zeger article:
http://sankhya.isical.ac.in/search/62b1/fpaper7.html
in which (among other things) they adopt a random intercept model for
pre-post designed trials, using a conditional likelihood
I am seeking an optimization routine that can deal with the following problem:
Maximize g(x), where x is a vector and g is nonlinear, subject to linear
constraints of the form h(x)0 and m(x)=0 and subject to the constraint that
all values of x are 0 or 1.
I can't find a nonlinear optimization
Hi,
searching for kalman filter R gives this paper that was published in
JSS. It may help.
@article{Tusell:2010:JSSOBK:v39i02,
author = Fernando Tusell,
title = Kalman Filtering in R,
journal = Journal of Statistical Software,
volume = 39,
number = 2,
pages =
On 11 Feb 2015, at 16:57 , anord andreas.n...@biol.lu.se wrote:
Dear R users,
We are working on a data set in which we have measured repeatedly a
physiological response variable (y)
every 20 min for 12 h (time variable; 'x') in subjects ('id') beloning to
one of five groups ('group';
I am using arima(x, order=c(p,0,q)) function for my project, which deals
with a set of large differenced time series data, data size varies from
8000 to 7. I checked their stationarity before applying arima.
Occasionally, arima(x, order=c(p,0,q)) gives me error like following (which
stops
Does anyone have code that uses a Kalman filter to impute time-series data? If
not, do you know of any software that can be used to impute time-series data?
Thank you,
John
John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of
Hello,
You're missing a dollar sign: 2$$$5, not 2$$5.
Hope this helps,
Rui Barradas
Em 11-02-2015 14:53, Tim Victor escreveu:
All,
Assume we have data in an ASCII file that looks like
Var1$Var2$Var3$Var4
1$2$3$4
2$$5
$$$6
When I execute
read.table( 'test.dat', header=TRUE, sep='$' )
I,
On 11/02/2015 19:38, Bert Gunter wrote:
Presumably you've checked out:
http://cran.r-project.org/web/views/Robust.html
If you can estimate the variance of parameter estimates, betahat, then
you can estimate the variance of a predicted value, X betahat; add the
estimated variance of individuals
All,
Assume we have data in an ASCII file that looks like
Var1$Var2$Var3$Var4
1$2$3$4
2$$5
$$$6
When I execute
read.table( 'test.dat', header=TRUE, sep='$' )
I, of course, receive the following error:
Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings,
:
line 2 did
Hi,
If x is a data frame, then x$getmean will try to get the vector named
getmean in x. You put () after x$getmean. I think r is confused about
it. It appears that you want to call a function named getmean().
--
View this message in context:
Dear R users,
We are working on a data set in which we have measured repeatedly a
physiological response variable (y)
every 20 min for 12 h (time variable; 'x') in subjects ('id') beloning to
one of five groups ('group'; 'A' to 'E'). Data are located at:
Hmmm.
Well, a youtube video is a file. Therefore search for R download file
and you will find the download.file() function.
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 2/11/15, 8:17 AM, Raoni Rodrigues
Petr,
My first attempt is to use the simple=TRUE argument to interaction2wt.
Then the bwplots in the item|item panel show the behavior of value over day
for each item. You get a plot similar to this panel with the growth curve plots
from nlme, for example,
bwplot(value ~ day | item,
Hi Rebecca,
It will be very helpful if you can provide a set of specific functions
g(x), h(x) and m(x).
--
View this message in context:
http://r.789695.n4.nabble.com/Nonlinear-integer-programming-question-tp4703122p4703128.html
Sent from the R help mailing list archive at Nabble.com.
I was playing with some examples of piecewise regression using lm() and have
come across a behavior I'm uncertain about.
Below is simple 3-segment dataset. I compare predicted output of a model
created by one call to lm() to that of 3 models created by 3 calls to lm().
In case A and B, the
anord andreas.nord at biol.lu.se writes:
[snip snip]
We are working on a data set in which we have measured repeatedly a
physiological response variable (y)
every 20 min for 12 h (time variable; 'x') in subjects ('id') beloning to
one of five groups ('group'; 'A' to 'E'). Data are
On 11/02/2015 4:30 PM, Goldschneider, Jill wrote:
I was playing with some examples of piecewise regression using lm() and have
come across a behavior I'm uncertain about.
Below is simple 3-segment dataset. I compare predicted output of a model
created by one call to lm() to that of 3 models
Hi all,
Simple question I should know: I'm unclear on the logic of why the sum of a
row of a data.frame returns a valid sum but the mean of a row of a
data.frame returns NA:
sum(rock[2,])
[1] 10901.05
mean(rock[2,],trim=0)
[1] NA
Warning message:
In mean.default(rock[2, ], trim = 0) :
Hi,
rock[2,] is a data frame and you should not use sum() on a data frame, first
google hit for the error message gives
http://stackoverflow.com/questions/19697498/r-beginner-argument-is-not-numeric-or-logical-returning-na
Otherwise I think you should use
?rowSums and ?rowMeans if you have
I am aware that it is possible to specify a subset with a single
logical operator when constructing a model, such as:
svyglm(formula, design=data, subset=variable==value).
What I can't figure out is how to specify a subset with two or more
logical operators:
svyglm(formula, design=data,
Hello
I tried use grofit package in our data set. We provide a subset of our
data with X iso, and 4 doses, and insect died was count each day for
long 5 days. We started with Y insects per dishes. When one is dead, it
was counted and removed. Died insect is cumulative in the next days.
i.e. day 1
Hi all,
In the attachment or this link (http://oi58.tinypic.com/35ic9qc.jpg) you'll
find the decision tree I made. I used the Rpart package to make the tree and
the rattle package using the fancyRpartPlot to plot it. The data in the tree
looks different than about every example I have seen
DeaR userRs,
I recently read this Liang-Zeger article:
http://sankhya.isical.ac.in/search/62b1/fpaper7.html
in which (among other things) they adopt a random intercept model for
pre-post designed trials, using a conditional likelihood approach
(I didn't think it possible with only two
First:
summary(ss.rpart1)
or summary(ss.rpart, file=whatever)
The printout will be quite long since your tree is so large, so the second form may be
best followed by a perusal of the file with your favorite text editor. The file name of
whatever above should be something you choose, of
gabx@hortensia [R] sessionInfo()
R version 3.1.2 (2014-10-31)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
LC_MONETARY=en_US.UTF-8
[6] LC_MESSAGES=en_US.UTF-8LC_PAPER=en_US.UTF-8
arnaud gaboury arnaud.gaboury at gmail.com writes:
Now when trying to update Rcpp:
gabx at hortensia [R] install.packages(Rcpp)
.
Error in unloadNamespace(pkg_name) :
namespace ‘Rcpp’ is imported by ‘reshape2’, ‘plyr’, ‘dplyr’ so
cannot be unloaded
..
hi brennan, survey design objects can be subsetted with the same subset()
syntax as data.frame objects, so following jeff's advice maybe you want
svyglm( formula , design = subset( surveydesign , variable %in% c( 'value
a' , 'value b' ) ) )
for some examples of how to construct a survey design
This seems like a fundamental misunderstanding on your part of how operators,
and in particular logical expressions, work in computer languages. Consider
some examples:
1+2 has a numeric answer because 1 and 2 are both numeric.
1+a has at the very least not a numeric answer because the values
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