Hi,
On Thu, Aug 22, 2013 at 9:49 PM, Jeff Newmiller
wrote:
> Please don't post in HTML format... it messes with code examples.
>
> Use character indexing (please read the Introduction to R... again if
> necessary).
>
> myf <- function(df, colname){
> df[ ,colname ]
> }
Or df[[colname]] for da
Hi all R mailing listers:
I am using the coda package. I tried to view the source of HPDinterval code
by typing fix(HPDinterval), it dispalys as follows:
function (obj, prob = 0.95, ...)
UseMethod("HPDinterval")
Then I search the answers about this case (see below), it still failed.
Thank you in
Please don't post in HTML format... it messes with code examples.
Use character indexing (please read the Introduction to R... again if
necessary).
myf <- function(df, colname){
df[ ,colname ]
}
colname <- "a"
myf(m,colname)
Until you learn simple R syntax, I strongly recommend avoiding wri
m=as.data.frame(outer(1:5,6:9))
colnames(m)=c('a','b','c','d')
tf=function(df, col){list(mean(eval(substitute(col),df,parent.frame())),col
)}
tf(m,a) will issue error: Error in tf(m, a) : object 'a' not found
How can I replace the col as char 'a' in the function?
Thank you
[[alternat
for example I have data frame m as below:
m=as.data.frame(outer(1:5,6:9))
colnames(m)=c('a','b','c','d')
and I define the function
myf=function(df, colname){
suppose colname is a, then:
how can I get the column 'a'
and how to get the colname as a string, 'a'
}
Thank you!
[[a
Hello,
I am confused with the design of experiment and I need your help. I study
macroinvertebrates from 40 lakes. Each lake was sampled employing stratified
random technique. Each lakes was divided into 3 different lake zones based on
light penetration: near-shore area, transitional, and the d
Dear all,
I have a couple of questions regarding the survival:::cch function.
1) I notice that Prentice and Self-Prentice functions are giving identical
standard errors (not by chance but by programming design) while their beta
estimates are different. My guess is they are both using the
Hello Joshua:
First of all, I thank you for the solution. The solution is quite
elegant and concise. I have one follow-up question/comments. Everything
seems ok but the problem comes if the last month stops somewhere in the
middle of the month. For instance, if I run the data as of yesterda
Dear all,
I have a couple of questions regarding the survival:::cch function.
1) I notice that Prentice and Self-Prentice functions are giving identical
standard errors (not by chance but by programming design) while their estimates
are different. My guess is they are both using the stan
On 22/08/13 21:57, Michael Weylandt wrote:
On Aug 22, 2013, at 7:39, Ben Harrison wrote:
No idea about the problem specifics but what are your OS and version of R? You
might be limited there.
I have 64-bit Ubuntu 12.04, R version 3.0.1.
More likely, however, is that your problem is j
On Aug 22, 2013, at 4:17 PM, Jim Lemon wrote:
> On 08/23/2013 04:32 AM, Jie wrote:
>> Dear All,
>>
>> I would like to draw a graph to illustrate the mapping between two vectors.
>> For instance,
>> a = c(2, 8, 5) ; mapped.a = c(8, 2, 5)
>>
>> I would like to get some picture as
>> http://www.fl
I tried your suggestion on the toy example that was posted by David Carlson
a while back:
set.seed(42)
x <- 1:15
y <- x/(1+x) + rnorm(15,0,0.02)
I found that I had to:
(a) Wrap the "1/x" inside "I()".
(b) Set the offset term to be rep(1,length(x)).
Bottom line:
fit <- glm(y ~
0+I(1/
On Wed, 21 Aug 2013 10:35:53 -0400
SH wrote:
It looks like your problem has already been answered, however, as a
rule of thumb anytime you see a peculiarity like this you should look
for minor variations between what you expected to export and what Excel
really exported as delimited text. Occasi
On 08/23/2013 04:32 AM, Jie wrote:
Dear All,
I would like to draw a graph to illustrate the mapping between two vectors.
For instance,
a = c(2, 8, 5) ; mapped.a = c(8, 2, 5)
I would like to get some picture as
http://www.flickr.com/photos/96546690@N02/9569526581/
Basically, plot all points of
Dear R-help list members,
Version 2.0-0 of the Rcmdr package is now on CRAN and should appear
presently on the various CRAN mirrors. As its number implies, this version
represents a milestone in the development of the package, which first
appeared on CRAN more than 10 years ago. The transition to
On 22.08.2013 19:52, Tal Galili wrote:
Dear Uwe,
Here is the updated code and error massage I get:
exe_URL = 'http://pandoc.googlecode.com/files/pandoc-1.11.1.msi'
exe_filename <- file.path(tempdir(), basename(exe_URL))
download.file(exe_URL, destfile = exe_filename, mode = "wb", method=
"int
The factor gm has nine levels, therefore you need to use a pch vector with
9 values
(duplicates are ok). Therefore this will work.
bwplot(var1 ~ gm, data=mydata, col="black",
pch=rep(1:3,3), ##par.settings=list(box.dot=list(pch=1:3)),
panel=panel.bwplot.intermediate.hh,
On Aug 22, 2013, at 9:23 AM, Garkuwa, Nuru Adamu wrote:
> I used the codes below trying to fit a model at .95 c.i. of size relationship
> but it fails.
>
> library(quantreg)
> range(len$length) # [1] 28 98
> range(len$preyl)
> x<-rq(len$preyl~len$length,tau=0.95)
> plot(x, type="b", xlab="le
Dear Henrik,
Thanks for getting back to me.
Apologies - have resolved the problem myself. The package in question
has quite a bit of legacy code written by others and as it appeared had
explicit library() calls inside the source code on the top level. After
removing those calls things started w
Dear R-users
I applied vegan's varpart function to partition the effects of
explanatory matrices. Adj. R square for the unique fraction [a] is
0.25. Does anyone know why the decomposition by hand using rda gives
me a different result for [a] (constrained proportion is 0.32)? I used
cbind()
*Issue:*
Usage of ETS R codes through RExcel macros in VBA
Given below is my command code:
Rinterface.runrcodefromrange Range(Sheet1!B2:D8)
Following are the codes written in the given cell reference:
#!rput
zz
'Sheet1'!$B$2:$B$22
library(forecast)
zz <- ts(zz,freq=365,start=c(20
Dear all,
I have a plot with two lines and I'm using labcurbe (package Hmisc) to show
the legend.
Everything works well, except that the line displayed in legend box should
have same style (plotting character) as the line in the plot. This is
achieved by pch parameter and looks like that it's only
On Thu, Aug 22, 2013 at 12:23 PM, Tal Galili wrote:
> Quick update - I think most of the problem is resolved.
> After more checks (following Uwe comments):
> 1) I see that the installer file runs properly when I set "mode='wb'", and
> fails when "mode='w'" (so that solves why some of the time the
It is not hard to create a basic function for that purpose:
map.vec <- function(x, y) {
dfa <- data.frame(a=x, x0=rep(1, length(x)),
y0=length(x):1)
dfamap <- data.frame(a=y, x1=rep(2, length(y)),
y1=length(y):1)
vec <- merge(dfa, dfamap, by=1)
plot
Hi, R.oo author here:
This looks like a possible namespace issues, which might have gone
under the radar if you have only used R v2.11.x until now. Could you
forward me your MyPackage test package so I can troubleshoot/reproduce
(I think the answer is in your DESCRIPTION/NAMESPACE file).
/Henrik
Quick update - I think most of the problem is resolved.
After more checks (following Uwe comments):
1) I see that the installer file runs properly when I set "mode='wb'", and
fails when "mode='w'" (so that solves why some of the time the installer
didn't run)
2) I still get the error massage mentio
I used the codes below trying to fit a model at .95 c.i. of size relationship
but it fails.
library(quantreg)
range(len$length)
range(len$preyl)
x<-rq(len$preyl~len$length,tau=0.95)
plot(x, type="b", xlab="length (cm)",ylab="preyl (cm)",ylim=c(5,35),
xlim=c(10,100), data=len)
Cheers,
Nuru
I have the following code and data
data.csv
"","Goal","Frequency","Weight","Group"
"1","Move",13,0.245283018867925,"Public"
"2","Create",10,0.188679245283019,"Public"
"3","Strengthen",30,0.566037735849057,"Public"
"4","Move",6,0.6,"Board"
"5","Create",0,0,"Board"
"6","Strengthen",4,0.4,"Board"
"
Dear All,
I would like to draw a graph to illustrate the mapping between two vectors.
For instance,
a = c(2, 8, 5) ; mapped.a = c(8, 2, 5)
I would like to get some picture as
http://www.flickr.com/photos/96546690@N02/9569526581/
Basically, plot all points of a in a vertical line, and mapped.a in
When you are using a package from CRAN, you should probably contact the
author of the package (in this case, that is me) with questions, instead of
asking on R-help.
The corrgram package DOES use base graphics. For instance, when using 3
variables, each one of the squares of the corrgram is using
Dear Uwe,
Here is the updated code and error massage I get:
exe_URL = 'http://pandoc.googlecode.com/files/pandoc-1.11.1.msi'
exe_filename <- file.path(tempdir(), basename(exe_URL))
download.file(exe_URL, destfile = exe_filename, mode = "wb", method=
"internal")
#
### Error massage:
tr
Free support for RExcel is available by subscribing to the mailing list at
rcom.univie.ac.at
and posting questions to the list.
On Aug 22, 2013, at 7:04 PM, Jeff Newmiller wrote:
> This is not on topic for this mailing list. See the Statconn website for
> support for RExcel.
__
This is not on topic for this mailing list. See the Statconn website for
support for RExcel.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Hi,
May be this helps:
dat1<- structure(...
dat1$ID<- row.names(dat1)
library(reshape2)
dat2<-melt(dat1,id.vars="ID")
dat2New<-dat2[order(as.numeric(dat2$ID)),]
head(dat2New)
# ID variable value
#1 1 Ramses 4
#11 1 Sheiks 5
#21 1 Trojans 7
#31 1 Unnamed 2
#2 2 R
Hi!
I am looking to choose a condom based on its pleasure score.
I received some summarised data from 10 individuals:
structure(list(Ramses = c(4, 4, 5, 5, 6, 3, 4, 4, 3, 4), Sheiks = c(5,
5, 6, 4, 7, 6, 4, 5, 6, 3), Trojans = c(7, 8, 7, 9, 6, 3, 2,
2, 2, 3), Unnamed = c(2, 1, 1, 3, 3, 4, 5, 4, 4
Hello,
I am trying to plot a few correlograms on the same figure, with the function
corrgram() from the package corrgram. However, the function does not seem to
use the base graphic system, as setting out the multiple figure layout with,
e.g., par(mfrow=c(2, 2,)) does not work.
Does anybody k
Hi,
Also, you could use:
library(data.table)
dt1<- data.table(dat1,key=c('SampDate','DepthM'))
unique(dt1)
# SiteID SiteName SampDate DepthM PDesc MAbbr Measure DNU
#1: 1 Big Platte Lake 2006-09-20 0.000 TP Grab 6.58 FALSE
#2: 1 Big Platte Lake 2006-09-20 2.286
HI Samuel,
Based on the output you wanted:
(It would be better to use ?dput() to show the example dataset)
dat1<- structure(list(SiteID = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L), SiteName = c("Big Platte Lake", "Big Platte Lake",
"Big Platte Lake", "Big Platte Lake", "Big
Issue:
Usage of ETS R codes through RExcel macros in VBA
Given below is my command code:
Rinterface.runrcodefromrange Range(“Sheet1!B2:D8”)
Following are the codes written in the given cell reference:
#!rput zz 'Sheet1'!$B$2:$B$22
library(forecast)
zz <- ts(zz,freq=365,start=c(2009
Genious!
Thank's a lot!
One final question: I am making black and white plots and will have to change
the pch=c(...) setting. The problem is that every time I change this setting I
get an error message and have to close RStudio and re-open it for anything to
work again. I can change the pch=c(..
This works:
expression(paste('Greek', italic('\uo3bc'))
thanks!
Jake
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thursday, August 22, 2013 2:01 AM
To: David Winsemius
Cc: Beaulieu, Jake; r-help@r-project.org
Subject: Re: [R] italic(mu)
On Aug 20, 2013
Hello,
I must tell you once again to address your questions to
r-help@r-project.org
And since there was no subject line, I've made up one.
Your question seems to be a question about excel, and I really can't
help you. Maybe someone else can.
But if you write.table(etszP) can't you read that
You specified 10 cuts to the dendrogram (all possible) so each
column represents the cluster membership for that number of
clusters. Column 1 is all ones because all rows belong to a
single cluster and in column 10 all the rows belong to different
clusters. For g9-9, 8 means that g9 belongs to clus
>From POSIXct to numeric and back with time zone
I am running regressions on data which has time series with different time
resolution. Some data has hourly resolution, while most has either daily or
weekly resolution. Aggregation is used to make the hourly data daily, while
liner interpolatio
Below are two ways to accomplish this. The last() function will not
pad with NA if you request more days than available, so you may need
to handle that if it's an issue. I prefer the period.apply method
because it keeps the data in an xts object, which I find easier to
work with.
getSymbols("YHO
On Aug 22, 2013, at 7:39, Ben Harrison wrote:
> I have a 70363 x 5 double matrix that I am playing with.
>
> > head(df)
>GR SP SN LN NEUT
> 1 1.458543 1.419946 -0.2928088 -0.2615358 -0.5565227
> 2 1.432041 1.418573 -0.2942713 -0.2634204 -0.5927334
> 3 1.4066
I have a 70363 x 5 double matrix that I am playing with.
> head(df)
GR SP SN LN NEUT
1 1.458543 1.419946 -0.2928088 -0.2615358 -0.5565227
2 1.432041 1.418573 -0.2942713 -0.2634204 -0.5927334
3 1.406642 1.418226 -0.2958296 -0.2652920 -0.6267121
4 1.382284 1.4188
Tal,
please quote the whole message!
I had to look for the original one now.
And then I found that code was not reproducible for me:
exe_URL = 'http://pandoc.googlecode.com/files/pandoc-1.11.1.msi'
exe_filename <- file.path(tempdir(), file.name.from.url(exe_URL))
Error in file.path(tempdir()
On 22/08/2013 09:24, Sachinthaka Abeywardana wrote:
Hi all,
I get the expected behaviour of getting a useful model if I do the following
This is an 'expectation problem', not a 'stepAIC problem'.
And stepAIC is not part of R and you are not crediting the tools you use.
fit<-lm(
expressions
Hi all,
I get the expected behaviour of getting a useful model if I do the following
fit<-lm(
expressions[,i]~expressions[,pa_all[1]]+expressions[,pa_all[2]]+expressions[,pa_all[3]]+expressions[,pa_all[4]]+expressions[,pa_all[5]])
step<-stepAIC(fit, direction="both")
Output:
Step: AIC=-78.75
ex
50 matches
Mail list logo
