Jim:
Did you forget about with() ?
Instead of:
by(lbtdat$latency,list(lbtdat$subject,
lbtdat$condition,lbtdat$state),mean)
##do
with(ibtdat,by(latency,list(subject,condition,state),mean))
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not w
Answered my own question.
In survey, summary does it.
On 2312//2013, 5:31 PM, "Nathan Pace" wrote:
>
>>The svycoxph function in the survey package loads the survival package
>>and
>>produces objects of class svycoxph and coxph.
>>
>>The print.coxph function - print(coxph.object, conf.int = 0.95
On 12/23/2013 11:31 PM, Laura Bethan Thomas [lbt1] wrote:
Hi All,
Sorry for what I imagine is quite a basic question. I have been trying to do is
create latency averages for each state (1-8) for each participant (n=13) in
each condition (1-10). I'm not sure what function I would need, or what
>The svycoxph function in the survey package loads the survival package and
>produces objects of class svycoxph and coxph.
>
>The print.coxph function - print(coxph.object, conf.int = 0.95) - in the
>survival package lists the values of the coxph object including the hazard
>ratios with 95% CIs.
>
I can't think of a straightforward way. You might be able to use
the image.plot() function in package fields by using
legend.only=TRUE to add the legend to your existing plot.
David
From: Morway, Eric [mailto:emor...@usgs.gov]
Sent: Monday, December 23, 2013 3:46 PM
To: dcarl...@tamu.edu
C
I believe you are right. We thank either Gmail or [[alternative HTML
version deleted]] for this. I think showNonASCII() is just irrelevant
here and pulling us to the wrong direction.
It is not reliable to paste code into Email due to the potentially
wrong text wrapping. Please consider an email at
Thanks David,
Instead of using terrain.colors or heat.colors, I went with:
library(colorRamps)
cv <- matrix(as.integer(cut(vals2, breaks=100)), dim(vals2))
pal <- blue2green2red(100) #a function from colorRamps
Do you happen to have any clever ideas for a legend? I could play around
with a ser
Thanks, I was not getting anything when I printed the x and so thought I was
doing something wrong. Instead I just didn't seem to have a non-ASCII
character.
John Kane
Kingston ON Canada
> -Original Message-
> From: murdoch.dun...@gmail.com
> Sent: Mon, 23 Dec 2013 15:52:32 -0500
> To
On 13-12-23 1:07 PM, John Kane wrote:
Thanks Duncan.
I had the feeling I was doing something wrong but did not realise it was that
stupid.
showNonASCII('ggplot(df, aes(x = x)) + geom_histogram(aes(y = ..density..)),
binwidth = 1, colour = "black", fill = "white")')
now runs
Hi,
You could either try:
#dat1 ##dataset
aggregate(latency~.,data=dat1,mean)
#or
library(data.table)
dt1 <- data.table(dat1,key=c('subject','conditionNo','state'))
dt1[,mean(latency),by=c('subject','conditionNo','state')]
A.K.
On Monday, December 23, 2013 2:20 PM, Laura Bethan Thomas [lbt
I would suggest using summaryBy()
library(doBy)
# sample data with you specifications
subject <- as.factor(rep(seq(13), each = 5))
state <- as.factor(sample(c(1:8), 65, replace = TRUE))
condition <- as.factor(sample(c(1:10), 65, replace = TRUE))
latency <- runif(65, min=750, max = 1100)
dat <- da
Waterloo Graphics is open-source and can be used from R.
Graphics can be copied and pasted in vector format to Word on Windows or
Mac.
There is also an SVG file save option that produces output with easy-to-use
object groupings for editing an Adobe Illustrator/Inkscape.
(as well as HTML5/Processi
Hi, chiming in.
Pasted the code in R studio and the format parser wouldn't mark the R code
chunks. It was because there were line breaks in the middle of chunk
options tags. Couldn't test if removing line breaks works, but maybe
that's the source of the problem?
On Mon, Dec 23, 2013 at 10:37 AM,
Hello, i'm using R for the exploration of a time series and i'm stuck in a
problem with the fitting of the distribution.
What's the difference between "fitdistr" and "mle"?
library(MASS)
fitting <- fitdistr(ret,densfun="normal")
print(c(mean(ret),sd(ret)))
> Hi All,
>
> Sorry for what I imagine is quite a basic question. I have been trying to do
> is create latency averages for each state (1-8) for each participant (n=13)
> in each condition (1-10). I'm not sure what function I would need, or what
> the most efficient ay of calculating this woul
Costas Vorlow gmail.com> writes:
> I am trying to write a code that executes an R command at specific time
> intervals.
>
> É want R to do that instead of the operating system.
>
> Any help/pointer extremely welcome.
Don't do it. Just rely on cron [if you're lucky enough to be on Linux
or OS
There is a fundamental problem with your
code, and there is the problem that you
have (sort of) identified.
The fundamental problem is that you are
only going to get the results of the last
call to 'ca.jo' that is done -- assuming
it were to run. You presumably want to
save some information from
Recode val into a set of integers equal to the number of color
levels you want and then use heat.colors(), terrain.colors(), or
a similar function to define a vector of continuous colors.
# cut makes it easy to split up the data but it creates a factor
# and loses the matrix dimensions so we have
Hi,
No problem.
If you have two columns and need the ratio, you could use ?transform
testframe$data1 <- c(2.24,6.5,4.34)
dcast(transform(testframe,ratio=data/data1),factor2~factor1,value.var="ratio",mean)
# factor2 a b
#1 1 1.491071 NaN
#2 2 0.483871 0.6461538
A
useRs,
The example code below is an attempt to plot some spatial data that is
associated with an irregularly spaced grid. The last thing I hope to do
with this example is assign the color of each polygon generated in the
nested for loop based on the value contained in "vals". The R code I'm
seek
Thanks Duncan.
I had the feeling I was doing something wrong but did not realise it was that
stupid.
showNonASCII('ggplot(df, aes(x = x)) + geom_histogram(aes(y = ..density..)),
binwidth = 1, colour = "black", fill = "white")')
now runs and does what the help page seems to im
On 13-12-23 12:40 PM, John Kane wrote:
Thanks Richard. I did not realise such a function existed.
Assuming I am using it correctly I do get an error though not where I was
expecting it. Anyway the code below returns an error
library(tools)
showNonASCII("ggplot(df, aes(x = x)) + geom_histogr
Thanks Richard. I did not realise such a function existed.
Assuming I am using it correctly I do get an error though not where I was
expecting it. Anyway the code below returns an error
library(tools)
showNonASCII("ggplot(df, aes(x = x)) + geom_histogram(aes(y = ..density..)),
binwidth = 1,
Does not seem to be. I 'think' I removed all the line breaks and it still is
not compiling. Thanks for the suggestion. I had not bothered to paste the
<<>>= text into RStudio and since TexMaker has an automatic wrap, I would never
have noticed it.
John Kane
Kingston ON Canada
-Original
Dear all,
I fit co-integration function between two integrated variables(y1 and y2)
over different grid points:
for(i in 1:N1){
for(j in 1:N2){
co<-ca.jo(data.frame(cbind(y2[i,j,],y1[i,j,])),type="trace", K=2,
spec="transitory",ecdet="const",season=NULL,dumvar=NULL)
}}
I have already extracted gr
If the problem seems to be non-ASCII characters, then the first investigation
step is to use the R functions
?tools::showNonASCII
?tools::showNonASCIIfile
On Mon, Dec 23, 2013 at 11:37 AM, John Kane wrote:
> Same result here with the same error message mentioned in my first post. I
> tried it
Same result here with the same error message mentioned in my first post. I
tried it in Texmaker which is my usual Latex editor, not that I do much in
Latex, and then tried it in RStudio and it is still choking.
Interestingly EMACS will process it and produce a pdf but it simply produces.
It
Rewarp gmail.com> writes:
> I am trying to install RcppEigen, which depends on Rcpp. Here's what the
> terminal says:
>
> > install.packages("RcppEigen")
[...]
> g++ -shared -o RcppEigen.so RcppEigen.o fastLm.o
> -L/home/rewarp/R/x86_64-pc-linux-gnu-library/3.0/Rcpp/lib -lRcpp
> -Wl,-rpath,/home/
Hello,
I am trying to write a code that executes an R command at specific time
intervals.
É want R to do that instead of the operating system.
Any help/pointer extremely welcome.
Thanks in advance,
Costas
[[alternative HTML version deleted]]
___
As I said, ?tapply gives you an answer (without using other packages) . Read it.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
H. Gilbert Welch
On Mon, Dec 23, 2013 at 6:2
HI,
I think this will be more appropriate.
dcast(testframe,factor2~factor1,value.var="data",mean)
factor2 a b
1 1 3.34 NaN
2 2 2.10 4.2
A.K.
On Monday, December 23, 2013 9:37 AM, arun wrote:
Hi,
You could try:
library(reshape2)
dcast(as.data.frame(as.table(by(testframe[,3],te
Hi,
You could try:
library(reshape2)
dcast(as.data.frame(as.table(by(testframe[,3],testframe[,-3],mean))),factor2~factor1,value.var="Freq")
# factor2 a b
#1 1 3.34 NA
#2 2 2.10 4.2
A.K.
On Monday, December 23, 2013 9:24 AM, Onur Uncu wrote:
Sure, here is a reproducible exam
Sure, here is a reproducible example:
testframe<-data.frame(factor1=c("a","b","a"),factor2=c(1,2,2),data=c(3.34,4.2,2.1))
splitframe<-split(testframe,list(factor1=testframe$factor1,factor2=testframe$factor2))
lapply(splitframe,function(x)mean(x[,"data"]))
The above lapply returns
$a.1
[1] 3.34
Book title: Data Mining Applications with R
Editors: Yanchang Zhao, Yonghua Cen
Publisher: Elsevier
Publish date: December 2013
ISBN: 978-0-12-411511-8
Length: 514 pages
URL: http://www.rdatamining.com/books/dmar
An edited book titled Data Mining Applications with R was released in
December 2013,
Hello,
See ?redfit from "dplR", for example.
HTH,
Pascal
On 23 December 2013 18:46, nuncio m wrote:
> Dear useRs,
>
> I have a time series of length approcimately 55. Is it possible to find the
> significance of fft spectral peaks with R?
>
> thank you
>
> --
> Nuncio.M
> Scientist
> National C
Hi Dan
I think you still have problems with embedded characters or some problems in
char code page conversion or the like.
Not knowing knitr but Sweave I cobbled the figures manually and ran the
sweave file to produce the latex file.
Latex was consistently stopping at the \caption and \ref
I am really sorry for posting a non-working example. It is running when I
cut the code from my previous mail into a clean session in RStudio (OSX).
However, I suspect that you are right. I did cut and paste some code from a
forum yesterday which had characters that had to be replaced. I gave emacs
Dear useRs,
I have a time series of length approcimately 55. Is it possible to find the
significance of fft spectral peaks with R?
thank you
--
Nuncio.M
Scientist
National Center for Antarctic and Ocean research
Head land Sada
Vasco da Gamma
Goa-403804
ph off 91 832 2525636
ph: cell 91 98903574
Hi
You have only single value for each participant in each gruppo and AFAIK you
can not do statistic on single value.
You can check differences among participants
fit<-lm(valor~participantes, data=df2)
> fit<-lm(valor~participantes, data=df2)
> anova(fit)
Analysis of Variance Table
Response: v
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