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:
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
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
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
Hello,
See ?redfit from dplR, for example.
HTH,
Pascal
On 23 December 2013 18:46, nuncio m nunci...@gmail.com 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
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,
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
$b.1
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 onuru...@gmail.com wrote:
Sure, here is a
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 smartpink...@yahoo.com wrote:
Hi,
You could try:
library(reshape2)
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 6:22
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]]
Rewarp rewarp at 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
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.
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 jrkrid...@inbox.com wrote:
Same result here with the same error message mentioned in my first
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 grid
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
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,
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)) +
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
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
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
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
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
Costas Vorlow costas.vorlow at 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
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 would be.
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, 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
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
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 -
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
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 and
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:
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
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
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
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.
When
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
Answered my own question.
In survey, summary does it.
On 2312//2013, 5:31 PM, Nathan Pace n.l.p...@utah.edu 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 =
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
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