Is there any way to train Weka classifiers using weighted instances in
RWeka? I have done it in Weka in Java, but I don't see any way to pass
the weights to the classifier in the RWeka documentation.
Thanks
__
R-help@r-project.org mailing list
https://s
Thanks Dennis,
Your code also produces same as Jim's but I am not looking that one, I need
to use "cbind" so that finally I will get the data frame of size 200X320
(i,e, 200 X(16X20)).
Thanks
On Tue, Dec 4, 2012 at 10:46 PM, Dennis Murphy wrote:
> In addition to Jim's reply, had you used the pl
Dear R community,
I am trying to fit an nlme model where I want to estimate the fixed effects of
two treatments on the parameters on the following equation
Photo~(a*(1-exp(-c*PARi/a)))-b
I was able to fit a simple model without covariates following the method
described in Mixed-Effects Methods
Hi,
p <- lapply(1:1e6, function(i)c(i, log2(i)))
system.time(z1 <- t(sapply(p,function(x)x)))
# user system elapsed
# 2.568 0.048 2.619
system.time(z1 <- do.call(rbind,p))
# user system elapsed
# 4.000 0.052 4.060
A.K.
- Original Message -
From: William Dunlap
Dear All,
I am trying to fit a generalized linear model with bigglm() (v. 0.8) on
a large dataset containing many zeros under a poisson distribution (in R
2.15.0, 64 bit Ubuntu 12.04). The bigglm() call gives an error message
(NA/NaN/Inf in foreign function call (arg 3)). If I understand
correctly
Hi,
In addition, you can also subset using:
births <- scan("http://robjhyndman.com/tsdldata/data/nybirths.dat";)
birthstimeseries <- ts(births, frequency=12, start=c(1946,1))
library(xts)
birthstimeseriesxts<-as.xts(birthstimeseries)
birthstsJanFeb<-birthstimeseriesxts[format(time(birthstimeserie
On 12/05/2012 04:24 AM, T Bal wrote:
Hi,
In the plot function I want to label x axis as the numbers between 1 and 12
(so 1, 2, 3, 4, 5, ..., 12). How should I do it? The range of x values are
different than this range. Thanks!
Hi T Bal,
You are probably troubled by missing labels. Try the staxl
On 12/05/2012 01:01 AM, anoumou wrote:
Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
...
But i do not get the results,i try by all means but i d'ont understant the
problem.
Hi anoumou,
Your function provides almost no indication of what two of its five
Thanks Jim but I need to use "cbind" otherwise I will not get the correct
information from the data. Also each data set has the dimension 200X365 but
I only need 200X20 each.
Thanks
On Tue, Dec 4, 2012 at 10:22 PM, jim holtman wrote:
> try::
>
> do.call(rbind, myfiles)
>
> On Tue, Dec 4, 2012 a
try::
do.call(rbind, myfiles)
On Tue, Dec 4, 2012 at 9:37 PM, Gyanendra Pokharel
wrote:
> Hi group,
>
> I imported 16 data frames using the function "list.files"
>
> temp <- list.files(path="...")
> myfiles = lapply(temp, read.table,sep = "")
>
> Now I have 16 data set imported in R win
You need to defined a svydesign object that contains all the variables you
want to use. This object is passed as the design= argument to all the
analysis functions in the survey package. The functions don't have a data=
argument, and many of them will not look outside the design= argument for
var
Hi group,
I imported 16 data frames using the function "list.files"
temp <- list.files(path="...")
myfiles = lapply(temp, read.table,sep = "")
Now I have 16 data set imported in R window.
I want to combine them by row and tried some thing like (Here I am
considering only 20 columns)
f
Fantastic - that does the trick, David, thanks for setting me straight!
Chris
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Tuesday, December 04, 2012 8:19 PM
To: Chris Solomon
Cc: r-help@r-project.org
Subject: Re: [R] control point size of superscript wh
Hi R users following this thread!
I evaluated both solutions given by Rui and Arun.
Both work very well.
Arun's is a little faster.
I did the following time measurements on a large matrix with 2 rows,
labelled "InterPro" and "GO", and 88664 columns, labelled with protein IDs.
I was interested in
On Dec 4, 2012, at 11:05 AM, Chris Solomon wrote:
> Hi-
>
> A journal has asked me to make all of my text annotations on a figure at
> 10-point size. For the most part this is easy, e.g. by creating figures with:
> pdf(..., family='Times', pointsize=10)
>
> But where I have superscripts (or
Hi all
Thanks for the attention and answers. I learned a lot I now I can go on my
work. I also tryed to you de command window().
I thought it would be possible to select one column of an ts object, like
we can do with a data.frame (plot(data[,2],data[,3]), to work. But as I saw
we need to extract
> No need for all this (see solutions including mine already given) --
> but even without those, this is silly. An identity map is a real waste
> if you just want the simplification bit of sapply() -- you'd be much
> better just using simplify2array()
You are right that simplify2array(p) does ever
Hello,
I have been creating many tileplots to try and illustrate the relative
abundance of fish through space and time. My issue is that the tiles that
border the plot are smaller than those in the center of the plot. In the
example I've provided the effect is pretty minor (I'm hoping this will be
On Tue, Dec 4, 2012 at 8:17 PM, arun wrote:
> Hi,
> Try this:
> list1<-list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17),
> c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25),
> c(45, 19), c(5e+05, 16))
>
>
> res<-t(sapply(list1,function(x) x))
Bah Humbug! (In Christ
Hello,
Try the following.
m[ , sapply(1:ncol(m), function(j) sapply("B", `%in%`, m[[1 , j]])),
drop=F ]
Hope this helps,
Rui Barradas
Em 04-12-2012 21:16, Asis Hallab escreveu:
Dear R users,
I have a matrix composed of lists:
m <- matrix( list(), nrow=1, ncol=3 )
m[[ 1, 1 ]] <- list("A",
HI,
Does this work for you?
mapply(function(x) x=="B",m)
[,1] [,2] [,3]
#[1,] FALSE FALSE FALSE
#[2,] TRUE FALSE TRUE
A.K.
- Original Message -
From: Asis Hallab
To: r-help@r-project.org
Cc:
Sent: Tuesday, December 4, 2012 4:16 PM
Subject: [R] How to find matching columns
Hi,
You can also do this:
dat1<-structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate =
c(4e-04,
5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04,
3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04,
5e-04, 7e-04, 7e-04)), .Names = c("group", "X3.Hy
Hi,
Not sure whether this helps:
library(reshape2)
dat1<-structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate =
c(4e-04,
5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04,
3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04,
5e-04, 7e-04, 7e-04)),
Hello,
m[ , sapply(1:ncol(m), function(j) sapply("B", `%in%`, m[[1 , j]])), drop=F
> ]
>
It indeed does.
Thank you very much!
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I am using R for hierarchical clustering of a number of documents.
I have a distance matrix on which I have applied hclust method. When I plot
the result of hclust method, I can see the dendogram plotted. What I need
now is the dendogram stored as a tree in a data structure. My goal is to
automati
Hi,
Try this:
list1<-list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17),
c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25),
c(45, 19), c(5e+05, 16))
res<-t(sapply(list1,function(x) x))
res
# [,1] [,2]
#[1,] 5 101
#[2,] 10 46
#[3,] 15 31
Thanks to both of you! I learned something new from both of your posts :-)
And you were both right, my example numbers were wrong! I accidentally
computed them based on 0.25 * mean instead of 0.2 * mean.
Thanks again!
Simon
On Wed, Dec 5, 2012 at 4:07 AM, arun wrote:
> HI,
>
> I am not sure t
HI,
I just wonder whether your code worked or not.
set.seed(8)
mat1<-matrix(sample(1:80,40,replace=TRUE),ncol=8)
set.seed(25)
mat2<-matrix(sample(1:160,40,replace=TRUE),ncol=8)
#Since the dimensions are the same,
m<-1:5
n<-1:8
sum1<-0
for(i in 1:length(m)){
for(j in 1:length(n)){
sum1<-sum1+
There is a book on MC with R:
Introducing Monte Carlo Methods with R by Robert/Casella:
http://www.springer.com/statistics/computational+statistics/book/978-1-4419-1575-7
On 4 December 2012 19:38, R. Michael Weylandt
wrote:
> replicate(1000, sum(rnorm(50)^2-rchisq(50, 3)))
>
> Or you know, many
That should read the "reshape" package -- not the "shape" package.
My apologies.
To: Charles Determan Jr ; "r-help@r-project.org"
Sent: Tuesday, December 4, 2012 4:36 PM
Subject: Re: [R] reformatting some data
It's not clear to me what it is you are att
It's not clear to me what it is you are attempting to do, as you switch from a
very specific example to some general example with the vague terms "var1"
"var2", and "var3."
It sounds like you might be trying to do something similar to what would be
available in the shape package using the melt
Hello,
I am trying to reformat some data so that it is organized by group in the
columns. The data currently looks like this:
group X3.Hydroxybutyrate X3.Hydroxyisovalerate ADP
347 4 4e-04 3e-04 5e-04
353 3 5e-04
Dear R users,
I have a matrix composed of lists:
m <- matrix( list(), nrow=1, ncol=3 )
m[[ 1, 1 ]] <- list("A", "B")
m[[ 1, 2 ]] <- list("A", "C")
m[[ 1, 3 ]] <- list("A", "B")
and want to get the sub-matrix where cells contain "B".
But
m[ , "B" %in% m[ 1, ], drop=F ]
as well as
m[ , "B" %in%
I don't think there's any reason for the calculation you're doing that
you must have the whole matrix in memory, is there?
Unless there's something more than what you've shown us, you're just
taking the sum of elementwise operations. You can read the matrix in
in manageable chunks, take the sum of
On Tue, Dec 4, 2012 at 8:14 PM, Charles Novaes de Santana
wrote:
> "Error in matrix(0, 48000, 48000) : too many elements specified"
>
> but I thought it was a machine limitation (and I was asking for access
> to a better machine in my labs...). Thanks for clarifying it.
>
> Well, when Sarah gave m
Thank you, Sarah! It is a wonderful new!!! :)
Now I need to solve the other question hehe How to allocate such large matrix :)
best,
Charles
On Tue, Dec 4, 2012 at 8:39 PM, Sarah Goslee wrote:
> Without a reproducible example it's hard to tell for certain, but what
> about simply (assuming nro
On Tue, Dec 4, 2012 at 8:09 PM, Sam Steingold wrote:
> How do I convert a list to a matrix?
>
> --8<---cut here---start->8---
> list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17),
> c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25),
> c
Try:
matrix(unlist(a), ncol=2, byrow=T)
--Mark Lamias
From: Sam Steingold
To: r-help@r-project.org
Sent: Tuesday, December 4, 2012 3:09 PM
Subject: [R] list to matrix?
How do I convert a list to a matrix?
--8<---cut here---start---
On Tue, Dec 4, 2012 at 8:43 PM, Peter Langfelder
wrote:
> On Tue, Dec 4, 2012 at 11:27 AM, Charles Novaes de Santana
> wrote:
>> Dear Michael,
>>
>> Thank you for your answer.
>>
>> I have 2 matrices. Each position of the matrices is a weight. And I
>> need to calculate the following sum of diffe
There is a book on MC with R:
Introducing Monte Carlo Methods with R by Robert/Casella:
http://www.springer.com/statistics/computational+statistics/book/978-1-4419-1575-7
On 4 December 2012 19:21, Adel ESSAFI wrote:
> Hello,
>
> How can I make a monte carlo simulation on R?
>
> Regards
> Adel
>
How do I convert a list to a matrix?
--8<---cut here---start->8---
list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17),
c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25),
c(45, 19), c(5e+05, 16))
as.matrix(a)
[,1]
[1,]
On Tue, Dec 4, 2012 at 11:27 AM, Charles Novaes de Santana
wrote:
> Dear Michael,
>
> Thank you for your answer.
>
> I have 2 matrices. Each position of the matrices is a weight. And I
> need to calculate the following sum of differences:
>
> Considering:
> mat1 and mat2 - two matrices (each of th
Without a reproducible example it's hard to tell for certain, but what
about simply (assuming nrows2 is actually columns):
sum((mat1/d1 - mat2/d2)^2)
R is smart enough to understand elementwise manipulation of a matrix:
you shouldn't need a loop at all.
Sarah
On Tue, Dec 4, 2012 at 2:27 PM, Cha
Dear Michael,
Thank you for your answer.
I have 2 matrices. Each position of the matrices is a weight. And I
need to calculate the following sum of differences:
Considering:
mat1 and mat2 - two matrices (each of them 48000 x 48000).
d1 and d2 - two constant values.
sum<-0;
for(i in 1:nrows1){
Hi-
A journal has asked me to make all of my text annotations on a figure at
10-point size. For the most part this is easy, e.g. by creating figures with:
pdf(..., family='Times', pointsize=10)
But where I have superscripts (or subscripts) in axis labels, the default seems
to be to shrink th
On Dec 4, 2012, at 6:47 PM, Charles Novaes de Santana
wrote:
> Dear all,
>
> I need to access data from a large matrix (48000 x 48000) and to do it
> I am trying to run two loops using "for" command. Surely it is been a
> very slow job.
>
> I heard that "for" is not the best option to perfor
Dear all,
I need to access data from a large matrix (48000 x 48000) and to do it
I am trying to run two loops using "for" command. Surely it is been a
very slow job.
I heard that "for" is not the best option to perform large loops in R,
but I don't really know what would be the best (fast) option
This is a BUGS problem, not an R problem, so would be better asked on the
BUGS list (a practical hint: if it's not working through R, run the code
directly in BUGS: the error reporting is better). Anyway, here BUGS manages
to tell R to tell you what's wrong:
variable n is not defined
And looking
replicate(1000, sum(rnorm(50)^2-rchisq(50, 3)))
Or you know, many other things...
Michael
On Tuesday, December 4, 2012, Adel ESSAFI wrote:
> Hello,
>
> How can I make a monte carlo simulation on R?
>
> Regards
> Adel
>
>
> --
> PhD candidate in Computer Science
> Address
> 3 avenue lamine, cité
On 04/12/2012 12:54 PM, David Romano wrote:
Hi everyone,
I normally include a call to browser() as I'm working out the kinks in my
scripts, and I am always able to step through each line by hitting
"Return", but for some reason, in the scripts I'm working on now, hitting
"Return" seems to cause
Hello,
How can I make a monte carlo simulation on R?
Regards
Adel
--
PhD candidate in Computer Science
Address
3 avenue lamine, cité ezzahra, Sousse 4000
Tunisia
tel: +216 97 246 706 (+33640302046 jusqu'au 15/6)
fax: +216 71 391 166
[[alternative HTML version deleted]]
__
Hi Arvin,
How are you preparing the data to be read: a spreadsheet?
How are you reading the data?
How have you verified that the CSV file is correct?
How have you verified that the data frame is incorrect?
Can you provide a reproducible example using a small portion of your dataset?
Sarah
On T
On 04/12/2012 1:02 PM, Torus Insurance wrote:
Hi list,
I am using read.csv to read data from csf files, but noticed that the
numeric data (those larger than 10 power 9) are rounded to the nearest
million (10 power 6). Any solution?
What makes you think that is happening? R rounds values for pr
Hi list,
I am using read.csv to read data from csf files, but noticed that the
numeric data (those larger than 10 power 9) are rounded to the nearest
million (10 power 6). Any solution?
Thanks
Arvin
--
Sent from my mobile device
__
R-help@r-project.org
Untested, I think it's the blank line in your script which exits the
debugger and then you're seeing regular execution.
MW
On Tue, Dec 4, 2012 at 5:54 PM, David Romano wrote:
> Hi everyone,
>
> I normally include a call to browser() as I'm working out the kinks in my
> scripts, and I am always a
Dear all,
I cannot reading a .xlsm file using read.xls.
I executed:
read.xls("resultados.xlsm",
colNames = TRUE,
sheet = 1,
type = "data.frame",
from = 1,
rowNames = NA,
colClasses = "character",
checkNames = TRUE,
dateTime = "numeric",
naStrings = NA,
stringsAsFactors = F)
Error:
Call("ReadXls
On 04-12-2012, at 18:50, Berend Hasselman wrote:
>
>
> You should return the vector y i.e. function values.
> But y has length 4 and x has length 4.
>
x has length 5 of course.
Berend
> So where is the fifth value for y?
>
>> }
>>
>> Xstart=c(1, 200, 0.5, 0.5, 12)
>> fstart= firstor
Hi everyone,
I normally include a call to browser() as I'm working out the kinks in my
scripts, and I am always able to step through each line by hitting
"Return", but for some reason, in the scripts I'm working on now, hitting
"Return" seems to cause execution of *all* the lines in my script. I'
On 04-12-2012, at 17:06, Alicia Ellis wrote:
> I'm solving 4 complex equations simultaneously. Code is below. The code
> returns only zero's for the solution though there should also be a non-zero
> result. I'm pretty confident that the equations are correct because they
> are straight from a
Hello,
You should provide sample data and code.
plot(3:10, xlim = c(0, 12), xaxt = "n")
axis(1, at = 1:12)
See the help page ?par for a description of the graphical parameters
'xlim' and 'xaxt', and of ?axis.
Hope this helps,
Rui Barradas
Em 04-12-2012 17:24, T Bal escreveu:
Hi,
In the pl
Hi,
In the plot function I want to label x axis as the numbers between 1 and 12
(so 1, 2, 3, 4, 5, ..., 12). How should I do it? The range of x values are
different than this range. Thanks!
Kind regards,
T. Bal
[[alternative HTML version deleted]]
___
What are you expecting?
What do you get?
What is the problem?
J
On 4 December 2012 06:01, anoumou wrote:
> Hello all,
> I need a help.
> I am modeling a disease and a create a R function like that:
>
> Lambda<-function (x,date1,r,h,a){
> ndate1 <- as.Date(date1, "%d/%m/%Y")
> t1 <- as.nume
HI,
I am not sure the output you wanted is correct:
"
sample1 sample2 sample3
1 1.0 0 0.5
"
because
0.2*colMeans(x[,-4])
sample1 sample2 sample3
# 28.40 24.08 21.36
This might help you:
apply(x[-4],2,function(y) length(y[y <0.2*mean(y) &
x$class=="a"])/length(x[x$class=="
John: I think you want the output from coef(fitRIRT). The
ranef(fitRIRT) will give you the subject specific random effect deviations
from the fixed effects means. The coef(fitRIRT) will give you the
combination of the fixed effect means with the subject specific random
effect deviations and
I'm solving 4 complex equations simultaneously. Code is below. The code
returns only zero's for the solution though there should also be a non-zero
result. I'm pretty confident that the equations are correct because they
are straight from a published paper and I checked them pretty thoroughly.
T
Dear all,
Do you know how to return all the R settings to original state? Other than
.Rdata and .Rhistory
Some weid thing happened to my machine. I was trying to get shaded confidence
band ploted using survplot from rms liberary.
It worked on one machine, but not on the other. I tried unsta
Yes, you are correct.
Thanks,
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605
I think the random effects represent the subject adjustments to the population
averages. You may have to do the addition yourself to get the subject specific
slopes and intercepts. Someone will hopefully correct me if I'm wrong.
On 12/04/12, John Sorkin wrote:
>
>
>
> Ken,
> Thank you for yo
Ken,
Thank you for your help. ranef(fitRIRT) does not give me what I expect. The
subject-specific slopes, and subject-specific intercepts are not anywhere close
to what I would expect them to be; the mean of the subject-specfic values
should be close to those reported by
summary(fitRIRT) and th
I am running a random intercept random slope regression:
fitRIRT <- lme(echogen~time,random=~
1+time|subject,data=repeatdata,na.action=na.omit)
summary(fitRIRT)
I would like to get the subject-specific slopes, i.e. the slope that the model
computes for each subject. If I have 10-subjects I s
Adam,
Getting the variance of MLE estimator when the true parameter is on the
boundary is a very difficult problem. It is known that the standard bootstrap
does not work. There are some sub-sampling approaches (Springer book:
Politis, Romano, Wolff), but I am not an expert on this.
However,
Hi,
I've got a new dataset which I don't know how to analyze with R. My knowledge
about R is limited for this kind of problem. I've tried to find a solution with
some spatio temporel packages and lme/lmer functions, but didn't find any
similar example.
I've got 10 locations on a coast on which
Thanks, Rui and David!
--
View this message in context:
http://r.789695.n4.nabble.com/Histogram-plot-help-tp4651958p4652065.html
Sent from the R help mailing list archive at Nabble.com.
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Without data to reproduce what you saw, we can only guess.
One possibility is due to tie-breaking. There are several places where ties
can occur and are broken at random, including at the prediction step. One
difference between the two ways of doing prediction is that when it's all done
withi
Hi,
If the frequency is >1, the error message will be gone.
For e.g.
birthstimeseriesJanFeb<-subset(birthstimeseries,cycle(birthstimeseries)==c(1,2))
birthstimeseriesJanFeb1<-ts(birthstimeseriesJanFeb,frequency=2,start=c(1946,1))
plot.ts(birthstimeseriesJanFeb1)
birthstimeseriesJanFebHW<-HoltWi
Hi,
Try this:
set.seed(15)
datos<-as.data.frame(matrix(sample(c(1:20,NA),30,replace=TRUE),ncol=6))
do.call(paste,c(na.omit(datos),sep=";"))
#[1] "5;18;14;10;17;3" "14;15;15;3;2;20" "8;18;19;17;12;11"
A.K.
- Original Message -
From: Dominic Roye
To: r-help@r-project.org
Cc:
Sent: Tu
HI,
I am getting an error message:
l.blist<-lapply(blist,HoltWinters)
#Error in decompose(ts(x[1L:wind], start = start(x), frequency = f), seasonal)
:
# time series has no or less than 2 periods
A.K.
- Original Message -
From: PIKAL Petr
To: Antonio Silva ; "R-help@r-project.org"
C
DT<-data.frame(time=c(0,1,5,24,36,48,72),DV=seq(0,60,10))
time DV
10 0
21 10
35 20
4 48 30
5 84 40
6 96 50
7 120 60
You want to add 24 to values that are >=24 in 'time'
DT[DT$time>=24,'time']<-DT[DT$time>=24,'time']+24
time DV
10 0
21 10
35 20
4 48 30
5 60
Are you using windows? If you are you may want to try to run your R code from a
batch file:
REM on Microsoft Windows (adjust the path to R.exe as needed)
"C:\Program Files\R\R-2.13.2\bin\x64\R.exe" CMD BATCH
"C:\Users\Frank\Documents\R\Projects\Heinrich\Heinrich.txt"
"C:\Users\Frank\Documen
Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
Lambda<-function (x,date1,r,h,a){
ndate1 <- as.Date(date1, "%d/%m/%Y")
t1 <- as.numeric(ndate1)
x[order(x$i),]
t <-x[,"t"]
i <-x[,"i"]
CONTAGIEUX <-x[,"CONTAGIEUX"]
while ( t1 < min(t) ){
for (i
Hi,
I am trying to covert a Winbugs code into R code. Here is the winbugs code
model{# models likelihoodfor (i in 1:n){time[i] ~ dnorm( mu[i], tau ) #
stochastic componenent# link and linear predictormu[i] <- beta0 + beta1 *
cases[i] + beta2 * distance[i]}# prior distributionstau ~ dgamma( 0.01
DT = data.frame(x=rep(c("a","b","c"),each=3), y=c(1,3,6), v=1:9, w=3:11,
z=LETTERS[1:9])
If I understand you right, and you want to select all rows where v>3 and
W<10
with(DT, DT[which(v>3 & w<10),])
x y v w z
4 b 1 4 6 D
5 b 3 5 7 E
6 b 6 6 8 F
7 c 1 7 9 G
you can use colSums, rowSums on th
There been, that done.
http://stackoverflow.com/questions/9968578/speeding-up-julias-poorly-written-r-examples/10712158#10712158
MW
On Tue, Dec 4, 2012 at 2:34 PM, Suzen, Mehmet wrote:
> Hello List,
>
> Probably many of you aware of the Julia language
> (http://julialang.org/), It is a promisin
Try the following:
set.seed(100)
rf1 <- randomForest(Species ~ ., data=iris)
set.seed(100)
rf2 <- randomForest(iris[1:4], iris$Species)
object.size(rf1)
object.size(rf2)
str(rf1)
str(rf2)
You can try it on your own data. That should give you some hints about why the
formula interface should be
Hello List,
Probably many of you aware of the Julia language
(http://julialang.org/), It is a promising project.
However it seems like R is very slow in their benchmarks. Very
important point they omit, they did not
use R's own JIT ! I had a feeling that R is mistreaded there :)
Also another imp
When you typed x as a command, R runs the command print(x). That function
produces a summary of the results which may include round off numbers to a
few decimal places to make them more readable. When you typed x$statistic,
you got the unrounded version of the result 5.6e-31 which I think you will
HI,
You can use either ?tapply(), ?aggregate(), ?ddply() from library(plyr)
dat1<-read.table(text="
0 12
1 10
1 4
1 6
1 7
1 13
2 21
2 23
2 20
3 18
3 17
3 16
3 27
3 33
4 11
4 8
4 19
4 16
4 9
",sep="",header=FALSE)
with(dat1,aggregate(dat1[,2],by=list(V1=dat1[,1]),sum))
# V1 x
Hello,
I have a problem with the "do.call-function". I would like to merge
the values of more than 30 columns, but not in all of the rows exist
values, so with this commando i get a lot of ";" or NA.
How get i only the merge of cells with a number?
datos$NEW <- do.call(paste, c(datos[,19:53], s
Thanks for the help,
Perhaps I should elaborate a bit, I am working on bioinformatics project in
which I am trying to run a forward selection algorithm for machine learning
classification of two biological conditions.
At each iteration I want to find the gene that in addition to those I have
fo
HI,
You can subset by:
birthstimeseriesJan<-subset(birthstimeseries,cycle(birthstimeseries)==1)
A.K.
- Original Message -
From: Antonio Silva
To: R-help@r-project.org
Cc:
Sent: Tuesday, December 4, 2012 4:26 AM
Subject: [R] partial analisys of a time series
Dear list members
I want
Hi
I am not an expert in time series. The problem is that resulting list is not
time series any more. So you need to convert it again to time series and you
need to give it frequency greater than 1.
something like
l.blist <- lapply(blist, function (x) HoltWinters(ts(x, frequency=2)))
But this
Try period.apply() from the xts package.
MW
On Tue, Dec 4, 2012 at 9:26 AM, Antonio Silva wrote:
> Dear list members
>
> I want to analyze separately the months of a time series. In other words, I
> want to plot and fit models for each month separately.
>
> Taking the example of
> http://a-littl
At 20:39 03/12/2012, Min Dong wrote:
Hi, I am a novice in R. It will be greatly appreciated if someone
can advise me with the following questions.
There are at least three packages available from CRAN (meta, metafor,
rmeta) which draw forest plots so it would help us if you had told us
which
Bonjour,
Je serais en congés jusqu'au Jeudi 6 Décembre.
Pour des raisons d'urgence, vous pourrez me contacter par téléphone au 06 46 34
81 03.
Cordialement,
--
Jérôme Boutet
Conservatoire d'espaces naturels de Picardie
1, place Ginkgo - village Oasis
80 044 AMIEN
You can use, 'sample' function for sampling and may consider using
partition clustering for selecting your regions, see Cluster task view:
http://cran.r-project.org/web/views/Cluster.html
On 4 December 2012 00:53, KoopaTrooper wrote:
> I am using package ks() to build 3D representations of bird
Hello, Simon,
see below!
On Tue, 4 Dec 2012, Simon wrote:
Hello all,
I have what feels like a simple problem, but I can't find an simple
answer. Consider this data frame:
x <- data.frame(sample1=c(35,176,182,193,124),
sample2=c(198,176,190,23,15), sample3=c(12,154,21,191,156),
class=c('a'
Thanks Petr
I thought there might be an equivalent for birthstimeseries[,1] if it were
a dataframe, but split function sounds great.
I could not reproduce the second line of your suggestion "l.blist <-
lapply(blist, HoltWinters)". I receive the message: Error in
decompose(ts(x[1L:wind], start = s
Hi,
Imagine the data you have is in a data frame, c, with columns a and b.
Then you can do this:
> res=tapply(b,c[,-2],sum)
> res[is.na(res)]<-0
> res
0 1 2 3 4
12 40 64 111 63
Hope it helps,
José
José Iparraguirre
Chief Economist
Age UK
-Original Message-
From: r-hel
On 04-12-2012, at 08:59, T Bal wrote:
> Hi,
> I have the following data:
>
> 0 12
> 1 10
> 1 4
> 1 6
> 1 7
> 1 13
> 2 21
> 2 23
> 2 20
> 3 18
> 3 17
> 3 16
> 3 27
> 3 33
> 4 11
> 4 8
> 4 19
> 4 16
> 4 9
>
>
> In this data file I would like to sum the numbe
Hi
That is quite usual. Excel is so widespread that almost everybody assumes it
shall not contain mistakes and behaves correctly. The contrary is true.
Spreadsheet often guess what user have on mind and "corrects" values to fit
such assumption, let aside mistakes in coded functions.
R expects
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