Dear all I have a few matrices that I would like to store alltogether under a
bigger object.
My matrixes with the same name were calculated inside a loop like
for (few times){
estimatedsr<- this is my matrix
savematrixasimagefile();
}
which means that I was losing all that instances.
Dear R colleagues,
is there an easier way to write R packages for the own use - without RTools and
TeX?
With R versions < 2.10.0 it was very easy to write a package. Under the path
with the package name you wrote a description file and built some directories
like "help", "html" and "R" with t
Hello,
I tried to fit a poisson distribution but looking at the function
fitdistr() it does not optimize lambda but simply estimates the mean
of the data and returns it as lambda. I'm a bit confused because I was
expecting an optimization of this parameter to gain a good fit...
If I would use mle(
Hello all,
Before getting to my question, I would like to apologize for asking this
question here. My question is not directly an R question, however, I still
find the topic relevant to R community of users - especially due to only *
partial* (current) support for interactive data visualization
You need to change models 2 and 3 to use ~ group
+ subject. You left subject out as a fixed factor.
At 05:17 PM 2/10/2011, array chip wrote:
Robert, thank you!
I tried all 3 models you suggested. Since each
subject only has one line of data in my dataset,
would including Subject as a factor
Hello
I have a time-comsuming program which need to simplify, I have tested
the annotated program as follow:
> #define function which will be call
> calsta <- function(c, n=10)
+ {
+ i <- seq(from=0, length=c)
+ logx <- lchoose(NT-n, CT-i) + lchoose(n, i)
+ logmax <- max(lo
On Feb 10, 2011, at 2:39 PM, sherri heck wrote:
Dear All -
I would like to make multi-page plots, but my plots are writing
over one
another - thus leaving just the last plot. Here is what I am doing:
pollutionRose(temp, "nox", type="hour")
update(trellis.last.object(), layout =c(3,3,3) )
Here is my primitive and computationally intensive solution to this problem.
Thank you to Michal (see above) for making me aware of the nice polygon
function. I am sure that there are better solutions but this one works for
me at work and for publication.
###
I would say it is quite tedious to process timestamps in such a mixed basis
unless it includes the daylight/standard time indicators in the data itself. Up
until you work with time in the repeated hour in the autumn you can skate by,
but those times require the extra indicator, and if it isn't i
Jeff Newmiller writes:
> Use Sys.setenv(TZ="Etc/GMT+8") before you read the files.
Thanks, Jeff and all for the suggestions (assuming this works, as it
seems it should, I may go with this one). I'll try them tomorrow.
After posting my query, I discovered a related problem which I have to
resea
Dear r-helpers,
I used lsfit to estimate regression coefficients from the matrices of
predictors and predicted values (i.e multivariate multiple regression
model). I wish, however, to compute rank-one matrices of predictor
values from the predicted ones.
How to do it in R? Could you suggest me app
Hi:
Here's one way:
plot(1~1,ylab=expression("Areas ("~mu*m^2~")"))
The tildes incorporate space between the math and text elements; they're
optional, but useful. Another way that also works is
plot(1~1,ylab=expression(paste("Areas (", mu*m^2, ")", sep = ' ')))
HTH,
Dennis
On Thu, Feb 10, 20
Hi:
And yet another: :)
library(plyr)
top2 <- function(x) {
y <- sort(x[-c(1, 2)], decreasing = TRUE)[1:2]
area <- rep(x[1], 2)
type <- rep(x[2], 2)
data.frame(area, type, value = y, variable = names(y))
}
v <- ldply(apply(df, 1, top2), rbind)
The last line does
Hi Alice,
You can use
pvals <- summary(myprobit)$coefficients[, "Pr(>|z|)"]
Notice that if the p-value is very small, the printed version is abbreviated,
but the object itself has full precision (not that it matters).
Bill Venables.
-Original Message-
From: r-help-boun...@r-project.o
I had a longer draft before but I'll just ask if you ever looked at your data?
cr=rainbow(3);
plot(df$success+df$failure,df$prop.success,col=cr[as.numeric(df$group)],cex=as.numeric(df$group))
you'd have to suspect that p is a function of the number of trials per subject
and that differences bet
!is.null(my.obj...@my.data.frame$my.var)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Kushan Thakkar
Sent: Friday, 11 February 2011 10:30 AM
To: r-help@r-project.org
Subject: [R] if a variable is defined
I have an object type m
Duncan Murdoch-2 wrote:
>
>
> There is no plotmath function. "plotmath" is the name of the help
> topic; it describes how various other functions plot text that includes
> math.
>
>
> grDevices is a base package, so if you've got R, you've got it.
>
> Duncan Murdoch
>
>
ok thanks.
I
I have an object type my.object. One of its slots contains a data frame
(i.e. my.obj...@my.data.frame) .. I want to check if one of the variables
exists in this data frame (i.e. my.obj...@my.data.frame$my.var)
I am trying to use the exists function but can't figure out how the
arguments work. Plea
Hi:
Try
summary(myprobit)$coefficients[, 4]
HTH,
Dennis
On Thu, Feb 10, 2011 at 3:46 PM, Allie818 wrote:
>
> I can get this summary of a model that I am running:
>
> summary(myprobit)
>
> Call:
> glm(formula = Response_Slot ~ trial_no, family = binomial(link = "probit"),
>data = neg_data,
A safer way to make sure you don't match the underscore:
> gsub("[^_]*_[^_]*_([^_]*).*", "\\1", "abcd_efgh_X_12ab3_dfsfd")
[1] "X"
On Thu, Feb 10, 2011 at 2:06 PM, Henrique Dallazuanna wrote:
> So, a way could be:
>
> gsub("(.*)_(.*)_(.*)_.*", "\\3", "abcd_efgh_X_12ab3_dfsfd")
>
>
I can get this summary of a model that I am running:
summary(myprobit)
Call:
glm(formula = Response_Slot ~ trial_no, family = binomial(link = "probit"),
data = neg_data, na.action = na.pass)
Deviance Residuals:
Min 1Q Median 3Q Max
-0.9528 -0.8934 -0.8418 1.44
Thank you Duncan. It works. It seems that NA is of great use.
Shiguo
On Thu, Feb 10, 2011 at 5:25 PM, Duncan Murdoch wrote:
> On 11-02-10 3:56 PM, Shiguo Jiang wrote:
>
>> Hello,
>>
>> I plot box plot, and add a point to the box indicating mean. I also add
>> some
>> range to the box width. I wa
Use Sys.setenv(TZ="Etc/GMT+8") before you read the files.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/So
On Thu, Feb 10, 2011 at 8:45 AM, Simon Gillings wrote:
> I am using bagging to perform Bagged Regression Trees on count data (bird
> abundance in Britain and Ireland, in relation to climate and land cover
> variables). Predictions from the final model are visually believable but I
> would reall
On 2011-02-10 09:40, mdc wrote:
Hi, I am a new R user and am trying to construct a palaeoenvironmental
transfer function (weighted averaging method) using the package rioja.
I've managed to insert the two matrices (the species abundance and the
environmental data) and have assigned them to the y
here is another way of doing it:
> set.seed(19)
>
> area<-c(rep(1,10),rep(2,10),rep(3,10),rep(4,10),rep(5,10))
> type<-c(rep(1:10,5))
> a<-rnorm(50)
> b<-rnorm(50)
> c<-rnorm(50)
> d<-rnorm(50)
> df<-cbind(area,type,a,b,c,d)
> df1 <- data.frame(df)
> require(reshape2)
> df.melt <- melt(df1, id=c('
On Thu, Feb 10, 2011 at 7:06 PM, Bill Harris
wrote:
> I'm reading in ~3 years worth of data that includes hourly timestamps.
> Presumably to avoid DST confusion, all the data is in PST time zone -- no
> discontinuities in the spring or fall.
>
> The data comes in a csv file, which I'm reading with
Hi:
It's not a perfect solution, but an 'easy' way out is to recognize that some
symbols are open and some are filled. If you want fill, use a filled symbol
and then change the color.
p <- rep(16:18, 4)
x<-c(1:12)
y<-c(rpois(12,4))
grp<-c(rep(c(3,4), each=6))
z<-c(rep(c(1,2), each=6))
dd <- data.
Željka biologija.unios.hr> writes:
>
> Dear all!
>
> I'm using R 2.12.1 in Ubuntu 10.04.
> Problem is that when using functions like interp (package akima) and
> akimaInterp (package fUtilities), I get different results every time for
> the same data set. I've checked the interpolated results,
I'm reading in ~3 years worth of data that includes hourly timestamps.
Presumably to avoid DST confusion, all the data is in PST time zone -- no
discontinuities in the spring or fall.
The data comes in a csv file, which I'm reading with
myvariable <- read.csv("my_data_file.csv",header=FALSE,
col
On 10/02/2011 4:08 PM, Lizblip wrote:
I get the same errors:
plotmath(area~cell,main="Myofibril areas by cell",ylab=expression("Areas
(mu*m^2)"),xlab="")
Error: could not find function "plotmath"
There is no plotmath function. "plotmath" is the name of the help
topic; it describes how var
My apologies to Eik - I didn't see his solution and essentially replicated
it. Sorry for the noise..
Dennis
On Thu, Feb 10, 2011 at 3:46 PM, Dennis Murphy wrote:
> Hi:
>
> On Thu, Feb 10, 2011 at 10:50 AM, Hui Du wrote:
>
>>
>> Hi all,
>>
>>I have a dataset. Each time I want to
Hi:
On Thu, Feb 10, 2011 at 10:50 AM, Hui Du wrote:
>
> Hi all,
>
>I have a dataset. Each time I want to sample N(i) elements
> from it and I want to repeated sampling M times. N(i) is varied from time to
> time. For example,
>
>dataset = 1:50;
>a
How about:
table(samp)
--
Robert Tirrell | r...@stanford.edu | (607) 437-6532
Program in Biomedical Informatics | Butte Lab | Stanford University
On Thu, Feb 10, 2011 at 13:54, Nipesh Bajaj wrote:
> > set.seed(100)
> > samp <- sample(c(1,-1,0), 20, replace=T); samp
>
[[alternative H
If you do use correlation, you should think about doing it on the log or sort
scale.
The train() function in the caret package can estimate performance using
resampling. There are examples in ?train that show how to define custom
performance measures (I think it shows how to do this with MAD es
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Marine Andersson
Sent: Thursday, February 10, 2011 3:53 PM
To: r-help@r-project.org
Subject: [R] Calculating rowMeans from different columns in each row?
Hello!
I have a dataset lik
x<-circular(dayofmax[,2]*2*pi/365)
res25 <- density(x, bw=25, control.circular=list(units="degrees"))
circularp(res25$x)
plot(res25, col=4, points.plot=TRUE, xlim=c(-1.5,1.5))
res50 <- density(x, bw=50, adjust=2)
lines(res50, col=2)
lines(res50, col=3, shrink=0.9)
this is my code, basically, it
I get the same errors:
> plotmath(area~cell,main="Myofibril areas by cell",ylab=expression("Areas
> (mu*m^2)"),xlab="")
Error: could not find function "plotmath"
> install.packages("grDevices")
Warning in install.packages("grDevices") :
argument 'lib' is missing: using
'/home/elizabeth/R/i686-
I would like to make a box plot with a dendogram on top that can be used
to indicate significant differences (with an asterisk or three at a node
depending on the p-value). I can supply the adjacency information a
priori. Is there an easy way to do this? Other ways of presenting the
same inform
Hi Ivan,
are you looking for quantiles of "type 1"? See ?quantile
x<-rnorm(50)
ecdfx<-ecdf(x)
ecdfx(quantile(x,.95,type=1)) #so this is the lowest x having ecdf(x)>=.95
ecdfx(quantile(x,.95,type=1)-1e-10) #left from that point ecdf<.95
so actually it is right-continuous (because its an inverse o
?rle
--
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360) 407-6815
PO Box 47600FAX:(360) 407-7534
Olympia, WA 98504-7600
Hi:
There are a couple of things here. Firstly, if you intend to predict from a
model in R, make sure your input to the model is a data frame. You can fit a
model piecemeal with a set of vectors, but there's a distinct possibility
you'll end up frustrated when trying to make predictions with new d
On 11-02-10 3:56 PM, Shiguo Jiang wrote:
Hello,
I plot box plot, and add a point to the box indicating mean. I also add some
range to the box width. I want to add legend for both the range (line) and
mean (point). However, I cannot add line legend and point legend together.
The code looks like t
Hi,
does
table(samp)
do what you want?
Am 10.02.2011 22:54, schrieb Nipesh Bajaj:
> Hello there, would you please look into my codes? Here I have following:
>
>> set.seed(100)
>> samp <- sample(c(1,-1,0), 20, replace=T); samp
> [1] 1 1 -1 1 -1 -1 0 -1 -1 1 -1 0 1 -1 0 0 1 -1 -1 0
Robert, thank you!
I tried all 3 models you suggested. Since each subject only has one line of
data
in my dataset, would including Subject as a factor in glm() or lm() lead to 0
df
for resaiduals?
Attached is my dataset and here is my version of the 3 models:
test<-read.table("test.txt",sep=
See ?.libPaths : as so often, all that is needed is attentive reading
of the documentation:
The library search path is initialized at startup from the
environment variable ‘R_LIBS’ (which should be a colon-separated
list of directories at which R library trees are rooted) followed
Marine -
Assuming your data frame is named "df", I think
apply(df,1,function(x)mean(x[x[7]:x[8]]))
will give you what you're looking for.
- Phil Spector
Statistical Computing Facility
Hello there, would you please look into my codes? Here I have following:
> set.seed(100)
> samp <- sample(c(1,-1,0), 20, replace=T); samp
[1] 1 1 -1 1 -1 -1 0 -1 -1 1 -1 0 1 -1 0 0 1 -1 -1 0
Here I want to calculate the length of each unique number for above vector.
How can I do that?
Given a dataset x, the ecdf is ecdf(x). Then I can use ecdf(x)(y) to find
the percentile of y. Given the ecdf is there a way to determine what is the
value of y that is the boundary of let's say 95 percentile? In other words,
is there a function I can call on the ecdf like:
fomeFunc( ecdf( x ), 0.
Hello,
I am trying to make a xyplot plot with points that are different symbols. I
want to call the symbol type (pch) from a column in my dataframe. Here is a
simplified example. In my real example I also have groups, which I have not
included here. This example doesn't change the symbols or
You could use the floating.pie function in the plotrix package, but even the
author of that function/package has stated in the past that pie charts are not
the best tool. Also look at the symbols function for some alternatives, I
would use the thermometers (and if that does not give enough opti
Hello!
I have a dataset like this:
X1 X2 X3 X4 X5X6X7X8
1 2 2 1 2 3 2 6
2 3 2 5 7 9 1 3
19 12 6 1 1 3 6
The columns X1-X6 contains ordinary numeric values.
X7 contains the numb
Hi,
try
ds<-1:50
M<-10
n<-sample(5:15,M,replace=T) #generate different n[i]
lapply(1:M,FUN=function(i) sample(ds,n[i]))
hth.
Am 10.02.2011 19:50, schrieb Hui Du:
>
> Hi all,
>
> I have a dataset. Each time I want to sample N(i) elements
> from it and I want to repeated sampli
1. If you use a random effects model, you should make Subject the
random factor. I.e., a random intercepts model with 1|Subject. Group
is a fixed effect: You have only 2 groups. Even if you had more than
2 groups, treating Group as random would return a standard deviation,
not a P-value as you
Hello,
I plot box plot, and add a point to the box indicating mean. I also add some
range to the box width. I want to add legend for both the range (line) and
mean (point). However, I cannot add line legend and point legend together.
The code looks like the following.
(1) First, I tried the follo
Do anyone know an R or bioconductor package to draw HGDP pie chart as in the
link below?
http://hgdp.uchicago.edu/cgi-bin/alfreqs.cgi?pos=118253964&chr=chr8&rs=rs12255372&imp=false
Many SNPs were missing from the link, so I would like to draw similar graphs in
R. I have all the allele frequency
Try
my.data.frame[my.data.frame$Buy==1 | my.data.frame$Sell ==1, ]
or
subset(my.data.frame,buy == 1 | sell == 1)
Then take a look at the help page for || i.e., help("||")
to see what you did wrong.
- Phil Spector
Hi, I am a new R user and am trying to construct a palaeoenvironmental
transfer function (weighted averaging method) using the package rioja.
I've managed to insert the two matrices (the species abundance and the
environmental data) and have assigned them to the y and x values
respectively. When I
Dear All -
I would like to make multi-page plots, but my plots are writing over one
another - thus leaving just the last plot. Here is what I am doing:
pollutionRose(temp, "nox", type="hour")
update(trellis.last.object(), layout =c(3,3,3) )
I would like 24 plots with 9 plots/page (the last pag
I have a data frame with the following columns:
Date Price Buy Sell
The Buy and Sell variables are binary. They are either zero or 1.
Additionally, they are mutually exclusive. Either Buy is 1, Sell is 1 or
they are both 0. But they are never both 1 for a given observation.
I want to see observa
Hi all,
I have a dataset. Each time I want to sample N(i) elements from
it and I want to repeated sampling M times. N(i) is varied from time to time.
For example,
dataset = 1:50;
a = list();
M = 1000;
I want to d
Hello!
Very sorry for a probably very simple question - I looked but did not
find an answer in the archives.
I have a table "counts" (below) that shows counts by Option within
each of my 2 groups. However, my groups have different sizes (N1=255
and N2=68). Table "prop" shows the resulting proporti
On 10 February 2011 12:01, Matt Shotwell wrote:
> On Thu, 2011-02-10 at 10:44 -0800, David Smith wrote:
>> The SAS import/export feature of Revolution R Enterprise 4.2 isn't
>> open-source, so we can't release it in open-source Revolution R
>> Community, or to CRAN as we do with the ParallelR pack
High,
on all of my computers the first entry in libPaths for root
was /usr/lib64/R/library or /usr/lib/R/library. Now it change at one
(lynx) anyway to
root@lynx:/root(2)# R
R version 2.12.1 (2010-12-16)
Copyright (C) 2010 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform:
On Thu, 2011-02-10 at 10:44 -0800, David Smith wrote:
> The SAS import/export feature of Revolution R Enterprise 4.2 isn't
> open-source, so we can't release it in open-source Revolution R
> Community, or to CRAN as we do with the ParallelR packages (foreach,
> doMC, etc.).
Judging by the language
On 02/10/2011 05:30 PM, Gabor Grothendieck wrote:
On Thu, Feb 10, 2011 at 10:49 AM, Nikhil Joshi wrote:
Gabor,
thanks for the suggestion.
I did as you have suggested- Started a clean session of R and set the java
option. Then I sourced the library and tried to read the xlsx file but got
the sa
On Feb 10, 2011, at 12:13 PM, Ben Bolker wrote:
> Simon Knos quantentunnel.de> writes:
>
>> Upfront: I am not sure if this is the right place to ask this question even
>> after reading through the FAQ, so apologies if I should have directed the
>> question elsewhere.
>
> I appreciate your con
On Feb 9, 2011, at 8:27 PM, Robert Baer wrote:
Is there an easy way to make the error bars the same color as the
points and lines they are plotted with.
My example
# fake data
x=sample(1:10, 100, replace =T)
y = rnorm(100) + runif(100)
df=data.frame(x,y)
# summarize data
m = aggregate(df,lis
On 2011-02-09 17:27, Robert Baer wrote:
Is there an easy way to make the error bars the same color as the points and
lines they are plotted with.
My example
# fake data
x=sample(1:10, 100, replace =T)
y = rnorm(100) + runif(100)
df=data.frame(x,y)
# summarize data
m = aggregate(df,list(x),mean)
Thank you guys, I've solved my problem with your precious help
--
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Sent from the R help mailing list archive at Nabble.com.
__
R
Dear All,
Could someone please suggest me the way to calculate the optimal threshold in
POT method via any available packages in R?
Thanks,
Fir
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the po
I have fitted the faults.data to glm.nb and to the function negbin from the
package aod. The output of both is the following:
summary(glm.nb(n~ll, data=faults))
Call:
glm.nb(formula = n ~ ll, data = faults, init.theta = 8.667407437,
link = log)
Deviance Residuals:
Min 1Q Media
If you want a random correlation matrix, why not just generate random data and
accept the correlation matrix that you get? The standard normal distribution
in k dimensions is (hyper)spherically symmetric. If you generate k standard
normal N(0,1) variates, you have a point in k-space with direc
Hi, all
I'm trying to solve an optimization problem with two variables, in wich I use
the following functions(they all are working perfectly):
FxMixedALT=function(x){calc=(-0.88*exp((-1.53655/1000)*(x^5.3)))-(0.12*exp(-0.06415*(x^2.5)))+0.88+0.12return(calc)}
RxMixedALT=function(x){calc=1-FxM
Hi Dennis,
Thanks for the solution, I really appreciate it!
Best,
Nathaniel
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Sent from the R help mailing list archive at Nabble.com.
I am using bagging to perform Bagged Regression Trees on count data (bird
abundance in Britain and Ireland, in relation to climate and land cover
variables). Predictions from the final model are visually believable but I
would really like a diagnostic equivalent to classification success that ca
Gabor,
thanks for the suggestion.
I did as you have suggested- Started a clean session of R and set the java
option. Then I sourced the library and tried to read the xlsx file but got
the same error. I am assuming that Java machine is loaded when the package
xlsx is sourced and not when the base
Hi,
I don't know how to make prediction with a factor in the linear model.
Say
yd=c(1,2,3,4,5)
> sl=c(2,3,4,5,6)
> sex=c("male","male","female","female","male")
> sex=factor(sex)
> m=lm(sl~yd+sex+sex:yd)
How to make a prediction with new data like yd=c(4,5,6,7,8) for male and
female s
So, a way could be:
gsub("(.*)_(.*)_(.*)_.*", "\\3", "abcd_efgh_X_12ab3_dfsfd")
On Thu, Feb 10, 2011 at 3:47 PM, Soumendra wrote:
> Hi Henrique,
>
> I believe your solution is wrong as it is fitted to find 12ab3,
> whereas Yan seems to be asking for the characters after the second
> unders
The SAS import/export feature of Revolution R Enterprise 4.2 isn't
open-source, so we can't release it in open-source Revolution R
Community, or to CRAN as we do with the ParallelR packages (foreach,
doMC, etc.).
It is, though, available for download free of charge to members of the
academic commu
HI,
I am interested in calculating hazard ratio of the stratified groups defined
by me.
library(survcomp)
binscores <-
cut(scores.train,c(-1000,-1,1,1000),c("low","intermediate","high"))
dd <- data.frame("surv.time"=OS, "surv.event"= status, "strat"=binscores)
km.coxph.plot(formula.s=Surv(surv.t
Simon Knos quantentunnel.de> writes:
> Upfront: I am not sure if this is the right place to ask this question even
> after reading through the FAQ, so apologies if I should have directed the
> question elsewhere.
I appreciate your confusion. I think this might be slightly more
appropriate for
prop.test() is applicable to a binomial experiment in each of two classes.
Your experiment is binomial only at the subject level. You then have
multiple subjects in each of your groups.
You have a random factor "Subjects" that must be accounted for.
The best way to analyze is a generalized li
Hello!
I am using rhierMnlRwMixture from bayesm package. I would like to use
it with a categorical covariate (Z).
I have 2 clariciation questions:
1. If the covariate is categorical, do I have to represent it as dummy
variable(s)? (e.g., 2 dummy variables for a 3-level categorical
variable)?
2. D
Alain -
Here's a reproducible data set:
set.seed(19)
area<-c(rep(1,10),rep(2,10),rep(3,10),rep(4,10),rep(5,10))
type<-c(rep(1:10,5))
a<-rnorm(50)
b<-rnorm(50)
c<-rnorm(50)
d<-rnorm(50)
df<-cbind(area,type,a,b,c,d)
First I'll make a helper function to operate on one
row of the data frame:
What you can do to find out is to type into your R session
RSiteSearch("multivariate fractional gaussian")
That seems to give some usefull results.
Kjetil
On Tue, Feb 8, 2011 at 1:51 PM, Wonsang You wrote:
>
> Dear R Helpers,
>
> I have searched for any R package or code for simulating multivar
Hi Henrique,
I believe your solution is wrong as it is fitted to find 12ab3,
whereas Yan seems to be asking for the characters after the second
underscore and before the third underscore.
For example, gsub(".*_.*_(.*)_.*", "\\1",
"abcd_efgh_X_12ab3_dfsfd") would still yield 12ab3 even though,
Thanks for the response, Rex. This is an interesting approach. The
Choleski decomposition approach that John suggested seems to be an
obvious and direct approach to this problem. Your approach is less
obvious to me but may be equal or superior to the Choleski
decomposition.
Are all possible cor
Hi, I initially posted this to the general R mailing list, but Bert Gunter
thought this may be a mixed model issue, so suggested me to post here.
I have a dataset that has 2 groups of subjects. For each subject in each group,
the response measured is the number of success (no.success) obatined w
Is there an easy way to make the error bars the same color as the points and
lines they are plotted with.
My example
# fake data
x=sample(1:10, 100, replace =T)
y = rnorm(100) + runif(100)
df=data.frame(x,y)
# summarize data
m = aggregate(df,list(x),mean)
se = aggregate(df,list(x),sd)/sqrt(10)
li
I can't see a good solution using a loop, but perhaps this
will get you what you want:
Dframe1$date = as.Date(Dframe1$sunrise)
Dframe2$date = as.Date(Dframe2$Logtime)
both = merge(Dframe1,Dframe2)
both$when = ifelse(both$Logtime >= both$sunrise &
+both$Logtime <= both$sunse
Dear All,
I am not that much into machine learning, but I know that R is wonderful
at that. My problem is the following: consider a set of N individuals
and an N by N symmetric matrix M.
This matrix stands for an interaction between individuals, i.e. the
element M[i,j]=M[j,i] is the (cumulative
Dear List Members
Upfront: I am not sure if this is the right place to ask this question even
after reading through the FAQ, so apologies if I should have directed the
question elsewhere.
My problem as follows: I need a modified version of some existing R package.
I am well able to make the mod
Baptiste Coulmont yahoo.com> writes:
>
> I want to create a randomly generated 25*15 binary matrix knowing what the
> colSums and rowSums of this matrix should be.
>
> colSums(a)
> [1] 3 7 2 5 0 4 6 5 5 4 3 4 4 5...
> rowSums(a)
> [1] 8 6 6 6 6 6 5 5 5 5 4 4 4...
>
This is a fairly ch
On Thu, Feb 10, 2011 at 10:49 AM, Nikhil Joshi wrote:
> Gabor,
> thanks for the suggestion.
> I did as you have suggested- Started a clean session of R and set the java
> option. Then I sourced the library and tried to read the xlsx file but got
> the same error. I am assuming that Java machine
Try using tabulate() instead of table(). E.g.,
compare your original
f0 <- function (x, u) {
tu <- table(u)
indices <- as.numeric(names(tu))
x[indices] <- x[indices] + tu
x
}
to
f1 <- function (x, u) {
x + tabulate(u, nbins = length(x))
}
I tried it for a 20-
On 10/02/2011 11:02 AM, Hosack, Michael wrote:
R experts,
I need to sample two rows without replacement from the following data frame
such that
neither row contains the same 'DOW'. For example, I cannot select both a Monday
morning
and a Monday afternoon. I am using STRATA_NUM as an index to r
I want to create a randomly generated 25*15 binary matrix knowing what the
colSums and rowSums of this matrix should be.
colSums(a)
[1] 3 7 2 5 0 4 6 5 5 4 3 4 4 5...
rowSums(a)
[1] 8 6 6 6 6 6 5 5 5 5 4 4 4...
. .
Baptiste Coulmont
coulm...@yahoo.com
R experts,
I need to sample two rows without replacement from the following data frame
such that
neither row contains the same 'DOW'. For example, I cannot select both a Monday
morning
and a Monday afternoon. I am using STRATA_NUM as an index to randomly select
rows from,
since this variable
Thanks all for your help. I fear text mining is an abstract little corner of
"R".
I have imported 3228 text (.txt) files, each a news story, into R using
[tm]:
textd <- Corpus(DirSource("other/docs"), readerControl = list(reader
=readPlain))
I can pre-process each individual document using tolo
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