Hi all,
I'm just wondering if there is a equivalent of SAS's FIRST. and LAST.
variables in R?
For example, suppose this is a snapshot of the data:
ClientCode CaseCode open closeImportant
1 37 28 2003-07-08 2003-09-021
2 37
> -Mensaje original-
> De: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] En nombre de Dipa Hari
> Enviado el: lunes, 12 de julio de 2010 22:19
> Para: r-help@r-project.org
> Asunto: [R] ed50
>
> I am using semiparametric Model
> library(mgcv)
> sm1=gam(y~x1+s(x2)
Dear R users,
I have panel data on the amount of money spent by travellers from 8 origin
countries in 4 destinations. I would like to carry out analysis for
destinations, origins and time. However, it seems to me that the package
"plm" can only esitmate two-way panel data (indexed by a two-dimensi
Hi,
Am using maanova package for doing anova.But am getting error like
this..plz, help me regarding this..
> TGR=read.madata("rmaexpr.dat",designfile="design.dat")
Reading one color array.
Otherwise change arrayType='twoColor' then read the data again
Warning messages:
1: In read.madata("rmaexp
strings <- replicate(1e5, paste(sample(letters, 100, rep = T), collapse = ""))
system.time(strings[-1] == strings[-1e5])
# user system elapsed
# 0.016 0.000 0.017
So it takes ~1/100 of a second to do ~100,000 string comparisons. You
need to provide a reproducible example that illustrates
I am asking this question because String comparison in R seems to be
awfully slow (based on profiling results) and I wonder if perhaps '=='
alone is not the best one can do. I did not ask for anything
particular and I don't think I need to provide a self-contained source
example for the question. S
You missed FAQ 7.22
7.22 Why do lattice/trellis graphics not work?
The most likely reason is that you forgot to tell R to display the graph.
Lattice functions such as xyplot() create a graph object, but do not display
it (the same is true of *ggplot2* graphics, and Trellis graphics in S-Plus).
T
Please get a copy of
R for SAS and SPSS Users
*by*
*Muenchen*, Robert A.
http://www.springer.com/statistics/computanional+statistics/book/978-0-387-09417-5
[[alternative HTML version deleted]]
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https://stat.e
I had tried that earlier and didn't work either, I probably have \Sexpr in the
wrong place. See example:
Column one header gets blank:
\documentclass[11pt]{article}
\usepackage{longtable,verbatim,ctable}
\usepackage{longtable,pdflscape}
\usepackage{fmtcount,hyperref}
\usepackage{fullpage}
\title
There is a standard notation for passwords in urls ... see for example
http://www.devx.com/tips/Tip/5604
"Cliff Clive" wrote:
>
>Hello everyone,
>
>Is it possible to download data from password-protected ftp sites? I saw
>another thread with instructions for uploading files using RCurl, but I
On 07/12/2010 07:16 PM, Roger Deangelis wrote:
Hi,
I am new to R.
I am trying to create an R function to do a SAS proc means/summary
proc.means ( data=bsebal;
class team year;
var ab h;
output out=BseBalAvg mean=;
Hi,
I am new to R.
I am trying to create an R function to do a SAS proc means/summary
proc.means ( data=bsebal;
class team year;
var ab h;
output out=BseBalAvg mean=;
run;)
I have a solution if I quote the the
Hello everyone,
Is it possible to download data from password-protected ftp sites? I saw
another thread with instructions for uploading files using RCurl, but I
could not find information for downloading them in the RCurl documentation.
I am using R 2.11 on a Windows XP 32-bit machine.
Thanks
Have a look at the Task View for spatial data...
http://cran.ms.unimelb.edu.au/web/views/Spatial.html
>>>
From: "chris howden"
To:,
Date: 13/Jul/2010 2:01p
Subject: [R] do the standard R analysis functions handle spatial "grid" data?
Hi everyone,
I'm doing a resource function analysis with r
Hi everyone,
I'm doing a resource function analysis with radio collared dingos and GIS
info.
The ecologist I'm working with wants to send me the data in a 'grid
format'...straight out of ARCVIEW GIS.
I want to model the data using a GLM and maybe a LOGISTIC model as well. And
I was planning on u
Hello,
I am working on an R package for storing order book data. I currently
have a display method that has the following output (ob is an S4 object):
display(ob)
Current time is 09:35:02
Price Ask Size
--
11.42 900
On Jul 12, 2010, at 6:26 PM, YANG, Richard ChunHSi wrote:
I ran the following script from xyplot Examples using Tin-R on
Windows and saw no plot produced.
EE <- equal.count(ethanol$E, number=9, overlap=1/4)
xyplot(NOx ~ C | EE, data=ethanol,
prepanel = function(x,y) prepanel.loess(x, y, spa
On Jul 12, 2010, at 6:18 PM, Josh B wrote:
Hi R sages,
Here is my latest problem. Consider the following toy example:
x <- read.table(textConnection("y1 y2 y3 x1 x2
indv.1 bagels donuts bagels 4 6
indv.2 donuts donuts donuts 5 1
indv.3 donuts donuts donuts 1 10
indv.4 donuts donuts donuts 10
The problem is that you have not pushed your viewport so it doesn't
exist in the plot. (You only pushed the layout viewport).
> grid.ls(viewports = TRUE)
ROOT
GRID.VP.82
Try this:
vp <- vplayout(2,2)
pushViewport(vp)
upViewport()
grid.ls(viewports = TRUE)
#ROOT
# GRID.VP.82
#GRID.VP.86
pr
On Jul 12, 2010, at 6:46 PM, David Winsemius wrote:
On Jul 12, 2010, at 6:03 PM, harsh yadav wrote:
Hi,
I have a function in R that compares two very large strings for
about 1
million records.
The strings are very large URLs like:-
http://query.nytimes.com/gst/sitesearch_selector.html
Hi R sages,
Here is my latest problem. Consider the following toy example:
x <- read.table(textConnection("y1 y2 y3 x1 x2
indv.1 bagels donuts bagels 4 6
indv.2 donuts donuts donuts 5 1
indv.3 donuts donuts donuts 1 10
indv.4 donuts donuts donuts 10 9
indv.5 bagels donuts bagels 0 2
indv.6 bagels
I ran the following script from xyplot Examples using Tin-R on
Windows and saw no plot produced.
EE <- equal.count(ethanol$E, number=9, overlap=1/4)
xyplot(NOx ~ C | EE, data=ethanol,
prepanel = function(x,y) prepanel.loess(x, y, span=1),
xlab="Compression Ratio", ylab="NOx (micrograms/J
Hi, Bert,
Thanks for your thoughtful explanation. I think the problem is quite
over my head and maybe I should leave "how" for experts :)
The situation is I have a group of sigmoid curves (let's say, they are
supposed to be the same) but occasionally you will see a few curves
kind of different. S
On Jul 12, 2010, at 6:03 PM, harsh yadav wrote:
Hi,
I have a function in R that compares two very large strings for
about 1
million records.
The strings are very large URLs like:-
http://query.nytimes.com/gst/sitesearch_selector.html?query=US+Visa+Laws&type=nyt&x=25&y=8
.
..
or of lar
On Jul 12, 2010, at 5:45 PM, Felipe Carrillo wrote:
Thanks for the quick reply Duncan.
I don't think I have explained myself well, I have a dataset named
"report" and
my column headers are run1,run2,run3,run4 and so on.
I know how to access the data below those columns with
\Sexpr{report[
Hi,
Is there a way to create a matrix in which the column names are not
checked to see if they are valid variable names?
I'm looking something similar to the check.names argument to
data.frame. If so, would such an approach work for the sparse matrix
classes in the Matrix package.
Many thanks!
Hi,
I have a function in R that compares two very large strings for about 1
million records.
The strings are very large URLs like:-
http://query.nytimes.com/gst/sitesearch_selector.html?query=US+Visa+Laws&type=nyt&x=25&y=8.
..
or of larger lengths.
The data-frame looks like:-
id url
1
http:/
Thanks Duncan... More appended at the bottom...
On 7/12/10 5:38 PM, "Duncan Murdoch" wrote:
> On 12/07/2010 5:25 PM, Bryan Hanson wrote:
>> Hello Wise Ones...
>>
>> I need a clever way around a problem with findInterval. Consider:
>>
>> vec1 <- 1:10
>> vec2 <- seq(1, 10, by = 0.1)
>>
>> x1
Thanks for the quick reply Duncan.
I don't think I have explained myself well, I have a dataset named "report" and
my column headers are run1,run2,run3,run4 and so on.
I know how to access the data below those columns with \Sexpr{report[1,1]} &
&\Sexpr{report[1,2]} and so on, but I can't access
How about this:
these = which(vec2 < x1[1] | vec2 > x1[2])
vec2[these]
# Or using logical indexation directly:
vec2[vec2 < x1[1] | vec2 > x1[2]]
>>>
From: Bryan Hanson
To:R Help
Date: 13/Jul/2010 9:28a
Subject: [R] findInterval and data resolution
Hello Wise Ones...
I need a clever way aroun
Greetings to all, and my apologies for a question that is mostly about
statistics and secondarily about R. I have just started a new job that
(this week, apparently) requires statistical knowledge beyond my training
(as an epidemiologist).
The problem:
- We have 57 food production facilities in
On 12/07/2010 5:25 PM, Bryan Hanson wrote:
Hello Wise Ones...
I need a clever way around a problem with findInterval. Consider:
vec1 <- 1:10
vec2 <- seq(1, 10, by = 0.1)
x1 <- c(2:3)
a1 <- findInterval(x1, vec1); a1 # example 1
a2 <- findInterval(x1, vec2); a2 # example 2
In the problem I'
Hello Wise Ones...
I need a clever way around a problem with findInterval. Consider:
vec1 <- 1:10
vec2 <- seq(1, 10, by = 0.1)
x1 <- c(2:3)
a1 <- findInterval(x1, vec1); a1 # example 1
a2 <- findInterval(x1, vec2); a2 # example 2
In the problem I'm working on, vec* may be either integer or n
On 12/07/2010 5:10 PM, Felipe Carrillo wrote:
Hi:
Since I work with a few different fish runs my column headers change everytime
I start a new Year. I have been using \Sexpr{} for my row and columns and now
I am trying to use with my report column headers. \Sexpr{1,1} is row 1 column 1,
what can
Hi:
Since I work with a few different fish runs my column headers change everytime
I start a new Year. I have been using \Sexpr{} for my row and columns and now
I am trying to use with my report column headers. \Sexpr{1,1} is row 1 column 1,
what can I use for headers? I tried \Sexpr{0,1} but swea
Hi everyone I dont know how to code in SAS but I do know how to code in R.
Can someone please be kind enough to translate this into R code for me:
proc mixed data = small method = reml;
class id day;
model weight = day/ solution ddfm = bw;
repeated day/ subject=id type = unstructured;
run;
==
Jun:
Short answer: There is no such thing as df for a nonlinear model (whether or
not mixed effects).
Longer answer: df is the dimension of the null space when the data are
projected on the linear subspace of the model matrix of a **linear model **
. So, strictly speaking, no linear model, no df.
Short course: Statistical Learning and Data Mining III:
Ten Hot Ideas for Learning from Data
Trevor Hastie and Robert Tibshirani, Stanford University
Georgetown University Conference Center
Washington DC,
October 11-12, 2010.
This two-day course gives a detailed overview of statistical m
Thanks Jim,
I acted on your suggestion and found the result unchanged. :-( Then I
noticed that fitdist doesn't like a sample size of 1 either.
If, then, "drop = TRUE" results in all empty combinations of m_id, year and
week being excluded, then (noticing the requirement is actually that the
sam
OK, here is a stripped down variant of my code. I can run it here
unchanged (apart from the credentials for connecting to my DB).
Sys.setenv(MYSQL_HOME='C:/Program Files/MySQL/MySQL Server 5.0')
> library(TSMySQL)
> library(plyr)
> library(fitdistrplus)
> con <- dbConnect(MySQL(), user="rejbyers
I am using semiparametric Model
library(mgcv)
sm1=gam(y~x1+s(x2),family=binomial, f)
How should I find out standard error for ed50 for the above model
ED50 =( -sm1$coef[1]-f(x2)) / sm1$coef [2]
f(x2) is estimated value for non parametric term.
Thanks
[[alternative HTML vers
try 'drop=TRUE' on the split function call. This will prevent the
NULL set from being sent to the function.
On Mon, Jul 12, 2010 at 3:10 PM, Ted Byers wrote:
> >From the documentation I have found, it seems that one of the functions from
> package plyr, or a combination of functions like split a
On Mon, Jul 12, 2010 at 2:25 PM, Bogaso Christofer
wrote:
> Hi all, can anyone please guide me how to create a sequence of months? Here
> I have tried following however couldn't get success
>
>> library(zoo)
>
>> seq(as.yearmon("2010-01-01"), as.yearmon("2010-03-01"), by="1 month")
>
There curren
I have the same problem and I wonder if there is any answer from the
community. Thanks.
--
View this message in context:
http://r.789695.n4.nabble.com/ggplot2-How-to-change-font-of-labels-in-geom-text-tp991579p2286671.html
Sent from the R help mailing list archive at Nabble.com.
___
Hello.
I've got a dataset that may have outliers in both x and y. While I am not
at all familiar with robust regression, it looked like the function lmrob in
package robustbase should handle this situation. When I try to use it, I
get:
Too many singular resamples
Aborting fast_s_w_mem()
Looki
Dear all,
I want to do a F test, which involves calculation of the degrees of
freedom for the residuals. Now say, I have a nlme object "mod.nlme". I
have two questions
1.How do I extract the degrees of freedom?
2.How is this degrees of freedom calculated in an nlme model?
Thanks.
Jun Shen
Some
Your code is not reproducible. Can you come up with a small example
showing the crux of your data structures/problem, that we can all run in
our R sessions? You're likely get much higher quality responses this way.
Ted Byers wrote:
From the documentation I have found, it seems that one of th
Hi Andrew,
Try
sapply(lst, "[", 2)
HTH,
Jorge
On Mon, Jul 12, 2010 at 3:12 PM, Andrew Leeser <> wrote:
> I am looking for a way to create a vector which contains the second element
> of
> every vector in the list. However, not every vector has two components, so
> I
> need to generate an NA f
I am looking for a way to create a vector which contains the second element of
every vector in the list. However, not every vector has two components, so I
need to generate an NA for those missing. For example, I have created the
following list:
lst <- list(c("a", "b"), c("c"), c("d", "e"), c("
>From the documentation I have found, it seems that one of the functions from
package plyr, or a combination of functions like split and lapply would
allow me to have a really short R script to analyze all my data (I have
reduced it to a couple hundred thousand records with about half a dozen
recor
I wanted to point out one thing that Ted said, about initializing the
vectors ('s' in your example). This can make a dramatic speed
difference if you are using a for loop (the difference is neglible
with vectorized computations).
Also, a lot of benchmarks have been flying around, each from a
diff
Hi There,
I get the following error from the code pasted below: Error in
storage.mode(test) <- "logical" :
object 'HGBmt12_Natl_Ave_or_Facility' not found
library(RODBC)
library(car)
setwd("c://temp//cms")
a07.connect <- odbcConnectAccess2007("DFC.accdb")
sqlTables(a07.connect) ##provides list
Am 12.07.2010 20:25, schrieb Bogaso Christofer:
>> library(zoo)
>
>> seq(as.yearmon("2010-01-01"), as.yearmon("2010-03-01"), by="1 month")
>
seq(as.Date("2010-01-01"), as.Date("2010-03-01"), by="1 month")
hth
Stefan
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R-help@r-project.org mailing
What was the question and answer here?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of pdb
Sent: Sunday, July 11, 2010 5:23 AM
To: r-help@r-project.org
Subject: Re: [R] eliminating constant variables
Importance: Low
Awsome!
It mad
Try fitdistr() in pkg MASS.
-Peter Ehlers
On 2010-07-12 11:17, Oscar Rodriguez wrote:
Dear R community:
Sorry if this question has a simple answer, but I am a new user of R.
Do you know a command, or package that can estimate the weibull distribution's
mean, standard deviation and variance
As in this example:
> seq(as.Date("2000/1/1"), as.Date("2003/1/1"), by="mon")
On 7/12/10 11:25 AM, "Bogaso Christofer"
wrote:
> Hi all, can anyone please guide me how to create a sequence of months? Here
> I have tried following however couldn't get success
>
>
>
>> library(zoo)
>
>> seq(
Hi all, can anyone please guide me how to create a sequence of months? Here
I have tried following however couldn't get success
> library(zoo)
> seq(as.yearmon("2010-01-01"), as.yearmon("2010-03-01"), by="1 month")
Error in del/by : non-numeric argument to binary operator
What is the cor
I am trying to recreate an analysis that has been done by another group
(in SAS I believe). I'm stuck on one part, I think because my stats
knowledge is lacking, and while it's OT, I'm hoping someone here can help.
Given this dataframe;
foo*<-*structure(list(OBS = structure(1:18, .Label = c("1
Thanks to you all. I stand corrected Ted and Manuela:) I am just an end user
and trying to pick up from such forums. Many thanks sirs.
On Mon, Jul 12, 2010 at 5:45 PM, Huso, Manuela wrote:
> Using Ted Harding's example:
>
> news1o <- runif(100)
> s2o<- runif(100)
>
> pt1 <- proc.tim
Hi Seth,
Can you recreate the example below using dbWriteTable?
Thanks!,
Erik
On Jul 11, 2010, at 6:13 PM, Seth Falcon wrote:
> On Sun, Jul 11, 2010 at 11:31 AM, Matt Shotwell wrote:
>> On Fri, 2010-07-09 at 20:02 -0400, Erik Wright wrote:
>>> Hi Matt,
>>>
>>> This works great, thanks!
>>>
Using Ted Harding's example:
news1o <- runif(100)
s2o<- runif(100)
pt1 <- proc.time()
s <- numeric(length(news1o))-1 # Set all of s to -1
s[news1o>s2o] <-1# Change to 1 only those values of s
# for which news1o>s2o
pt2<-
Dear R community:
Sorry if this question has a simple answer, but I am a new user of R.
Do you know a command, or package that can estimate the weibull distribution's
mean, standard deviation and variance? or can direct me to where to find it?
Thanks in advance,
Oscar Rodriguez Gon
There is no constraint on the magnitude of probability density values, though
the area under the curve must be equal to one. You may be thinking of
cumulative probability distributions? If so, take a look at smoothed.df() in
library (cwhmisc).
"Katja Hillmann" wrote:
>Hello,
>
>I used the com
Richardson, Patrick vai.org> writes:
>
> I'm trying to use multiple plotting colors in my code. My first "ifelse"
statement successfully does what I
> want. However, now I want anything less than -4.5 to be green and the rest
black. I want another "col"
> argument but can only use one. How could
Steve,
That worked perfectly. Thank You!
Best regards,
Patrick
-Original Message-
From: Steve Lianoglou [mailto:mailinglist.honey...@gmail.com]
Sent: Monday, July 12, 2010 12:55 PM
To: Richardson, Patrick
Cc: r-help@r-project.org
Subject: Re: [R] Multiple Ploting Colors
Hi,
On Mon,
Ugh:
> Depending on what you want the plot for, perhaps you might consider
> changing your color palette from green -> black -> red to something
> like blue -> black -> yellow, since many folks who are color can not
> differentiate green from red all that well.
... folks who are color *blind* can
chen jia fisher.osu.edu> writes:
>
Check out the ?par(). Specifically mgp.
HTH,
Ken
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and
Hello,
I used the command kdensity in order to calculate the density of
fractions/ratios (e.g. number of longterm unemployed on total
unemployment). Thus I try to calculate the denisty of values less than
1. However, the values of the kernel densitiy R provided (y-scale) are
all greater than
One more thing:
On Mon, Jul 12, 2010 at 12:55 PM, Steve Lianoglou
wrote:
> Hi,
>
> On Mon, Jul 12, 2010 at 12:02 PM, Richardson, Patrick
> wrote:
>> I'm trying to use multiple plotting colors in my code. My first "ifelse"
>> statement successfully does what I want. However, now I want anything
Hi,
On Mon, Jul 12, 2010 at 12:02 PM, Richardson, Patrick
wrote:
> I'm trying to use multiple plotting colors in my code. My first "ifelse"
> statement successfully does what I want. However, now I want anything less
> than -4.5 to be green and the rest black. I want another "col" argument but
Hi everybody!
I have the next code which makes a reduction of the *a *variable in two
slaves, using the Rmpi package.
library(Rmpi)
mpi.spawn.Rslaves(nslaves=2)
reduc<-function(){
a<-mpi.comm.rank()+2
mpi.reduce(a,type=2, op="prod")
return(paste("a=",a))
}
mpi.bcast.Robj2slave(reduc)
mpi
Richardson, Patrick vai.org> writes:
>
> I'm trying to use multiple plotting colors in my code. My first "ifelse"
statement successfully does what I
> want. However, now I want anything less than -4.5 to be green and the rest
black. I want another "col"
> argument but can only use one. How could
On Mon, Jul 12, 2010 at 9:17 AM, Erik Wright wrote:
> Hi Seth,
>
> Can you recreate the example below using dbWriteTable?
>
Not sure if that is possible with the current dbWriteTable code (don't
have time to explore that right now). You are welcome to poke around.
You could wrap the example in a
On 12-Jul-10 14:09:30, Raghu wrote:
> When I just run a for loop it works. But if I am going to
> run a for loop every time for large vectors I might as well
> use C or any other language.
> The reason R is powerful is becasue it can handle large vectors
> without each element being manipulated? Pl
On Jul 12, 2010, at 10:09 AM, Raghu wrote:
When I just run a for loop it works. But if I am going to run a for
loop
every time for large vectors I might as well use C or any other
language.
The reason R is powerful is becasue it can handle large vectors
without each
element being manipula
Hi,
one comment: Claeskens and Hjort define AIC as 2*log L - 2*p for a model
with likelihood L and p parameters; consequently, they look for models
with *maximum* AIC in model selection and averaging. This differs from
the vast majority of authors (and R), who define AIC as -2*log L + 2*p
and sea
In an if statement, you can use only elements. In your example, news1o
and s2o are vectors so there is a warning saying the two vectors have a
bigger length than one.
If you don't send two messages about the same problem in two minutes,
you can see what people answer you... For example, I adv
You probably want to use "ifelse"
s <- ifelse(news1o>s2o, 1, -1)
'if' only handle a single logical expression.
On Mon, Jul 12, 2010 at 10:02 AM, Raghu wrote:
> I know the following may sound too basic but I thought the mailing list is
> for the benefit of all levels of people. I ran a simple i
> The reason R is powerful is becasue it can handle large vectors without
each
> element being manipulated? Please let me know where I am wrong.
>
> for(i in 1:length(news1o)){
> + if(news1o[i]>s2o[i])
> + s[i]<-1
> + else
> + s[i]<--1
> + }
You might give ifelse() a shot here.
s <- ifelse(news
I don't know what is wrong with your code but I believe you should use
ifelse instead of a for loop:
s <- ifelse(news1o > s2o, 1 , -1 )
Alain
On 12-Jul-10 16:09, Raghu wrote:
When I just run a for loop it works. But if I am going to run a for loop
every time for large vectors I might as wel
I'm trying to use multiple plotting colors in my code. My first "ifelse"
statement successfully does what I want. However, now I want anything less than
-4.5 to be green and the rest black. I want another "col" argument but can only
use one. How could I go about getting separate colors for anyth
Hi,
I have a problem converting a XML file, via the XML package, to a data.frame.
The XML file looks like this:
I don't usually use XML files, but I have searched for an answer for quite a
while. I have tried "xmlToDataFra
Hi there,
I place a vector of strings as labels at the tick points by using
axis(1,at=seq(0.1,0.7,by=0.1),
labels=paste(seq(10,70,by=10),"%",sep=""), tick=FALSE)
However, there is a large space between those labels and the boundary
of plot box. I want to reduce this space so that the labels appe
I know the following may sound too basic but I thought the mailing list is
for the benefit of all levels of people. I ran a simple if statement on two
numeric vectors (news1o and s2o) which are of equal length. I have done an
str on both of them for your kind perusal below. I am trying to compare t
When I just run a for loop it works. But if I am going to run a for loop
every time for large vectors I might as well use C or any other language.
The reason R is powerful is becasue it can handle large vectors without each
element being manipulated? Please let me know where I am wrong.
for(i in 1
Dear all,
I share the problem Linda Garcia and Ram Kumar Basnet described; I have a
biclust object, containing several clusters. For drawing a heatmap, it is
possible to specify the cluster to be plotted. However, I'd like to extract
the clusters in this manner:
Cond.1 Cond.2
Gene - v
If you want to use the mondate package you will need to specify the day of
the month. Dates "March-2010" and "May-2010" are ambiguous. Recommend you
choose last day of both months as "representative days." Then those days
will be integer months apart.
> a<-mondate("March 31, 2010", displayFormat="
I encourage all authors and maintainers of packages that they use
"findFn" in the "sos" package to search for other uses of a name you
want to use. The "findFn" function searches for matches in the help
pages of contributed packages, including all of CRAN plus some
elsewhere. The "grepF
Juliet,
I've been corrected off list. I did not read properly that you are on 64bit.
The calculation should be :
53860858 * 4 * 8 /1024^3 = 1.6GB
since pointers are 8 bytes on 64bit.
Also, data.table is an add-on package so I should have included :
install.packages("data.table")
requi
On Jul 12, 2010, at 11:19 AM, Duncan Murdoch wrote:
On 12/07/2010 11:10 AM, g...@ucalgary.ca wrote:
I would like to sum/mean/min a list of lists and numbers to return
the
related lists.
-1+2*c(1,1,0)+2+c(-1,10,-1) returns c(2,13,0) but
sum(1,2*c(1,1,0),2,c(-1,10,-1)) returns 15 not a list.
require("grid")
require("lattice")
fred = data.frame(x=1:5,y=runif(5))
vplayout <- function (x,y) viewport(layout.pos.row=x, layout.pos.col=y)
grid.newpage()
pushViewport(viewport(layout=grid.layout(2,2)))
p = xyplot(y~x,fred)
print( p,newpage=FALSE,draw.in=vplayout(2,2)$name)
On Mon, Jul 12, 20
I am trying to create a model using the Quantmod package in R. I am
using the following string of commands:
> ema<-read.csv(file="ESU0 Jul 7 1 sec data.csv")
> Bid=(ema$Bid)
> twentysell=EMA(Bid,n=1200)
> fortysell=EMA(Bid,n=2400)
> sigup<-ifelse(twentysell>fortysell,1,0)
> sigdn<-ifelse(twentysell
Hi,
After searching the archives and Google and not turning up anything, I
thought I'd ask here.
Has anyone done an R package for calculating Gwet's AC1 statistic and variance?
K.L. Gwet. Computing inter-rater reliability and its variance in the
presence of high agreement. Br J Math Stat Psychol
On Jul 12, 2010, at 11:10 AM, g...@ucalgary.ca wrote:
I would like to sum/mean/min a list of lists and numbers to return the
related lists.
You will advance in your understanding faster if you adopt the correct
terminology:
-1+2*c(1,1,0)+2+c(-1,10,-1) returns c(2,13,0) but
... which is
On 12/07/2010 11:10 AM, g...@ucalgary.ca wrote:
I would like to sum/mean/min a list of lists and numbers to return the
related lists.
-1+2*c(1,1,0)+2+c(-1,10,-1) returns c(2,13,0) but
sum(1,2*c(1,1,0),2,c(-1,10,-1)) returns 15 not a list.
Using the suggestions of Gabor Grothendieck,
Reduce('+',l
Dear Deepayan,
Thank you for taking the time to look into this issue.
I have a data object called "Data", please find it at the end of the
message. Then I can run the code below separately in the console.
#Construct the nlme object
mod.nlme<-nlme(RESP~E0+(Emax-E0)*CP**gamma/(EC50**gamma+CP**gamm
I would like to sum/mean/min a list of lists and numbers to return the
related lists.
-1+2*c(1,1,0)+2+c(-1,10,-1) returns c(2,13,0) but
sum(1,2*c(1,1,0),2,c(-1,10,-1)) returns 15 not a list.
Using the suggestions of Gabor Grothendieck,
Reduce('+',list(-1,2*c(1,1,0),2,c(-1,10,-1))) returns what we
Hi Juliet,
Thanks for the info.
It is very slow because of the == in testData[testData$V2==one_ind,]
Why? Imagine someoone looks for 10 people in the phone directory. Would
they search the entire phone directory for the first person's phone number,
starting
on page 1, looking at every single
You might want to do 'object.size' on myData to see how big it is and
then if you do try to run reshape again take a look and see if there
is any paging happening on your system which may be an indication that
you don't have enough memory. Also with 53M observations, it may take
a lot of time to d
Dear R-listers,
I have a set of plant multivariate morphological data (including
thousands of individuals from several species) which may be identified
as several units of morphological variation.
Firstly, I want to defined the optimal number of such units
Then, I want to choose proper discrimina
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