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
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
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,
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
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
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$Bid)
twentysell=EMA(Bid,n=1200)
fortysell=EMA(Bid,n=2400)
sigup-ifelse(twentysellfortysell,1,0)
require(grid)
require(lattice)
fred = data.frame(x=1:5,y=runif(5))
vplayout - function (x,y) viewport(layout.pos.row=x, layout.pos.col=y)
grid.newpage()
pushViewport(viewport(layout=grid.layout(2,2)))
p = xyplot(y~x,fred)
print( p,newpage=FALSE,draw.in=vplayout(2,2)$name)
On Mon, Jul 12, 2010
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.
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)
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 grepFn can
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=%B %d, %Y)
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 -
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
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
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 appear
Hi,
I have a problem converting a XML file, via the XML package, to a data.frame.
The XML file looks like this:
Transaction
ID value=0044/
Var1 value=XYZ159/
Var2 value=_/
Var3 value=AMR1.0-INT-1005/
Var4 value=2010-05-25 10:44:16:673/
Var5 value=1/
Var6 value=0/
/Transaction
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 well
You probably want to use ifelse
s - ifelse(news1os2o, 1, -1)
'if' only handle a single logical expression.
On Mon, Jul 12, 2010 at 10:02 AM, Raghu r.raghura...@gmail.com wrote:
I know the following may sound too basic but I thought the mailing list is
for the benefit of all levels of people.
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(news1o s2o, 1,
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
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
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
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? Please
On Mon, Jul 12, 2010 at 9:17 AM, Erik Wright eswri...@wisc.edu 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
Richardson, Patrick Patrick.Richardson at 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
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)
Hi,
On Mon, Jul 12, 2010 at 12:02 PM, Richardson, Patrick
patrick.richard...@vai.org 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
One more thing:
On Mon, Jul 12, 2010 at 12:55 PM, Steve Lianoglou
mailinglist.honey...@gmail.com wrote:
Hi,
On Mon, Jul 12, 2010 at 12:02 PM, Richardson, Patrick
patrick.richard...@vai.org wrote:
I'm trying to use multiple plotting colors in my code. My first ifelse
statement successfully
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
chen jia chen_1002 at fisher.osu.edu writes:
Check out the ?par(). Specifically mgp.
HTH,
Ken
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
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 not
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
Richardson, Patrick Patrick.Richardson at 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
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 katja.hillm...@wiso.uni-hamburg.de
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
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 shotw...@musc.edu 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[news1os2o] -1# Change to 1 only those values of s
# for which news1os2o
pt2-
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 manuela.h...@oregonstate.edu
wrote:
Using Ted Harding's example:
news1o - runif(100)
s2o-
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,
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 correct
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 bogaso.christo...@gmail.com
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
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
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
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
__
R-help@r-project.org mailing list
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 of
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
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
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(f, g))
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 for those
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
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
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()
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.
On Mon, Jul 12, 2010 at 2:25 PM, Bogaso Christofer
bogaso.christo...@gmail.com 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)
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 r.ted.by...@gmail.com wrote:
From the documentation I have found, it seems that one of the functions from
package plyr, or a combination of
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
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,
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
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
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
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;
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
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
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'm
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
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 han...@depauw.edu
To:R Help r-h...@stat.math.ethz.ch
Date: 13/Jul/2010 9:28a
Subject: [R] findInterval and data resolution
Hello Wise
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 my
Thanks Duncan... More appended at the bottom...
On 7/12/10 5:38 PM, Duncan Murdoch murdoch.dun...@gmail.com 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)
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+Lawstype=nytx=25y=8.
..
or of larger lengths.
The data-frame looks like:-
id url
1
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!
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
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+Lawstype=nytx=25y=8
.
..
or of
I ran the following script from xyplot Examples using Tin-R on
Windows and saw no plot produced.
EE - equal.count(ethanol$E, number=9, overlap=1/4)
xyplot(NOx ~ C | EE, data=ethanol,
prepanel = function(x,y) prepanel.loess(x, y, span=1),
xlab=Compression Ratio, ylab=NOx (micrograms/J),
Hi R sages,
Here is my latest problem. Consider the following toy example:
x - read.table(textConnection(y1 y2 y3 x1 x2
indv.1 bagels donuts bagels 4 6
indv.2 donuts donuts donuts 5 1
indv.3 donuts donuts donuts 1 10
indv.4 donuts donuts donuts 10 9
indv.5 bagels donuts bagels 0 2
indv.6 bagels
On Jul 12, 2010, at 6:46 PM, David Winsemius wrote:
On Jul 12, 2010, at 6:03 PM, harsh yadav wrote:
Hi,
I have a function in R that compares two very large strings for
about 1
million records.
The strings are very large URLs like:-
The problem is that you have not pushed your viewport so it doesn't
exist in the plot. (You only pushed the layout viewport).
grid.ls(viewports = TRUE)
ROOT
GRID.VP.82
Try this:
vp - vplayout(2,2)
pushViewport(vp)
upViewport()
grid.ls(viewports = TRUE)
#ROOT
# GRID.VP.82
#GRID.VP.86
On Jul 12, 2010, at 6:18 PM, Josh B wrote:
Hi R sages,
Here is my latest problem. Consider the following toy example:
x - read.table(textConnection(y1 y2 y3 x1 x2
indv.1 bagels donuts bagels 4 6
indv.2 donuts donuts donuts 5 1
indv.3 donuts donuts donuts 1 10
indv.4 donuts donuts donuts 10 9
On Jul 12, 2010, at 6:26 PM, YANG, Richard ChunHSi wrote:
I ran the following script from xyplot Examples using Tin-R on
Windows and saw no plot produced.
EE - equal.count(ethanol$E, number=9, overlap=1/4)
xyplot(NOx ~ C | EE, data=ethanol,
prepanel = function(x,y) prepanel.loess(x, y,
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
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
Have a look at the Task View for spatial data...
http://cran.ms.unimelb.edu.au/web/views/Spatial.html
From: chris howden tall.chr...@yahoo.com.au
To:r-help@r-project.org, r-sig-ecology-requ...@r-project.org
Date: 13/Jul/2010 2:01p
Subject: [R] do the standard R analysis functions handle
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
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
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=;
There is a standard notation for passwords in urls ... see for example
http://www.devx.com/tips/Tip/5604
Cliff Clive cliffcl...@gmail.com wrote:
Hello everyone,
Is it possible to download data from password-protected ftp sites? I saw
another thread with instructions for uploading files using
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}
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
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R-help@r-project.org mailing list
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).
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.
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 have 30 files in the current directories, i would like to perform the
cbind(fil1,file2,file3,file4file30)
how could i do this in a for loop:
such as:
file2 - list.files(pattern=.out3$)
for (j in file2) {
cbind(j)...how to implement cbind here
}
Thanks.
On Mon, Jul 12, 2010 at 2:51 AM, Jun Shen jun.shen...@gmail.com wrote:
Dear all,
When I construct an nlme model object by calling nlme(...)-mod.nlme,
this object can be used in xyplot(). Something like
xyplot(x,y,..
..
ind.predict-predict(mod.nlme)
..
) is working fine in
how to use boxplot on all the columns from he date frame instead of manually
entering the columns like below
bhtest1 - read.table(bhtest1.txt, header=TRUE)
boxplot (bhtest1[,2], bhtest1[,3], bhtest1[, 4], bhtest1[,5], bhtest1[,6],
bhtest1[,7])
please help, Thanks,
--
View this
This is also mentioned in FAQ 7.31
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
Also if you search the R-help archives for 'precision' you can find a
lot of threads discussing the issue in further depth.
On Sun, Jul 11, 2010 at 9:02 PM, Wu
Actually,
boxplot (bhtest1)
Should do what you want...
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com
Hi,
Assuming that you have read the files into R,
and that their names (in R) are held in some object
(e.g., 'file2'), then this works
do.call(what = cbind, args = mget(x = file2, envir = .GlobalEnv)
Here is a reproducible example:
x1 - data.frame(x = 1:10)
x2 - data.frame(y = 1:10)
file.names
On Sun, 11 Jul 2010, Spencer Graves wrote:
Hi, Richard and Duncan:
Thank you both very much. You provided different but workable
solutions.
1. With Rgui 2.11.1 on Vista x64, the escape worked, but neither
ctrl-c nor ctrl-C worked for me.
Why did you expect them too?
On Fri, Jul 9, 2010 at 9:11 PM, Gene Leynes gleyne...@gmail.com wrote:
I thought the apply functions are faster than for loops, but my most
recent test shows that apply actually takes a significantly longer than a
for loop. Am I missing something?
Check Rnews for an article discussing proper
On 12/07/10 08:16, Liviu Andronic wrote:
On Fri, Jul 9, 2010 at 9:11 PM, Gene Leynesgleyne...@gmail.com wrote:
[...]
Check Rnews for an article discussing proper usage of apply and for.
Liviu
I am guessing you are referencing
@article{Rnews:Ligges+Fox:2008
Thanks a lot for the reply, some comments below
On 07/10/2010 04:11 AM, Steve Lianoglou wrote:
Hi,
On Fri, Jul 9, 2010 at 12:15 PM, manuel.martin
manuel.mar...@orleans.inra.fr wrote:
Dear all,
after having calibrated a svm model through the svm() command of the e1071
package, is there a
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