On Sep 17, 2010, at 2:54 AM, C.H. wrote:
Dear R users,
I have a very simple question and I've tried to search for the answer.
(But failed.)
there should be a function (func) that work like
abc <- c(1,2,3,4)
func(abc)
"abc"
> func <- function(xyz) deparse(substitute(xyz))
> func(abc)
[1]
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 17/09/10 08:38, Peter Dalgaard wrote:
> On 09/17/2010 12:14 AM, Jeremy Miles wrote:
>>> I know from organizing a conference in Germany that the only really good way
>>> was and is ordinary money transfer via BIC and IBAN numbers. Unfortunately,
>>>
On Fri, Sep 17, 2010 at 7:54 AM, C.H. wrote:
> Dear R users,
>
> I have a very simple question and I've tried to search for the answer.
> (But failed.)
>
> there should be a function (func) that work like
>
>> abc <- c(1,2,3,4)
>> func(abc)
> "abc"
>
> I would like to know the name of that functio
On Fri, 17 Sep 2010, Peter Dalgaard wrote:
On 09/17/2010 12:14 AM, Jeremy Miles wrote:
As far as I remember, this requires that a real person opens
the account, and takes on the associated tax issues. It may
be different for US-based projects.
Nope -- all around the world, the governments
First, I don't have the correct lingo for this topic, so I can't really find a
solution for my problem. And
maybe I formulate it incorrectly, so bear with me.
How would I calculate a 'constant transition matrix' if I know a given value at
a given time?
Let's say I know that my value is 54,0
On 09/16/2010 11:40 PM, Josh B wrote:
Hi all,
Please consider the following code:
require(plotrix)
l<- list(rnorm(50),rnorm(50,sd=2),rnorm(50,mean=3))
multhist(l)
I have two things I need help with:
(1) In the output, there are empty spaces on the x-axis. How would I eliminate
these spaces?
On 17-Sep-10 08:19:20, Stefan Petersson wrote:
> First, I don't have the correct lingo for this topic, so I can't
> really find a solution for my problem. And maybe I formulate it
> incorrectly, so bear with me.
>
> How would I calculate a 'constant transition matrix' if I know a
> given value at
Hi,
> I would GLADLY donate 30-50$ each year just to see R develop, but there
> needs to be a way for me to do it in a civilized manner. If the userbase of
> R is over 2 million there will surely be 100,000 users who, like myself,
> will happily fork out 40$ a year - would that help? you can d
I would like to thank you again for your help.
But it seems that the floor function (ceiling, round too) create more dots in
the matrix that line really "touches".
unique( floor( cbind( seq(2,62, by=0.1), linefn(seq(2,62, by=0.1)) ) ) )
You can see that in the picture below
http://yfrog.com/5
Hi Jim,
First, I think it's better if you reply to the list, other users might
be interested and have better answers.
Second, as other replies showed you, using a list is actually way easier
than creating new objects every time. Check especially the reply from
Joshua Wiley, which gives you g
I think that's because of your definition for "touches".
If I read it correctly, David's code assumes that a "cell" is a square
1 unit wide and then followed the normal geometric definition of
"touches", ie. the line intersects with at least one point on the
cell's boundary.
You seem to have a di
Hi Jan
Jan private napsal dne 17.09.2010 12:43:40:
> Hello Petr,
>
> > but I think this is how your code really works. Did you try it?
>
> it does, but the R documentation says somewhere:
> "Warning: for() loops are used in R code much less often than in
> compiled languages. Code that takes a
Hello everyone,
please consider the following lines of a matrix
[574,] 59 32
[575,] 59 32
[576,] 59 32
[577,] 59 32
[578,] 59 32
[579,] 59 32
[580,] 59 32
[581,] 60 32
[582,] 60 33
[583,] 60 33
[584,] 60 33
[585,] 60 33
[586,] 60 33
[587,] 60
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 17/09/10 14:19, Alaios wrote:
> Hello everyone,
> please consider the following lines of a matrix
>
>
> [574,] 59 32
> [575,] 59 32
> [576,] 59 32
> [577,] 59 32
> [578,] 59 32
> [579,] 59 32
> [580,] 59 32
> [581,] 6
Hello Petr,
> but I think this is how your code really works. Did you try it?
it does, but the R documentation says somewhere:
"Warning: for() loops are used in R code much less often than in
compiled languages. Code that takes a `whole object' view is likely to
be both clearer and faster in R."
Try this:
aggregate(rep(1, nrow(x)), x, sum)
On Fri, Sep 17, 2010 at 9:19 AM, Alaios wrote:
> Hello everyone,
> please consider the following lines of a matrix
>
>
> [574,] 59 32
> [575,] 59 32
> [576,] 59 32
> [577,] 59 32
> [578,] 59 32
> [579,] 59 32
> [580,] 59 3
On 17-Sep-10 12:19:10, Alaios wrote:
> Hello everyone,
> please consider the following lines of a matrix
>
> [574,] 59 32
> [575,] 59 32
> [576,] 59 32
> [577,] 59 32
> [578,] 59 32
> [579,] 59 32
> [580,] 59 32
> [581,] 60 32
> [582,] 60 33
> [583,] 60 33
>
Dear list,
I have a question I'm trying to use the following command in R, but it gives
me an error message.The command is:
data<-ddply(data,c("year","name"), transform, check1=ifelse(check1==1 &
check2==1, 1,NULL))
so in my data frame I already have the check1 variable, if the conditions
(che
Help!
I am unsure if I can analyze data from the following experiment.
Fish were placed in a tank at (t=0)
Measurements of Carbon Dioxide were taken each day for 120 days
(t=0,...120)
A few fish were then randomly pulled out of the tank at different days,
killed and examined for the presence of a
Hi,
Not sure since I've never done it, but shouldn't it be NA instead of NULL?
Ivan
Le 9/17/2010 15:23, n.via...@libero.it a écrit :
> Dear list,
> I have a question I'm trying to use the following command in R, but it gives
> me an error message.The command is:
>
> data<-ddply(data,c("year","
>
setClass("person",representation(name="character",age="numeric",kids="list"))
[1] "person"
> bob <- new("person")
> length(b...@name)
[1] 0
> length(b...@age)
[1] 0
> length(b...@kids)
[1] 0
> is.na(b...@kids)
logical(0)
> is.na(b...@age)
logical(0)
> is.na(b...@name)
logical(0)
> b...@kids <- l
Alaios,
Try
as.data.frame(table(x[,1], x[,2))
where x is your matrix.
HTH,
Jorge
On Fri, Sep 17, 2010 at 8:19 AM, Alaios <> wrote:
> Hello everyone,
> please consider the following lines of a matrix
>
>
> [574,] 59 32
> [575,] 59 32
> [576,] 59 32
> [577,] 59 32
> [578,] 5
Dear Felipe,
maybe the data (which I can't see through the digest) do not have
variability, maybe something else.
Try sending me the data.frame, I'll see what happens.
Giovanni
--
Message: 72
Date: Thu, 16 Sep 2010 11:49:19 -0500
From: Luis Felipe Parra
To: r-help@r-
Just to confirm,
this one worked for me
as.data.frame(table(temp[]))
Var1 Freq
1 1 11
2 2 29
3 3 29
4 4 29
5 5 29
6 6 29
7 7 29
8 8 29
9 9 29
10 10 29
11 11 29
12 12 29
13 13 29
14 14 29
15 15 29
16 16 29
17 17
Hello Greg,
Thank you for your comment.
I had a chance to use Windows GUI R and I found several interesting
differences between Mac and Windows GUI R.
1. txtStart() command
- Windows GUI R - worked perfectly
- Mac GUI R version.string R version 2.11.1 (2010-05-31) (both
32bit and 64bit) - o
I'm attempting to create an array of treatment comparisons for modelling data
generation. This involves comparison of one treatment (c2) with another (c3),
relative to a common comparator (c1).
Attached code gives me the correct array but need to remove duplicates.
Duplicates relate only to c
On Fri, Sep 17, 2010 at 4:46 AM, Ivan Calandra
wrote:
> Hi Jim,
>
> First, I think it's better if you reply to the list, other users might
> be interested and have better answers.
>
> Second, as other replies showed you, using a list is actually way easier
> than creating new objects every time.
My bad :(
unfortunately does not work correct.
This is some of the output of the table
..
[494,] 50 27
[495,] 50 27
[496,] 50 27
[497,] 50 28
[498,] 50 28
[499,] 50 28
[500,] 50 28
[501,] 51 28
[502,] 51 28
[503,] 51 28
[504,] 51 28
[505,]
So try this :
aggregate(rep(1, nrow(x)), as.data.frame(x), sum)
On Fri, Sep 17, 2010 at 11:54 AM, Alaios wrote:
> My bad :(
> unfortunately does not work correct.
>
> This is some of the output of the table
> ..
> [494,] 50 27
> [495,] 50 27
> [496,] 50 27
> [497,] 50
On 9/16/10 5:00 PM, Rolf Turner wrote:
On 17/09/2010, at 8:51 AM, Duke wrote:
Hi Duncan,
On 9/16/10 3:47 PM, Duncan Murdoch wrote:
On 16/09/2010 3:40 PM, Duke wrote:
Hi all,
I am writing a function (fun.R), but I dont know how to code the
function so that the Help Text will be shown u
Dear R users
I have tried to install the optimx but met problems.
I have gone to the website you suggested:
https://r-forge.r-project.org/R/?group_id=395
and tried to install it with the following method:
install.packages("optimx", repos="http://R-Forge.R-project.org";)
I have received the
You are pretty good.
Worked nicely :)
Thanks!
Alex
From: Henrique Dallazuanna
Cc: Rhelp
Sent: Fri, September 17, 2010 4:58:50 PM
Subject: Re: [R] count frequency
So try this :
aggregate(rep(1, nrow(x)), as.data.frame(x), sum)
My bad :(
>unfortunately
Dear R-experts,
I was wondering if anyone know of a library in R that has functions
for calculating the Banzhaf and Shaply-Shubik indeces?
Best, Thomas
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
Here it is...
> str(Males$BMXHT)
num [1:2801] 168 161 180 182 169 ...
> str(Females$BMXHT)
num [1:3440] 162 159 164 165 159 ...
> str(Males$yourWeight)
num [1:2801] 1148 788 10298 25115 8691 ...
> str(Females$myWeight)
num [1:3440] 9169 4964 2608 2806 907 ...
I want to combine Males$BMXHT wi
Hi,
I want to know whether there is any function in R to find STRESS after using
cmdscale and estimating the coordinates, I have written these functions to find
stress (for p =1,2,3,4,5)
stress<-rep(0,5)
for(p in 1:5)
{
datahat<-cmdscale(d,p)
deltahat<-as.matrix(dist(datahat))
a<-0
b<-0
for(i i
Sorry,
Rather, it should be:
wtd.quantile(Everybody$BMXHT, weights=Everybody$ourWeight, 0.05)
Thanks again,
Brian.
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Sent from the R help mailing list archive at Nabble.com.
_
Hello,
I am having a problem configuring a Redhat server to run R as a shared library
called from Perl.
The error message I get is:
Can't find 'boot_R' symbol in libR.so
Has anyone run across something similar?
Any help would be greatly appreciated.
Thank you,
Kenneth Buck
Emory Winship Cance
I know Matlab's M file can be converted to a stand-alone executable file. I
wonder if there is a project aimed at compiling R scripts into stand-alone
executable file. I think it will be very promising for R to be more widely
used in different fields.
--
View this message in context:
http://
I get an error message:
Error in file(file, ifelse(append, "a", "w")) : invalid 'open' argument
How do I resolve this?
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___
On Thu, Sep 16, 2010 at 5:29 PM, Yihui Xie wrote:
> Hi all,
>
> We know old.packages() can check for updates of add-on packages, but
> is there a way to check updates of R itself? "go to R homepage" is a
> way, of course, but I hope this can be done by R.
>
> I'm not sure about the "reliable" plac
I made simulation with Weibull and create Matrix,
How can I create mean/min/max/stdev on column or rows of matrix?,
Thanks,
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-p
More context would be useful, but my first guess is that you are
running into problems because both file and append are functions.
If you didn't explicitly set them to something else, R is using the
function.
What happens if you instead use:
filename <- "myfile.csv"
toappend <- FALSE
file(filename
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of lord12
> Sent: Friday, September 17, 2010 9:02 AM
> To: r-help@r-project.org
> Subject: Re: [R] convert to csv file
>
>
> I get an error message:
>
> Error in file(file, ifelse
On Sep 17, 2010, at 7:22 AM, Alaios wrote:
I would like to thank you again for your help.
But it seems that the floor function (ceiling, round too) create
more dots in the matrix that line really "touches".
You said "cells" not "dots". Are you trying to change the problem now?
My concern
I am making a heatmap. In the dataset, I have couple of rows with zero value
in all the columns. When I cluster data using hierarchical clustering, R
gives an error message. When I get rid of those particular rows, the script
runs smoothly with no problem. I have to include those rows somehow
Good News!
I think I figured out my problem.
The command I was looking for was append().
Thank you for your help,
Brian
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Sent from the R help mailing list archive at Nabble.com
If having the definition of the function appear when you enter the name of the
function elicits a "hate" response, then perhaps you should not be using R.
That characteristic of R is fundamental and unlikely to change: in R everything
is an object, and the result of evaluating an expression is d
On Sep 17, 2010, at 11:15 AM, btpagano wrote:
Here it is...
str(Males$BMXHT)
num [1:2801] 168 161 180 182 169 ...
str(Females$BMXHT)
num [1:3440] 162 159 164 165 159 ...
str(Males$yourWeight)
num [1:2801] 1148 788 10298 25115 8691 ...
str(Females$myWeight)
num [1:3440] 9169 4964 2608
DearR Users,
I have a problem which I think you might be able to help. I have a dataframe
which I'm trying to "filter" following different groups I specified. It's a
little hard to explain, so here is an example:
My dataframe:
ESS DHP
1 EPB 22
2 SAB 10
3 SAB 20
4 BOJ 14
5 ERS 2
Hi:
Here's a simple example that you can tune to your needs:
m <- matrix(rpois(100, 10), nrow = 10)
# Function to compute summary statistics:
f <- function(x) c(min = min(x), med = median(x), mean = mean(x), max =
max(x))
# Apply to rows (index 1)
apply(m, 1, f)
[,1] [,2] [,3] [,4] [,5] [,
Hi:
The first argument of ifelse is a logical statement that must evaluate to
either TRUE or FALSE. I'm guessing that you're using the append function
here, which is not a logical statement. Perhaps it would help if you
described what you wanted to do with a reproducible example to illustrate
the
I might add that ifelse() is a vectorized function - is the operation you're
trying to perform vectorized?
D.
On Fri, Sep 17, 2010 at 10:21 AM, Dennis Murphy wrote:
> Hi:
>
> The first argument of ifelse is a logical statement that must evaluate to
> either TRUE or FALSE. I'm guessing that you'
Hi Bastien,
You can use match(), or the convenience function %in%, like this
(assuming your data.frame is named "dat"):
subset(dat, ESS %in% c("EPB","SAB"))
dat[dat$ESS %in% c("EPB","SAB"), ]
best,
Ista
On Fri, Sep 17, 2010 at 1:02 PM, Bastien Ferland-Raymond
wrote:
> DearR Users,
>
> I have
On Sep 17, 2010, at 12:42 PM, btpagano wrote:
Good News!
I think I figured out my problem.
The command I was looking for was append().
Er, maybe.
It does appear that you could use append() but c() is much more the
typical R way of concatenating objects like vectors and lists. The
only
Bastien -
In what way did
subset(yourdataframe,ESS %in% softwood)
not work?
- Phil Spector
Statistical Computing Facility
Department of Statistics
Although, I can fix this, I am trying to sort out something as
straighforward as possible for my students, and I have some questions
that hopefully someone can help me with.
My data is:
Species DistanceCount
A 5 0
A 10 5
A 15 5
On 16.09.2010 22:40, Jack Siegrist wrote:
I am saving several plots to different formats. I want to be able to change
the font size for all of them at once. Is there a way to do this without
having to have a call to par every time a new device is opened?
No.
At least you will need some wrap
Hi,
You have not provided a reproducible example, so I can only make an educated
guess. This is likely a numerical problem with trying to solve a linear
system of equations in the Lagrange-Newton method (sequential quadratic
programming or SQP) for solving nonlinearly programming problems. I am
So it looks like the problem is with the mac GUI. Here are a couple more
things to test.
Try it with just sink and see if the output comes through. Also try with some
other functions besides date (maybe summary(rnorm(100))).
If there are other Mac users out there reading this, can some of you
Perhaps, the "smacof" package on CRAN might be useful to you.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Na.Ebrahimi
Sent: Friday, September 17, 2010 11:08 AM
To: r-help@r-project.org
Subject: [R] How to find STRESS crite
Graham Smith myotis.co.uk> writes:
>
> Although, I can fix this, I am trying to sort out something as
> straighforward as possible for my students, and I have some questions
> that hopefully someone can help me with.
>
> My data is:
>
> Species DistanceCount
> A 5
On Sep 17, 2010, at 2:04 PM, Greg Snow wrote:
So it looks like the problem is with the mac GUI. Here are a couple
more things to test.
Try it with just sink and see if the output comes through. Also try
with some other functions besides date (maybe summary(rnorm(100))).
If there are ot
yehengxin hotmail.com> writes:
> I know Matlab's M file can be converted to a stand-alone executable file. I
> wonder if there is a project aimed at compiling R scripts into stand-alone
> executable file. I think it will be very promising for R to be more widely
> used in different fields.
Dear friends,
thank you
Liina Pilv
On Jan 24, 2010, at 6:56 PM, Jim Holtman wrote:
where did you save it to; what directory? did you restart in the
same directory? most likely a procedural error.
What is the problem you are trying to solve?
Sent from my iPhone.
On Jan 24, 2010, at 9:46,
Hi, I asked this on mixed model mailing list, but that list is not very active,
so I'd like to try the general R mailing list. Sorry if anyone receives the
double post.
Hi, I have a dataset of animals receiving some eye treatments. There are 8
treatments, each animal's right and left eye was
Hi all,
Is there a built in or easier way to work with dates that are *just* Year-Month?
Right now I paste() on a day as a work around, but it ultimately needs to be
in Year-Month form, so then I use format() to get rid of the day.
x <- c("2006-December", "2006-July")
x <- paste(x, "-01", sep='')
Try this:
library(zoo)
as.yearmon(x, "%Y-%B")
On Fri, Sep 17, 2010 at 4:23 PM, Joshua Wiley wrote:
> Hi all,
>
> Is there a built in or easier way to work with dates that are *just*
> Year-Month?
> Right now I paste() on a day as a work around, but it ultimately needs to
> be
> in Year-Month fo
Hello R masters,
I have sent this same message to other lists and none so far could give some
light. I was trying to use odfWeave to generate a report from R and Im
getting an error that I think is related to latin characters. I looked
around and did find some stuff related to this problem about S
If uniroot is not a bisection method, then what function in R does use
bisection?
Thanks.
--- On Fri, 9/10/10, David Winsemius wrote:
From: David Winsemius
Subject: Re: [R] Is there a bisection method in R?
To: "huang min"
Cc: "Gregory Gentlemen" , r-help@r-project.org
Received: Friday, Sept
I had started thinking of writing my own class...this is much nicer. Thanks!
Josh
On Fri, Sep 17, 2010 at 12:27 PM, Henrique Dallazuanna wrote:
>
> Try this:
>
> library(zoo)
> as.yearmon(x, "%Y-%B")
>
>
> On Fri, Sep 17, 2010 at 4:23 PM, Joshua Wiley
> wrote:
>>
>> Hi all,
>>
>> Is there a bu
On 9/17/10 12:46 PM, Jeff Newmiller wrote:
If having the definition of the function appear when you enter the name of the function
elicits a "hate" response, then perhaps you should not be using R. That
characteristic of R is fundamental and unlikely to change: in R everything is an object,
a
On 09/17/2010 09:14 PM, array chip wrote:
> Hi, I asked this on mixed model mailing list, but that list is not very
> active,
> so I'd like to try the general R mailing list. Sorry if anyone receives the
> double post.
>
>
> Hi, I have a dataset of animals receiving some eye treatments. There
Why does it matter what method is used? The uniroot function used a quicker
method than bisection, but solves the same problem, so why do you need
bisection? The only reason I can think is this is a homework problem, in which
case we don't do homework.
In fact I have assigned this as homework
On 09/17/2010 09:28 PM, Gregory Gentlemen wrote:
> If uniroot is not a bisection method, then what function in R does use
> bisection?
>
Why do you assume that there is one? uniroot contains a better algorithm
for finding bracketed roots.
It shouldn't be too hard to roll your own if you need on
I think what is going on (and someone is likely to correct me otherwise) is
that formulas have an associated environment that gets passed along with them
while character strings do not.
This means that when you pass the object to another function which passes it to
another function, etc. that e
I can't see the graphs (my work has a rather strict policy about "file sharing"
websites and apparently imageshack is redlisted).
But try adding outer=TRUE to the axis function calls and see if that fixes the
problem. If not, play with the par(xpd= ) setting.
--
Gregory (Greg) L. Snow Ph.D.
S
Hello
I am at the moment trying to get to grips with a data cube in R, and I am
increasingly wondering whether I am actually making things unnecessarily
difficult for myself.
The idea is that I have a data cube with three dimensions (so a 3D matrix):
companies, figures, and years.
So along the z a
Ben,
Thanks for your help on this. Obviously a bit of a mental block on my
part, as it seems painfully obvious now.
Graham
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PLEASE do read the posting guide http://www.
Hello,
This is a semi-complicated question about comparing two datasets,
probably using merge, but I am open to other ideas. I have a large
frame of information about companies. It's over 30,000 rows and looks
something like...
df1 <-
identifier1 identifier2nameother_name
Thank you Peter. Actually 3 people from mixed model mailing list tried my code
using lmer(). They got the same results as what I got from lme4(). So they
couldn't replicate my lmer() results:
Random effects:
Groups NameVariance Std.Dev.
eye:id (Intercept) 3.59531 1.89613
id
I have been tinkering around with this for a bit, and I am proud to
share navel gazer 1.0.
If no arguments are passed, it will look up the top 50 authors on the
r-help list, for the given month in the given year. You can also
specify one or more months as a character vector (e.g., "August" or
c("A
I confirm John's problems with lmer. I'm using R 2.11.1. on Windows XP.
R> m4 <- lmer(score~trt+(1|id/eye),dat)
R> m4
Linear mixed model fit by REML
Formula: score ~ trt + (1 | id/eye)
Data: dat
AIC BIC logLik deviance REMLdev
446.7 495.8 -212.4430.9 424.7
Random effects:
Groups
(This is my first post -- I hope I am doing this right)
Why not use an array?
An array can be indexed in as many dimensions as you would like, and do not
require any extra packages.
x = array(1:27, dim = c(3, 3, 3))
x
x[1, , ]
x[ , 1, ]
x[ , , 1]
-David A. Johnston
--
View this message in c
Here is something simple (does not have any checks for bad input), yet
should be adequate:
bisect <- function(fn, lower, upper, tol=1.e-07, ...) {
f.lo <- fn(lower, ...)
f.hi <- fn(upper, ...)
feval <- 2
if (f.lo * f.hi > 0) stop("Root is not bracketed in the specified interval
\n")
chg <- uppe
See also the matrixStats package on CRAN.
/Henrik
On Fri, Sep 17, 2010 at 10:17 AM, Dennis Murphy wrote:
> Hi:
>
> Here's a simple example that you can tune to your needs:
>
> m <- matrix(rpois(100, 10), nrow = 10)
>
> # Function to compute summary statistics:
> f <- function(x) c(min = min(x),
For some reason, I have trouble pasting commands into the R Console (XP Pro and
R v 2.11.1). This occurs mainly when copying from a PDF document and then
pasting into the R Console. It will work for a while and then stop. It is
difficult to identify what the sequence is prior to it not allowing
Dear All,
I am in a trouble with reading data.
It is in txt file looking like this.
0.00632 18.00 2.310 0 0.5380 6.5750 65.20 4.0900 1 296.0 15.30
396.90 4.98 24.00
0.02731 0.00 7.070 0 0.4690 6.4210 78.90 4.9671 2 242.0 17.80
396.90 9.14 21.60
0.02729 0.0
Dear R experts,
I'm new to R. It seems to be a simple question but I just can't find a way to
do it. Please help me.
I have two data sets x and y as shown in the following. I want to compare the
first two columns in x and y, find the matched ones and assign the relative
value from column 2 of
Hi Troy,
If you are using the R GUI for Windows, I highly recommend just
copying and pasting everything into the accompanying script editor,
and then you can run the commands you want directly from there.
The Windows clipboard can be finicky sometimes, so I would not be too
hasty to think the pro
Soyeon -
I think scan() (combined with matrix and data.frame) is the
easiest way.
Suppose your text file is called "data.txt". Then
data.frame(matrix(scan('data.txt'),byrow=TRUE,ncol=14))
should give you what you want.
- Phil Spector
On Sep 17, 2010, at 11:47 AM, yehengxin wrote:
I know Matlab's M file can be converted to a stand-alone executable
file. I
wonder if there is a project aimed at compiling R scripts into stand-
alone
executable file. I think it will be very promising for R to be more
widely
used in diff
Hi:
df <- data.frame(x = 1:10, y = rep(0, 10))
dg <- data.frame(x = c(7, 4, 2, 8, 12, 15), y = letters[1:6])
merge(df, dg, by = 'x', all.x = TRUE)
x y.x y.y
1 1 0
2 2 0c
3 3 0
4 4 0b
5 5 0
6 6 0
7 7 0a
8 8 0d
9 9 0
10 10 0
HTH,
Denn
Hi all,
I have major heterogeneity in variances across labs (100-fold). There is no
apparent variance heterogeneity across y-hat. By using lme4 in the following
way, am I accounting for the variance differences in labs?:
lmer(y ~ fixed1 + covariates + (fixed1|labs))
I'm not sure that it is - I t
I should have been more careful in my description. When I say there is a
100-fold difference in variance across labs, I mean that variance in fixed1
(the predictor) varies 100-fold across labs, not the variance in y is
100-fold difference across labs.
Can lme4 account for such variance differences
Dear R users,
Some of you may be interested in the following announcement:
The Vim-R-plugin now works on Windows too. With the Vim-R-plugin we
can send commands to R from the text editor Vim:
http://www.vim.org/scripts/script.php?script_id=2628
Notes:
On Windows, the plugin copies the command
Please give me some help, many thanks.
I graphed a expected CDF line of a binomial distribution on a graph,
And I have some observed points (observed CDF) from 4 groups fall on the
smooth CDF line.
I cannot really get the legend I want
legend ('topleft', c('a, 'b', 'c', 'd', 'expected CDF'), c
Try this:
apply(m, 1, summary) # 1 or 2
On Fri, Sep 17, 2010 at 1:08 PM, Halabi, Anan wrote:
> I made simulation with Weibull and create Matrix,
> How can I create mean/min/max/stdev on column or rows of matrix?,
> Thanks,
>
> __
> R-help@r-project.or
On Sep 17, 2010, at 7:57 PM, aegea wrote:
Please give me some help, many thanks.
I graphed a expected CDF line of a binomial distribution on a graph,
And I have some observed points (observed CDF) from 4 groups fall on
the
smooth CDF line.
I cannot really get the legend I want
legend ('
Hi folks,
Debian 504 64-bit
Emacs Version 22.1.1
I have Emacs+ESS running on the box. R can work on ESS. But the fonts on the
menu bar (top) of Emacs are NOT clear, difficult to read, grey foreground. I
have been
googling around for solution without result. Please help.
TIA
B.R.
Stephen
Try emacs 23
Cheers,
Simon.
Simon Blomberg, BSc (Hons), PhD, MAppStat, AStat
Lecturer and Consultant Statistician
School of Biological Sciences
The University of Queensland
St. Lucia Queensland 4072
Australia
T: +61 7 3365 2506
email: S.Blomberg1_at_uq.edu.au
http://www.uq.edu.au/~uqsblomb/
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