On 04/02/2010 7:33 PM, Mark Huberty wrote:
Greetings -
Does CRAN or someone similar make a splint / lint-like syntax checker
for R? I realize that both ESS and the debug package and similar tools
have debugging and error analysis features, but those appear to require
running the code through
Hi all,
Thanks for your answers, it worked, but still can't get the time scale on the x-axis, probably has
to do with the unit in the viewport or something like that. But following your recommendation I've
prepared some dummy data to go with the scripts. As before we have two graphs, one that ha
Hello,
In a study exploring transgenerational transmission of anxiety
disorder we investigate whether infants react to experimentally
induced mood changes of their mothers. We measured the time that an
infant needed to cross a cliff (=crossing time) depending on whether
his mother had prev
Dear all,
I've found this nice code fragment on the web
(from Jan T. Kim if I'm correct):
hitReturn <- function(msg)
{
invisible(readline(sprintf("%s -- hit return", msg)));
}
But it does not seem to work in non-interactive mode
( I mean, when I start a script like this: R --vanilla < sc
Hi,
is there a way to receive a total summary of a numerical matrix?
I'm trying to get a total overview of numerical matrix. Using
summary(nummat) only hands back the Min, 1st Qu, Median, Mean, 3rd Qu
and Max for each column. What I actually need is all the Information
for the whole matrix.
Hi
r-help-boun...@r-project.org napsal dne 05.02.2010 10:33:38:
> Hi,
>
> is there a way to receive a total summary of a numerical matrix?
> I'm trying to get a total overview of numerical matrix. Using
> summary(nummat) only hands back the Min, 1st Qu, Median, Mean, 3rd Qu
> and Max for each
You have not told us your version of R. In earlier versions of R
readline() reads a line from the script (and input buffering affects
which line: a number of packages fail their example checks on some
platforms because of this). In R-devel it returns "" immediately.
To make this work you nee
On 05/02/2010 4:31 AM, Jonne Zutt wrote:
Dear all,
I've found this nice code fragment on the web
(from Jan T. Kim if I'm correct):
hitReturn <- function(msg)
{
invisible(readline(sprintf("%s -- hit return", msg)));
}
But it does not seem to work in non-interactive mode
( I mean, when I
Thanks, that helped!
--
Anne Skoeries
Am 05.02.2010 um 10:40 schrieb Patrick Burns:
summary(c(the.matrix))
should work for you. Or perhaps more telling:
summary(as.vector(the.matrix))
On 05/02/2010 09:33, Anne Skoeries wrote:
Hi,
is there a way to receive a total summary of a numerical
On 05.02.2010 00:24, casperyc wrote:
Yes, that is pretty much what I want.
However, there was slightly a mistake.
we need to use ''rate=rate[i]"" and "shape=shape[i]" because the default is
==
dgamma(x, shape, rate = 1, scale = 1/rate, log = FALSE
Dear All
Subject : Multiple Lines in Graph
Could you please help me to draw two lines in a graph.
plot(f1, t1, type ="b") and plot(f2,t2, type="b") where t1, t2, f1,and f2
are single dimensional matrix.
I have these two graph. How can I draw these two lines in a single window?
Thanks in advanc
Thank you very much, works perfectly!
I understand why as well now (thank you too Duncan ;))
I have been away from this mailinglist for a while, but I'm happy to
see that you are both still this active :)
Sorry for forgetting to mention my R version, it was
$ rpm -q R
R-2.10.1-1.fc12.i686
Jonne.
Dear all,
I'm attemping to find the overall range of values from column 5 of a series of
data frames, on a per-row basis, and assign the results to a new object. At
present, I'm only able to receive the overall range of all values, whereas I'm
intending to get the results of the range for each
On 02/05/2010 09:13 PM, wesley mathew wrote:
Dear All
Subject : Multiple Lines in Graph
Could you please help me to draw two lines in a graph.
plot(f1, t1, type ="b") and plot(f2,t2, type="b") where t1, t2, f1,and f2
are single dimensional matrix.
I have these two graph. How can I draw these
Hi
r-help-boun...@r-project.org napsal dne 05.02.2010 11:13:23:
> Dear All
>
> Subject : Multiple Lines in Graph
>
> Could you please help me to draw two lines in a graph.
> plot(f1, t1, type ="b") and plot(f2,t2, type="b") where t1, t2,
f1,and f2
> are single dimensional matrix.
With your
Hi
r-help-boun...@r-project.org napsal dne 04.02.2010 17:44:05:
> Hi everybody!
>
> I would like to export the results of a test statistic in a *.csv file,
> but get an error.
> The code is below.
> The data are attached as .txt with tab as separator. I tried to get a
> sample dataset, but for
Dear all,
I have some trouble using the "id"-argument with aftreg (accelerated
failure time regression analysis from the eha library).
As far as I understand it, the id argument is used to group
individuals together if there are time-varying covariates and the
data is arranged in counting pr
Wesley,
As has already been pointed out, it's a good idea to
provide _reproducible_ code. I'll just add that it's
also a good idea to peruse the help pages, where you
would find that ?plot points you to ?lines (under See Also).
Making frequent use of the help pages is pretty well
an R-requiremen
On 02/05/2010 09:21 PM, Steve Murray wrote:
Dear all,
I'm attemping to find the overall range of values from column 5 of a series of
data frames, on a per-row basis, and assign the results to a new object. At
present, I'm only able to receive the overall range of all values, whereas I'm
inte
Thanks to all for your help.
So to answer all questions:
- "test" is indeed a list. With my real data:
> str(test)
List of 3
$ output : num [1:15, 1:6] 1 2 3 4 5 6 7 8 9 10 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:6] "con.num" "psihat" "p.value" "p.crit" ...
$ c
In case this time-dependent covariate is an internal time-dependent
covariate (aka endogenous time-dependent covariate), you can use the
jointModel() function from package JM, with the option "weibull-AFT-GH"
for the 'method' argument.
For more information you may have a look at:
http://rwiki
Dimitris,
thanks for the hint, I'll definitely give it a shot later.
But for now it would be great to understand the aftreg arguments
first :)
All the best
Philipp
Dimitris Rizopoulos wrote:
In case this time-dependent covariate is an internal time-dependent
covariate (aka endogenous time-d
I don't really know qcc, but it seems to me that you might
have to provide information about within-group variability.
But maybe I'm completely out to lunch on this.
-Peter Ehlers
vikrant wrote:
Thanks Bart and Peter for your help and the example is working for c chart as
well withut any error
On Fri, Feb 5, 2010 at 11:30 AM, Philipp Rappold
wrote:
> Dear all,
>
> I have some trouble using the "id"-argument with aftreg (accelerated failure
> time regression analysis from the eha library).
>
> As far as I understand it, the id argument is used to group individuals
> together if there are
Hi,
I dont know if this a coding rule or a program bug.
When I write this function below it work:
a <- c(1,2)
for(i in a){
print(i)
}
[1] 1
[1] 2
But if I write this function with line space it dont work.
a <- c(1,2)
for(i in a){
print(i)
}
Error: unexpected '}' in "}"
I remember that
On 05/02/2010 7:46 AM, Ronaldo Reis Júnior wrote:
Hi,
I dont know if this a coding rule or a program bug.
When I write this function below it work:
a <- c(1,2)
for(i in a){
print(i)
}
[1] 1
[1] 2
But if I write this function with line space it dont work.
a <- c(1,2)
for(i in a){
print(
Folks:
Can anyone throw some light on this? Thanks.
Satish
-
Satish Vadlamani
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Hi,
Could you guide me to upload Rgraphviz library. I have installed Graphviz
2..20.3.1 , even followed the instructions in Read me, but I cannot load the
package.
There is a error message "This application has failed to start becaues
libcdt-4.dll was not found. Re-installing the application
Where should be shine it? No information provided on operating
system, version, memory, size of files, what you want to do with them,
etc. Lot of options: put it in a database, read partial file (lines
and/or columns), preprocess, etc. Your option.
On Fri, Feb 5, 2010 at 8:03 AM, Satish Vadlama
Hi,
I'm using the function "mlogit" from the package "mlogit" in order to make a
multinomial model, with random and nested effect*. But, currently, even a basic
model as
> mlogit(c ~ lma + poids , MC, shape = "long", alt.var = "N")
Erreur dans drop(.Call("La_dgesv", a, as.matrix(b), tol, PACKAG
On Feb 5, 2010, at 1:50 AM, Bert Gunter wrote:
Folks:
You can make use of matrix subscripting and avoid R level loops and
applys
altogether. This will end up being many times faster.
Here's your original code:
Z=matrix(rnorm(20), nrow=4)
index=replicate(4, sample(1:5, 3))
P=4
tmpr=list()
Using pointLabel in the maptools package will plot text labels for points to
avoid overlap of text.
But is there a way to avoid/minimize overlap of text labels with the outline of
an underlying polygon?
For example, when plotting location of cities, one would like the text labels
to not fall o
the csv files are downloaded from a database and it looks like some
character fields contain the CR-LF sequence within them.
This causes R to see a new record/row and the number of rows it sees
is different (usually higher) from the number of rows actually
extracted.
Any suggestions?
Thanks.
__
On 05/02/2010 8:13 AM, Deepak Manohar wrote:
Hi,
Could you guide me to upload Rgraphviz library. I have installed Graphviz
2..20.3.1 , even followed the instructions in Read me, but I cannot load the
package.
There is a error message "This application has failed to start becaues libcdt-4.dll
This is very weird!!
I know the 'lty=1' command and I tried yesterday , many times!!
didnt work.
However, today, it worked!!!
Maybe the university's computer is stupid!
Thanks!
--
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http://n4.nabble.com/legend-help-tp1461992p1470216.html
Sent from the R help mailin
On Fri, Feb 5, 2010 at 10:23 AM, analys...@hotmail.com
wrote:
> the csv files are downloaded from a database and it looks like some
> character fields contain the CR-LF sequence within them.
>
> This causes R to see a new record/row and the number of rows it sees
> is different (usually higher) fr
Carrie Li wrote:
Dear r-helpers,
I have a small dataset (n<50), and I want to compute the Hodges Lehmann
exact confidence interval.
So far, I know that "pairwiseCI" has the function "HL.diff". The description
is as follows :
HL.diff calculates the Hodges-Lehmann confidence interval for the diff
Dear All,
I would like to count or list the names of the existing worksheets
within an .xls file. Any hints?
Thaks,
Gabor
--
Pozsgai Gábor
www.coleoptera.hu
www.photogabor.com
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Hi:
Here's another approach: use apply to do the sampling inline. Using Carrie's
original problem,
apply(Z, 1, function(x) x[sample(1:length(x), 3)])
[,1] [,2][,3] [,4]
[1,] 0.6236041 -0.7554920 0.58903794 0.8390664
[2,] 0.8291094 0.4041808 0.07874168 1.2790384
[
See ?sheetCount and ?sheetNames in the gdata package.
On Fri, Feb 5, 2010 at 9:13 AM, Gábor Pozsgai wrote:
> Dear All,
>
> I would like to count or list the names of the existing worksheets
> within an .xls file. Any hints?
>
> Thaks,
>
> Gabor
>
> --
> Pozsgai Gábor
> www.coleoptera.hu
> www.pho
I agree with Jim. The term "do analysis" is almost meaningless, the posting
guide makes reference to statements such as that. At least he tried to
define large, but inconsistenly (first of all 850MB, then changed to
10-20-15GB).
> Satish wrote: "at one time I will need to load say 15GB into R"
I have a function to read xls files that tells me the name of the available
sheets. See below.
Kevin Wright
read.xls = function (file, sheet, condition)
{
if (missing(file))
stop("No file specified.")
if (!file.exists(file))
stop("File ", file, " does not exist in direct
On 02/05/2010 05:54 AM, Duncan Murdoch wrote:
> On 05/02/2010 8:13 AM, Deepak Manohar wrote:
>> Hi,
>>
>> Could you guide me to upload Rgraphviz library. I have installed
>> Graphviz 2..20.3.1 , even followed the instructions in Read me, but I
>> cannot load the package.
>>
>> There is a error mess
I find this odd because it doesn't appear to happen in larger datasets. I
have the following data set ENV with the first column set as row.names:
> ENV
TPlog
001S29H 0.601
002S42H 0.602
003S43S 0.779
004S43S 0.702
005S51H 0.978
006S52P 2.718
If I apply > ENV <- ENV[-1,] # remove f
You're not only removing a row of data, you are invoking the default
behavior of subset, which is to collapse the subsetted result to the
smallest possible type, which in this case is a vector. Vectors have
no rows, and thus no row names.
You need the drop=FALSE argument, as in
ENV <- ENV[-1, , dr
> str(MC)
'data.frame': 21130 obs. of 10 variables:
$ male : int 1 1 1 1 1 1 1 1 1 1 ...
$ pop : int 2 2 2 2 2 2 2 2 2 2 ...
$ lma : num 4.9 4.9 4.9 4.9 4.9 4.9 4.9 4.9 4.9 4.9 ...
$ poids : num 3.03 3.03 3.03 3.03 3.03 3.03 3.03 3.03 3.03 3.03 ...
$ ventre: int 4 4 4 4
Thank you both.
Gabor, do I need perl to be installed onj my computer to use those functions?
Gabor
2010/2/5 Kevin Wright :
> I have a function to read xls files that tells me the name of the available
> sheets. See below.
>
> Kevin Wright
>
>
> read.xls = function (file, sheet, condition)
> {
>
Thank you Sarah.I'm glad it was a quick fix:
On Fri, Feb 5, 2010 at 8:50 AM, Sarah Goslee <> wrote:
> You're not only removing a row of data, you are invoking the default
> behavior of subset, which is to collapse the subsetted result to the
> smallest possible type, which in this case is a vecto
I am currently attempting to split a long list of strings (let's call it
"string.list") that is of the format:
"1234567.z3.abcdef-gh.12"
I have gotten it to:
"1234567" "z3" "abcdef-gh" "12"
by use of the strsplit function.
This leaves me with each element of "string.list" having a split stri
Kevin
This function is very valuable to me as well. Can you explain what the
function is looking for with regards to the arguments sheet and condition?
Thanks
Steve
Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead,
Yes. Get it from here:
http://www.activestate.com/activeperl/
On Fri, Feb 5, 2010 at 10:10 AM, Gábor Pozsgai wrote:
> Thank you both.
> Gabor, do I need perl to be installed onj my computer to use those functions?
>
> Gabor
>
> 2010/2/5 Kevin Wright :
>> I have a function to read xls files that
Does this help:
> x <-
> c("1234567.z3.abcdef-gh.12","1234567.z3.abcdef-gh.12","1234567.z3.abcdef-gh.12")
> y <- strsplit(x, '[.]')
>
> y
[[1]]
[1] "1234567" "z3""abcdef-gh" "12"
[[2]]
[1] "1234567" "z3""abcdef-gh" "12"
[[3]]
[1] "1234567" "z3""abcdef-gh" "12"
> y
On Fri, Feb 5, 2010 at 9:29 AM, jim holtman wrote:
> Does this help:
>
>> x <-
>> c("1234567.z3.abcdef-gh.12","1234567.z3.abcdef-gh.12","1234567.z3.abcdef-gh.12")
>> y <- strsplit(x, '[.]')
Here's another way with the stringr package:
library(stringr)
x <-
c("1234567.z3.abcdef-gh.12","1234567.
Hi,
I'm building a graph (barplot) in which the X axis label
disappears.
I tried to use the option mgp of par() and I could not get
the desired result.
Note that want the axis labels horizontally.
caes = c(37,20,19,16,75,103)
names(caes) = c("Pinscher", "Pastor \n Alemão", "Poodle",
"Rottwe
Dear all,
I am attempting to perform a calculation which counts the number of positive
(or negative) values based on the sample mean (on a per-row basis). If the mean
is>0 then only positive values should be counted, and if the mean is <0 then
only negative values should be counted. In cases w
Hello there,
I spent all day yesterday trying to get a small data set from Splus into R,
no luck! Both, Splus and R, are run on a 64-bit RedHat Linux machine, the
versions of the softwares are 64-bit and are as what follows:
Splus:
TIBCO Software Inc. Confidential Information
Copyright (c) 1988
Fran,
The trick is to use box.width, not box.ratio.
xyplot(Perc ~ as.POSIXct(hora,format="%d-%m-%Y %H:%M"),
data=digrate, groups=Drate, ## key=leg,
xlab="time of the day",
horizontal=FALSE,
scales=list(alternating=FALSE,
tck=c(1,0),
x=list(at=seq(r
Hey all,
So I'm manually conducting a sliding window test, and I would like to tack on
results to some variable as the results are outputted. I'm using a for loop,
and I am currently using an array as my 'output collector' variable, though I
know I should be using a dataframe of some sort. My q
Hi,
I have been working with LMMs for a while now and convergence has proved
to be a very common problem.
Right now, your model has two random effect terms, that is, one for the
intercept and one for the coefficient of the "treat" variable. This
implies that the default variance-covariance mat
Hello all,
I hate to add to the daily queries regarding R's handling of large
datsets ;), but...
I read in an online powerpoint about the ff package something about
the "length of an ff object" needing to be smaller than
.Machine$integer.max. Does anyone know if this means that the # of
elements
Have you tried dput/dget or dump/source?
On the S-Plus side, you need to tell it to use the older format.
Rich
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PLEASE do read the posting guide http://www.R-project.org/p
Yes, that was perfect! Thank you so much!
Just to clarify, since I'm kind of new to string manipulation-- is that '[['
in the sapply function what is designating splits/elements within the
string? So that's the part that says "I want this particular element" and
the "1" or "2" or "number" is what
On Sun, Jan 31, 2010 at 10:24 PM, Anton du Toit wrote:
> Dear R-helpers,
>
> I’m writing for advice on whether I should use R or a different package or
> language. I’ve looked through the R-help archives, some manuals, and some
> other sites as well, and I haven’t done too well finding relevant in
On 05/02/2010 10:48 AM, Silvano wrote:
Hi,
I'm building a graph (barplot) in which the X axis label
disappears.
I tried to use the option mgp of par() and I could not get
the desired result.
Note that want the axis labels horizontally.
caes = c(37,20,19,16,75,103)
names(caes) = c("Pinscher"
Hi Steve,
as far as i understood, you're trying to do this:
direction_func <- function(combdframe) {
ifelse(mean(combdframe==0), -9,
sum((sign(mean(combdframe))*combdframe)>0))
}
direction<-apply(combdframe, 1, direction_func)
direction
cheers,
thomas
--
Thomas Liebig
Fraunhofe
On Feb 5, 2010, at 10:48 AM, Steve Murray wrote:
Dear all,
I am attempting to perform a calculation which counts the number of
positive (or negative) values based on the sample mean (on a per-row
basis). If the mean is>0 then only positive values should be
counted, and if the mean is <0
Before being helpful let me raise a couple of questions:
1. "I know I'm looking at longevity data (which is believed to have a
Gompertz distribution for mammals dying from 'old age')".
I'm not as convinced. The Gompertz is a nice story, but is
confounded by individual risk or 'frailty'. B
It was exactly what I needed, thank you.
--
Silvano Cesar da Costa
Departamento de Estatística
Universidade Estadual de Londrina
Fone: 3371-4346
--
- Original Message -
From: "Duncan Murdoch"
To: "Silvano"
Cc:
Dear all,
I would like to ask how to extract the p-value for the whole model from
summary(lm).
This didn't help a lot summary.lm
summary(lm(speed~dist, cars))
Thanks a lot!
[[alternative HTML version deleted]]
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1. I am stuck with a copy of S-PLUS 4.x. At that time I used dump() in
S-PLUS and source() to get things into R afterwards ...
2. Why do you think that 32-bit vs. 64-bit issues matter? The file
format does not change (well, this is guessed since I do not have any
64-bit S-PLUS version availabl
The '[[' is just the index access to an object. type:
?'[['
to see the help page.
Actually I should have used '[' in this case:
> sapply(y, '[', 1)
[1] "1234567" "1234567" "1234567"
is equivalent to:
> sapply(y, function(a) a[1])
[1] "1234567" "1234567" "1234567"
>
So set a value based o
On Feb 5, 2010, at 11:54 AM, Trafim Vanishek wrote:
Dear all,
I would like to ask how to extract the p-value for the whole model
from
summary(lm).
This didn't help a lot summary.lm
I disagree. The help page tells you that "coefficients" has the p-
values sogo ahead and get them:
> summ
You should be able to access the p-value using the $coefficients variable,
which is part of summary.
Try:
results <- summary(lm(speed~dist, cars))
results$coefficients
and then:
results$coefficients[x]
where x is the location of particular p-value, or coefficient supplied, you
want, from the
use a list:
result <- list()
for (i in 1:limit){
computation
# store results
result[[i]] <- yourData
}
On Fri, Feb 5, 2010 at 11:14 AM, Turchin, Michael
wrote:
> Hey all,
>
> So I'm manually conducting a sliding window test, and I would like to tack on
> results to some var
Thanks Jim. If I want to store multiple columns of data in my list, how would I
do that? result[j][[i]] where j could be 1, 2 or 3? And out of curiosity, why
would I use a list over a dataframe in the instance were I only collecting one
column of data? Is list more efficient than a dataframe at
On Feb 5, 2010, at 12:00 PM, Turchin, Michael wrote:
You should be able to access the p-value using the $coefficients
variable, which is part of summary.
Try:
results <- summary(lm(speed~dist, cars))
results$coefficients
and then:
results$coefficients[x]
The "coefficients" element of th
The element that you store in the list can be anything. If you have a
matrix, or dataframe or anything else, it handles it just fine. Also
each of the elements can have different dimensions. A list in this
instance is probably more efficient than a dataframe that you would be
adding to. Also if
> I would like to ask how to extract the p-value for the whole model
> from
> summary(lm).
If you mean the p-value given at the end of the summary() printout, it
isn;t held in the summary object. But information to get it is. Using
the ?lm example:
ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,
Dear all,
I have a table like this:
> eds
R.ID Region Gender Agegr Time nvisits
11 A F 60--64 1:00 1
22 OF 55--591:20 1
33 OF 55--59 3:45 3
44 S
Dear all,
I want to use the histogtam as a density estimator, with the binwidths
calculated using scott's formula which is
binwidth = 3.49*ST.dev.*n^(-1/3)
for the following data (30 data points)
12-9-3-6-1-23-21-7-18-16-15-4-19-22-20-2-3-18-8-10-1-7-5-4-11-12-3-9-19-7
so first,I' ve tried this
Hi all,
I have encountered a problem which appears to have defeated my (admittedly
nascent) R skills. I want to draw a spider plot with many cases (just over
300). I am primarily interested in the difference between 4 categories of
cases, and want to display them as different colors. the col.star
Hello! I have a fairly simple (I guess) question. Suppose I have a table
like this:
Score Group
1 Ctrl
2 Ctrl
.......
10Treat
11Treat
.. ... ...
25Treat
What is(are) the simplest command(s) in R to perform an ordinary t-test for
significant d
Uwe Ligges a écrit sur 2010/02/05
11:04:44 :
> 1. I am stuck with a copy of S-PLUS 4.x. At that time I used dump() in
> S-PLUS and source() to get things into R afterwards ...
>
> 2. Why do you think that 32-bit vs. 64-bit issues matter? The file
> format does not change (well, this is guessed s
Hi,
I'm using a GLM with a quasi binomial error distribution and I would like
to do a model selection method similar to step(AIC) to carry out a
restricted search for the "best" model. I would like to know which of my 5
predictor variables would be included in the "best" model if I start with
a 'f
See the identical example in ?t.test
ErikRH wrote:
Hello! I have a fairly simple (I guess) question. Suppose I have a table
like this:
Score Group
1 Ctrl
2 Ctrl
.......
10Treat
11Treat
.. ... ...
25Treat
What is(are) the simplest command(
For a data.frame with only numeric and factor
columns using dump() on the S+ end and source()
on the R end ought to work. If you have timeDate
columns you will need to convert them to character
data before exporting and convert them to your
favorite R time/date class after importing them.
If you
Hey,
So I found workaround for what I'm now pretty sure is a bug.
the reason I'm sure it's a bug is that i found that I found that
linebreaks are handled in a wonky but regular way. There's a rule,
and it's a bad rule, ie a bug.
The wonky rule is as follows: text(x,y,expression(paste("thing1
\n
On Thu, Feb 4, 2010 at 4:53 PM, Wade Wall wrote:
> Hi all,
>
> Is there a function to open a script file from the command line? I have
> several students who are Mac users and when they open up a script file it
> does not send commands to the console, and unfortunately I don't know how
> to
> so
>
>
>> On Linux and Unix machines, such as OS X, a the following hashbang line
> could be added to the top of the script:
>
> #!/usr/bin/env Rscript
>
> Then the script can be run from a terminal using:
>
> cd path/to/script/files
>
I forgot to mention that in order to use the hashbang method
On Fri, 2010-02-05 at 17:44 +0100, ErikRH wrote:
> Hello! I have a fairly simple (I guess) question. Suppose I have a table
> like this:
>
> Score Group
> 1 Ctrl
> 2 Ctrl
> .......
> 10Treat
> 11Treat
> .. ... ...
> 25Treat
>
> What is(are) t
I vote to 'fortunize' Doug Bates on
Hierarchical data sets: which software to use?
"The widespread use of spreadsheets or SPSS data sets or SAS data sets
which encourage the "single table with a gargantuan number of columns,
most of which are missing data in most cases" approach to organization
Hi,
I have two numeric columns (dat1 and dat2) to bwplot
data=expand.grid(dat1=rnorm(20,10,5),class=1:5,school=letters[1:3]))
data$dat2=rnorm(300,13,6)
What I want is to plot two boxes side by side for class,
bwplot((dat1+dat2)~as.factor(class)|school,data=data)
but the code does not work thi
On Fri, Feb 5, 2010 at 9:41 AM, George Locke
wrote:
>
> Hey,
>
> So I found workaround for what I'm now pretty sure is a bug.
>
> the reason I'm sure it's a bug is that i found that I found that
> linebreaks are handled in a wonky but regular way. There's a rule,
> and it's a bad rule, ie a bug.
Has anybody used RJDBC to read tables from an MS Access mdb file? I know how
to do it with RODBC, and I have used RJDBC with SQL Server. I am, however
interested now in this particular combination. Is it possible?
--
View this message in context:
http://n4.nabble.com/RJDBC-with-MS-Access-tp1470
Hi everyone,
I am trying to construct a glm and am running into a couple of questions.
The data set I am using consists of 6 categories for the response and 6
independent predictors representing nutrient concentrations at sample point
locations. Ultimately I'd like to use the probabilities for
Hi All,
I'm fitting a linear (multiple) regression model with 3 predictors + their
interactions.
Can anyone suggest some test in R which can help me know whether I need a
non-linear (regression) model or some transformation? I'm mostly concerned
about finding a way to know whether I should fit a
On Fri, 2010-02-05 at 13:10 -0500, steve_fried...@nps.gov wrote:
> Hi everyone,
>
> I am trying to construct a glm and am running into a couple of questions.
>
> The data set I am using consists of 6 categories for the response and 6
> independent predictors representing nutrient concentrations a
Also, a general suggestion about errors like this: google them!
Searching for "libcdt-4.dll was not found" brings up
http://n4.nabble.com/Rgraphviz-install-td878526.html as the first hit,
and appears to describe the solution.
Best,
Ista
On Fri, Feb 5, 2010 at 2:44 PM, Martin Morgan wrote:
> On 0
Thanks a lot for your help, the examples worked fine, just have to
change the colours to produce a b&w plot, and add the legend
On 05/02/2010 17:11, RICHARD M. HEIBERGER wrote:
Fran,
The trick is to use box.width, not box.ratio.
xyplot(Perc ~ as.POSIXct(hora,format="%d-%m-%Y %H:%M"),
d
On 05/02/2010 12:21 PM, maram salem wrote:
Dear all,
I want to use the histogtam as a density estimator, with the binwidths calculated using scott's formula which is
binwidth = 3.49*ST.dev.*n^(-1/3)
for the following data (30 data points)
12-9-3-6-1-23-21-7-18-16-15-4-19-22-20-2-3-18-8-10-1-7-5
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