Hello,
I have a probem with dbWriteTable method of package RPostrgreSQL
The table is well added in the database but R doesn't succeed in
inserting rows
But If I send the "COPY FROM" as an sql query in R, the rows are then
well added
I think it is a problem with the temp file create by dbWriteT
A possible work around would be to append the selection.r file to the .Rhistory
file and then reload the history, e.g.
file.append(".Rhistory", .trPaths[5])
loadhistory(file= ".Rhistory")
You can then access the code on the console, skipping the last 2 lines.
[[alternative HTML
If it is enough to have two y labels on the left side and two x label
along the bottom, you can specify xlab/ylab as vectors. Rather than
drawing different xlabs or ylabs for all panels in a grid, you would
typically use strips to label the panels.
By the way, you could use the 'split' or 'positio
On 05/07/2010 06:07 AM, (Ted Harding) wrote:
...
How about an R "device" called xml? This would implement the XML
"extensible markup language" which is basically capable of
encapsulating any formatted material.
Ted, this is perhaps a better idea than you think. Much of the latest
and greatest
Dear users,
I have monthly station data (44) stations data for 45 years which I have
read in R using read.table. the data is in the format:
Year Month Station1 Station2 ..
Station 44 (i.e the column names in the 1st row), I also have the
latitude and l
Hi,
Check the functions merge_all and merge_recurse from the reshape package.
I think you're looking for the latter function.
Cheers!!
Albert-Jan
~~
All right, but apart from the sanitation, the medicine, education, wine, pu
Dear Felix,
the split and position argument can indeed be used (see below). But one has to
manually adjust the position argument for the last plot (bottom right) in order
to be aligned with the second (top right). This issue does not appear in the
"panel function" approach.
As you can see from
Thanks.
On Thu, May 6, 2010 at 7:49 PM, Gabor Grothendieck
wrote:
> See ?rollapply in the zoo package.
>
> On Thu, May 6, 2010 at 6:20 PM, Dipankar Basu wrote:
> > Hi All,
> >
> > I am using R 2.11.0 on a Ubuntu machine. I have a time series data set
> and
> > want to run rolling regressions wit
On Fri, May 7, 2010 at 12:44 AM, Frank E Harrell Jr
wrote:
> Ted I can't resist offering my $.02, which is that I'm puzzled why LaTeX,
> being free, flexible, and powerful, is used only by millions of people and
> not tens of millions.
I think it's the whole geek/suit divide thing. Back when onl
Puzzling question. You install R, you click on "install packages", you
select a mirror, you select hmisc, and done. There is a 64bit version of R,
but a 32bit runs smooth on a Windows 7 64bit as well. if you love the
command line, look at ?install.packages.
I can't see why you would like to compil
Hi,
I have some data on the effect of cycle shape (categorical) and
frequency (continuous) on the efficiency of muscle contraction. My
minimum adequate model is:
m15<-glm(efficiency~cycle.shape*freq, family=quasipoisson)
However, I wish to know where significant differences lie between
On 05/07/2010 07:00 PM, Barry Rowlingson wrote:
...
I think it's the whole geek/suit divide thing. Back when only geeks
had computers we had LaTeX, and FORTRAN, and S, and Unix (and
VAX/VMS). But then the suits thought 'hey, we can make money out of
this', and computers started working their wa
Short answer : No, and you shouldn't.
Longer answer:
R is built under the GNU GPL license version 2. So the idea is that whenever
you make something for R and spread it around, you give the source code as
well. If you don't want functions to be loaded in the general environment,
you can eg create
See also this thread:
https://stat.ethz.ch/pipermail/r-help/2007-December/148116.html
Cheers
Joris
On Thu, May 6, 2010 at 11:24 PM, Michael Dykes wrote:
> When trying to install Rcmdr, I get the following error messages. I am not
> aware of how to fix the problem, i.e. how to remove the lock.
>
I need to compute the mean and the standard deviation of a data set and would
like to have the results in one table/data frame. I call tapply() two times
and do then merge the resulting tables to have them all in one table. Is
there any way to tell tapply() to use the functions mean and sd within
hi all ,
has anyone tried to predict a univariate time series by a neural networks
packages ? please help me in this problem . I am new in R and I did not
found any document that explains this problem.
thanks in advance
David
[[alternative HTML version deleted]]
___
Dear R users,
When I was trying to use the function *optim* to get the MLEs, quite a few
warning messages showed up as below:
Warning in log(psi * lam) : NaNs produced
I am just wondering what does this mean? Was it something wrong with my
likelihood function or was it sth wrong with the data?
Hi Kevin,
The obvious work-around is to start with a model that can be fitted without
giving the start-parameters. If you have to specify the start parameters,
that usually means there is too much parameters or too much dependence in
your data for the algorithm to converge. Meaning that the outcom
hello,
i searched archives for post-hocs on mixed models and wondered why the
method which intuitively came up to me as first, namely reordering the
levels of the fixed factors and to examine the t-statistics (my model
contains only categorical factors, one with 2 and one with 4 levels), does
n
chrish...@psyctc.org wrote:
I've changed the subject line a bit here as Max is asking such a
fundamental question.
Max Kuhn sent the following at 01/05/2010 19:22:
Chris,
...
Why is it R Core's job to fulfill your wants and desires? I have a
hard time thinking that very busy peo
Hi Nathalie,
depending on the hypotheses you want to test, you can set the contrasts
either by using
options(contrasts=c("unordered contrasts","ordered contrasts")) (see
?contrast for the possible choices)
or you can specify the contrasts in the glm function itself (contrasts
argument)
If you w
Well, there's always RExcel to get all your R stuff into something M$
Ruffice can understand. And they're even working on a Word link if I got it
right.
Cheers
Joris
On Fri, May 7, 2010 at 12:24 PM, Duncan Murdoch wrote:
> chrish...@psyctc.org wrote:
>
>> I've changed the subject line a bit here
Look at the reshape package - it allows that when using "cast"
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
Dear R experts,
I have four panels plotted with relation="free" and I would like to have labels
for both axis for each of the four plots.
How can this be done? Below is a minimal example. There are two problems:
1) How to determine the current panel being plotted inside the panel function.
2) Ho
Take a look at the package reshape, the functions ?melt and ?cast. Take a
look at their website as well : http://had.co.nz/reshape/
Cheers
Joris
On Fri, May 7, 2010 at 10:21 AM, Zablone Owiti wrote:
> Dear users,
>
> I have monthly station data (44) stations data for 45 years which I have
> rea
Robert A LaBudde wrote:
At 01:40 PM 5/6/2010, Joris Meys wrote:
On Thu, May 6, 2010 at 6:09 PM, Greg Snow wrote:
Because if you use the sample standard deviation then it is a t test not a
z test.
I'm doubting that seriously...
You calculate normalized Z-values by substractin
just check whether some values in the parameter space forcing the log()
function to apply logarithm on negative values !!!
--
View this message in context:
http://r.789695.n4.nabble.com/problem-in-using-optim-tp2133938p2133942.html
Sent from the R help mailing list archive at Nabble.com.
__
Hi,
I am a student and I am starting to use R. I have a question regardint
the Data mining - rattle package.
Once I create a decission tree (predictive) I want to evaluate it.
Under the "evaluate" option I can choose between:
"Pr v Ob": I do not know how to interpret the Pseudo R square va
Hi all,
what are the books that i can read them for begenning with R ?
thanks in advance
David
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting gui
Hi I am analysing from a behavioural experiment comparing left and right
turns of snails.
Basically I end up with a list of twenty values like this:
R
R
R
L
R
L
L
L
L
R
R
R
R
L
L
L
R
L
R
L
I want to analyse runs of turns but would like to automate how this is
processed and end up with a table th
Maximilian Kofler wrote:
Hi all!
I¹m just implementing the Ullmann¹s algorithm for searching subgraph
isomorphisms in graphNEL objects. The algorithm is running with smaller
graphs, but when I¹m calling it i get an R error message saying that
functions are nested too deeply in source code.
I
http://www.r-project.org/ : see under "Manuals" and "FAQ"
On Fri, May 7, 2010 at 11:44 AM, david sabine wrote:
> Hi all,
> what are the books that i can read them for begenning with R ?
> thanks in advance
>
> David
>
>[[alternative HTML version deleted]]
>
> _
Dear Thomas,
I have been running simulations in order me to understand this problem! I
have found something online where the absolute median difference is computed
and permutations are ran to compute a p-value. Is such a test (if I can call
it a test) tests the null hypothesis that median group 1
Hi,
What you can do is define your own function which takes a vector of
values, computes the statistics you want and then returns a string
which displays the output the way you want it. Then use this function
in your tapply call.
like (untested)
mySummary <- function(x) {
paste(mean(x),sd(x),
Dear Dario Sacco,
> "DS" == Dario Sacco
> on Thu, 06 May 2010 17:45:30 +0200 writes:
DS> Dear Dr. Maechler,
DS> I am an agronomist and a researcher at the University of Turin. I am
DS> also teaching "Applied statistics", then I have some knowledge in
DS> Statistics,
Duncan Murdoch gmail.com> writes:
> I doubt if that was the error message. More likely you saw
>
> Error: evaluation nested too deeply: infinite recursion /
> options(expressions=)?
Exactly this was the error message I recieved
>This isn't a case of the source being nested to deeply, but r
Thank you all for your help, this has solved my problem. My main
problem with using gsubfn was that i was getting confused by the
square brackets in
[^]]+[^]
but I now have a much better understanding of what this means.
Cheers!
Tony Breyal
On 6 May, 19:38, Gabor Grothendieck wrote:
> This is
For simply doing tables xtable has done some nice work for me.
--- On Fri, 5/7/10, Joris Meys wrote:
> From: Joris Meys
> Subject: Re: [R] What is the best way to have "R" output tables in an MS Word
> format? (shaping R core)
> To: "Duncan Murdoch"
> Cc: r-h...@stat.math.ethz.ch
> Received:
data.table is an enhanced data.frame with fast subset, fast
grouping and fast merge. It uses a short and flexible syntax
which extends existing R concepts.
Example:
DT[a>3,sum(b*c),by=d]
where DT is a data.table with 4 columns (a,b,c,d).
data.table 1.4.1 :
* grouping is now 10+ times faster
That's a robust way of obtaining a p-value, and can be classified as a test.
The important trick here is to take into account that you have two tails in
a distribution. If the p-value is calculated taking both tails into account,
then it indeed tests the null hypothesis that median group1 - median
Let me see. I open a Word document and type,
Dear Dr.Harrell,
I open a new LaTeX document and type something like:
\documentclass[10pt,a4paper]{letter}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\address{your name and address}
\signature{your s
tapply does handle functions with vector outputs, e.g. using the built
in CO2 data set the following data frame is returned:
> f <- function(x) data.frame(mean = mean(x), sd = sd(x))
> do.call(rbind, tapply(CO2$uptake, CO2$Type, f))
mean sd
Quebec 33.5 9.67
Mississippi 20.9 7.82
On May 6, 2010, at 9:01 PM, LeandroTV wrote:
Hi,
I have a barchart very similar to the example on the function
documetation,
however, I want to sort the bars according one group in one panel.
Reminding:
library(lattice)
barchart(yield ~ variety | site, data = barley,
groups = year
On 05/07/2010 07:45 AM, John Kane wrote:
Let me see. I open a Word document and type,
Dear Dr.Harrell,
I open a new LaTeX document and type something like:
\documentclass[10pt,a4paper]{letter}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\addres
It probably means that your parameters are taking on values that are
infeasible. You may have to impose constraints on your parameters to get a
solution (e.g. bounds constraints). If you need better help, you have to
send us a minimal, self-contained, reproducible example.
Ravi.
-Original Me
On May 7, 2010, at 1:04 AM, Mohan L wrote:
Hi all,
I have data like this:
sample <- read.csv(file="sample.csv",sep=",",header=TRUE)
sample
stdate Domainsex age Login
1 01/11/09xxx FeMale 25 2
2 01/11/09xxx FeMale 35 4
3 01/11/09xxx Male 1830
4 01/11
There is the contrast package
On Fri, May 7, 2010 at 5:15 AM, Natalie Holt wrote:
> Hi,
>
> I have some data on the effect of cycle shape (categorical) and frequency
> (continuous) on the efficiency of muscle contraction. My minimum adequate
> model is:
>
> m15<-glm(efficiency~cycle.shape*freq, f
You have not given enough information to reproduce your problem, so it
is difficult to say. Most of the time results such as you display will
be due to a data error. It is possible to get a crossing result,
however.
subject 1 2 3 4 5
-
Death 10 - -
Hello,
i have a dataframe with the GDP for different Country (in the columns) and
Years (in the rows).
Now i want for every year the best three values, if possible with name of
the countries (columnnames).
For the best it's no problem but for the other two values.
Thanks,
Hi,
I have a very simple request (I think).
I have a vector/array,
> c <- c(1.3,1.2,1,3.3,3.3,5.1,1.1,1.2,0)
I produce a reverse sorted table/histogram with it,
> t = rev(sort(table(c))
> t
c
3.3 1.2 5.1 1.3 1.1 1 0
2 2 1 1 1 1 1
I would now like to get the labe
That was a superb answer to a question that has already appeared in
various forms on r-help at least 4 times that I can remember just this
week. I think your text could be appended without much editing into
the FAQ. It could then have its own hyperlink, and you wouldn't need
to type it aga
Maximilian Kofler wrote:
Duncan Murdoch gmail.com> writes:
I doubt if that was the error message. More likely you saw
Error: evaluation nested too deeply: infinite recursion /
options(expressions=)?
Exactly this was the error message I recieved
This isn't a case of the sourc
Dear list,
in the following loop im generating objects of type table. What I would like to
do is to put all those objects together in a list (that i called cc).I did this
but the result is not what i espect to get:
cc=list()
d=1
for (i in data) {
cc=list(cc,assign(paste("n",d,sep=""),table(i,
Rolf Turner wrote:
I recently tried to install the latest version of spatstat, from CRAN,
using the install.packages() function. It proceeded to install version
1.17-5 of spatstat, although the current version is 1.18-4.
One reason this could happen is that the package depends on an R versi
Hi Zablone,
I have a few questions about your data, but think the reshape package is
ultimately what you want. So just look at it and see if you can get it
to do what you want.
Marsh Feldman
On Fri, 7 May 2010 01:21:09 -0700 , Zablone Owiti wrote:
> Dear users,
>
> I have monthly station
On May 7, 2010, at 9:42 AM, SHANE MILLER, BLOOMBERG/ 731 LEXIN wrote:
Hi,
I have a very simple request (I think).
I have a vector/array,
c <- c(1.3,1.2,1,3.3,3.3,5.1,1.1,1.2,0)
I produce a reverse sorted table/histogram with it,
t = rev(sort(table(c))
t
c
3.3 1.2 5.1 1.3 1.1 1 0
Duncan Murdoch wrote:
Maximilian Kofler wrote:
Duncan Murdoch gmail.com> writes:
I doubt if that was the error message. More likely you saw
Error: evaluation nested too deeply: infinite recursion /
options(expressions=)?
Exactly this was the error message I recieve
Dear list,
in the following loop im generating objects of type table. What I would like to
do is to put all those objects together in a list (that i called cc).I did this
but the result is not what i espect to get:
cc=list()
d=1
for (i in data) {
cc=list(cc,assign(paste("n",d,sep=""),table(i
Hmm...
set.seed(17*11)
d<-data.frame(africa=sample(50, 10),
europe= sample(50, 10),
n.america= sample(50, 10),
s.america= sample(50, 10),
antarctica= sample((1:50)/20, 10)
)
#Get three top from each row
t(apply(d,1
Am 07.05.2010 16:02, schrieb Duncan Murdoch:
Rolf Turner wrote:
I recently tried to install the latest version of spatstat, from CRAN,
using the install.packages() function. It proceeded to install version
1.17-5 of spatstat, although the current version is 1.18-4.
One reason this could happ
Hello,
It depends a bit on what 'data' is, but look at...
cc <- list()
d <- 1
data <- expression(d1, d2, d3)
for(i in data) {cc[[paste(i)]] <- table(i, subsample$vD31NADD}
if 'data' is a character vector or of numbers, you would just need
for(i in data) {cc[[i]] <- table(i, subsample$vD31NADD}
Zach,
The R-gurus will correct me when I'm wrong, but as far as my very limited
experience goes, the 64bit version only gives you an advantage when throwing
around huge datasets or doing very memory-intensive tasks. For most of the
things I do with R, there is no difference at all. Now the differe
Dear R-users,
I am pleased to announce the release of scaRabee, a new R package which
constitutes a toolkit for modeling and simulation in the field of
pharmacometrics. It was designed as a user-friendly interface to create and
modify model scripts, define system observations and inputs, and run
s
Dear R users,
I am pleased to announce the release of three new R packages: optimbase,
optimsimplex, and neldermead.
- optimbase provides a set of commands to manage an abstract optimization
method. The goal is to provide a building block for a large class of
specialized optimization methods. This
Hi,
I have a very simple request (I think).
I have a vector/array,
> c <- c(1.3,1.2,1,3.3,3.3,5.1,1.1,1.2,0)
I produce a reverse sorted table/histogram with it,
> t = rev(sort(table(c))
> t
c
3.3 1.2 5.1 1.3 1.1 1 0
2 2 1 1 1 1 1
I would now like to get the labe
Actually I have never mentioned the majority of the items in this post
in that form. remix only appeared yesterday on CRAN and as pointed
out to me offline the binaries are still not there (but should be
automatically built shortly).
On Fri, May 7, 2010 at 9:49 AM, David Winsemius wrote:
> That
I am trying to format data for the Geneland package
I need to write a series of paired numbers and paired "000"
to a plain ascii file without surrounding the numerals with quotes,
the original data is in a matrix formatted paired strings,
and written to file using
write.table(x,"output filename",
As pointed out to me offline, data.table should be added to the list
of relevant packages as well. Its primary advantage is for large data
sets as it is very fast. Its interface does take some getting used to
but its most recent version on CRAN does have several vignettes which
should ease learni
Hi Nevli,
Take a look at the "quote" argument under ?write.table.
In your case
write.table(x, "output filename", col.names = F, row.names = F, quote =
FALSE)
should do what you want.
HTH,
Jorge
On Fri, May 7, 2010 at 11:18 AM, Nevil Amos <> wrote:
> I am trying to format data for the Genela
dat = c(1.3,1.2,1,3.3,3.3,5.1,1.1,1.2,0)
tt = table(dat)
sum(as.numeric(names(tt))*tt)/ sum(tt)
- Phil Spector
Statistical Computing Facility
Department of Statistics
Here is one possibility:
tmp <- sample(c('R','L'), 100, replace=TRUE)
tmp2 <- rle(tmp)
tmp3 <- sapply( 1:length(tmp2$values),
function(i) paste( rep(tmp2$values[i], tmp2$lengths[i]), collapse='') )
table(tmp3)
the rle function computes the run lengths, the sapply line converts those
results i
Permutation tests are real tests (if done properly), but one subtle but
important note: The null hypothesis being tested is that the 2 distributions
are identical, the medians being equal is part of that, but the null includes
more than just that assumption.
--
Gregory (Greg) L. Snow Ph.D.
St
Hello Alfred,
I found the solution from S. Ellison
(https://stat.ethz.ch/pipermail/r-help/2010-May/238158.html) really
inspiring.
Here I am using tail and the library 'plyr':
set.seed(17*11)
d<-data.frame(africa=sample(50, 10),
europe= sample(50, 10),
n.americ
> As pointed out to me offline, data.table should be added to the list
> of relevant packages as well. Its primary advantage is for large data
> sets as it is very fast. Its interface does take some getting used to
> but its most recent version on CRAN does have several vignettes which
> should e
At 07:10 AM 5/7/2010, Duncan Murdoch wrote:
Robert A LaBudde wrote:
At 01:40 PM 5/6/2010, Joris Meys wrote:
On Thu, May 6, 2010 at 6:09 PM, Greg Snow wrote:
Because if you use the sample standard deviation then it is a t test not a
z test.
I'm doubting that seriously...
You calculate n
Using d from the post of S Ellison, try
val <- apply(d, 1, function(x) head(sort(x, decreasing = TRUE), 3))
co <- apply(d, 1, function(x) head(names(d)[order(x, decreasing = TRUE)], 3))
data.frame(values = c(val), countries = c(co), row = c(col(val)))
The first line gives a matrix of the three hi
-Ursprüngliche Nachricht-
Von: S Ellison
Gesendet: 07.05.2010 16:44:08
An: r-help@r-project.org,alfred-schu...@web.de
Betreff: Re: [R] Find the three best values in every row
>Hmm...
>
> set.seed(17*11)
> d<-data.frame(africa=sample(50, 10),
> europe= sample(50, 10
See my recent reply under the subject "bar order using lattice barchart()"
Running this code before doing your plot:
barley$variety <- reorder(barley$variety, barley$yield, function(x) x[2] )
will cause the bars in all the plots to be reordered such that 1931 Waseca is
increasing, is that what
Hi, trying to get ODBC to work with MySQL on my Mac running Leopard.
I followed the instructions here: www.stats.ox.ac.uk/pub/bdr/RODBC-manual.pdf -
setting up iODBC, creating a connection to MySQL (previously
installed) which I tested with the iODBC tester as well as my MySQL
GUI client. So far,
On Fri, 7 May 2010, cheba meier wrote:
Dear Thomas,
I have been running simulations in order me to understand this problem! I
have found something online where the absolute median difference is computed
and permutations are ran to compute a p-value. Is such a test (if I can call
it a test) test
depends on how you interprete "absolute median difference". Is that the
absolute difference of the medians, or the median of the absolute
differences. Probably the latter one, so you would be right. If it's the
former one, then it is testing whether the difference of the medians is
zero.
Cheers
Jo
Hi - I'm relatively new to R, and would appreciate a bit of advice on
how to get off the ground.
I'd like to install RCmdr to take advantage of the GUI.
I've installed the package, and all the dependencies. I started X11.
In my R-GUI window, I have checked the box next to RCmdr in the
Package Ma
Please let me quote an eminently sensible person, who observed that ...
"p-values are dangerous, especially large, small, and in-between ones."
- Frank E Harrell Jr., Prof. of Biostatistics and Department Chair,
Vanderbilt University
Charles Annis, P.E.
charles.an...@statisticalengineering.co
-- Forwarded message --
From: Vivek Ayer
Date: Fri, May 7, 2010 at 10:17 AM
Subject: Re: [R] custom metric for dist for use with hclust/kmeans
To: Greg Snow
Hi Greg,
The pam function is exactly what I needed. I can now create my own
distance matrix, run as.dist on it and pass i
Which fortune is this? :)
Dennis
On Fri, May 7, 2010 at 10:08 AM, Charles Annis, P.E. <
charles.an...@statisticalengineering.com> wrote:
> Please let me quote an eminently sensible person, who observed that ...
>
> "p-values are dangerous, especially large, small, and in-between ones."
> - Frank
Don't use the assign function (it is not even clear what you are trying to do
with it).
Here is a better approach based on your code:
> cc <- list()
> d <- 1
> for (i in data) {
+ cc[[d]] <- table( ...
+ d <- d + 1
+ }
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Cent
Robert A LaBudde wrote:
At 07:10 AM 5/7/2010, Duncan Murdoch wrote:
Robert A LaBudde wrote:
At 01:40 PM 5/6/2010, Joris Meys wrote:
On Thu, May 6, 2010 at 6:09 PM, Greg Snow wrote:
Because if you use the sample standard deviation then it is a t test not a
z test.
David Petersen wrote:
Hi - I'm relatively new to R, and would appreciate a bit of advice on
how to get off the ground.
I'd like to install RCmdr to take advantage of the GUI.
I've installed the package, and all the dependencies. I started X11.
In my R-GUI window, I have checked the box next to
RSQLite is an R package conforming to the R DBI interface that allows
for interaction with SQLite.
Version 0.9-0 highlights:
* Support for SQLite BLOBs using raw vectors in R
* New memory model for db connections allows for more familiar R
semantics and no predefined limit to the number of co
Hi R-users,
I try to put image legend via image.plot function, but I do not know put
this in log scale. The last line of code put legend in linear scale, but I
want put this in log scale. This is a image of the ocean color web site
8-day composite L3 4KM. Anybody colud get me advise please.
This i
Dear all,
I constructed this function called my.boxplot.stats by replacing fivnum()
with quantile() in function boxplot.stats(). So I can try different quantile
methods in bwplot(). The problem is I couldn't pass different values to the
"type" argument to my.boxplot.stats, which in turn is an argu
Hi all,
Thank you for your reply.
if done properly! What does this mean? The R-code I have is using the
R-function sample without replacement. Am I doing this properly?
median of the differences is zero! Does this mean if I run 1000 permutation
and for each permutation I compute the median diffe
On Fri, May 7, 2010 at 9:45 PM, cheba meier wrote:
> Hi all,
>
> Thank you for your reply.
>
> if done properly! What does this mean? The R-code I have is using the
> R-function sample without replacement. Am I doing this properly?
>
> median of the differences is zero!
This is only valid if the
Perhaps this might help clarify:
sample A: 10 15 20
sample B: 12 15 22
Median of sample A = 15; median of sample B = 15. Sample medians are =.
But: B-A differences are 2,0,2 with a median of 2. So the median difference
does not equal the difference of the medians.
Clarity in what yo
On May 7, 2010, at 3:30 PM, Jun Shen wrote:
Dear all,
I constructed this function called my.boxplot.stats by replacing
fivnum()
with quantile() in function boxplot.stats(). So I can try different
quantile
methods in bwplot(). The problem is I couldn't pass different values
to the
"type"
David,
my.boxplot.stats is modified from boxplot.stats (package grDevices) as
follows. x is the original argument, I guess it's the data object for
processing. I only added "type". Thanks.
Jun
=
> boxplot.stats
function (x, coef = 1.5,
Max Kuhn wrote:
Can you send the rest of the sessionInfo() results? I'd like to see your locale.
Thanks,
Max
Hi Max,
I've put the locale info below. I wanted to let you know that I tried
some things again today, starting from scratch, and found that if I just
ran the odfWeave command,
Paul,
That line defines the bullet character. I don't think that you'll be
missing much by excluding it.
I think that to avoid these issues, you should use a utf-8 locale. The
one that I get in my sessions is:
locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
I'm sure that there a
I use the ci.examp function (TeachingDemos package), but your question is so
vague that I don't know if that will help you or not.
If you can give more detail on what you want to accomplish then we have a
better chance of being able to help.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data C
On May 7, 2010, at 4:18 PM, Jun Shen wrote:
David,
my.boxplot.stats is modified from boxplot.stats (package grDevices)
as follows. x is the original argument, I guess it's the data object
for processing. I only added "type". Thanks.
Right, but you should be assigning stats=my.boxplot.sta
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