Yea,i am very new to R. Thanks a lot Jim Lemon! I appreciate it very much!!
Hi Andrew,
As you seem to be an R newbie and some of the replies may have been
cryptic to a newbie, try this:
boxplot(a$Life.Expectancies.at.Birth)
To explain the above a little bit, when reading in a text file, R
Russ Abbott russ.abb...@gmail.com [Sun, Mar 20, 2011 at 06:46:04AM CET]:
I'm afraid I disagree.
I don't think there is much disagreement. You have been shown that
there is a functional way to rephrase your example. I suppose you
don't disagree that the structure() expression has no side
My aplogies:
The Table of effects did not come through as I had intended them to. HEre they
are reformatted:
Again I would like to see someone actually run mahalanobis() for this data set
to arrive at ?1 and ?2. I do not know what exactly (after reading the manual)
goes in for x,center,or
On Mar 20, 2011, at 06:46 , Russ Abbott wrote:
I'm afraid I disagree. As a number of people have shown, it's certainly
possible to get the end result
pH - c(4.5,7,7.3,8.2,6.3)
names(pH) - c('area1','area2','mud','dam','middle')
pH
area1 area2muddam middle
4.57.07.3
On 11-03-19 10:21 PM, Kenn Konstabel wrote:
On Sun, Mar 20, 2011 at 4:13 AM, Kenn Konstabellebats...@gmail.com wrote:
you can omit the list and do the following:
/.../
(but you don't really need this in this case as you can use balance
instead of this$balance)
P.S. using this would
Hi:
Here's a lattice version:
y - rbeta(1, 2, 5)
densityplot( ~ y, xlim = c(0, 1), plot.points = FALSE,
panel = function(x, y, ...) {
p - seq(0.001, 0.999, by = 0.001)
panel.densityplot(x, col = 'blue', ...)
panel.xyplot(p, dbeta(p, 2,
On Sun, Mar 20, 2011 at 12:43 PM, Duncan Murdoch
murdoch.dun...@gmail.comwrote:
On 11-03-19 10:21 PM, Kenn Konstabel wrote:
On Sun, Mar 20, 2011 at 4:13 AM, Kenn Konstabellebats...@gmail.com
wrote:
you can omit the list and do the following:
/.../
(but you don't really need this in
Hello,
I found an interesting program on Pierre Legendre's webpage:
http://www.bio.umontreal.ca/casgrain/en/labo/corr_permute.html
With this program one can compute a Pearson correlation coefficient matrix
with permutation test.
This is exactly what I need as an R-package because so far I
Thanks for your answer, Peter. I solved it. The problem actually seemed to be
in tha data format in the SPSS file I imported into R.
C.
--
View this message in context:
http://r.789695.n4.nabble.com/Problems-with-package-npmc-tp3390472p3390915.html
Sent from the R help mailing list archive at
Hi,
I' am interested in differences between sample's result when samples consist
of full elements and consist of only distinct elements. When sample consist
of full elements it take about 120 sec., but when consist of only distinct
elements it take about 4.5 or 5 times more sec. I expected that
HI,
While I was working on R , I had a pop up from mcafee asking me if I wasih to
block or allow something to connect. I did not see r in the question, so I had
clicked on the option block. since then, when I start R through excel , I
only get R console and I do nto get R script anymore.
Is there a function similar to pkgDepends() that returns the packages that a
particular package imports or imports from? I believe I can get this
information from the matrix returned from installed.packages() but something
like pkgDepends() would be more convenient.
I did not find anything
On Sun, Mar 20, 2011 at 10:45 AM, herbert8...@gmx.de wrote:
Hello,
I found an interesting program on Pierre Legendre's webpage:
http://www.bio.umontreal.ca/casgrain/en/labo/corr_permute.html
With this program one can compute a Pearson correlation coefficient matrix
with permutation test.
On 20.03.2011 17:38, Derek Ogle wrote:
Is there a function similar to pkgDepends() that returns the packages that a
particular package imports or imports from? I believe I can get this
information from the matrix returned from installed.packages() but something
like pkgDepends() would be
1) If you think McAfee is interfering, then you should look toward your support
options for that software to help you remove the setting that relates to R or
rcom. You may be able to test your theory by disabling that software
temporarily.
2) If R is working at the console, and RGui is
Dear List,
I'm aware that this has been brought up before (e.g.
http://tolstoy.newcastle.edu.au/R/e6/help/09/03/7365.html
http://tolstoy.newcastle.edu.au/R/e6/help/09/03/7365.html ;
https://stat.ethz.ch/pipermail/r-help/2009-March/190902.html
After pondering all the pros and cons regarding the usefulness of a
rating/review system for R packages, don't you think it would make sense
to implement such a thing?
Or to look what is there, and how little it is filled:
http://crantastic.org/
Dieter
--
View this message in context:
Hi, R users,
I there a way that I can control the position of the legend while using
barplot function?
For example:
a-matrix(c(0,0,0.5,0.8,0.9,0.9),2,3)
colnames(a)-c(X,Y,Z)
rownames(a)-c(A,B)
a
barplot(a,width = 0.2,legend =
TRUE,axes=FALSE,col=c(coral3,steelblue3),beside=TRUE)
In this case,
Hello Gabor,
Thanks for your response. There are certainly several offers in R to construct
confidence envelopes.
But here I am specifically interested in sort of a summary output such as
provided by the function cor().
I have more than a thousand correlations and I just need for each
Sorry, the last part of code does not work when uniqu() are used, the true
version;
e - rnorm(n=50, mean=0, sd=sqrt(0.5625))
x0 - c(rep(1,50))
x1 - rnorm(n=50,mean=2,sd=1)
x2 - rnorm(n=50,mean=2,sd=1)
x3 - rnorm(n=50,mean=2,sd=1)
x4 - rnorm(n=50,mean=2,sd=1)
y - 1+ 2*x1+4*x2+3*x3+2*x4+e
x2[1] =
On 20.03.2011 18:52, Hongwei Dong wrote:
Hi, R users,
I there a way that I can control the position of the legend while using
barplot function?
For example:
a-matrix(c(0,0,0.5,0.8,0.9,0.9),2,3)
colnames(a)-c(X,Y,Z)
rownames(a)-c(A,B)
a
barplot(a,width = 0.2,legend =
Hey List,
I did a multiple regression and my final model looks as follows:
model9-lm(calP ~ nsP + I(st^2) + distPr + I(distPr^2))
Now I tried to predict the values for calP from this model using the
following function:
xv-seq(0,89,by=1)
yv-predict(model9,list(distPr=xv,st=xv,nsP=xv))
The
Dieter Menne dieter.menne at menne-biomed.de writes:
After pondering all the pros and cons regarding the usefulness of a
rating/review system for R packages, don't you think it would make sense
to implement such a thing?
Or to look what is there, and how little it is filled:
Dear all, can somebody guide me how to install the package RQuantLib, for
which windows binary is not available. I have tried installing it with R CMD
INSTALL in my windows vista machine (I have Rtools installed), however it
stopped due to an error saying: compilation failed for package
I like Sweave, which I consider to be a great contribution. I have just
written a .Rnw document that comes to about 6 pages of mixed code and
mathematical explanation. Now I want to turn the R code into a function. My
R code currently contains statements like N-1000 and theta- pi/10. In the
next
On Fri, 2011-03-18 at 06:21 -0700, bra86 wrote:
Hello, everybody,
I hope somebody could help me with a dist() function.
I have a data frame of size 2*4087 (col*row), where col corresponds to the
treatment and rows are
So you have 4087 species? If yes, normally, you'd have the species in
Hi Anna,
On Sun, Mar 20, 2011 at 2:54 PM, Anna Lee ana-...@web.de wrote:
Hey List,
I did a multiple regression and my final model looks as follows:
model9-lm(calP ~ nsP + I(st^2) + distPr + I(distPr^2))
Now I tried to predict the values for calP from this model using the
following
Dear Ista!
Thank you for replying. The point you made is exactly what's the
problem: I want to predict the values at different points in space.
calP stands for the water content at each sampling point (n=90) but I
don't quite understand what R does. calP is my vector of measured data
and I
Hi David --
On 03/20/2011 12:19 PM, David.Epstein wrote:
I like Sweave, which I consider to be a great contribution. I have just
written a .Rnw document that comes to about 6 pages of mixed code and
mathematical explanation. Now I want to turn the R code into a function. My
R code currently
Hello folks - I have been trying to figure this out. I have a set of very
large files that are of this format
, , , ,
1/4/1999,9:31:00 AM,blah, blah, blah
1/4/1999,9:32:00 AM,blah, blah, blah
1/4/1999,9:33:00 AM,blah, blah, blah
I want to write R code that reads only that data between a start
Hi all,
I am having a problem using xyplot from lattice library. I am plotting one
variable against another one, conditioning on two other variablesm one of
which has 3 levels and the other 6 levels. The output is thus a 6x3 panel
plot. Here's my code:
print((xyplot(CloDurPercent[CloDurNA !=
Dear friends,
Sorry for this somewhat generically titled posting but I had a question
with using contrasts in a manova context. So here is my question:
Suppose I am interested in doing inference on \beta in the case of the
model given by:
Y = X %*% \beta + e
where Y is a n x p matrix of
One thing that it would be good to learn is how to use Rprof to look
at where time is being spent in your script. Most of the time (over
80%) is spent in the 'unique' call which is doing absolutely nothing
in your code since you are not assigning its output to an object.
Here is what I saw when
How big is the file? WHy not read the entire file in and then use
'subset' to extract only the data that you want? If the file is too
large to be able to read in, then you could put it in a database and
use SQL to extract what you want. You could also create a 'perl'
script to filter the data
On 3/19/2011 8:32 PM, Sally Luo wrote:
Also, is it possible to draw a historical map of European countries using R?
If you want a map of France ~ 1830, see
http://www.datavis.ca/gallery/guerry/guerrymap.html
__
R-help@r-project.org mailing list
Are Pos and Consld -- which CANNOT be in Stops -- the same length
unsubscripted as the subscripted variables? Do all the combinations of
levels of these variables actually occur (you can check with table()
)?
You would also probably do better to use the subset argument of xyplot
to subset the
On Mar 20, 2011, at 3:56 PM, Anna Lee wrote:
Dear Ista!
Thank you for replying. The point you made is exactly what's the
problem: I want to predict the values at different points in space.
calP stands for the water content at each sampling point (n=90) but I
don't quite understand what R
It's unclear to me, why the rating/review system should relate to
entire packages. Would it not be more informative, if single specific
functions would be rated and reviewed?
I would like to see if + is rated better than -, or if more
difficulties are reported for * than for /. I could then
On Mar 20, 2011, at 4:06 PM, askh wrote:
Hi all,
I am having a problem using xyplot from lattice library. I am
plotting one
variable against another one, conditioning on two other variablesm
one of
which has 3 levels and the other 6 levels. The output is thus a 6x3
panel
plot. Here's
Thanks Jim for the reply. The file has 1,183,318 rows and there are 20 such
files.
Too big for R to handle?
--
View this message in context:
http://r.789695.n4.nabble.com/read-file-part-way-through-based-on-start-and-end-date-first-column-tp3391769p3392005.html
Sent from the R help mailing
On Mar 20, 2011, at 21:05 , Ranjan Maitra wrote:
Dear friends,
Sorry for this somewhat generically titled posting but I had a question
with using contrasts in a manova context. So here is my question:
Suppose I am interested in doing inference on \beta in the case of the
model given by:
Hi Anna,
Maybe you can start again and tell us what you are trying to
accomplish. If you are trying to calculate predicted values for the
cases in the data set used to fit the model you don't need the newdata
argument at all. Just do
predict(model9)
If you are trying to do something else, please
Dear Peter and Ranjan,
In addition to Anova(), linearHypothesis() in the car package handles
multivariate linear models, including those for repeated measures.
Best,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department of Sociology
Depends on what version of R you are using. If you are running a 32
bit version and if all the columns were numeric, if you had about 20
columns, I would guess that might require 300MB for a single copy of
the object and for the reading in and then subsetting, you might
require 3-4X that space.
Subscription options are available at the link at the bottom of this
(and every r-help) email.
Best,
Ista
On Sun, Mar 20, 2011 at 11:25 PM, Dan Davis dan7a7da...@gmail.com wrote:
take me off the R help mailing list.
[[alternative HTML version deleted]]
PROBLEM How can I concatenate the following lists into ONE LIST WITHOUT
the unhelpful message operator is invalid for atomic vectors? Combine
as a data frame?
EXAMPLE
Birth_Date - NULL
Birth_Date[1:3] - c(01/17/1939,01/17/1949, 01/17/1959)
Later_Date - NULL
Later_Date[1:3] -
First of all, you are working with 'vectors' not 'lists'. Look in the
following script what str(family) is on your original input.
What you probably want is a 'data.frame'. May be time to go back and
re-read the Intro to R to understand the differences between
vectors, lists and dataframes.
thank you ! I will try it !
--
View this message in context:
http://r.789695.n4.nabble.com/confirmatory-factor-analysis-program-in-R-tp3386133p3392279.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Search for lavaan, sem, and OpenMx.
On Mon, Mar 21, 2011 at 12:34 AM, rvohen bingbingzhan...@126.com wrote:
thank you ! I will try it !
--
View this message in context:
http://r.789695.n4.nabble.com/confirmatory-factor-analysis-program-in-R-tp3386133p3392279.html
Sent from the R help
THanks Jim. Eventually I do want to store the records in a database... so
mySQL. But right now I want to run some analytics on the data so I'm looking
for a quick and dirty solution that can give me the flexibility to extract
data on various time periods. I think a perl script that uses
Try:
family - list(Birth_Date=Birth_Date, Later_Date=Later_Date, Names=
Names)
family$Birth_Date
Sent from my iPhone ( so untested)
On Mar 20, 2011, at 9:58 PM, jim holtman jholt...@gmail.com wrote:
First of all, you are working with 'vectors' not 'lists'. Look in the
following
On Sun, Mar 20, 2011 at 3:47 PM, algotr8der algotr8...@gmail.com wrote:
Hello folks - I have been trying to figure this out. I have a set of very
large files that are of this format
, , , ,
1/4/1999,9:31:00 AM,blah, blah, blah
1/4/1999,9:32:00 AM,blah, blah, blah
1/4/1999,9:33:00 AM,blah,
Dear R Users,
I am working with gsub for the first time. I am trying to remove some
characters from a string. I have hit the problem where the period is the
shorthand for 'everything' in the R language when what I want to remove is
the actual periods. In the example below, I simply want to
Hi John,
Try
gsub([.],,txt)
See Extended Regular Expressions in ?regex.
HTH,
Jorge
*
*
On Mon, Mar 21, 2011 at 12:49 AM, Sparks, John James wrote:
Dear R Users,
I am working with gsub for the first time. I am trying to remove some
characters from a string. I have hit the problem
I think it is as simple as adding the argument fixed=TRUE to the function call.
txt = this. is. a. test.
gsub(., , txt, fixed = TRUE)
result
[1] this is a test
HTH,
Peter
On Sun, Mar 20, 2011 at 9:49 PM, Sparks, John James jspa...@uic.edu wrote:
Dear R Users,
I am working with gsub for the
55 matches
Mail list logo
