tapply(x, y, min)
where x is the vector of numbers and y the vector of class labels.
Best,
--
Wolfgang Viechtbauerhttp://www.wvbauer.com/
Department of Methodology and StatisticsTel: +31 (0)43 388-2277
School for Public Health and Primary Care Office Location:
Maast
I am trying to find the root of the following function. Basically, I am
trying to find n1, n2, prob1, and prob2 from a mixture of two negative
binomials.. I differentiated the log likelihood with respect to n and am
trying to solve this equation. zij and y are both a column of vectors. y is
the dat
I have searched help topics but don't know exactly what to search for.
Need to use my_num to find any matching STREPs in my_df
my_num <-
c("101","102","103","104","105","107","108","112","113","114","115")
## "my_df" has 8,000 different STREPS in it.
## I have a statement where I can select
Hi,
I have a matrix, first column is of certain values, second column is
the class labels or a factor.
e.g.
1.2 1
1.3 1
1.3 1
1.5 1
2.1 2
2.0 2
9.9 2
1.4 3
1.8 3
1.9 3
I want to find out what is the min values of column 1 for each
corresponding class (column 2). For the above example, I want to
Hi all,
Could anybody please shed some lights on me about good books/literature
about modeling and forecasting financial time series in the commodity space?
Thanks so much!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
DOES NOBODY KNOW?
HELP!
lybaomc wrote:
>
> Hi,all
>
> with my data,there are more than 1000 quantitative results of antibody
> concentrations, there may be 2 components(positive and negative), or 3
> components (may be strong positive, positive, and negative), or 4-6
> components. Could you
Here is one way of doing it:
> a <-
+ data.frame(id=c(c("A1","A2","A3","A4","A5"),
+ c("A3","A2","A3","A4","A5")),loc=c("B1","B2","B3","B4","B5"),
+ clm=c(rep(("General"),6),rep("Life",4)))
> # split the indices based on 'id' & 'loc'
> a.indx <- split(seq(nrow(a)), paste(a$id, a$loc))
> #
On Oct 28, 2009, at 9:30 PM, Steven Kang wrote:
Dear R users,
Basically, from the following arbitrary data set:
a <-
data
.frame
(id
=
c
(c
("A1
","A2
","A3
","A4
","A5
"),c
("A3
","A2
","A3
","A4","A5")),loc=c("B1","B2","B3","B4","B5"),clm=c(rep(("General"),
6),rep("Life",4)))
a
id
Dear R users,
Basically, from the following arbitrary data set:
a <-
data.frame(id=c(c("A1","A2","A3","A4","A5"),c("A3","A2","A3","A4","A5")),loc=c("B1","B2","B3","B4","B5"),clm=c(rep(("General"),6),rep("Life",4)))
> a
id loc clm
1 A1 B1 General
2 A2 B2 General
3 A3 B3 General
4 A
I'm sure this is a simple problem, however I need a bit of help to sort it
out.
I have a multivariate data set of elements. Some of the observations have
some amount of certain elements but are lacking in others. Therefore, I
want to quickly group/cluster the observations based on a presence/ab
Thanks to Dr. Ripley and Dr. Murdoch for the workaround
and the solution to the problem with sub() and gsub() memory problem.
Now, with the perl=TRUE option added it works perfect (with the full
database)!
alumnos$AL_NUME_ID<-gsub("(^ +)|( +$)","",alumnos$AL_NUME_ID,perl=TRUE)
I am going to test
OOPS!
("Why can't the English learn to speak? ... Why in America, they haven't
spoken it for years!" --Prof. Henry Higgins in MY FAIR LADY. )
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: Rolf Turner [mailto:r.tur...@auckland.ac.nz]
Sent: Wednesday, Oc
On Wed, 2009-10-28 at 19:03 -0400, Duncan Murdoch wrote:
> > Second, install.packages("intervals") produced the usual line about
> > selecting a mirror, but no selection list popped up. It just sat
> there
> > until I interrupted the session and reentered the command with a
> > pre-specified repos
On Oct 28, 2009, at 7:06 PM, David Winsemius wrote:
On Oct 28, 2009, at 6:22 PM, Mark W. Miller wrote:
I am guessing that your is not really an NA_character, but
rather a factor with a level of "".
See if str(s1) confirms my suspicions.
df1 <- read.table(textConnection(" Firstname
On Oct 28, 2009, at 6:22 PM, Mark W. Miller wrote:
I am guessing that your is not really an NA_character, but
rather a factor with a level of "".
See if str(s1) confirms my suspicions.
df1 <- read.table(textConnection(" Firstname Lastname Age
+ 1Bob Smith 20
+ 2 John
On 28/10/2009 5:54 PM, Ross Boylan wrote:
I just installed 2.10 on XP; ESS is my primary interface.
I seem to be able to access all the help files; under 2.8.1 I was having
seemingly random problems accessing some help topics (including one day
I could and a few days later I couldn't, and vice-v
Thanks!
Rolf Turner-3 wrote:
>
>
> ?get
>
> On 29/10/2009, at 11:25 AM, skyjo wrote:
>
>>
>> Hello. I am trying to write an interactive function that asks the
>> user for a
>> vector of observations. Unfortunately, if a user inputs a vector, R
>> treats
>> the vector name as a string in
?get
On 29/10/2009, at 11:25 AM, skyjo wrote:
Hello. I am trying to write an interactive function that asks the
user for a
vector of observations. Unfortunately, if a user inputs a vector, R
treats
the vector name as a string instead of a variable. Here is an example:
vector.input<-fun
Hello. I am trying to write an interactive function that asks the user for a
vector of observations. Unfortunately, if a user inputs a vector, R treats
the vector name as a string instead of a variable. Here is an example:
vector.input<-function(){
k<-as.integer(readline("Input number of vect
On 29/10/2009, at 10:48 AM, Bert Gunter wrote:
Folks:
If generalities -- with the attendant risk of occasional specific
caveats
and violations -- can be tolerated, then George Box's (paraphrased)
comments
of circa 40-50 years ago seem apropos: why do statisticians obsess
over
normality,
I am guessing that your is not really an NA_character, but
rather a factor with a level of "".
See if str(s1) confirms my suspicions.
> df1 <- read.table(textConnection(" Firstname Lastname Age
+ 1Bob Smith 20
+ 2 John Clark NA
+ 3 Andy 40"), header=T)
The obvious search strategy within Thompson's excellent compendium of
R methods to accompany Agresti succeeds:
https://home.comcast.net/~lthompson221/Splusdiscrete2.pdf
In particular pages 260-267.
On Oct 28, 2009, at 5:54 PM, emilio.vir...@urjc.es wrote:
dear colleagues,
I want to use hier
On Oct 28, 2009, at 5:22 PM, Carl Witthoft wrote:
Just wondering if it's possible to have an item with no content at
all. Here's what I was hoping to do, inside a larger function that
acts on a 3-D array.
I want to enter, as one of the arguments to the main function, the
index over whic
dear colleagues,
I want to use hierarchical partitioning methods to evaluate variance
contributions of differente predictors within a generalized linear
mixed model framework. I don't know if this is possible, because I can
not found anything about this issue. My error is poisson and I, at
I just installed 2.10 on XP; ESS is my primary interface.
I seem to be able to access all the help files; under 2.8.1 I was having
seemingly random problems accessing some help topics (including one day
I could and a few days later I couldn't, and vice-versa). That's good.
I noticed a few glitch
On Oct 28, 2009, at 5:19 PM, Mark Miller wrote:
I have a small Excel data file with two columns of character
variables, one column with a numeric variable and three rows. One
of the character cells is blank and one of the numeric cells is blank.
I read the data file with the following cod
Folks:
If generalities -- with the attendant risk of occasional specific caveats
and violations -- can be tolerated, then George Box's (paraphrased) comments
of circa 40-50 years ago seem apropos: why do statisticians obsess over
normality, to which most analyses -- i.e. inference (especially from
Just wondering if it's possible to have an item with no content at all.
Here's what I was hoping to do, inside a larger function that acts on
a 3-D array.
I want to enter, as one of the arguments to the main function, the index
over which I'm going to do some action. For example, if the actio
I have a small Excel data file with two columns of character variables, one
column with a numeric variable and three rows. One of the character cells is
blank and one of the numeric cells is blank.
I read the data file with the following code:
library(RODBC)
channel <- odbcConnectExcel('u:/t
Kjetil Halvorsen wrote:
On Wed, Oct 28, 2009 at 7:25 AM, David Scott wrote:
Karl Ove Hufthammer wrote:
On Tue, 27 Oct 2009 18:06:02 -0400 Ben Bolker wrote:
If transforming your data brings you closer to satisfying
the assumptions of your analytic methods and having a sensible
analysis, then
Hi All:
I am trying to use read.fwf and encountering the following error below. Any
ideas on what I can do?
I tried to use read.table (and the default for read.table is space) and it
works. I am not sure why read.fwf is not working
test_data_frame = read.fwf(file="small.txt",widths=width_vec,he
Hi David,
>> Now when I turn on R again the script is now completely blank.
This happened to me about 4--5 months ago under Vista. I cannot quite
remember what I did but I think I got the script working by opening it in
another editor (a hex editor would do) and removing either the first few
byt
Call for Papers:
R/Finance 2010: Applied Finance with R
April 16 and 17, 2010
Chicago, IL, USA
The second annual R/Finance conference for applied finance using R
will be held this spring in Chicago, IL, USA on April 16 and 17, 2010.
The two-day conference will cover topics including portfolio
ma
Righto!
Thanks to Henrik and Martin for pointing this out. unlist() wins by a mile!
So this _was_ instructive ... for me!
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: henrik.bengts...@gmail.com [mailto:henrik.bengts...@gmail.com] On
Behalf
S Ellison wrote:
"Lathouri, Maria" 10/28/09 6:02 PM >>>
I want to have as label in y-axis Temperature (oC).
First, look at ?plotmath and find the 'degree' symbol...
Then look at the symbol for spacing.
Then try
ylab=expression(Temperature~degree*C)
and then perhaps
ylab=expression(Temper
On 10/28/2009 11:21 AM, trz wrote:
Hi There,
I'm attempting to plot 10 values on a three-dimensional PCA with text labels
next to each point. While i have no trouble doing this on 2D plots using the
'text' or 'textxy' function, I cannot find a function to do this on a 3D
plot.
I am using princom
unlist(..., use.names=FALSE) is heaps faster than the default
unlist(..., use.names=TRUE), cf.
> z <- split(sample(1000,1e6,rep=TRUE),rep(1:1e5,10))
> system.time(y1 <- Reduce(union,z))
user system elapsed
5.980.005.89
> system.time(y2 <- unique(unlist(z)))
user system elapsed
Bert Gunter writes:
> ... and just for amusement: unique(do.call(c,l))
>
> The do.call and unlist approaches should be faster than Reduce; do.call
> _may_ be marginally faster than unlist. Here's a timing comparison:
For large named lists, unlist(l, use.names=FALSE) can have important
performanc
... and just for amusement: unique(do.call(c,l))
The do.call and unlist approaches should be faster than Reduce; do.call
_may_ be marginally faster than unlist. Here's a timing comparison:
> z <- split(sample(1000,1e6,rep=TRUE),rep(1:1e5,10))
> length(z)
[1] 10
## the comparisons:
> system
>>> "Lathouri, Maria" 10/28/09 6:02 PM >>>
>I want to have as label in y-axis Temperature (oC).
First, look at ?plotmath and find the 'degree' symbol...
Then look at the symbol for spacing.
Then try
ylab=expression(Temperature~degree*C)
***
Peng Yu wrote:
>
> Suppose that I have a list of vectors. I want to compute the union of
> all the vectors in the list. I could use 'for' loop to do so. But I'm
> wondering what would be a better solution that does not need a 'for'
> loop.
>
> l=list(a=c(1,3,4), b=c(1,3,6), c=c(1,3,7), )
>
or
unlist(l)
and possibly you want a unique() on that...
b
On Oct 28, 2009, at 5:15 PM, Jorge Ivan Velez wrote:
Hi Peng,
Here is a suggestion:
unique(do.call(c, l))
# [1] 1 3 4 6 7
Best regards,
Jorge
On Wed, Oct 28, 2009 at 3:07 PM, Peng Yu <> wrote:
Suppose that I have a list of ve
William Dunlap wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Erik Iverson
Sent: Wednesday, October 28, 2009 9:22 AM
To: Gurpal Kalsi; r-help@r-project.org
Subject: Re: [R] Selecting rows according to a column
Hello,
He
Hi Peng,
Here is a suggestion:
unique(do.call(c, l))
# [1] 1 3 4 6 7
Best regards,
Jorge
On Wed, Oct 28, 2009 at 3:07 PM, Peng Yu <> wrote:
> Suppose that I have a list of vectors. I want to compute the union of
> all the vectors in the list. I could use 'for' loop to do so. But I'm
> wonder
Suppose that I have a list of vectors. I want to compute the union of
all the vectors in the list. I could use 'for' loop to do so. But I'm
wondering what would be a better solution that does not need a 'for'
loop.
l=list(a=c(1,3,4), b=c(1,3,6), c=c(1,3,7), )
_
Achim Zeileis wrote:
On Wed, 28 Oct 2009, David Croll wrote:
Am I paranoid, or am I in search of an exactness statistics cannot
deliver?
Paranoid. ;-) With your sample size all solutions should be sufficiently
close to exact.
Well, the world could still be out to get him... It is required
As does Muenchen in RforSASSPSSusers.pdf and in the book that grew out
of that effort:
http://rforsasandspssusers.googlepages.com/RforSASSPSSusers.pdf
http://www.amazon.com/SAS-SPSS-Users-Statistics-Computing/dp/0387094172/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1217456813&sr=8-1
http://rforsasan
My OS is Windows XP.
"Writing R Extensions", Section 1.1 states "‘README’ or ‘ChangeLog’ will be
ignored by R, but may be useful to
end-users." I see examples of README files in packages grDevices folder afm
and nlme folder mlbook.
The README file was placed at the top level of the packag
Alzola and Harrell discuss some of these issues in "An introduction to
S and the Hmisc and Design Libraries".
-ista
On Wed, Oct 28, 2009 at 1:27 PM, Jacob Wegelin wrote:
>
> Often it is useful to keep a "codebook" to document the contents of a
> dataset. (By "dataset" I mean
> a rectangular stru
I find that Harrell's describe ( Hmisc) provides some of that desired
functionality. When I am creating a paper codebook I will print the
results of describe function fro a dataframe to create an overview
snapshot and will post a copy of str(dfname) on the wall.
As his help page says:
"desc
Here's a simple example that might help you get what you want:
set.seed(1)
x <- matrix(rnorm(30), ncol=3, dimnames=list(NULL, letters[1:3]))
(xc <- cor(x))
a b c
a 1.000 -0.3767034 -0.7158385
b -0.3767034 1.000 0.6040273
c -0.7158385 0.6040273 1.000
Dear all
I am doing some plots in R.
I want to have as label in y-axis Temperature (oC). I have used
ylab=expression(paste({Temperature} ^o*C)) but what I get is TemperatureoC.
How can I have a space between Temperature and the units and also the units to
be in brackets?
Many thanks
Maria
So is this factor ordered or not. Is it just in your model to create
unordered strata? What does the Time variable look like now?
summary(Time)
Depending on how Time and Treatment are associated with the subjects
you may choose different options. And are they sitting in a dataframe?
--
Dav
On Wed, 28 Oct 2009, David Croll wrote:
Dear Achim,
let me thank you for this assurance!
The sample size is too large (~4000 observations per group) to solve this
problem exactly. The error message could maybe be improved, but the message
is clear: This is too large to deal with.
However
On Oct 28, 2009, at 1:17 PM, David Winsemius wrote:
It appearred to be available from within my Mac OSX system so I went
to CRAN. Appears it fails the automated checks for Windows devices:
http://cran.r-project.org/web/checks/check_results_RandomFields.html
And looking at that compile chec
On 10/28/2009 12:41 PM, morp...@comcast.net wrote:
The package list was accessed October 28, 10:13 am MST using Windows XP and
mirror USA (CA 1).
That list comes from CRAN. You can go there and follow the links
through contributed packages until you get to it at
http://cran.r-project.org/w
Mark Kimpel wrote:
I am analyzing an experiment in which time is a factor, represented by
numbers indicating time since last treatment, but in this particular case
there is no reason to think that time has a numeric meaning in the sense
that 24 would be greater than 6. We have no idea which genes
Often it is useful to keep a "codebook" to document the contents of a dataset. (By
"dataset" I mean
a rectangular structure such as a dataframe.)
The codebook has as many rows as the dataset has columns (variables, fields).
The columns (fields)
of the codebook may include:
• va
On Wed, Oct 28, 2009 at 1:08 PM, David Winsemius wrote:
>
> On Oct 28, 2009, at 12:21 PM, Val wrote:
>
>
>
> On Wed, Oct 28, 2009 at 11:59 AM, David Winsemius
> wrote:
>
>>
>> On Oct 28, 2009, at 11:46 AM, Val wrote:
>>
>> Val, please take it slow, you are missing basic stuff here.
>>>
It appearred to be available from within my Mac OSX system so I went
to CRAN. Appears it fails the automated checks for Windows devices:
http://cran.r-project.org/web/checks/check_results_RandomFields.html
You are supposed to direct such questions to the package maintainer.
On Oct 28, 2009,
I am analyzing an experiment in which time is a factor, represented by
numbers indicating time since last treatment, but in this particular case
there is no reason to think that time has a numeric meaning in the sense
that 24 would be greater than 6. We have no idea which genes will be
increasing o
On Oct 28, 2009, at 12:21 PM, Val wrote:
>
>
> On Wed, Oct 28, 2009 at 11:59 AM, David Winsemius > wrote:
>
> On Oct 28, 2009, at 11:46 AM, Val wrote:
>
> Val, please take it slow, you are missing basic stuff here.
>
> (1) Windows Explorer may hide extensions; the 'Type' column should
> read 'R
Here is a faster way of doing the replacement: (provide reproducible
data next time)
> x <- matrix(sample(6:9, 64, TRUE), 8)
> x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]87767879
[2,]77867677
[3,]777696
Use 'rle':
> x <- rle(sample[,2])
> x
Run Length Encoding
lengths: int [1:9] 10 3 3 1 1 1 1 17 3
values : num [1:9] 0 1 0 1 0 1 0 1 0
> which.max(x$lengths[x$values==1])
[1] 4
> which.max(x$lengths * x$values) # makes use of the fact you are only using 0
> & 1
[1] 8
> cumsum(c(1, x$lengths))
#Mdarts is a matrix 2343x788
#frequencia is a vector 2343x1
# 9 in Mdarts[fri,frj] stands for my missing values which i want to replace
by the value in the vector frequencia
Mdarts<-t(matrix(scan("C:/GWS/CNB/dartg.txt"),ncol=nindT,nrow=nm, byrow=T))
frequencia <- matrix(scan("C:/GWS/CNB/freq.txt
Hi Martin,
Unfortunately, the error is coming on the data set that I have right now. I
was successfully able to display any field in the matrix and even the whole
matrix when I tried the example code provided by you. However, it is failing
on the dataset I am working on.I can share the file with y
Hadley, thanks - that was a permutation that I did not try (but should have
thought of it). But...
Now, when some observations are removed, you get the count on the plot
(previously one did not), however, alas, a new problem: the value of
.count.. includes the NA's or something similar. Revised
Thanks very much!
Gurpal
Centrica Energy.
On Wed, Oct 28, 2009 at 4:31 PM, William Dunlap wrote:
> > -Original Message-
> > From: r-help-boun...@r-project.org
> > [mailto:r-help-boun...@r-project.org] On Behalf Of Erik Iverson
> > Sent: Wednesday, October 28, 2009 9:22 AM
> > To: Gurpal
The package list was accessed October 28, 10:13 am MST using Windows XP and
mirror USA (CA 1).
Bill Morphet, Ph.D.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.
Your example is too complicated for me. But few points:
1) What do you mean by "instrument"? Do you mean variable?
2) diff(demand) is identical to demand[-1] - demand[-204]
3) system() is a built-in R function, so avoid using it as variable name
4) The variable "yd" is in the eqInvest formula
I thought I'd share this with the list since it appears to provide a
quick fix to some memory problems, and I haven't see it discussed in
relation to R.
To reallocate virtual memory from kernel-mode to user-mode in 32-bit
Vista or Windows 7 one can use the increaseuserva boot option value.
See
htt
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Erik Iverson
> Sent: Wednesday, October 28, 2009 9:22 AM
> To: Gurpal Kalsi; r-help@r-project.org
> Subject: Re: [R] Selecting rows according to a column
>
> Hello,
>
> Here's a
I need to determine the length of the longest series of consecutive numbers
(1's to be specific) and the start time of that series. For example, in the
following sample, the first column is "time" and the second column indicates
the presence of the target behavior.
I would like a function that wou
Not very elegant but try:
z <- data.frame(a = 1:5, b=10*(1:5), c = c("a", "a", "b", "b", "b") )
z[ cbind( 1:nrow(z), match( as.character(z$c) , colnames(z) ) ) ]
If you have very few columns, you can use ifelse() too.
Regards, Adai
Gurpal Kalsi wrote:
Hi,
With a data such as:
z = data.f
There is a package in beta testing now that looks interesting:
http://openmx.psyc.virginia.edu/installing-openmx
-Ista
On Wed, Oct 28, 2009 at 10:37 AM, Robert Terwilliger wrote:
> Dear R-help,
>
> I am interested in using structural equation modeling.
>
> Just getting started with it, but I'm
Hello,
Here's an idea:
ifelse(z$c == "a", z$a, z$b)
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Gurpal Kalsi
> Sent: Wednesday, October 28, 2009 11:15 AM
> To: r-help@r-project.org
> Subject: [R] Selecting rows accordin
On Wed, Oct 28, 2009 at 11:59 AM, David Winsemius wrote:
>
> On Oct 28, 2009, at 11:46 AM, Val wrote:
>
> Val, please take it slow, you are missing basic stuff here.
>>
>>>
>>> (1) Windows Explorer may hide extensions; the 'Type' column should
>>> read 'R file' anyway.
>>>
>>>
>> * Yes I looked
Hello Jim
Thanx a lot.
Actually since long I was trying
> dts = c("1989-09-28", "2001-01-15", "2004-08-30", "1990-02-09")
> dts
[1] "1989-09-28" "2001-01-15" "2004-08-30" "1990-02-09"
> GMT = timeDate(dts, zone = "GMT", FinCenter = "GMT")
> GMT
GMT
[1] [1989-09-28] [2001-01-15] [2004-08-30] [199
How about this:
> x <- c("2008-09-12T23:48:07.747Z", "2008-09-12T20:35:07.747Z")
> x.t <- as.POSIXct(x, format="%Y-%m-%dT%H:%M:%OS")
>
> x.t
[1] "2008-09-12 23:48:07 EDT" "2008-09-12 20:35:07 EDT"
> diff(x.t)
Time difference of -3.216667 hours
>
On Wed, Oct 28, 2009 at 12:11 PM, Sunita22 wrote:
Hi,
With a data such as:
> z = data.frame(a = 1:5, b=10*a, c = c("a", "a", "b", "b", "b") )
* a b c*
*1* 10 *a*
*2* 20 *a*
3 *30* *b*
4 *40* *b*
5 *50* *b*
Can anyone suggest a way to select [1, 2, 30, 40, 50],
ie. using column "c" to specify which column is selected for each row.
Many th
Hello
I have a data set which contains a column of Standard Time Stamps
(2008-09-12T23:48:07.747Z, 2008-09-12T20:35:07.747Z, etc)
I need to find differences in consecutive time stamps and then need to plot
a graph using it, can anyone guide me as to how to handle this type of data?
Thanks in ad
To stop in Rgui mode, you can try pressing the ESC key. If you are using
within emacs, change to R buffer and try C-c C-c to stop it.
I am not sure how to recover the script (emacs usually makes a .R~
backup). Maybe if you still have the output printed to screen or
terminal make a copy of it
The gamlss package, by Mikis Stasinopoulos and available at
http://www.gamlss.com/ as well as from CRAN, is also very flexible,
allowing shape and scale adjustment.
Steve E
>>> David Winsemius 28/10/2009 14:18 >>>
You might want to take a look at this article by WEI, PERE, KOENKER,
AND HE.
On Oct 28, 2009, at 11:46 AM, Val wrote:
Val, please take it slow, you are missing basic stuff here.
(1) Windows Explorer may hide extensions; the 'Type' column should
read 'R file' anyway.
* Yes I looked at it and it only shows type. To check I downloaded
another script with R extens
On 10/28/2009 11:37 AM, David Young wrote:
Hi all,
I just had a rather unpleasant experience. After considerable work I
finally got a script working and set it to run. It had some memory
allocation problems when I came back so I used Windows to stop it.
During that process it told me that the
David Winsemius wrote:
On Oct 28, 2009, at 10:55 AM, Val wrote:
The working directory is
getwd()
[1] "C:/Documents and Settings/Val/My Documents"
The data file(Rossi.dat) and the script(Rossi.R) are in
"C:/Documents and Settings/Val/My Documents/R_data/prd"
So you are not giving a prope
Thanks a lot. Have a nice day!
Best,
Pat
On Wed, Oct 28, 2009 at 10:29 AM, Max Kuhn wrote:
> There are a few. I'm partial to the function in the caret package:
> createDataPartition. Also, there are functions there for
> pre-processing on training sets and applying it to new data sets.
>
> For
Hi There,
I'm attempting to plot 10 values on a three-dimensional PCA with text labels
next to each point. While i have no trouble doing this on 2D plots using the
'text' or 'textxy' function, I cannot find a function to do this on a 3D
plot.
I am using princomp for my PCA:
>PCA<-princomp(eucdat
-- begin included message ---
> (basehazzft.ln$stra[285])
[1] stra=2
134 Levels: stra=1 stra=10 stra=100 stra=101 stra=102 ... stra=99
> c(basehazzft.ln$stra[285])
[1] 47
while the desired value is 2, I get a 47. What am I doing wrong? I tried
the as.numeric function but I have the same problem..
Dear all,
I am using the Kruskal-Wallis test in R (kruskal.test()) to compare
non-normally distributed observations for 5 different groups. I now want to
perform multiple comparisons to identify the groups with significant
differences in the mean ranks. On searching the forum I found a number of
Val, please take it slow, you are missing basic stuff here.
>
> (1) Windows Explorer may hide extensions; the 'Type' column should
> read 'R file' anyway.
>
* Yes I looked at it and it only shows type. To check I downloaded
another script with R extension "test.R" and the type column shows th
The general way to create a title for multiple plots on the same page is to
first create some room for the title by setting the outer margins (using
par(oma=...)), then use mtext or title with the outer=TRUE argument to place
the overall title.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data
Hi all,
I just had a rather unpleasant experience. After considerable work I
finally got a script working and set it to run. It had some memory
allocation problems when I came back so I used Windows to stop it.
During that process it told me that the script had been changed and
asked if I wanted
On Wed, Oct 28, 2009 at 9:23 AM, Tom Gottfried wrote:
> ?curve
>
> regards,
> Tom
and I was in the process of writing a curve example when I noticed Tom
sent this. Here it is:
set.seed(777)
x <- runif(100, 0, 100)
y <- 10*x + x^2 - .01*x^3 + rnorm(100, 0, 500)
fit <- lm(y ~ x + I(x^2) + I(x^3))
On Oct 28, 2009, at 10:55 AM, Val wrote:
The working directory is
getwd()
[1] "C:/Documents and Settings/Val/My Documents"
The data file(Rossi.dat) and the script(Rossi.R) are in
"C:/Documents and Settings/Val/My Documents/R_data/prd"
So you are not giving a proper path when you issue the
There are a few. I'm partial to the function in the caret package:
createDataPartition. Also, there are functions there for
pre-processing on training sets and applying it to new data sets.
For a somewhat dated summary of the packages, see:
http://www.jstatsoft.org/v28/i05
also:
http://c
Do you mean like regexpr() (on the same help page)?
Depending on your locale, you might actually prefer the character
offset: if you want to match in a MBCS and have byte offsets you will
need to work a bit harder if useBytes=TRUE is not sufficient for you.
On Wed, 28 Oct 2009, Johannes Graum
?curve
regards,
Tom
Ken Ervin schrieb:
> I have a data set of 6 or so ordered pairs, and I've been able to graph
> them and have decided to use a high-order polynomial regression. I've
> used the following piece of code:
>
> regression <- function(x,y) {
>x <- c(insert_numbers_here)
>y
Hi, Users,
I am a new user. I am trying to partition data into training and test. Is
there any R package or function that can partition dataset? Also, is there
any package do crossvalidation? Any help will be appreciated.
Best,
Pat
[[alternative HTML version deleted]]
___
Hi All:
Bear with me on this longer e-mail.
Questions:
1) Can you share with me on any example code that you may have that calculates
bias of a statistical forecast in a time series?
2) Supposed I have the file in the fixed width format (details below).
1-62 character key
63-76 sales data point
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