Hello!
I'm a college undergrad desperately trying to finish up my thesis. I
have a dataset on the distribution of a grassland bird from the
Breeding Bird Survey. I have a very straightforward and simple version
of the logistic growth model to describe changes in this bird's
abundance over time.
Giovanni,
You can use the '...' for that, as in:
loocv - function(data, fnc, ...) {
n - length(data.x)
score - 0
for (i in 1:n) {
x_i - data.x[-i]
y_i - data.y[-i]
yhat - fnc(x=x_i,y=y_i, ...)
score - score + (y_i - yhat)^2
}
score - score/n
return(score)
}
scoreks -
Hi Daniel,
Thanks for taking the time to go through this.
You wrote:
I don't see the advantage over using the R_LIBS environment variable
You are correct - there is no advantage *over *doing that, because that is *
exactly* what my code is doing :)
You wrote:
You need to initially copy
Hello Jan,
On Apr 26, 2010, at 8:56 AM, Jan van der Laan wrote:
You can use the '...' for that, as in:
loocv - function(data, fnc, ...) {
n - length(data.x)
score - 0
for (i in 1:n) {
x_i - data.x[-i]
y_i - data.y[-i]
yhat - fnc(x=x_i,y=y_i, ...)
score - score + (y_i - yhat)^2
Hi, Dear R community,
These is a paper of Wheat Grain Yield Response to N Application
Evaluated through Canopy Reflectance.
This paper of Materials and Methods said A single regression equation
relating A and B was derived for each location using PROC NLIN (SAS
Inst. 1990). Slopes and intercepts
inputfille
snpid indid genotypegvariable probeid genegeneexpression
rs1040480 CHB_NA18524 C/T 2 GI_19743926-I PTPRT 5.850586
rs1040480 CHB_NA18526 C/C 1 GI_19743926-I PTPRT 6.028641
rs1040480 CHB_NA18529 C/C 3
Hi
r-help-boun...@r-project.org napsal dne 26.04.2010 06:52:55:
Having some difficulties with understanding how tapply works and getting
return values I expect
Data: dataframe. DF DF$Id $D $Year...
Id D Year Jan Feb Mar Apr May Jun Jul Aug Sep
Oct
Nov
I've tried both mean and colMean.
I did success with one attempt using mean, however if only have 1 year and
its a NA
then I get NaN ( which I can replace). I'll keep trying.
On Mon, Apr 26, 2010 at 12:26 AM, Petr PIKAL petr.pi...@precheza.cz wrote:
Hi
r-help-boun...@r-project.org napsal
Hi Charlotte ,
I can't reproduce your code, but skimming through it -
It would appear that:
1) in
eqn1- function(K1, bird)
you didn't define bird (you did define it before the function, so I'd
suggest just removing it from the function call like this:
eqn1- function(K1)
2) you didn't return and
mm...
I also noticed the function you wrote didn't use parenthesis, mixed b and c
and used different names for K.
Your code is a great exercise in debugging (no offense intended :) )
Try using:
bird-bird.density[0] # I assume this exists
eqn- function(K1, b1 = 1.22, c1 = .55) {
Have a look here:
http://www.r-bloggers.com/a-free-book-on-geostatistical-mapping-with-r/
http://www.r-bloggers.com/a-free-book-on-geostatistical-mapping-with-r/
Contact
Details:---
Contact me: tal.gal...@gmail.com |
That fails:
The manual says:
tapply(X, INDEX, FUN = NULL, ..., simplify = TRUE)
ArgumentsXan atomic object, typically a vector.INDEXlist of factors, each of
same length as X. The elements are coerced to factors by
as.factorhttp://127.0.0.1:31214/library/base/help/as.factor
.
my error says:
Great! Thank you for your help!
-Charlotte
On Mon, Apr 26, 2010 at 1:12 AM, Tal Galili tal.gal...@gmail.com wrote:
mm...
I also noticed the function you wrote didn't use parenthesis, mixed b and c
and used different names for K.
Your code is a great exercise in debugging (no offense
Hi Peter
Thanks. This was my problem
I saw this term in the next document
http://cran.r-project.org/doc/Rnews/Rnews_2008-2.pdf
Best regards
--
View this message in context:
http://r.789695.n4.nabble.com/Library-tm-Error-could-not-find-function-TermDocMatrix-tp2062464p2064843.html
Sent from
Hi:
Use of ddply() in the plyr package appears to work.
library(plyr)
ddply(df[, -1], .(Year), colwise(mean), na.rm = TRUE)
D Year Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
1 1.00 1980 NaN NaN NaN NaN NaN 212 203 209 228 237 NaN NaN
2 0.50 1981 NaN 251 243 246 241 NaN NaN
Hi
steven mosher mosherste...@gmail.com napsal dne 26.04.2010 10:21:37:
That fails:
The manual says:
tapply(X, INDEX, FUN = NULL, ..., simplify = TRUE)
Arguments
X
an atomic object, typically a vector.
INDEX
list of factors, each of same length as X. The elements are
Dear list,
There is a new rwiki (http://rwiki.sciviews.org) section for 'large scale
data' at http://rwiki.sciviews.org/doku.php?id=large_scale_data. This is a
massive topic of interest to many. I could use some help filling in the
content so that this can develop into a powerful resource for the
hi everybody,
How can you plot some dates? I mean how can i have the tickmarks with the label
of each date on my x axis?
The dates i use are in POSIXct format, i don't know if it matters.
thanks a lot
Karine
Dear all,
this is probably a very silly question, but could anyone tell me what the
different parameters in a coxph model with a frailty.gamma term mean?
Specifically I have two questions:
(1) Compared to a normal coxph model, it seems that I obtain two standard
errors [se(coef) and se2].
What
On 04/26/2010 08:05 PM, karine heerah wrote:
hi everybody,
How can you plot some dates? I mean how can i have the tickmarks with the label
of each date on my x axis?
The dates i use are in POSIXct format, i don't know if it matters.
Hi Karine,
Here's one way:
plot(1:3,xaxt=n)
On 26/04/2010 6:05 AM, karine heerah wrote:
hi everybody,
How can you plot some dates? I mean how can i have the tickmarks with the label
of each date on my x axis?
The dates i use are in POSIXct format, i don't know if it matters.
If your x variable is a POSIXct variable, it
Did you check what comes out x$fstatistic[1L] etc... ? Seems to me at least
one of those is not a number.
Please give ready-to-use code, nobody here will transform your mail to a
text file, try to read it in and so on. Also specify which package you use,
lmList is not a standard function in R and
Dear all,
I have a multiline plot with each line labeled with a different letter.
But I'm not able to make the legend display the same kind of pattern
'-a-', instead the letter is overwritten by the line. A simpler legend
with only the letter is not very visible and the pt.bg does nothing with
Try this:
aggregate(DF[c('data', 'data2')], DF[ 'years'], FUN = sum, na.rm = TRUE)
aggregate(DF[c('data', 'data2')], list(as.character(factor(DF[, 'years'],
labels = c('5-7', '5-7', '5-7', 8, FUN = sum, na.rm = TRUE)
On Sun, Apr 25, 2010 at 3:29 AM, steven mosher mosherste...@gmail.comwrote:
Dear R users and developers,
Due to a failure of the electrical system, the IT services (also hosting
R-project related websites) of the WU (Vienna University of Economics and
Business) are currently unavailable.
Affected websites and services are: CRAN master (cran.r-project.org), the
Please check the installation guide for Tinn-R. It might be you simply
forgot to configure Tinn-R (RConfigurepermanent). I use it for 2 years
now, and find it a blessing.
Mind that there is a problem sometimes with the connection between Tinn-R
and R, and the problem is Windows-related. If you
Google goes a long way...
I would start here
http://www-stat.stanford.edu/~tibs/ElemStatLearn/
On Apr 26, 2010, at 12:44 AM, Changbin Du changb...@gmail.com wrote:
Hi, Dear R community,
Does anyone know how to constructdecision tree with boosting? Is any
tutorial I can read?
--
On 26.04.2010 13:32, Mario Valle wrote:
Dear all,
I have a multiline plot with each line labeled with a different letter.
But I'm not able to make the legend display the same kind of pattern
'-a-', instead the letter is overwritten by the line. A simpler legend
with only the letter is not very
You can try this:
plot(1:10,10:1,lty=1,type='b', lwd=2,pch='a')
text(2, pretty(10:1)[2] - 0.3, paste(\u2013 a \u2013, ds1, sep =),
col = 'black', font = 2)
text(2, pretty(10:1)[2] - 0.6, paste(\u2013 b \u2013, ds2, sep =),
col = 'red', font = 2)
rect(0, 0, 3, 2)
On Mon, Apr 26, 2010 at
Thanks Uwe and Henrique!
Both solution works and both have the same niceness/work.required ratio :-)
Thanks again!
mario
On 26-Apr-10 14:02, Uwe Ligges wrote:
On 26.04.2010 13:32, Mario Valle wrote:
Dear all,
I have a multiline plot with each line labeled with a
Try this kludge:
legend(left, c(-a-, -b-, ds1 , ds2 ), ncol = 2, text.col = 1:2)
On Mon, Apr 26, 2010 at 7:32 AM, Mario Valle mva...@cscs.ch wrote:
Dear all,
I have a multiline plot with each line labeled with a different letter.
But I'm not able to make the legend display the same kind of
I am using apply to run exact permutation t-tests by columns
using the coin package. For example:
library(coin)
dat - matrix(rnorm(7*35),7,35)
fun - function(x) {
pvalue( independence_test(x~f,
data=data.frame(x, f=factor(c(rep(a,4),rep(b,3,
Hello everyone!
My data is composed of 277 individuals measured on 8 binary variables
(1=yes, 2=no).
I did two similar cluster analyses, one on SPSS 18.0 and one on R 2.9.2. The
objective is to have the means for each variable per retained cluster.
1) the R analysis ran as followed:
call
Please specify reproducible code that show the problem.
When I do it, ewverything is fine on my data with the most recently
released versions of R, R2WinBUGS and coda.
But perhaps you have another package loaded with another
summary.mcmc.list method?
On 26.04.2010 06:31, Chris Fonnesbeck
(Resending, forgot to cc. r-help)
On Apr 26, 2010, at 1:47 PM, Stefan Peter Theußl wrote:
Dear R users and developers,
Due to a failure of the electrical system, the IT services (also hosting
R-project related websites) of the WU (Vienna University of Economics and
Business) are
Hi Jeoffrey,
How stable are the results in general ?
If you repeat the analysis in R several times, does it yield the same
results ?
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me:
I'm not sure why you'd expect Euclidean distance and squared Euclidean
distance to
give the same results.
Euclidean distance is the square root of the sums of squared
differences for each variable, and that's exactly what dist() returns.
http://en.wikipedia.org/wiki/Euclidean_distance
On a map,
So I have Transactions as a dataframe and Transactions$Symbol is a column in
the frame
I simply want to run sub over all elements in that column where the new value
replaces the old value
this doesn't seem to work
Transactions$Symbol = lapply(Transactions$Symbol, function(x) sub('[
Try this:
Transactions$Symbol - gsub('[ -]*','INST',Transactions$Symbol)
On Mon, Apr 26, 2010 at 10:14 AM, Robert Nicholson
robert.nichol...@gmail.com wrote:
So I have Transactions as a dataframe and Transactions$Symbol is a column
in the frame
I simply want to run sub over all elements
Yeah I finally got there thanks.
On Apr 26, 2010, at 8:22 AM, Henrique Dallazuanna wrote:
Try this:
Transactions$Symbol - gsub('[ -]*','INST',Transactions$Symbol)
On Mon, Apr 26, 2010 at 10:14 AM, Robert Nicholson
robert.nichol...@gmail.com wrote:
So I have Transactions as a
On Apr 26, 2010, at 12:55 AM, Rob James wrote:
What causes the error report:
logical(0)
to arise in the rms function lrm?
Here's my data:
But both the dependent and the independent variable seem fine...
str(AABB)
'data.frame':1176425 obs. of 9 variables:
$ sex : int 1 1 0 1 1 0
Chris Fonnesbeck fonnesbeck at gmail.com writes:
I am trying to get summary statistics from WinBUGS/JAGS output in the
form of mcmc.list objects, using the summary() function. However, I
get odd warning messages:
Warning messages:
1: In glm.fit(x = X, y = Y, weights = weights, start =
Dear R community,
The only working mirror under a separate subdomain in the zone r-project.org is
cran.us.r-project.org. Some mirrors are just virtualhosts on the master server
and thus cannot be resolved via DNS. These are:
cran.au.r-project.org
cran.br.r-project.org
cran.ca.r-project.org
On 4/25/2010 19:39:52, Tal Galili wrote:
*c) R core implementation ?!*
I hope I am not being rude (or jumping into any open doors) in asking this
but...
What do you think about implementing this strategy into the R basic
installation?
Tal,
As a general rule, I think R should make
Question: How do I isolate the p-value from the serial.test function
(portmanteau test)?
Details: I tried the following set of commands
testResult - serial.test(vec2varModelFit)
thePValue - testResult$p.value
When i type
1) thePValue
I get:
NULL
2) testResult
I get
Portmanteau Test
I do get the following error message:
*Error in lookup.svc[i, j] - svc[svc$st == unique(svc$st)[i] svc$vc ==
: *
* replacement has length zero*
I also thought it might be because of how R treats NA, but then I would
expect the loop to stop at the place of error (i=1, j=2) and not continue to
Try this:
testResult$serial$p.value
On Mon, Apr 26, 2010 at 11:17 AM, Andrew Hill andy.bob.h...@gmail.comwrote:
Question: How do I isolate the p-value from the serial.test function
(portmanteau test)?
Details: I tried the following set of commands
testResult -
Hello,
I have
JRip(Species ~ ., data = iris)
JRIP rules:
===
(Sepal.Width = 2.9) and (Petal.Width = 1.6) = Species=versicolor
(39.0/5.0)
(Petal.Length = 4.1) and (Petal.Width = 1.7) = Species=versicolor
(17.0/2.0)
(Petal.Length = 4.8) = Species=virginica (46.0/1.0)
= Species=setosa
In a .Rnw file I want to insert the R command
pairs(mydataframe)
and achieve the following effects
1. the command itseld is echoed into the tex document generated by Sweave
fig=TRUE,echo=TRUE=
2. The graphics generated appears in the tex document, with the graphics
centred.
3. The R command
Hi,
I would like to compute the autocorrelation time from my 2-D time series,
but I cannot find how to do using R. I have found the fucntion acf and acf
plot, but how can find the autocorrelation time?
thanks
--
View this message in context:
Try this;
xtabs(y ~ st + vc, data = x)
On Mon, Apr 26, 2010 at 11:35 AM, Robin Jeffries rjeffr...@ucla.edu wrote:
I do get the following error message:
*Error in lookup.svc[i, j] - svc[svc$st == unique(svc$st)[i] svc$vc ==
: *
* replacement has length zero*
I also thought it might
Hi everybody,
How can I set the seed for my simulations ? Is set.seed(123456) alright ?
Thanks a lot,
Jimmy
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
This should get you close:
x - read.table(textConnection(st vc y
+ A Z .2
+ B Z .4
+ B Y .3
+ C Y .1
+ C X .8), header=TRUE)
closeAllConnections()
x
st vc y
1 A Z 0.2
2 B Z 0.4
3 B Y 0.3
4 C Y 0.1
5 C X 0.8
x.1 - melt(x)
Using st, vc as id variables
x.1
Hi,
On Mon, Apr 26, 2010 at 10:45 AM, Jimmy Söderly jimmy.sode...@gmail.com wrote:
Hi everybody,
How can I set the seed for my simulations ? Is set.seed(123456) alright ?
Hmm ...
R set.seed(123456)
R rnorm(5)
[1] 0.83373317 -0.27604777 -0.35500184 0.08748742 2.25225573
R rnorm(5)
[1]
Seriously! That easy!
I kept thinking that xtab would just give me frequencies of how many times
the combination occurred, and not the values themselves.
Thanks!
-Robin
On Mon, Apr 26, 2010 at 7:40 AM, Henrique Dallazuanna www...@gmail.comwrote:
Try this;
xtabs(y ~ st + vc, data = x)
Here is a kludge:
fm - JRip(Species ~., iris)
as.numeric(sub(.*:, , grep(Number of Rules, capture.output(fm), value =
TRUE)))
[1] 4
On Mon, Apr 26, 2010 at 7:23 AM, fascob samto...@yahoo.com wrote:
Hello,
I have
JRip(Species ~ ., data = iris)
JRIP rules:
===
(Sepal.Width =
Thanks,
I was trying to stick with the base package and figure out how the base
routines worked. I looked at plyer and it was very appealing. I guess i'll
give in and use it
On Mon, Apr 26, 2010 at 2:33 AM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
Use of ddply() in the plyr package
I guess my problem was seeing a bunch of examples where they pulled a
variable from a dataframe..
tapply(df$data, index=list(..
and I
assumed that the df$data was just generalizable to a collection of vectors
a vector of vector being a vector
Thanks.
On Mon, Apr 26, 2010 at 2:43 AM, Petr
Thanks Steve, it seems to work !
2010/4/26 Steve Lianoglou mailinglist.honey...@gmail.com
Hi,
On Mon, Apr 26, 2010 at 10:45 AM, Jimmy Söderly jimmy.sode...@gmail.com
wrote:
Hi everybody,
How can I set the seed for my simulations ? Is set.seed(123456) alright
?
Hmm ...
R
Dear list,
I have a big data frame which looks like this:
variable YEAR VAR
EC01 2006 100
EC01 2007 200
EC02 2006
I would like to select rows if a row contains any one of several values. I can
do the selection as follows:
result[,Subject]==JEFF | result[,Subject]==BG
But this is very unwieldily if one wishes to select many, many rows as one has
to continuously repeat the source:
result[,Subject]==JEFF |
Do you mean
ccf
?
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
On Apr 26, 2010, at 11:12 AM, John Sorkin wrote:
I would like to select rows if a row contains any one of several
values. I can do the selection as follows:
result[,Subject]==JEFF | result[,Subject]==BG
But this is very unwieldily if one wishes to select many, many rows
as one has to
On Apr 26, 2010, at 10:12 AM, John Sorkin wrote:
I would like to select rows if a row contains any one of several values. I
can do the selection as follows:
result[,Subject]==JEFF | result[,Subject]==BG
But this is very unwieldily if one wishes to select many, many rows as one
has to
See ?%in%
result$Subject %in% c(JEFF, BG)
John Sorkin wrote:
I would like to select rows if a row contains any one of several values. I can
do the selection as follows:
result[,Subject]==JEFF | result[,Subject]==BG
But this is very unwieldily if one wishes to select many, many rows as one
Hi all,
I need an R expression, expression() to get the follows Latex code:
$P^\uparrow_{SF}$.
Imposible, for me!!.
Thank you in advance
/salva
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Here are four ways:
# state.name comes with R
DF - data.frame(state = state.name, num = seq_along(state.name))
DF[DF$state %in% c(Iowa, Utah),]
subset(DF, state %in% c(Iowa, Utah))
subset(DF, grepl(Iowa|Utah, state))
library(sqldf) # see http://sqldf.googlecode.com
sqldf(select * from DF
On Apr 26, 2010, at 11:27 AM, Gabor Grothendieck wrote:
Here are four ways:
# state.name comes with R
DF - data.frame(state = state.name, num = seq_along(state.name))
DF[DF$state %in% c(Iowa, Utah),]
subset(DF, state %in% c(Iowa, Utah))
subset(DF, grepl(Iowa|Utah, state))
John;
The
Hi guys,
I'm trying to install Rmpi on the slave nodes of my cluster. The
installation seems to work with no errors or warnings using this
command:
R CMD INSTALL
--configure-args=--with-Rmpi-include=/usr/lib64/openmpi/1.3.2-gcc/include
--with-Rmpi-libpath=/usr/lib64/openmpi/1.3.2-gcc/lib
Hi there,
I am working on a project that requires me to find the point at which values
become negative in a sequence about the max.
For example, say I have some sequence:
x=c(-12, -2,-19, 0, -14, -2, 9,10,20,35,56,89,60,39,12,8,-5,-2,0,10)
In this sequence, the max is 89, and I need to
HI, Dear Greg,
I AM A NEW to GBM package. Can boosting decision tree be implemented in
'gbm' package? Or 'gbm' can only be used for regression?
IF can, DO I need to combine the rpart and gbm command?
Thanks so much!
--
Sincerely,
Changbin
--
[[alternative HTML version deleted]]
Try this:
func - function(x)
{
which.negative - which(x0)
index.to.return - which.negative[which.negative which.max(x)][1]
return(index.to.return)
}
func(x)
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com |
On Apr 26, 2010, at 12:29 PM, Su Chu wrote:
Hi there,
I am working on a project that requires me to find the point at
which values
become negative in a sequence about the max.
For example, say I have some sequence:
x=c(-12, -2,-19, 0, -14, -2, 9,10,20,35,56,89,60,39,12,8,-5,-2,0,10)
Hi everybody,
I wanted to use Sweave but I do not have the Latex package:
LaTeX Error: File 'Sweave.sty' not found.
I cannot find it with MiKTeX :-(
Can somebody help me ?
Thanks a lot,
Jimmy
[[alternative HTML version deleted]]
__
Dear users,
I am trying to access a Microsoft Access database from R using RODBC
package
but I have had little success. The setup works with isql, RODBC seems
to
connect to the database, but RODBC does not recognize the data in the
database. Can anybody advise where I am going wrong?
I am
If you have the package you have this file. Maybe check FAQ A.12
http://www.stat.uni-muenchen.de/~leisch/Sweave/FAQ.html
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jimmy Söderly
Sent: Monday, April 26, 2010 1:10 PM
To:
You'll find that this is a very common question, and 30 seconds with google
will get you many nice explanations and solutions.
The quickest is to copy Sweave.sty from the package directory to your
working directory.
Sarah
On Mon, Apr 26, 2010 at 1:10 PM, Jimmy Söderly jimmy.sode...@gmail.com
I am new to R and have tried for a good while to figure out how to code this
in R. The dataset below:
FTIStandKey
State
County
FTITract
CoverType
Ver_CT
V_Origin
V_SpGrp
NAH6005-001
Texas
Jasper
NAH6005
PPLB-2000-U
PPLB-2000-U
P
P
NAH6005-002
Texas
Try this:
sweep(with(x, tapply(VAR, list(variable, YEAR), FUN = prod)), 2, with(x,
tapply(VAR, YEAR, FUN = prod)), FUN = /)
On Mon, Apr 26, 2010 at 12:07 PM, n.via...@libero.it n.via...@libero.itwrote:
Dear list,
I have a big data frame which looks like this:
variable YEAR
Its at this path (this command is issued from within R):
file.path(R.home(), share, textmf, Sweave.sty)
You can add it to your MiKTeX installation or just put it in the same
directory as your Rnw file.
On Mon, Apr 26, 2010 at 1:10 PM, Jimmy Söderly jimmy.sode...@gmail.com wrote:
Hi
Thanks so much, Greg!
On the demo(bernoulli), I FOUND the following information: IT is used for
logistic regression.
My question is: when I define a decision tree, can I still use the formula
Y~X1+X2+X3,# formula, even though I dont know the detailed
formula of decision tree.
Sorry, typo. Try this:
file.path(R.home(), share, texmf, Sweave.sty)
On Mon, Apr 26, 2010 at 1:18 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Its at this path (this command is issued from within R):
file.path(R.home(), share, textmf, Sweave.sty)
You can add it to your MiKTeX
I have created a multivariate regression tree using mvpart, with 3-4
responses. Though the plot shows bargraphs for each response, I would
like to have the VALUES of the responses
printed or indicated (via a scale or something) alongside the bargraph.
Is this possible ??
Thanks,
Manjunath
I am new to R and have tried for a good while to figure out how to code this
in R. The dataset below:
FTIStandKey
State
County
FTITract
CoverType
Ver_CT
V_Origin
V_SpGrp
NAH6005-001
Texas
Jasper
NAH6005
PPLB-2000-U
PPLB-2000-U
P
P
NAH6005-002
Texas
you can do this:
a - tapply(VAR, YEAR, prod)
The use merge to create a new variable of the length of your original VAR,
and just do
VAR/prod.VAR
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me:
On Apr 26, 2010, at 12:11 PM, boris.vasil...@forces.gc.ca wrote:
Dear users,
I am trying to access a Microsoft Access database from R using RODBC
package
but I have had little success. The setup works with isql, RODBC seems
to
connect to the database, but RODBC does not recognize the
Is this what you want:
x - read.table(textConnection(variable
YEAR VAR
+ EC01 2006 100
+
+ EC01 2007 200
+
+ EC02 2006 500
GBM implements boosted trees. It works for 0/1 outcomes, count outcomes,
continuous outcomes and a few others. You do not need to combine rpart
and gbm. You're best bet is to just load the package and run a demo
demo(bernoulli).
From: Changbin Du
On 26/04/2010 7:07 AM, David.Epstein wrote:
In a .Rnw file I want to insert the R command
pairs(mydataframe)
and achieve the following effects
1. the command itseld is echoed into the tex document generated by Sweave
fig=TRUE,echo=TRUE=
2. The graphics generated appears in the tex document,
On Apr 26, 2010, at 12:40 PM, Marc Schwartz wrote:
On Apr 26, 2010, at 12:11 PM, boris.vasil...@forces.gc.ca wrote:
Dear users,
I am trying to access a Microsoft Access database from R using RODBC
package
but I have had little success. The setup works with isql, RODBC seems
to
Hi all,
I am getting different results from ccf and cor,
Here is a simple example:
set.seed(100)
N - 100
x1 - sample(N)
x2 - x1 + rnorm(N,0,5)
ccf(x1,x2)$acf[ccf(x1,x2)$lag == -1]
cor(x1[-N], x2[-1])
Results:
ccf(x1,x2)$acf[ccf(x1,x2)$lag == -1]
[1] -0.128027
cor(x1[-N], x2[-1])
[1]
Thanks !!!
2010/4/26 Gabor Grothendieck ggrothendi...@gmail.com
Sorry, typo. Try this:
file.path(R.home(), share, texmf, Sweave.sty)
On Mon, Apr 26, 2010 at 1:18 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Its at this path (this command is issued from within R):
What I want to do is to be able to select certain FTIStandKey's and change
values in the dataset. For example, I would like to select by
FTIStandKey==NAH6253-003, and change the entries for CoverType=
XSOP--C and change the V_SpGrp==T, and also change V_Origin==T. I
only want to change
Hi Randy,
The SAS code for what I want to do is:
/*
if FTIStandKey= NAH6253-003 then do;
CoverType= XSOP--C ;
V_SpGrp=T;
V_Origin=T;
end;
I admit I didn't read all the sample code you added, but this is what
I *think* you're trying to accomplish.
Hi,
I am trying to connect to RServe across a network.
I had put RServe on my local machine and it worked just fine. When am try to
connect to it across a network - it is able to go through the handshake and
get in. However , when it gets to the 'request' method in the RTalk class,
it hangs. Any
The dear mark leeds markle...@verizon.net
Has pointed me that the answer to my question was in the MASS book, in page
390
Where it is said that acf works by dividing the covariance with N instead of
N-t
so to insure that the covariance sequence is positive definite.
Although I am not sure if to
On Fri, 23 Apr 2010 15:22:45 +0300, Tal Galili tal.gal...@gmail.com
wrote:
Due to the new R 2.11 release, I want to implement Dirk's suggestion
herehttp://stackoverflow.com/questions/1401904/painless-way-to-install-a-new-version-of-r
.
So for that I am asking - How can I (permanently) change R's
On Thu, 22 Apr 2010 12:00:13 -0700 (PDT), Paul Miller
pjmiller...@yahoo.com wrote:
I was just wondering if anyone could give me some advice about the wisdom or
folly of trying to use both [R and S-Plus].
I suspect that trying to use both will give you heartburn. When I
switched from S-Plus to
Hi Dennis,
Thanks for the idea, but the order of the rowSums does not
necessarily correspond to the order of rows that maximizes the rank
of the matrix.
Ex:
a[1:9]-c(1,1,30,50,1,1,1,20,1)
a
[,1] [,2] [,3]
[1,]1 501
[2,]11 20
[3,] 3011
I think it makes more sense for most users to have a global library
(as you call it), rather than put the library under the current
installation. I have been doing that for years, and it saves a lot of
trouble.
When I have helped people learn R, the need to copy the library when
updating is a
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