Hi,
I need to do factor analysis with non-constant variance. Is there a package
that contains Welch ANOVA ?
Thanks,
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-projec
package: psych
function: error.bars()
R version: 2.9.1
OS: both linux and windows
a<-c(1,2,3,4,5)
b<-c(1,2,3,4,4)
c<-c(1,2,3,NA,5)
data<-data.frame(a,b,c)
error.bars(data[,1:2],ylim=c(0,6)) looks fine
however
error.bars(data,ylim=c(0.6))
shows multiple error bars for each vector and the num
try this:
set.seed(123)
dat <- data.frame(x = round(rnorm(10)), y = round(rnorm(10)))
dat$Min <- pmin(dat$x, dat$y)
dat$Max <- pmax(dat$x, dat$y)
dat
ind <- dat$Min != dat$Max
dat[ind, ]
I hope it helps.
Best,
Dimitris
kxk wrote:
I have two variables in a data frame, I want to generate tw
Hi John,
Thank you for the easily cut-and-pastable example.
You are providing syntactically complete code on the first line so the
second line is not seen as a continuation. Try moving the '+' on the
second line to the first. E.g,
pb1 <- p + geom_point(aes(Area, Pcode)) + geom_segment() +
yes, you need to use the 'drop' argument; have a look at the help file
?"[" and then try this:
z <- matrix(rnorm(20), ncol = 5)
z[1, , drop = FALSE]
z[, 1, drop = FALSE]
I hope it helps.
Best,
Dimitris
Jim Nemesh wrote:
Is there any way to force a slice of a matrix to stay a matrix? R tends
I have two variables in a data frame, I want to generate two additional
variables. For every observations (i.e. every row), I want the first new
variable 'min' to carry the minimum of the two existing variables, and I
want the second new variable 'max' to carry the maximum of the two existing
vari
Is there any way to force a slice of a matrix to stay a matrix? R
tends to convert a single row of a matrix into a vector.
Example:
z<-matrix (rnorm(20), ncol=5)
zz<-z[1,]
is.matrix(zz) #FALSE
I usually resort to:
zz<-matrix(z[1,], ncol=dim(z)[2], dimnames=list(rownames(z)[1],
colnames(z))
Thanks a lot for the reply.
1. How I compare the lars v.s. lm
Your understand is correct.
2. Answer to my question.
I guess in each step of lars, the coeff are not the same as the lm function
returns using the same "model", because the coeff of lars are the
"accumulated steps" towards the s
I have a quick question, and I apologize in advance if, in asking, I
expose my woeful ignorance of R and its packages. I am trying to use
the bootcov function to estimate the standard errors for some
regression quantiles using a cluster bootstrap. However, it seems that
bootcov passes arguments tha
Hi Greg,
Thanks, very encouraging: with my example, this is 10x more efficient
than my loop:
utilisateur système écoulé
13.819 5.510 20.204
utilisateur système écoulé
156.206 44.859 202.150
In real life, I did some work on each file before doin
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Gabor Grothendieck
Sent: Thursday, July 23, 2009 7:33 PM
To: Nair, Murlidharan T
Cc: r-h...@stat.math.ethz.ch
Subject: Re: [R] computing the radius of an arc
See ?draw.arc in the plo
Hello,
I have been trying to plot correctly a graph for 2 month now, with no success.I
want to put a grid, adjusted on the X axis tickers.
Here is the way I build my X-Axis and my grid:
grid(11, NULL, col="grey40")
axis(2)
# build the tickers on the beginning of each monthticks.at <-
seq(IS
Hi
Mr Derik wrote:
Thank you for your help.
I've read all the documentation I can find and I still can't get this to
work.
postscriptFonts()
in my console produces a list of fonts already mapped yes?
one of which is:
$ComputerModernItalic
$family
[1] "ComputerModernItalic"
$metrics
[1] "C
What you are reporting is that there are two duplicated dates in your
data. 'duplicated' returns a logical vector that is TRUE for the
second and subsequent duplicates. Notice what is returned:
> x <- c(1,2,2,3,4,4,5,6,4,7,3)
> x[duplicated(x)]
[1] 2 4 4 3
>
On Thu, Jul 23, 2009 at 8:50 PM, T
If by "problem" you mean the problem of determining how, in general,
you should proceed perhaps you need an introductory guide on R
and time series such as Cowpertwait's book.
On Thu, Jul 23, 2009 at 10:37 PM, Hongwei Dong wrote:
> Hi, Gabor, it seems ARIMA model does not have that problem. For ex
Hi, Gabor, it seems ARIMA model does not have that problem. For example:
set.seed(123)
y<-ts(c(1:20))
x = ts(rnorm(20))
z = ts(rnorm(20))
tt<-ts(cbind(x, lag(x,-1),lag(x,-2),z))
fit <- arima(y[1:15],order=c(1,0,0),xreg=tt[(1:15),])
fit
pred <- predict(fit, n.ahead=5,tt[(16:20),])
pred
What do you
See ?draw.arc in the plotrix package to draw an arc.
On Thu, Jul 23, 2009 at 10:13 PM, Nair, Murlidharan T wrote:
> Hi!!
>
> I am interesting in computing the radius of an arc that best approximates a
> curve. Is there an R function that I can use to draw an arc?
> Nothing useful came up when I s
Hi!!
I am interesting in computing the radius of an arc that best approximates a
curve. Is there an R function that I can use to draw an arc?
Nothing useful came up when I searched help.search. Does anyone have any
suggestion to do this?
Thanks ../Murli
___
Thanks very much :)
Daniel Malter wrote:
>
> length(datasetname$Events[datasetname$Length<28 &
> datasetname$Method%in%c(1,2,3) ...additional conditions... , ]
> returns
> the number of rows
> sum(datasetname$Events[datasetname$Length<28 &
> datasetname$Method%in%c(1,2,3) ...additional con
Hi,
On Jul 23, 2009, at 7:30 PM, Lars Bishop wrote:
Dear experts,
I'm new in R and trying to learn by writing a version of the
Perceptron
Algorithm. How can I tell in the code below to stop the iteration
when the
condition in the "for loop" is not satisfied for all training
examples?
T
Try this:
library(dyn)
set.seed(123)
tz <- zoo(cbind(Y = 0, x = rnorm(10), z = rnorm(10)))
# simulate values
for(i in 2:10) {
tz$Y[i] <- with(tz, 2*Y[i-1] + 3*z[i] +4* x[i] + 5*x[i-1] + rnorm(1))
}
# keep copy of tz to compare later to simulated Y's
tz.orig <- tz
# NA out Y's that are to be p
What I want R to do is to use the estimated Y at t-1 to be the lag(Y,-1) in
the forecast equation for time t. Is there anyway I can realize this with R?
For example, when the Y value for year 18 is forecast, the estimated Y for
year 17 is used, not the actual Y for year 17 already in the data.
Than
length(datasetname$Events[datasetname$Length<28 &
datasetname$Method%in%c(1,2,3) ...additional conditions... , ] returns
the number of rows
sum(datasetname$Events[datasetname$Length<28 &
datasetname$Method%in%c(1,2,3) ...additional conditions... , ] returns
the sum of the Events variable
>
> Here's a way to get to your solution, but it's not very pretty:
>
> testdfr <- data.frame(POB=c("Oregon","Oregon","Oregon","New
> York","California","California"))
>
> nstates <- length(unique(testdfr$POB))
> testdfr$ POBR <- c(nstates:1)[table(testdfr$POB)][testdfr$POB]
Hmm I will hav
Dear list,
I just had a function (as.ltraj in Adehabitat) give me the following error:
"Error in as.ltraj(xy, id, date = da) : non unique dates for a given burst"
I checked my dates and got the following:
> dupes<-mydata$DateTime[duplicated(mydata$DateTime)]
> dupes
[1] (07/30/02 00:00:00) (
You can't remove Y since its in the rhs of your model.
On Thu, Jul 23, 2009 at 8:25 PM, Hongwei Dong wrote:
> Thanks, Gabor. Here are the problems I'm trying to solve.
> FIRST, I run this to simulate a 20 years time series process. The data from
> 1-15 years are used to estimate the model, and thi
Thanks, Gabor. Here are the problems I'm trying to solve.
*FIRST*, I run this to simulate a 20 years time series process. The data
from 1-15 years are used to estimate the model, and this model is used to
predict the year from 16-20. The following script works.
set.seed(123)
tt <- ts(cbind(Y = 1:2
Another situation would be if you have comment characters in strings that
are intended to be content.
On Thu, Jul 23, 2009 at 5:35 PM, Greg Snow wrote:
> Some programs quote everything to be "safe", others only quote when needed.
> The only case that I know of that read.table and friends require
Dear R-Users,
I have a huge data set (172,696 rows) that is set out in the following way:
Area | Length | Species | Method | Date | Events
I want to sum the number of Events if:
Length < 28AND
Species is NOT "X" ,"Y", or "Z" AND
Method is either "1
Try
enter at the R console: help.start()
and then when the help comes up in your browser click on Packages
and then click on the package you want
and then click on the help file you want
On Thu, Jul 23, 2009 at 3:30 PM, Steven McKinney wrote:
>
> Hi all,
>
> Is there a way to navigate directly to
Perfectly explained Ted. One might, at first reflection, consider that
simply repeating the values 7 through 12 and sampling (w/o replacement)
from among the 18 resulting values, would be similar to just doubling the
selection probabilities for 7 through 12 and then sampling. That would
clearly
Dear experts,
I'm new in R and trying to learn by writing a version of the Perceptron
Algorithm. How can I tell in the code below to stop the iteration when the
condition in the "for loop" is not satisfied for all training examples?
Thanks in advance for your help!
## Generate a linearly separa
I have just started using ggplot2 and I seem to be doing something stupid
in writing ggplot2 commands on more than one line.
In the example below the commands on one line are working fine, but as
soon as I put them on two lines I get an error. Can any one point out
what I am doing wrong? It must
On 23-Jul-09 22:16:39, Thomas Lumley wrote:
> On Thu, 23 Jul 2009 ted.hard...@manchester.ac.uk wrote:
>
>> The general problem, of sampling without replacement in such a way
>> that for each item the probability that it is included in the sample
>> is proportional to a pre-assigned weight ("sampli
Another approach that might be worth trying is to
create an empty data frame with lots and lots of
rows before looping, and then replace rather than
append. Of course, this requires knowing at least
approximately how many rows total you will have.
This suggestion comes from the help page for
Here's a way to get to your solution, but it's not very pretty:
testdfr <- data.frame(POB=c("Oregon","Oregon","Oregon","New
York","California","California"))
nstates <- length(unique(testdfr$POB))
testdfr$ POBR <- c(nstates:1)[table(testdfr$POB)][testdfr$POB]
greetings,
Remko
---
Hi all,
I am using ACS micro data (PUMS) with one of the columns as a
factor for the place of birth (POBPF). I would like to create
a column (POBR) containing a rank
corresponding to the place of the observation
in the POBPF rankings. For example,
if a person is from Oregon, Oregon is
the
After you fix your data frame and if you don't using 2 packages, you might
try something like:
lib(plyr) #for 'by' processing
lib(Hmisc) # for its wtd.mean function
d=data.frame(x=c(15,12,3,10,10),g=c(1,1,2,2,3),w=c(2,1,5,2,5)) ; d
ddply(d,~g,function(df) wtd.mean(df$x,df$w))
milton ruser wrot
On Thu, 23 Jul 2009 ted.hard...@manchester.ac.uk wrote:
The general problem, of sampling without replacement in such a way
that for each item the probability that it is included in the sample
is proportional to a pre-assigned weight ("sampling with probability
proportional to size") is quite tr
Indeed, Jim! And that's why I said to read carefully what is said about
"prob" in '?sample':
If 'replace' is false, these probabilities are applied
sequentially, that is the probability of choosing the
next item is proportional to the weights amongst the
remaining items.
Whereas, if you r
Something like this (untested) may work for you:
> for (i in 1:132) {
+ a <- clim[ , , i]
+ nm <- sprint('clim%03d.txt',i)
+ write.table(a,nm)
+ }
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Origi
On Jul 23, 2009, at 4:18 PM, Alexis Maluendas wrote:
Hi R experts,
I need know how calculate a weighted mean by group in a data frame.
I have
tried with aggragate() function:
data.frame(x=c(15,12,3,10,10),g=c(1,1,1,2,2,3,3),w=c(2,3,1,5,5,2,5))
-> d
aggregate(d$x,by=list(d$g),weighted.mea
On 7/23/2009 5:18 PM, Alexis Maluendas wrote:
> Hi R experts,
>
> I need know how calculate a weighted mean by group in a data frame. I have
> tried with aggragate() function:
>
> data.frame(x=c(15,12,3,10,10),g=c(1,1,1,2,2,3,3),w=c(2,3,1,5,5,2,5)) -> d
> aggregate(d$x,by=list(d$g),weighted.mean,
try your first "reproducible line" first :-)
On Thu, Jul 23, 2009 at 5:18 PM, Alexis Maluendas wrote:
> Hi R experts,
>
> I need know how calculate a weighted mean by group in a data frame. I have
> tried with aggragate() function:
>
> data.frame(x=c(15,12,3,10,10),g=c(1,1,1,2,2,3,3),w=c(2,3,1,
I already posted this on another forum
I kept having the same problem. Turns out the issue was - I set my
pathwrong.
Make sure you have both, the path to R.dll and jri.dll in your path
CORRECTLY, and make sure that you don't put any spaces inbetween values in
the path.
madhura wrote:
>
> The p
Some programs quote everything to be "safe", others only quote when needed.
The only case that I know of that read.table and friends require quotes for is
when a separator is inside of a string, for example if you are using spaces as
the separator and have some names with spaces in them (e.g. "
> -Original Message-
> From: Marc Schwartz [mailto:marc_schwa...@me.com]
> Sent: Thursday, July 23, 2009 2:04 PM
> To: Steven McKinney
> Cc: R-help@r-project.org
> Subject: Re: [R] Navigate to Index page of a package from R command
> prompt
>
> On Jul 23, 2009, at 3:30 PM, Steven McKinney
Try something like (untested):
> mylist <- lapply(all.files, function(i) read.csv(i) )
> mydf <- do.call('rbind', mylist)
If all the csv files are conformable that rbind works on them (if the loop
method works then that should be the case) then this will read in each file,
store the data frames
On Jul 23, 2009, at 3:30 PM, Steven McKinney wrote:
Hi all,
Is there a way to navigate directly to the "Index" page of help
for a package?
Here's my connundrum:
I download and install package "foo".
I don't know what functions are in package "foo",
so I can't invoke the help for package "foo
Hi R experts,
I need know how calculate a weighted mean by group in a data frame. I have
tried with aggragate() function:
data.frame(x=c(15,12,3,10,10),g=c(1,1,1,2,2,3,3),w=c(2,3,1,5,5,2,5)) -> d
aggregate(d$x,by=list(d$g),weighted.mean,w=d$w)
Generating the following error:
Error en FUN(X[[1L]
Having given a lecture on "Numerical Derivatives" just a short time ago, I
would like to mention the following:
Many functions, especially in engineering, are not available as formulas
built simply from arithmetical operators and elementary functions. They are
provided as intricate procedures, a
You are absolutely correct Ted. When no weights are applied it doesn't
matter if you sample with or without replacement, because the probability
of choosing any particular value is equally distributed among all such.
But when they're weighted unequally that's not the case.
It is also interestin
Hi Mary,
Something such as
> plot(1,1)
> title(expression(paste("LBAuo = -", infinity)))
Is this what you're after?
Best
Steve McKinney
From: Mary A. Marion [mms...@comcast.net]
Sent: July 23, 2009 4:23 PM
To: Steven McKinney
Subject: Re: [R] mathemati
Dear all,
I have in my possession a netcdf from which I want to extract some data files.
I have used the "ncdf" package to read the netcdf file and used the
"get.var.ncdf" function to identify the variable i wish to use. The data is in
the form of a time-series of geographical data points that
Hi all,
Is there a way to navigate directly to the "Index" page of help
for a package?
Here's my connundrum:
I download and install package "foo".
I don't know what functions are in package "foo",
so I can't invoke the help for package "foo" via
> ?someFunction
help(package = "foo")
pops up so
Dear Alberto,
Try this:
Res <- apply( bm[,colnames(bm) %in% d], 1, function(x) sum( as.numeric(x) )
)
cbind(bm, Res)
HTH,
Jorge
On Thu, Jul 23, 2009 at 3:31 PM, Alberto Lora M wrote:
> Dear R project group
>
> I have the following problem
>
> Let suppose the following data:
>
>
>
> b<-c("0",
OOPS! The result of a calculation below somehow got omitted!
(325820+326140+325289+325098+325475+325916)/
(174873+175398+174196+174445+173240+174110)
# [1] 1.867351
to be compared (as at the end) with the ratio 1.867471 of the
expected number of "weight=2" to expected number of "weight=1"
Thank you very much for the reply. I checked the rows and it was the
unbalanced " quote marks in some of the rows that caused the problem. Once I
disabled quoting altogether, the problem is solved.
I have one more basic question. I disabled quoting when loading the file to
R, and all the columns
Hi,
On Jul 22, 2009, at 6:18 PM, lulu9797 wrote:
The returned values of lars function include R squares along the
variable
selection path.
Correct.
However, such values are always slightly different from the
R squares returned by the regression function lm using the same
models.
Anyone
Try:
> cbind( bm, res=apply(bm[,d],1 , sum) )
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Alberto Lora M
Hi,
I often have to do this:
select a folder (directory) containing a few hundred data files in csv
format (up to 1000 files, in fact)
open each file, transform some character variables in date-tiime format
make into a dataframe (involves getting rid of a few variables I don't
need
conc
Dear R project group
I have the following problem
Let suppose the following data:
b<-c("0","1","1","1","0","1","0","0","1","0","1","1","1","0","1","1","0","1","1","0")
bm<-matrix(b,ncol=4)
colnames(bm)<-c("F1", "F2", "F3", "F4")
d<-c("F1","F4")
For the matrix bm i need to create a fifth colum
On 23-Jul-09 17:59:56, Jim Bouldin wrote:
> Dan Nordlund wrote:
> "It would be necessary to see the code for your 'brief test'
> before anyone could meaningfully comment on your results.
> But your results for a single test could have been a valid
> "random" result."
>
> I've re-created what I did
This problem is due to a bug in the iterators package.
It has been fixed in version 1.0.2 of iterators, which is
on CRAN now.
- Daya
Daya Atapattu
Revolution Computing
Olivier ETERRADOSSI wrote:
Many thanks David for making the connection betweeen the two reports.
If I understand, other frenc
Marc,
Thanks,
sapply(ls(pat = "^name"),get)
was exactly what I was after. The default behavior for vectors of
equal length is nice too, but I was most interested in the ragged
case, which produces a list.
url:www.econ.uiuc.edu/~rogerRoger Koenker
emailrkoen...@uiu
Best thing is to test it out on simulated data for which you
already know the answer.
library(dyn)
set.seed(123)
DF <- data.frame(x = rnorm(10), z = rnorm(10))
DF$Y <- 0
for(i in 2:10) {
DF$Y[i] <- with(DF, 2*Y[i-1] + 3*z[i] +4* x[i] + 5*x[i-1] + rnorm(1))
}
DF.zoo <- do.call(merge, lapply(DF, z
There are a couple of options:
The help page for lapply also includes the help for sapply and sapply has a
USE.NAMES argument that may do what you want (specify simplify=FALSE to force
the same behavior as lapply).
You can post specify the names like:
> names(mylist) <- vector.of.names
Do eit
On Jul 23, 2009, at 9:19 AM, roger koenker wrote:
This is an attempt to rescue an old R-help question that apparently
received
no response from the oblivion of collective silence, and besides I'm
also
curious about the answer
From: Griffith Feeney (gfee...@hawaii.edu)
Date: Fri 28 Jan 2000
Well one quick way (for non-generics) is the 'args' function:
> args(sample)
function (x, size, replace = FALSE, prob = NULL)
NULL
A similar line appears near the top of the help page when you do '?sample'.
The "replace = FALSE" in the line above means that false is the default (with
the assu
On Thu, Jul 23, 2009 at 11:04 AM, Mark Knecht wrote:
> Hi,
> I'm having trouble within my function CalcPos to get it to call
> CalcHorz with values from each row. I *think* it's calling CalcHorz
> with the final values of the inputs and not the values from each row.
> How can I do this properly i
This is an attempt to rescue an old R-help question that apparently
received
no response from the oblivion of collective silence, and besides I'm
also
curious about the answer
From: Griffith Feeney (gfee...@hawaii.edu)
Date: Fri 28 Jan 2000 - 07:48:45 EST wrote (to R-help)
Constructing lis
Hi Jean-Paul,
>> However, I've tried both solutions on my model, and I got different
>> residuals :...
>> What could be the difference between the two?
There is no difference. You have made a mistake.
##
tt <- data.frame(read.csv(file="tt.csv", sep="")) ## imports your data set
T.aov <- aov(PH
On Thu, Jul 23, 2009 at 12:36 PM, Kingsford
Jones wrote:
>
> As you might have suspected from the lmer t-value close to 0, the
> associated p-value is about .5.
you can ignore the sentence above -- it's a two sided test and the
t-value is not that close to 0...
>
> hth,
>
> Kingsford Jones
>
>>
Thanks Greg, that most definitely was it. So apparently the default is
sampling without replacement. Fine, but this brings up a question I've had
for a bit now, which is, how do you know what the default settings are for
the arguments of any given function? The HTML help files don't seem to
ind
Hi, Gabor, got it. Thanks a lot.
I have one more question about how the "predict" function works here,
especially for the lag(Y,-1) part.
In my model, I assume I know predictors x and z in the next two years, and
use them to predict Y. For each forecast step at time t, the lag(Y,-1) in
the model s
How's this?
f2 <- function(v) {
n <- length(v)
s <- matrix(0,n,n)
for(i in 1:n)
for(j in 1:i)
s[j,i] <- ifelse(i>1, s[j,i-1]+v[i]*ifelse(j>1, s[j-1,i-1], 1),
v[i])
c(1,s[,n])
}
> system.time(f2.result <- f2(1:20))
user system elapsed
0 0 0
> system.time(f
Mark,
My example is essentially identical to Jorge's. This is a good opportunity to
compare two solutions to a problem, one using "for" loops, and one using the
apply family of functions. Compare this with Daniel's solution.
## BEGIN EXAMPLE
## sample list of data.frames, different number o
On Thu, Jul 23, 2009 at 9:02 AM, John Sorkin wrote:
> R 2.8.1
> Windows XP
>
> I am trying to analyze repeated measures data (the data are listed at the end
> of this Email message) and I need help to make sure that I have properly
> specified my model, and would like to know why lmer does not re
Try adding replace=TRUE to your call to sample, then you will get numbers
closer to what you are expecting.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailt
Hi Erin,
On Jul 23, 2009, at 2:12 PM, Erin Hodgess wrote:
Dear R People:
I'm having trouble with something that should be very simple.
I'm setting up a matrix outside of a loop and writing items into it
during the loop.
Here is the output:
glob3b("sites.info")
dim 27 3
[1] "/raid1/osg-app
Hi Mark, study the following example. The two simulated dataframes are put
in a list called listdata. The loop iterates throught the elements in the
list and multiplies the last column "listdata[[i]][,length(listdata[[i]])]"
by the column before the last "listdata[[i]][,length(listdata[[i]])-1]".
T
Dear Mark,
Try this:
lapply(yourlistofdataframes, function(d){
k <- ncol(d)
d$product <- d[,k-1] * d[,k]
d
}
)
HTH,
Jorge
On Thu, Jul 23, 2009 at 2:05 PM, Mark Na wrote:
> Hi R-helpers,
>
> I have a list
Dear R People:
I'm having trouble with something that should be very simple.
I'm setting up a matrix outside of a loop and writing items into it
during the loop.
Here is the output:
> glob3b("sites.info")
dim 27 3
[1] "/raid1/osg-app"
Error in xy[i, ] : incorrect number of dimensions
Here is
Hi R-helpers,
I have a list containing 10 elements, each of which is a dataframe. I wish
to add a new column to each list element (dataframe) containing the product
of the last two columns of each dataframe.
I'd appreciate any pointers, thanks!
Mark Na
[[alternative HTML version deleted
Hi,
I'm having trouble within my function CalcPos to get it to call
CalcHorz with values from each row. I *think* it's calling CalcHorz
with the final values of the inputs and not the values from each row.
How can I do this properly in R?
The values aa,bb,cc,dd are inputs. CalcPos first calc
Dan Nordlund wrote:
"It would be necessary to see the code for your 'brief test' before anyone
could meaningfully comment on your results. But your results for a single
test could have been a valid "random" result."
I've re-created what I did below. The problem appears to be with the
weighting
Please provide your code in a reproducible form (as requested previously).
Here we fit with first 9 points and then add a point for prediction. (Of
course your model can only predict the current value of Y so you may
have to rethink your model even aside from the implementation if you
really want
Hi all
Can anybody help me with this? I am trying to include in an automatic way
the argument in arg.names in a barplot. I generate the labels I want to
appear below the bars with a for loop, and they contain subscripts, so I
need to use expression
anch<-0.05
esp<-4
for (i in 1:dim(Ntot)[1
I vote for it.
--- On Thu, 7/23/09, Marc Schwartz wrote:
> From: Marc Schwartz
> On Jul 23, 2009, at 8:59 AM, Greg Snow wrote:
>
> > Doing the computations in R then the graphs in Excel
> reminds me of the maxim:
> >
> > Measure with a micrometer
> > Mark with chalk
> > Cut with an ax
>
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Jim Bouldin
> Sent: Thursday, July 23, 2009 9:49 AM
> To: ted.hard...@manchester.ac.uk; ted.hard...@manchester.ac.uk; r-h...@r-
> project.org
> Subject: Re: [R] error message: .Ran
Hi, Gabor and Other R users,
I'm re-posting my script and the results I got.
here is the dynamic model I used to estimate in-sample model (1996-2006)
and it works:
fit<-dyn$lm(Y~lag(Y,-1)+z+x+lag(x,-1)+lag(x,-2)+lag(x,-3)+lag(x,-4))
Then I used this model to do out sample forecast with t
>
> Hi all,
>
> I need some help about how to calculate w in a SVR in package e1071.
>
> I have a regression y_i=f(x_i)+e
>
> where f(*x*)=(w,phi(x))+b
>
> then go on with the SVR calculation I know that w*=Sum_i=1^n [(á_i -
> á*_i)K(x,x_i) ] where á_i and á*_i are the lagrangian multipliers of the
Thanks much Ted. I actually had just tried what you suggest here before
you posted, and resolved the problem. Thanks also for the other tips. I
wrote x = as.vector(c(1:12)) because I thought that the mode of x might be
the problem, the error message pointing to .Random.seed notwithstanding.
On
I am not sure that I fully understand what all you want to do (and I don't
understand why you need the correlation and if a correlation based on 3 pairs
is even meaningful), but here is a first stab at what you are trying to do:
tmp <- "Species Control_CR Damage_DR
A 10 2
A 9 3
A 7 4
A 9 2
A 8 3
Hello,
I am trying to create a heatmap that clusters based on a k-means scheme
rather than a hierarchical clustering scheme.
Suppose I have the following input data, located in sample.table:
x1 x2 x3 x4
x1 17.198 16.306 16.806 16.374
x2 14.554 10.866 15.780 14.596
x3 14.374 14.118 14.569 17.352
x
here is the solution
x <- subset(read.csv("foo.csv"), VALUE>0)
On Thu, Jul 23, 2009 at 11:28 AM, stephen sefick wrote:
> I have a csv file that I am trying to read in and know that values <0
> are erroneous - is there a way to read only value grater than 0.
>
> thanks,
>
> --
> Stephen Sefick
>
>
I have a csv file that I am trying to read in and know that values <0
are erroneous - is there a way to read only value grater than 0.
thanks,
--
Stephen Sefick
Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us
On 23-Jul-09 16:08:17, Jim Bouldin wrote:
>> Jim Bouldin wrote:
>> > Thank you. However, when I tried that, I got this message:
>>
>> > Warning message:
>> > In rm(.Random.seed) : variable ".Random.seed" was not found
>>
>> In that case, have you attached some package that has its own
>> .Random
On 23-Jul-09 15:30:05, Jim Bouldin wrote:
> I'm trying to run this simple random sample procedure and keep
> getting the error message shown. I don't understand this; I've
> Thanks.
>
>> x = as.vector(c(1:12));x
> [1] 1 2 3 4 5 6 7 8 9 10 11 12
>> mode(x)
> [1] "numeric"
>> sample(x, 3)
>
>
> Jim Bouldin wrote:
> > Thank you. However, when I tried that, I got this message:
> >
> > Warning message:
> > In rm(.Random.seed) : variable ".Random.seed" was not found
>
>
> In that case, have you attached some package that has its own
> .Random.seed?
> Try to find where the current
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