Hi,
I am wondering if anyone knows of an easy way to fit a saturated model
using the sem package on raw data? Say the data were:
mtcars[, c(mpg, hp, wt)]
The model would estimate the three means (intercepts) of c(mpg,
hp, wt). The variances of c(mpg, hp, wt). The covariance
of mpg with hp
On 07/12/2012 01:39 AM, Duncan Murdoch wrote:
On 12-07-11 2:34 PM, Jonas Stein wrote:
Take a look at the predicted values at your starting fit: there's a
discontinuity at 0.4, which sure makes it look as though overflow is
occurring. I'd recommend expanding tanh() in terms of exponentials and
Dear list,
I'm quite confused when interpreting results from a mixed linear model.
For example, working on Iris data frame, I want to know the effect of
species on slope of the model Petal.Length~Sepal.Length
I write this :
data(iris)
reg01 - lm(Petal.Length~Sepal.Length +
Hi,
Try this: (a variant of andrija)
testct-table(test)
subset(test,!is.na(match(test,as.integer(names(testct[testct%in%max(testct)])
[,1]
[1,] 1
[2,] 1
[3,] 1
[4,] 1
[5,] 7
[6,] 7
[7,] 7
[8,] 7
A.K.
- Original Message -
From: Amanduh320
Hi,
Try this:
func1-function(x,y,z)
{ifelse(is.na(y[[x]]),z[[x]],y[[x]])}
dat3-data.frame(lapply(colnames(df1),function(x) func1(x,df1,df2)))
colnames(dat3)-colnames(df1)
dat3
cola colb colc cold cole
1 1.4 5.0 9.00 1.6 17.0
2 1.4 6.0 0.02 14.0 0.6
3 3.0 0.8 11.00 15.0 19.0
4
Hi,
Here i have a matrix like this,
OLDMatrix -
X1 X2X3
- ----
22 24 23
25 27 27
10 13 15
the thing is,
im running two function(SUM,COUNT) to get output in another matrix called
NEWMatrix
NEWMatrix -
On 12/07/2012 07:10, Dimitris.Kapetanakis wrote:
Dear all,
I created a function which it has many arguments i.e. Mu(x, Y, B, X, P)
where Mu is the function. This function works perfect but when I am trying
to use it in the apply() function in the manner apply(Matr, 1, Mu, Y=y, B=b,
X=x, P=p) it
Dear Mr. Holtman and especially dear Rui,
thank you VERY much.
You helped me a lot!
I've just added the following:
rsort - ratios[order(ratios$vpNum),]
Now the test subjects are arranged according to their vpNum.
Thanks a lot again!
--
View this message in context:
Hi all, may i know is it possible to plot a graph by first letter?
for example:
Name: Age:
Angel20
Amelia 20
Bernard 19
Stephanie 20
Vanessa 22
Angeline 23
Camel
Michael, Mikhail
Many thanks for your helpful comments. My faith in community support continues
to grow.
Michael: I'm looking to use some sort of flexible spline-like fit
(smooth.spline, lowess etc).
Many thanks for sharing your expertise. I actually cross posted this on to the
manipulatr
I have developed a package for conjoint analysis in R, you may use package
faisalconjoint.
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Sent from the R help mailing list archive at Nabble.com.
Hi,
I need to read source code of function as a string vector or matrix.
if i am using get method it return as closure type.
example
readcsv-function(filepath)
{
output-read.csv(filepath)
}
how to get as string vector like this
test[1]=readcsv-function(filepath)
test[2]={
On 7/12/2012 11:00 AM, purushothaman wrote:
Hi,
I need to read source code of function as a string vector or matrix.
if i am using get method it return as closure type.
example
readcsv-function(filepath)
{
output-read.csv(filepath)
}
how to get as string vector like this
On 07/11/2012 05:38 PM, Rantony wrote:
Hi,
Here i have an matrix like this,
ABCPQRXYZ MNO
-- --- --
367 15
2 122415
20 5 1 2
25 50 15 35
i need to get the
Hi,
readLines read a file and retrun as vector.
i need to read particular function in Rfile.
Thanks
B.Purushothaman
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Sent from the R help mailing
On Jul 11, 2012, at 20:34 , Jonas Stein wrote:
Take a look at the predicted values at your starting fit: there's a
discontinuity at 0.4, which sure makes it look as though overflow is
occurring. I'd recommend expanding tanh() in terms of exponentials and
rewrite the prediction in a way
Hi everybody,
I try to simulate random numbers from a trivariate nested Archimedean
copula. My aim is to correlate two processes with, e.g. theta2, as the so
called child pair and then to correlate these two processes with a third one
with theta1 (parent). This figure tries to capture what I am
This looks a bit like homework, and makes no attempt whatsoever to follow
the posting guide or present previous work. But still, I'd start with
?grepl, and move on to use rseek.org to look for functions for whatever
kind of plot you need to make. The R Graph Gallery is also a good resource.
And
Hi Jim,
Thanks alots for helping me. Is it possible to get the any percentile
without using library?
Thanks
-Antony.
From: Jim Lemon [via R]
[mailto:ml-node+s789695n4636272...@n4.nabble.com]
Sent: Thursday, July 12, 2012 2:46 PM
To: Akkara, Antony (GE Energy, Non-GE)
Subject: Re:
I want to use the caret package and found out about the timingSamps
obtion to obtain the time which is needed to predict results. But, as
soon as I set a value for this option, the whole model generation fails.
Check this example:
-
library(caret)
Dear R users,
if all my numerical variables in my datasets having the same units, may I
leave them unnormalized, just do cv.glmnet
directly(cv.glmnet(data,standardize=FALSE))?
i know normally if there is a mixture of numerical and categorical , one has
to standardize the numerical part before
Dear Eik,
yes, with the memisc package I was able to import only a few variable of
the SPSS dataframe!
Thank you very much for your help!
Marion
2012/7/5 Eik Vettorazzi e.vettora...@uke.de
Hi Marion,
package memisc does what you want, just have a look at ?importers.
Hth
Eik
Am 05.07.2012
Hello!
I'm having some problems with
pit function in ensembleBMA. When I run on mu data I have the
same warnings as below, when I type the example(pit).
Any suggestion on how to resolve it?
library(ensembleBMA)
Loading required package: chron
example(pit)
pit data(ensBMAtest)
pit
Eli,
I think that this should be possible. If your integration allows you to
express the salinity as a weighted sum of evaluated values of the
spatial smooth function (where the weights are known), then you could
use the summation convention for smooths described in
?linear.functional.terms.
Hi,
How to get all list item to one string variable.
example
a[1]=abc
b[1]=def
output=abc def
Thanks
B.Purushothaman
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Sent from the R help mailing list archive at
it's working
Thank You so much
B.purushothaman
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__
With paste().
On Thursday, July 12, 2012, purushothaman wrote:
Hi,
How to get all list item to one string variable.
example
a[1]=abc
b[1]=def
output=abc def
Thanks
B.Purushothaman
--
View this message in context:
hi,
sorry it's not 2 different list all item in same list like this
a[1]=abc
a[2]=def
...
output =abc def ...
Thanks
B.Purushothaman
--
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Sent from the R help
Gary Dong pdxgary163 at gmail.com writes:
Dear R users,
I have created a Loess surface in R, in which x is relative longitude by
miles, y is relative latitude by miles, and z is population density at the
neighborhood level. The purpose is to identify some population centers in
the region.
Try using the rbind() function to combine the two vectors from colSums()
into a matrix. Then assign row names and get rid of column names using
the dimnames() function. For example:
OLDMatrix - matrix(c(22, 25, 10, 24, 27, 13, 23, 27, 15),
ncol=3, dimnames = list(NULL, c(X1, X2,
Try this...
a-c(abc,def,ghi,klm,nop,qrs,tuv,wxyz)
b-data.frame()
for (i in 1:length(a)){
b-paste(b,a[i],sep=)
}
print(b)
Thanks
Arun
---
--
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Sent from the R help
Hi,
What you did is equivalent to, but more complicated (and slower as well
I guess) than:
paste(a, collapse=)
Ivan
--
Ivan CALANDRA
Université de Bourgogne
UMR CNRS/uB 6282 Biogéosciences
6 Boulevard Gabriel
21000 Dijon, FRANCE
+33(0)3.80.39.63.06
ivan.calan...@u-bourgogne.fr
On Thu, Jul 12, 2012 at 05:22:45AM -0700, purushothaman wrote:
hi,
sorry it's not 2 different list all item in same list like this
a[1]=abc
a[2]=def
...
output =abc def ...
Hi.
Try this
a - list(abc, def, ghi)
paste(a, collapse= )
[1] abc def ghi
Hope this helps.
Petr
Hi!
I would like to post the following question:
I was trying to figure out how to do the simulation shown in Fig 10.6 of
John Verzani's book 'Using R for Intro Statistics'. It is on page 290, with
a description on the previous page. It seems like a simple thing... Just
needing to duplicate a
Thanks jean.
From: Jean V Adams [via R]
[mailto:ml-node+s789695n4636289...@n4.nabble.com]
Sent: Thursday, July 12, 2012 6:14 PM
To: Akkara, Antony (GE Energy, Non-GE)
Subject: Re: Add row into a Matrix witout headers from Function
Try using the rbind() function to combine the two vectors
Hello,
I am using read.csv.sql first time for reading the large data file.If I am
ran this code that showns warning “closing unused connection”.
Is it I am missing any argument from my command or how to comeout from this
warning?.
R code:-
Library(sqldf)
ip_dir_path -
Hello,
The reason you are only getting one row is because each time you are replacing
the NEWMatrix with new output.
Try this:
Oldmatrix-read.table(text=
X1 X2 X3
22 24 23
25 27 27
10 13 15
,sep=,header=TRUE)
NewMatrix1-rbind(
Hi,
If I understand it correctly, probably this is what you want:
a-list(abc,def,ghi)
a1-unlist(a)
paste(a1[1],a1[2],a1[3],sep= )
[1] abc def ghi
A.K.
- Original Message -
From: purushothaman purushothama...@ge.com
To: r-help@r-project.org
Cc:
Sent: Thursday, July 12, 2012 8:22 AM
Hi Jean,
If we convert the NEWMatrix to dataframe,then headers do appear.
NewMatrix1-data.frame(NewMatrix)
NewMatrix1
X1 X2 X3 X4
SumMatrix SUM 57 64 65
CountMat COUNT 3 3 3
#assigning headers to NULL
colnames(NewMatrix1)-NULL
NewMatrix1
NA NA NA NA
Hi Manning,
There are two obvious mistakes:
- close the brackets for replicate() and
- do the things in the correct order
replicate(100, {
x=rep(1:10,10) ## first define x
y=rnorm(100,x,5) ## and then y because it depends on x
plot(y~x) ## then plot, because it depends on x
... and you are unlikely to get a helpful reply until you follow the
posting guide and post code that shows what you did. Unless there's a
claiRvoyant package out there somewhere to figure it out.
-- Bert
On Thu, Jul 12, 2012 at 3:43 AM, yan y.j...@ucl.ac.uk wrote:
Dear R users,
if all my
Post on the r-sig-mixed-models list, not here.*
-- Bert
* Or consult a local statistician, as your problem appears to be
insufficient statistical knowledge, rather than R-related. In particular,
fitting a (mixed effect) linear model without an intercept is almost always
unwise.
On Thu, Jul 12,
Thanks! For some reason, example 5 in that documentation doesn't work for me.
It runs the query, but I'm getting zero results. If I substitute the
variable with the actual value, I do get the results I want. Is there a
reason for this?
--
View this message in context:
Hi,
I have a question on testing spatial autocorrelation on raster data
including NA-values. In particular, I like to calculate Moran´s I and
Geary´s C indices by using inverse distance weighting matrices.
Calculating Moran´s I with moran.test works fine, because the function
contains the
Duncan has given a indication of why nls() has troubles, and you have found a
way to work
around the problem partially. However, you may like to try nlmrt (from R-forge
project
R-forge.r-project.org/R/?group_id=395
It is intended to be very aggressive in finding a solution, and also to deal
I think that you have the statements out of order and I know that you are
lacking the last ) to match the replicate(
Try this
replicate(100, {
x=rep(1:10,10)
y=rnorm(100,x,5)
plot(y~x)
abline(lm(y~x))
})
John Kane
Kingston ON Canada
-Original Message-
From: aldoushuxle...@gmail.com
I really am not sure of the question but perhaps ?order for a start?
John Kane
Kingston ON Canada
-Original Message-
From: jubil...@live.com.sg
Sent: Thu, 12 Jul 2012 01:15:26 -0700 (PDT)
To: r-help@r-project.org
Subject: [R] plot graph by first letter
Hi all, may i know is it
?subset
John Kane
Kingston ON Canada
-Original Message-
From: jubil...@live.com.sg
Sent: Wed, 11 Jul 2012 20:53:51 -0700 (PDT)
To: r-help@r-project.org
Subject: [R] plot only a same variable timing graph
hi all.
for example :
Table 1 variable:
BERTH TIME
A
On Thu, Jul 12, 2012 at 9:41 AM, scstrein scstr...@ncsu.edu wrote:
Thanks! For some reason, example 5 in that documentation doesn't work for me.
It runs the query, but I'm getting zero results. If I substitute the
variable with the actual value, I do get the results I want. Is there a
reason
You could read the .png (or some other formats) in then use the
rasterImage function to put that into the current plot.
On Mon, Jul 9, 2012 at 8:25 PM, Jie Tang totang...@gmail.com wrote:
hi R-users
Now I have a figure in emf or png or tiff format that have been drawn
by other tool and I
?deparse
And please include context -- relatively few of us use Nabble.
Michael
On Jul 12, 2012, at 4:16 AM, purushothaman purushothama...@ge.com wrote:
Hi,
readLines read a file and retrun as vector.
i need to read particular function in Rfile.
Thanks
B.Purushothaman
--
View
Dear Joshua,
If I understand correctly what you want to do, the sem package won't do it.
That is, the sem() function won't do what often is called FIML estimation
for models with missing data. I've been thinking about implementing this
feature, and don't think that it would be too difficult, but
Hello,
There's a package, lavaan, that implements FIML as an option of function
sem(). I have never used it, though, so I can't say much about it.
Hope this helps,
Rui Barradas
Em 12-07-2012 16:20, John Fox escreveu:
Dear Joshua,
If I understand correctly what you want to do, the sem
Hi David,
please make sure that you have the 0.6 version of igraph installed.
You might need to upgrade your R installation to install the 0.6
version of igraph.
Best,
Gabor
On Wed, Jul 11, 2012 at 10:36 AM, David Marx dm...@soundexchange.com wrote:
Hi,
I've installed the igraph package and
Dear R users,
I have a lot of experience with traditional R graphics, but I decided to turn
to trellis as
it was recommended for spatial graphs by the sp package. In traditional R
graphics
I always first set the size of the device region absolute units (e.g. mm) and
then I
firmly fix the
It is not clear to me what you want in the lattice context, but perhaps
?print.trellis
may be relevant.
Note that inner and outer margins are not directly translatable in
lattice, as there are multiple levels of plots involved, all determined by
viewport contexts. Ergo my confusion about what
You can use the locator function to retrieve the user coordinates of a
point that you click on in the plot, then use those points with
rasterImage to add the image. So replace the last 2 lines of
Michael's answer with something like:
barplot(VADeaths, border = dark blue)
tmp - locator(1)
If you want something other than an arrow (or an arrow that looks
different from those produced by the arrows function) then look at the
my.symbols function in the TeachingDemos package.
On Mon, Jul 9, 2012 at 9:37 PM, Manish Gupta mandecent.gu...@gmail.com wrote:
Hi,
I am working on stacked
If you are going to be doing a lot of this then you might want to
consider using logspline density estimates (logspline package) instead
of kernel density estimates.
On Wed, Jul 11, 2012 at 8:33 AM, firdaus.janoos fjan...@bwh.harvard.edu wrote:
Hello,
I wanted to know if there is a simple way
Hi:
Using nls how can I increase the numbers of iterations to go beyond 50.
I just want to be able to predict for the last two weeks of the year.
This is what I have:
weight_random - runif(50,1,24)
weight - sort(weight_random);weight
weightData - data.frame(weight,week=1:50)
See ?nls.control (referenced from ?nls).
On 12/07/2012 18:47, Felipe Carrillo wrote:
Hi:
Using nls how can I increase the numbers of iterations to go beyond 50.
I just want to be able to predict for the last two weeks of the year.
This is what I have:
weight_random - runif(50,1,24)
Read the Help file!
?nls ## Note the control argument
?nls.control
-- Bert
On Thu, Jul 12, 2012 at 10:47 AM, Felipe Carrillo
mazatlanmex...@yahoo.comwrote:
Hi:
Using nls how can I increase the numbers of iterations to go beyond 50.
I just want to be able to predict for the last two weeks
try this:
readcsv-function(filepath)
+ {
+ output-read.csv(filepath)
+ }
readcsv
function(filepath)
{
output-read.csv(filepath)
}
x - capture.output(readcsv)
x
[1] function(filepath) {
output-read.csv(filepath)
[4] }
On Thu, Jul 12, 2012 at 5:00 AM, purushothaman
Hi,
I'm trying to run a permutation test on paired samples.
First I tried the package exactRankTests:
require(exactRankTests)
x - c(1.83,0.50,1.62,2.48,1.68,1.88,1.55,3.06,1.30)
y - c(0.878,0.647,0.598,2.05,1.06,1.29,1.06,3.14,1.29)
wilcox.test(x,y,paired = TRUE,alternative = greater)
Dear R-helpers,
I am puzzled by the following observation:
On my home dual core Windows desktop computer, I am used to running two R
sessions in parallel. These do very well in using the full CPU of the
computer (half each) and don't seem to slow each other down.
Today I have started some large
Hello!
I'm doing a svar and when I make the estimation the next error message
appears:
In SVAR(x, Amat = amat, Bmat = bmat, start = NULL, max.iter = 1000, :
The AB-model is just identified. No test possible.
Could you help me to interpret it please.
Also I have the identification assumption
Hi!
Would be grateful if somebody helped me understand this error message after
trying to run a panel data:
PanelData - read.xls(/Users/Bahman/Desktop/Taylor
rule/Data/PanelData2.xls)
attach(PanelData)
# Panel Data regresson for all countrys.
Answer - plm(NOM.INT.RATE ~ INFL + TARGET.INFL +
Thanks Simon! I can use the functional approach to bridge the gap between
this and more traditional forms of data assimilation.
At the moment, though, what I am doing is related to your work with soap
films, in which your response comes from linearly combine solutions of a PDE
evaluated at the
Hi,
Try this:
dat1-read.table(text=
Name Age
Angel 20
Amelia 20
Bernard 19
Stephanie 20
Vanessa 22
Angeline 23
Camel 21
,sep=,header=TRUE)
I am trying to write a loop to forecast realized volatility over successive
days for the purpose of VaR prediction using the HAR-RV-CJ model which is as
follows:
log(RV_t+1) = β_0 + β_CD log(CV_t) + β_CW log(CV_t-5) + β_CM
log(CV_t-22) + β_JD log(J_t) + β_JW J_t-5 + β_JM J_t-22 + e_t
where
I have a graph of residuals and I am attempting to get a list of the indexes
of each time the residual is greater than 2 standard deviations or less than
-2 standard deviations, but only the first point of the section. And then
I'd also need the first point where the point returns to the range
I am trying to write a loop to forecast realized volatility over successive
days for the purpose of VaR prediction using the HAR-RV-CJ model which is as
follows:
log(RV_t+1) = β_0 + β_CD log(CV_t) + β_CW log(CV_t-5) + β_CM
log(CV_t-22) + β_JD log(J_t + 1) + β_JW log(J_t-5 + 1) + β_JM
Hi William,
Glad to know that.
So, I guess difftime() did the trick if both befong to the class Date.
A.K.
From: William Mabe billm...@gmail.com
To: arun smartpink...@yahoo.com
Sent: Thursday, July 12, 2012 1:12 PM
Subject: Re: [R] unable to subtract dates
Hi Gabor,
I updated my RStudio installation to 0.96.316 (which did not modify my R build
of 2.14.0) and that did it. Thanks for the help!
David Marx
Please note our office has moved:
David Marx | Data Analyst Specialist, Claims Dept | SoundExchange, Inc.
733 10th Street, NW | 10th Floor |
I have a spatial salinity field s and a model g(s) ~ Xb where the X comes from
slightly modified GAM basis functions.
I am trying to deal with the following set of requirements:
1. The underlying physics are linear, and plain salinity (the identity link) is
the correct response to my
Thank you,
I am sorry but I am still trying to figure out how to make the function
work.
I have a column called tUnitsort$BlockNumber which can range from 0 to 6.
I have another two columns with the date and the hour ending for the given
day.
Example
DateHour BlockNumber MyTo
I have a text file like this
2.5 3.6 7.1 7.9
100 3 4 2 3
200 3.1 4 3 3
300 2.2 3.3 2 4
I used r - read.table(a.txt, header=T)
The row names becomes X2.5, X3.6... What I need is the row names are
numeric, so I can use the row names as numbers on
HI,
Much more simplified code that my previous one.
a-list(abc,def,ghi, jkl, mno, pqr)
paste(a[],sep= )
[1] abc def ghi jkl mno pqr
A.K.
- Original Message -
From: purushothaman purushothama...@ge.com
To: r-help@r-project.org
Cc:
Sent: Thursday, July 12, 2012 8:22 AM
Subject: Re:
Thanks Bert, I increased the number of iterations:
M_model - nls(weight ~ alpha +
beta*exp(gamma*week),control=nls.control(maxiter=200), weightData,
start = c(alpha = 0.0, beta = 1, gamma = 0.2), trace = TRUE)
But now the 'start' argument seems to be the problem.
Looking at the
I can reproduce the errors. I'll take a look.
Thanks,
Max
On Thu, Jul 12, 2012 at 5:24 AM, Dominik Bruhn domi...@dbruhn.de wrote:
I want to use the caret package and found out about the timingSamps
obtion to obtain the time which is needed to predict results. But, as
soon as I set a value
On Thu, Jul 12, 2012 at 3:02 PM, Felipe Carrillo
mazatlanmex...@yahoo.com wrote:
Thanks Bert, I increased the number of iterations:
M_model - nls(weight ~ alpha +
beta*exp(gamma*week),control=nls.control(maxiter=200), weightData,
start = c(alpha = 0.0, beta = 1, gamma = 0.2),
I get a different error now:
nls(weight ~ cbind(1, exp(gamma*week)), weightData, start = list(gamma=
0.2), alg = plinear)
Error in nls(weight ~ cbind(1, exp(gamma * week)), weightData, start =
list(gamma = 0.2), :
step factor 0.000488281 reduced below 'minFactor' of 0.000976562
The help
On Thu, Jul 12, 2012 at 3:40 PM, Felipe Carrillo
mazatlanmex...@yahoo.com wrote:
I get a different error now:
nls(weight ~ cbind(1, exp(gamma*week)), weightData, start = list(gamma=
0.2), alg = plinear)
Error in nls(weight ~ cbind(1, exp(gamma * week)), weightData, start =
list(gamma = 0.2),
On Tue, 10-Jul-2012 at 11:19PM +0200, Erdal Karaca wrote:
| german Null == english zero :-)
German Gift == English poison :-(
|
| 2012/7/10 Rolf Turner rolf.tur...@xtra.co.nz
|
|
|
| In addition to taking cognisance of Richard Heiberger's reply you
| should also learn to
Hi all!
I have built a model to predict interactions with turtles and the model
includes an offset for effort:
ZIP-zeroinfl(Sturgeon~fMesh+fSeason+offset(LogEffort),dist=poisson,link=logit,data=data)
I wasn't clear about one aspect of the response to a similar question I
recently posted...I
Hi,
Try this:
dat1-read.table(text=
2.5 3.6 7.1 7.9
100 3 4 2 3
200 3.1 4 3 3
300 2.2 3.3 2 4
,sep=,header=TRUE)
#Either
colnames(dat1)-c(2.5,3.6,7.1,7.9)
#or
colnames(dat1)-c(2.5,3.6,7.1,7.9)
#produce character column names
Another new question:
I want to be able to subset the data based on whether or not that data point
was recorded on a holiday. The is.holiday() function from the chron package
would be perfect for this. However, when I try it, the following happens
(I'm also using the timeDate package):
To add to Gabor's remarks:
This has nothing to do per se with R -- it is your insufficient
understanding of the underlying mathematical issues.
It is pretty trivial to do the plinear algorithm by hand. Fit linear
regressions
weight ~ lm(weight ~z)
where z is exp(gamma*week) for a suitable
just do this:
colnames(r)-substr(colnames(r),2,nchar(colnames(r)))
This will remove the X.
Later when you want to use the headed to plot something, cast it as numeric:
plot(colMeans(r)~as.numeric(colnames(r)))
-
Yasir Kaheil
--
View this message in context:
P.S.: I should have mentioned: The operating system is Windows XP.
--
View this message in context:
http://r.789695.n4.nabble.com/Two-R-sessions-on-multicore-computer-seem-to-inhibit-each-other-tp4636336p4636355.html
Sent from the R help mailing list archive at Nabble.com.
Dear all
I am using lars package to do lasso in R. I dont undesrtand what max.steps
do?and how I can understand from the outputs to obtain the last steps in this
packagethanks for your helpbest
[[alternative HTML version deleted]]
__
If you don't understand the methodology, why are you using it??
Use the glmnet package instead and read the supporting documents. If it's
too technical for you, consult a statistician or do something else.
I do NOT do brain surgery.
-- Bert
On Thu, Jul 12, 2012 at 1:04 PM, SHIR SHIRY
Thank you all for your help.
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish Wildlife Service
California, USA
http://www.fws.gov/redbluff/rbdd_jsmp.aspx
From: Gabor Grothendieck ggrothendi...@gmail.com
To: Felipe Carrillo
I have independent event sequences for example as follows :
Independent event sequence 1 : A , B , C , D
Independent event sequence 2 : A, C , B
Independent event sequence 3 :D, A, B, X,Y, Z
Independent event sequence 4 :C,A,A,B
Independent event sequence 5 :B,A,D
I want to able to
I wrote a little function called first() to help with situations like
this. It returns a 1 every time an element of a vector is different from
the previous element, and a 0 otherwise.
first - function(x) {
L - length(x)
c(1, 1-(x[-1]==x[-L]))
}
sd - 1
residuals - c(1, 2.1, 3, 4,
On Thu, 12 Jul 2012, Lee, Laura wrote:
Hi all!
I have built a model to predict interactions with turtles and the model
includes an offset for effort:
ZIP-zeroinfl(Sturgeon~fMesh+fSeason+offset(LogEffort),dist=poisson,link=logit,data=data)
Note that this includes the offset both in the
Hello,
I've not been following this thread but this seems ndependent from
previous posts. Try the following.
url - http://r.789695.n4.nabble.com/file/n4636337/BR3_2011_New.csv;
tUnitsort - read.csv(url, header=TRUE)
cols - sapply(c(Date, Hour, BlockNumber, MyTo), function(x)
Hi,
I managed to use the attached data set and figure out the following:
flies - read.table(example12_1.dat,header=TRUE,sep=\t)
boxplot(long ~ group,
data = flies,
horizontal = TRUE,
col = red)
I'm very new to R and would like some help with the following:
1. Change the
The example you gave had only one split. If your real situation has three
splits, you'll have to take a look at testtree$csplit matrix and decide
how you want to define the new grouping variable. Here's one way to do it
...
Jean
library(rpart)
library(rpart.plot)
test_set - data.frame(
Looks like data was not attached. Here is is.
longgroup
40 1
37 1
44 1
47 1
47 1
47 1
68 1
47 1
54 1
61 1
71 1
75 1
89 1
58 1
59 1
62 1
79 1
96 1
58 1
62 1
70 1
72 1
74 1
96
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