-- Forwarded message --
From: alexander russell ssv...@gmail.com
Date: Tue, Oct 20, 2009 at 4:34 PM
Subject: Re: [R] When modeling with negbin from the aod package...
To: Matthieu Lesnoff matthieu.lesn...@gmail.com
Hello again,
It seems that, though we have a simple estimate of
Karl Ove Hufthammer wrote:
In article 4addc1d0.2040...@yahoo.de, niederlein-rs...@yahoo.de
says...
In every list entry is a data.frame but the columns are only partially
the same. If I have exactly the same columns, I could do the following
command to combine my data:
do.call(rbind, myList)
Dear Rlers,
in the following dataset I would like to insert a new column that refers to
the data of the previous row.
My question is whether the probability of a female (Id) changes if she had
given birth to a pup in the previous year. So my dataframe consists of the
column Id, year (2003-2007
Hi Mark,
After reviewing the IDE/Script Editors article at sciviews.org, I
wanted to pose a quick question here to see if anyone can offer an
opinion or commentary about GUI editors that can be installed in a
Windoze environment that allow editing/saving of remote .R files and
running R
Hi Giovanni,
Thanks for your reply. I can make the function work after parsing the code
directly into R. The problem arise after compiling the function into a
package and then calling the function, because the files inside source()
seems to be missing.
I tried to include the sourced files in the
Dear all
I am new R user, and trying to learn more.
I am doing linear regression analysis in R with my data. I am trying to find
the way to calculate the slope value (coefficient of x) to degree of slope.
Please give me idea on this.
Thanking you in anticipation
Warm regardMS
summary(ma.dati2$death.status)- censoring
Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
0.0 0.0 0.0 0.05332 0.0 1.0 39.0
##
summary(ma.dati2$time.death)--- time
Min. 1st Qu. MedianMean 3rd Qu.Max.NA's
Only a confusion with two dataset with similar names...
Sorry
Giovanni
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PLEASE do read the posting guide
On 10/21/2009 01:30 AM, jcano wrote:
betav-c(0.78,0.94,0.88,0.41,0.59,4.68)
etav-c(235.6,59.5,31.2,8.7,3.2,1174)
Hi Javier,
Maybe not exactly what you want, but try:
addtable2plot(2,8,rbind(betav,etav),bty=o,
display.colnames=FALSE,display.rownames=TRUE)
using your own x and y
I have three time series, x, y, and z, and I want to analyse the
relations between them. However, they have vastly different
resolutions. I am writing to ask for advice on how to handle this
situation in R.
x is a stimulus, and y and z are responses.
x is a rectangular pulse 4 sec long. Its
On 10/21/2009 01:59 AM, hadley wickham wrote:
It currently works (because I can't figure out how to make it an
error) but you really should not do it.
Please pardon my nomination, Hadley, but that is just too good to pass up.
Jim
__
Hi R users
I used R to get the results of a linear regression
reg-lm(y~x)
here are the results:
# Call:
# lm(formula = donnees$txi7098 ~ donnees$txs7098)
#
# Residuals:
# Min
On 10/21/2009 09:40 AM, mnstn wrote:
Hello All,
My question is regarding the attached plot. I would like to have multiple
transparent green bands running the length (yaxis) of the plot the width of
which is determined by the green lines at y=0 in the plot. Can you suggest a
way to do it?
PS I think one way to get the average waveforms I want from the
analysis is using cross-correlation, but again the multiple scale
problem is present.
On Wed, Oct 21, 2009 at 9:56 AM, William Simpson
william.a.simp...@gmail.com wrote:
I have three time series, x, y, and z, and I want to analyse
Hello,
We are a group of PhD students working in the field of toxicology. Several
of us have small data sets with N=10-15. Our research is mainly about the
association between an exposure and an effect, so preferrably we would like
to use linear regression models. However, most of the time our
I have a dataset that contains numbers between -10 and 0. E.g. x =
c(-9.23, -9.56, -1.40, ...)
If I no do a
qplot(x, geom=histogram)
I get the error:
Error: position_stack requires non-overlapping x intervals
Strangely, the following both work:
qplot(x * -1, geom=histogram)
qplot(x+100,
Hi all R Users,
I am using R in batch mode to do an automatic reporting, I'm saving all the
picture in wmf. When I'm launching manually the batch file(.bat) it's
working but when I'm lauching the batch file from the server I have in the
outputfile the following erro message:
Error in
Dear Maling list,
I have a data.frame with three columns. I want to produce a fourth
column which is result of eliminating the characters present in the
second and third colum from the first.
Example:
a b c
1 f f
2 j h
3
Hi
I am using Ubuntu Hardy, and I installing many packages from source. I am
keeping my R packages fairly up to date.
My question is: is there a way, of keeping the source packages, so that when
I am installing a new version of R, an update.packages(checkBuilt=TRUE) will
only fetch the packages
Hi there,
I am having trouble getting the plotting of multiple time series to work.
I have used RBloomberg to download data, which I then convert to a data
frame. After I have calculated my new index values, I would like to plot the
new index.
My problem is that I can't get the plot feature to
Dear all,
Lets say I have the following data frame:
set.seed(1)
col1 - c(rep('happy',9), rep('sad', 9))
col2 - rep(c(rep('alpha', 3), rep('beta', 3), rep('gamma', 3)),2)
dates - as.Date(rep(c('2009-10-13', '2009-10-14', '2009-10-15'),6))
score=rnorm(18, 10, 3)
df1-data.frame(col1=col1,
In article OF1D427087.01614A71-ONC1257656.003BFAB8-C1257656.003C88F8
@basf-c-s.be, sebastian.roh...@basf.com says...
I have a dataset that contains numbers between -10 and 0. E.g. x =
c(-9.23, -9.56, -1.40, ...)
If I no do a
qplot(x, geom=histogram)
I get the error:
Error: position_stack
aves = aggregate(df1$score, by=list(col1=df1$col1, col2=df1$col2), mean)
results = merge(df1, aves)
b
On Oct 21, 2009, at 9:03 AM, Tony Breyal wrote:
Dear all,
Lets say I have the following data frame:
set.seed(1)
col1 - c(rep('happy',9), rep('sad', 9))
col2 - rep(c(rep('alpha', 3),
On 10/21/2009 7:03 AM, Tony Breyal wrote:
Dear all,
Lets say I have the following data frame:
set.seed(1)
col1 - c(rep('happy',9), rep('sad', 9))
col2 - rep(c(rep('alpha', 3), rep('beta', 3), rep('gamma', 3)),2)
dates - as.Date(rep(c('2009-10-13', '2009-10-14', '2009-10-15'),6))
In article 800acfc0-2c3c-41f1-af18-3b52f7e43...@jhsph.edu,
bcarv...@jhsph.edu says...
aves = aggregate(df1$score, by=list(col1=df1$col1, col2=df1$col2), mean)
results = merge(df1, aves)
Or, with the 'plyr' package, which has a very nice syntax:
library(plyr)
ddply(df1, .(col1, col2),
John Fox has a nice explanation here:
http://cran.r-project.org/doc/contrib/Fox-Companion/appendix-bootstrapping.pdf
-Ista
On Wed, Oct 21, 2009 at 6:38 AM, Charlotta Rylander z...@nilu.no wrote:
Hello,
We are a group of PhD students working in the field of toxicology. Several
of us have
Dear R:ers,
I'm using the svm from the e1071 package to train a model with the
option probabilities = TRUE. I then use predict with probabilities
= TRUE and get the probabilities for the data point belonging to either
class. So far all is well.
My question is why I get different results
Hi all!
I am a grad stat student and is fairly new in using R. I am doing a
regression tree in one of my problem sets. I have already identified the
cut-points (using Stata) for the regression. My problem is how to graph the
fitted values of z against the independent variables x and y. Basically,
Iff dd is your data frame then:
dd$prev - ave(dd$of, dd$Id, FUN = function(x) c(NA, head(x, -1)))
On Wed, Oct 21, 2009 at 2:55 AM, clion birt...@hotmail.com wrote:
Dear Rlers,
in the following dataset I would like to insert a new column that refers to
the data of the previous row.
My
Hi MS,
I think it is simple trigonometry:
atan(beta)
or in degrees instead of radians
atan(beta)*360/(2*pi)
hth.
ms.com schrieb:
Dear all
I am new R user, and trying to learn more.
I am doing linear regression analysis in R with my data. I am trying to find
the way to calculate the slope
I wasn't clear: x and z are pulse *trains* with irregular gaps between pulses.
x is a rectangular pulse 4 sec long. Its onset and offset are known
with sub-millisecond precision. The onset varies irregularly -- it
doesn't fall on neat 1/2 sec or sec boundaries for example.
y is a sampled
On 10/21/2009 3:53 AM, Johan Lassen wrote:
Hi Giovanni,
Thanks for your reply. I can make the function work after parsing the code
directly into R. The problem arise after compiling the function into a
package and then calling the function, because the files inside source()
seems to be missing.
Hi CE.KA,
Take a look at the following:
# Data
set.seed(123)
x - rnorm(100)
y - 2 + 1.5*x + rnorm(100)
# Regression model
reg - lm(y ~ x)
# The summary
summary(reg)
# Objects present in the summary()
names(summary(reg))
# Extracting the coefficients
summary(reg)$coeff
HTH,
Jorge
On Wed,
Thank you all for your responses, i have now achieved the desired
output for my own real data using your suggestions. I will also have
to look into this 'plyr' package as i have noticed that it gets
mentioned a lot.
On 21 Oct, 13:33, Karl Ove Hufthammer k...@huftis.org wrote:
In article
On 10/21/2009 9:03 AM, Michael Ralph M. Abrigo wrote:
Hi all!
I am a grad stat student and is fairly new in using R. I am doing a
regression tree in one of my problem sets. I have already identified the
cut-points (using Stata) for the regression. My problem is how to graph the
fitted values of
Hello All..
Please consider the following:
y - rnorm(20, mean = 10)
f1 - as.factor(rep(c(A, B, B, A), 5))
f2 - as.factor(rep(c(C, D), 10))
testdata - data.frame(y, f1, f2)
testFunc - function(formula, data, ...) {
#mf - model.frame(formula, data)
kw.res - kruskal.test(formula, data)
Suppose I have the following function
myFun - function(formula, data){
f - formula(formula)
dat - model.frame(f, data)
dat
}
Applying it with this sample data yields a new dataframe:
qqq - data.frame(grade = c(3, NA, 3,4,5,5,4,3), score = rnorm(8), idVar =
Here is one way:
fo - y ~ f1 * f2
one.x - lapply(all.vars(fo[[3]]), function(x) { fo[[3]] - as.name(x); fo })
one.x
[[1]]
y ~ f1
[[2]]
y ~ f2
On Wed, Oct 21, 2009 at 11:29 AM, Bryan Hanson han...@depauw.edu wrote:
Hello All..
Please consider the following:
y - rnorm(20, mean = 10)
f1
Greetings!
As part of my research project I am using R to study temperature data
collected by a network. Each node (observation point) records temperature of
its surroundings throughout the day and generates a dataset. Using the
recorded datasets for the past 7 days I need to build a prediction
Hi Anders,
On Oct 21, 2009, at 8:49 AM, Anders Carlsson wrote:
Dear R:ers,
I'm using the svm from the e1071 package to train a model with the
option probabilities = TRUE. I then use predict with
probabilities = TRUE and get the probabilities for the data point
belonging to either class.
Good day, i imported some data into R from Excel. By using the edit()
function, this is what one of the dates looks like in R:
x - structure(1254351600, class = c(POSIXt, POSIXct), tzone = )
[1] 2009-10-01 BST
However, when i do the following, the date changes:
as.Date(x, formate=%Y-%m-%d )
Dear R-Help-Team,
I would like to cluster my data using the ward-method. In several papers I
read (e.g. Bahrenberg) that it is neccesary to use the squared euclidean
distance with the ward-method. Unfortunatelly I cannot find this term in r
as a method for measuring the distance.
Does anybody
Hello Megha and Jim,
Thanks for your comments. The green bands actually correspond to data that
looks like:
13 0
42 0
183 0
186 0
187 0
192 0
194 0
and so on. I plotted them using:
plot(v[,1],3+v[,2],type=h,col=gray,lwd=5) and the rest of the data using
points. The result is here:
Hi,
On Oct 21, 2009, at 12:31 PM, Aneeta wrote:
Greetings!
As part of my research project I am using R to study temperature data
collected by a network. Each node (observation point) records
temperature of
its surroundings throughout the day and generates a dataset. Using the
recorded
Hi again, and thank you Steve for your reply!
Hi Anders,
On Oct 21, 2009, at 8:49 AM, Anders Carlsson wrote:
Dear R:ers,
I'm using the svm from the e1071 package to train a model with the
option probabilities = TRUE. I then use predict with
probabilities = TRUE and get the
Good afternoon
Using R 2.9.2 on a machine running Windows XP
I have a longitudinal data set, with data on schools and their test scores over
a four year period. I have centered year, and run the following
m1.mod1 - lme(fixed = math_1 ~ I(year-2007.5)*TFC_,
data = long,
Dear All,
Let have 10 pair of observations, as shown below.
##
x - 1:10
y - c(1,3,2,4,5,10,13,15,19,22)
plot(x,y)
##
Two fitted models, with ranges of [1,5] and [5,10], can be easily fitted
separately by lm function as shown below:
Hello
Running R2.9.2 on Windows XP
I am puzzled by the performance of LME in situations where there are missing
data. As I understand it, one of the strengths of this sort of model is how
well it deals with missing data, yet lme requires nonmissing data.
Thus,
m1.mod1 - lme(fixed =
On Wed, Oct 21, 2009 at 12:09 PM, FMH kagba2...@yahoo.com wrote:
Dear All,
Let have 10 pair of observations, as shown below.
##
x - 1:10
y - c(1,3,2,4,5,10,13,15,19,22)
plot(x,y)
##
Two fitted models, with ranges of [1,5] and [5,10], can be
Thanks Gabor, you taught me two useful things: all.vars, and the fact that a
formula object always has length 3. Problem solved. Bryan
On 10/21/09 11:57 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote:
Here is one way:
fo - y ~ f1 * f2
one.x - lapply(all.vars(fo[[3]]), function(x) {
You can calculate the Euclidean distance with dist() and then square it.
Sarah
On Wed, Oct 21, 2009 at 11:35 AM, Caro B. carolin.w...@googlemail.com wrote:
Dear R-Help-Team,
I would like to cluster my data using the ward-method. In several papers I
read (e.g. Bahrenberg) that it is
Emulate lm.
myFun -
function(formula, data, na.action, id, ...){
mf - match.call(expand.dots = FALSE)
m - match(c(formula, data, id, na.action), names(mf), 0L)
mf - mf[c(1L, m)]
mf$drop.unused.levels - TRUE
mf[[1L]] - as.name(model.frame)
mf - eval(mf,
On Wed, 2009-10-21 at 17:35 +0200, Caro B. wrote:
Dear R-Help-Team,
I would like to cluster my data using the ward-method. In several papers I
read (e.g. Bahrenberg) that it is neccesary to use the squared euclidean
distance with the ward-method. Unfortunatelly I cannot find this term in r
Hi,
In order to have more eyes on this, I'm CCing this back to lease
(please try to keep further correspondence here, since most mail to
*this* address of mine probably gets lost if it's not coming in from a
list to begin with) ...
I'm not really sure I have much to say about your
Howdy,
On Oct 21, 2009, at 1:05 PM, Anders Carlsson wrote:
snip
Yes, exactly that. In your example, though, the variation seems to
be a lot smaller. I'm guessing that has to with the data.
If I instead output the decision values, the whole procedure is
fully reproducible, i.e. the exact
Hi all,
I have a matrix of correlation values between all pairwise comparison in
an experiment. For example, I have 2 time points (1,2) each in
triplicate. Thus, I have the following matrix
1-1
1-2
1-3
2-1
2-2
2-3
1-1
NA
...
...
...
...
...
1-2
...
NA
...
...
...
...
David:
Do you mean inappropriate or embarrassing?
How would we R-ians know what has happened at REVolution were it not for
Ajay's note? Were you planning a press release? Something like, 47% of
Revolution summarily fired. Nobody left with more than a year of
experience...?
Charles Annis,
On 22/10/2009, at 7:20 AM, David M Smith wrote:
It is clearly inappropriate for me or anyone else from REvolution to
comment on matters pertaining to any employee or ex-employee. I would
further add that this is a highly inappropriate use of this list.
Oh, I dunno. I have the
Peter Flom wrote:
I am puzzled by the performance of LME in situations where there are
missing data. As I
understand it, one of the strengths of this sort of model is how well it
deals with missing
data, yet lme requires nonmissing data.
You are confusing missing data with an
Hi,
To generate random numbers between 0 and 1, do you use rnorm followed by dnrom?
for ex, for 10 variables
a = rnorm(10)
a
[1] -0.87640764 -0.95842391 -1.33434559 -0.63844932 -1.69829393 0.80010865
[7] -0.01026882 -0.23887516 2.29912600 -1.38352143
dnorm(a)
[1] 0.27171985 0.25202507
On 10/21/2009 3:09 PM, Charles Annis, P.E. wrote:
David:
Do you mean inappropriate or embarrassing?
How would we R-ians know what has happened at REVolution were it not for
Ajay's note? Were you planning a press release?
I think he already did post a note on his blog mentioning that
Well I do not think that this is highly inappropriate use of this list. I am
sure you new designation of VP makes you feel all more powerful but your
authority does not extend to this list. It should continue to rest with Core R
team. It seems had this list been under your control (i.e
Uniformly distributed random numbers between 0 and 1? Try ?runif
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of carol white
Sent: Wednesday, October 21, 2009 2:26 PM
To: r-h...@stat.math.ethz.ch
Subject: [R] random numbers
On 22/10/2009, at 8:27 AM, kulwinder banipal wrote:
Well I do not think that this is highly inappropriate use of this
list. I am sure you new designation of VP makes you feel all more
powerful but your authority does not extend to this list. It should
continue to rest with Core R team.
I would suggest to use the generator at
http://submoon.freeshell.org/pix/valium/dilbert_rng.jpg
and subtract 8.5.
Best,
Gabor
On Wed, Oct 21, 2009 at 9:25 PM, carol white wht_...@yahoo.com wrote:
Hi,
To generate random numbers between 0 and 1, do you use rnorm followed by
dnrom? for ex, for
On Wed, Oct 21, 2009 at 8:09 PM, Charles Annis, P.E.
charles.an...@statisticalengineering.com wrote:
David:
Do you mean inappropriate or embarrassing?
How would we R-ians know what has happened at REVolution were it not for
Ajay's note? Were you planning a press release? Something like,
I apologize for the previous post using HTML. Haven't posted for a while and
e-mail client default.
Best,
Ken
Hi all,
I have a matrix of correlation values between all pairwise comparison in an
experiment. For example, I have 2 time points (1,2) each in triplicate. Thus,
I have the
On Wed, Oct 21, 2009 at 8:37 PM, Gábor Csárdi csa...@rmki.kfki.hu wrote:
I would suggest to use the generator at
http://submoon.freeshell.org/pix/valium/dilbert_rng.jpg
and subtract 8.5.
You may laugh (indeed I did) but some medical trials have used (and
poss still do) telephone-a-human
On 22/10/2009, at 8:25 AM, carol white wrote:
Hi,
To generate random numbers between 0 and 1, do you use rnorm
followed by dnrom? for ex, for 10 variables
a = rnorm(10)
a
[1] -0.87640764 -0.95842391 -1.33434559 -0.63844932 -1.69829393
0.80010865
[7] -0.01026882 -0.23887516
On 22/10/2009, at 8:37 AM, Barry Rowlingson wrote:
On Wed, Oct 21, 2009 at 8:09 PM, Charles Annis, P.E.
charles.an...@statisticalengineering.com wrote:
David:
Do you mean inappropriate or embarrassing?
How would we R-ians know what has happened at REVolution were it
not for
Ajay's note?
Hi,
snip
If I instead output the decision values, the whole procedure is
fully reproducible, i.e. the exact same values are returned when I
retrain the model.
By the decision values, you mean the predict labels, right?
The output of decision values can be turned on in the predict.svm,
The data that I use has been collected by a sensor network deployed by Intel.
You may take a look at the network at the following website
http://db.csail.mit.edu/labdata/labdata.html
The main goal of my project is to simulate a physical layer attack on a
sensor network and to detect such an
I wrote
I am puzzled by the performance of LME in situations where there are
missing data. As I
understand it, one of the strengths of this sort of model is how well it
deals with missing
data, yet lme requires nonmissing data.
Mark Difford replied
You are confusing missing data with
I am sorry.
I will refrain from this in the future (using the list which is a technical
resource and not a forum inappropriately).
My blog has my views on it -http://decisionstats.com so I wont cut and paste
on that.
I would like to applaud David's team at REvolution for finally releasing an
Hi
This is on WinXP with regional settings as EST (we are now on DST but I run
EST) R2.9.2
x - structure(1254351600, class = c(POSIXt, POSIXct), tzone = )
x
[1] 2009-10-01 09:00:00 EST
as.POSIXlt(x)
[1] 2009-10-01 09:00:00 EST
as.Date(x, formate=%Y-%m-%d )
[1] 2009-09-30
I had a similar
Basically I need to use the following data to calculate a squared error for
each Sample based on the expected Survival for the zone.
Basically, this code has Live/Dead for each sample, and I need to calculate
the square error based on the Expected Mean (ie, Survival). The code looks
up the
Hello R user community,
can anyone tell me how to increase the size of a filled box in a legend?
thanx,
Janet
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PLEASE do read
Hi,
Am am plotting aggregated frequency data (extracted from an RDBMS)
relating to DNA sequence features for each of the human chromosomes as
described in the table chromosomes below (the frequency data is in a
table 'hits' that has a value (or not) for each of a set of bins across
each
Dear R People:
Suppose I have the following output:
table(xx)
xx
A C G T
13 12 10 15
I would like to have the output sorted in descending order by height
or frequency.
But when I do the following:
rev(table(xx))
xx
T G C A
15 10 12 13
the output is sorted by the names rather than
I would like to have the output sorted in descending order by height
or frequency.
But when I do the following:
rev(table(xx))
xx
T G C A
15 10 12 13
Err, I guess you meant to write
sort(table(xx))
here?
Cheers,
Stefan
__
On 22/10/2009, at 11:16 AM, Erin Hodgess wrote:
Dear R People:
Suppose I have the following output:
table(xx)
xx
A C G T
13 12 10 15
I would like to have the output sorted in descending order by height
or frequency.
But when I do the following:
rev(table(xx))
xx
T G C A
15 10
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of chipmaney
Sent: Wednesday, October 21, 2009 2:58 PM
To: r-help@r-project.org
Subject: [R] How do I vectorize this
Dear R users,
I'd like to ask u whether you know how to sort out the following:
I'm trying to reproduce a dataset of clusters, and for that I need to build
up a cluster index (inside a function) using the command replicate as
follows:
dataset- function(
clusters=100,
cluster.size=50,
I am running into a problem using 'mix' for multiple imputation (over
continuous and categorical variables).
For the way I will be using this I would like to create an imputation
model on some training data set and then use this model to impute
missing values for a different set of individuals
Hello,
I'm using R to run a acoustic analysis software called Seewave. I ask the code
to extract a list of variables from my recording, and the program give ONE
table for each of these. The tables consist of a two column data.frame with the
time in column 1 and the frequency in column 2.
Hi everybody,
I noticed a strange behavior when using loops versus apply() on a data frame.
The example below explicitly computes a distance matrix given a
dataset. When the dataset is a matrix, everything works fine. But when
the dataset is a data.frame, the dist.for function written using
Hello all,
I am new to the list serve but hope to contribute what I can. For now,
however, I hope to tap into all your knowledge about mixed effects models.
Currently, I am running a mixed effects model on a time series panel data.
For this model I want to find out the fixed and random effect
Hi,
Below is the code that is giving me this error message:
#===
#The code below results in error: Error: NA/NaN/Inf in foreign function #call
(arg 1)
#===
good day everyone!
i have a time series (andong.ts) and fitted and AR() model using the
following code
andong.ts - ts(read.table(D:/.../andong.csv, header = TRUE), start =
c(1966,1), frequency = 1)
ar(andong.ts)
Call:
ar(x = andong)
Coefficients:
1 2 3
0.3117 0.0607
good day everyone!
i have a time series (andong.ts) and fitted and AR() model using the
following code
andong.ts - ts(read.table(D:/.../andong.csv, header = TRUE), start =
c(1966,1), frequency = 1)
ar(andong.ts)
Call: ar(x = andong)
Coefficients:
1 2
On Wed, Oct 21, 2009 at 11:06 AM, Peter Flom
peterflomconsult...@mindspring.com wrote:
...
I have a longitudinal data set, with data on schools and their test scores
over a four year period. I have centered year, and run the following
m1.mod1 - lme(fixed = math_1 ~ I(year-2007.5)*TFC_,
Hi - a simple question. I know that on plot one can set log=xy to
set scale on both axes to log scale. I use this to plot
autocorrelation of a slow decaying autocorrelated process. It has some
points that are zero or slightly negative and cause this plot to
return error. I'm wondering if there is
Mixed models based on likelihood methods can often handle missing
observations within subjects, but they not do well with missing
individual elements in the design matrices (think unit nonresponse vs
item nonresponse in the survey world). Continuing with the example I
recently sent to you
try running Rprof on the two examples to see what the difference is.
what you will probably see is a lot of the time on the dataframe is
spent in accessing it like a matrix ('['). Rprof is very helpful to
see where time is spent in your scripts.
Sent from my iPhone
On Oct 21, 2009, at
Hi Peter,
See e.g. Hedeker and Gibbons, Longitudinal Data Analysis, which
repeatedly stresses that
mixed models provide good estimates if the data are missing at random.
This may be true. However, one of the real strengths of LME is that it
handles unbalanced designs, which is a different
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