Hi Caitlin,
You just need to set the ylim argument to plot (you already did but
g_range is 0, 13 not 0, 15. So...
On Thu, Mar 17, 2011 at 10:49 PM, Caitlin wrote:
> Hi all.
>
> I'm working on an assignment for a psychology class and I am not sure how to
> adjust the y-axis so it displays the ra
Hi all.
I'm working on an assignment for a psychology class and I am not sure how to
adjust the y-axis so it displays the range: 0, 5, 10, 15
The code below is almost ideal:
lsd = c(3, 5, 13)
mar = c(1, 2, 3)
g_range <- range(0, lsd, mar)
plot(lsd, type="o", col="blue", ylim=g_range,
axes=FALSE,
Dear List,
First off, this is completely off topic, but I thought others might
find it interesting. The good people at IBM are apparently providing
written interpretations of results now. I was working with a student
the other day with simple nonparametric tests using SPSS, and under
the nonpara
Does someone have confirmatory factor analysis program in R,which includes
factor loading and some tests?thank you!
--
View this message in context:
http://r.789695.n4.nabble.com/confirmatory-factor-analysis-program-in-R-tp3386133p3386133.html
Sent from the R help mailing list archive at Nabble.
Dear R community members
I have been struggling on this simple question, but never get appropriate
solution. So please help.
# my data, though I have a large number of variables
var1 <- rnorm(500, 10,4)
var2 <- rnorm(500, 20, 8)
var3 <- rnorm(500, 30, 18)
var4 <- rnorm(500, 40, 20)
datafr1 <- da
Have timestamp in format HH:MM:SS.MMM.UUU and need to remove the last "." so
it is in format HH:MM:SS.MMMUUU.
What is the fastest way to do this, since it has to be repeated on millions
of rows. Should I use regex ?
Currently doing it with a string split, which is slow:
>head(ts)
[1] 09:30:00
Thanks
> which.max(table(x))
129.46.71.19
10
How do I get only "129.46.71.19" back as a str...
-Original Message-
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Thursday, March 17, 2011 4:39 PM
To: Khanvilkar, Shashank
Cc: r-help
Subject: Re: [R] Histograms with strings,
Hi all.
I'm having a little trouble with the function "texteval" (session package).
I have used "texteval" in the construction of the function "ExpandData1".
"ExpandData1" does not work as expected. However, when I run only the inside
code
of "ExpandData1" I get the right result. Apparently "te
Hi,
I have been learning the quantmod package over the last several days. I
went to check some of my data pulls against other sources and was
surprised to find that a few tickers that have single characters do not
successfully scrape from Google Finance using getFin(). Particularly
require(quan
Hi everyone,
Is there any command to identify the pattern of responses of a database
with this format:
year id
20081
20091
20082
20092
20083
20093
20084
20094
20104
I just need the frequency of the patterns grouped by id:
2008 2009 2010 = 80
2009
I read it but it said PDF file and Ps, didn´t specify which other files, so I
attached a csv file, which I thought would work.I have uploaded the file in
rapidhare (second option was putting on the
web):http://rapidshare.com/files/453101614/Coordinates_and_values.csv
Hope this works.Thanks for
On Mar 17, 2011, at 6:33 PM, Pamela Allen wrote:
Hi All,
I'm trying to plot data that is a time series of flows that are
associated
with a specific level, and I would like each level to represent a
colour
in a line plot. Here is some data that approximates what I'm using:
date=c(1:300)
On Mar 17, 2011, at 6:06 PM, Khanvilkar, Shashank wrote:
Hello,
Thanks in advance for any help,
I have read a CSV file in which there is a column for an IP addr as
in:
tmpInFile$V2
[1] "74.125.224.38" "74.125.224.38" "129.46.71.19" "129.46.71.19"
[5] "129.46.71.19" "129.46.71.19" "129
On Mar 17, 2011, at 5:44 PM, ufuk beyaztas wrote:
Hi dear all,
It may be a simple question, i have a list output with different
number of
elements as following;
[[1]]
[1] 0.86801402 -0.82974691 0.3974 -0.98566707 -4.96576856
-1.32056754
[7] -5.54093319 -0.07600462 -1.34457280 -1.04
Taby,
At the end of your note are you referring to the bootstrap confidence
intervals in the "external validation" case, i.e., not corrrected for
overfitting? If so you can get that without the bootstrap (e.g., Hmisc
package rcorr.cens function).
You can get bootstrap overfitting-corrected ROC a
names(which.max(table(x)))
This is a 'named' vector
On Thu, Mar 17, 2011 at 8:58 PM, Khanvilkar, Shashank
wrote:
> Thanks
>
>> which.max(table(x))
> 129.46.71.19
> 10
>
> How do I get only "129.46.71.19" back as a str...
>
>
> -Original Message-
> From: jim holtman [mailto:jholt
This mostly seemed like a condensation of what had been said
before ... threads from the past which were readily available through
searching.
http://finzi.psych.upenn.edu/R/Rhelp02/archive/20370.html
http://finzi.psych.upenn.edu/R/Rhelp02/archive/73750.html
http://finzi.psych.upenn.edu/R/Rhe
On 2011-03-17 16:37, Axel Urbiz wrote:
Dear List,
This is an embarrassing question, but I can seem to make this work…How do I
change the font size on the xlab and on the numbers shown in the x-axis on
the time series plot below. The arguments cex.lab and cex.axis do not seem
to be 'passing' to
Indeed, I forgot about the segments function.
with(d,plot(date,flow,type="n"))
with(d,segments(start,start.y,end,end.y,col=colour))
> Hi,
>
> because each colour is defined on non-consecutive points, you'll
> probably need to cut the intervals to define segments around each
> point. One approa
Hi:
Try this:
plot(ts(rnorm(100), start = 2004, freq = 12), xaxt = 'n', yaxt = 'n',
xlab = '', ylab = '')
axis(1, at = c(2004:2012), cex.axis = 0.7)
axis(2, cex.axis = 0.7)
title(xlab = 'My X lab', ylab = 'RQI', cex.lab = 0.1)
The illegible dots in the region where the axis labels would nor
Hi,
because each colour is defined on non-consecutive points, you'll
probably need to cut the intervals to define segments around each
point. One approach might be the following,
d = transform(data, start = date - c(0, diff(date)/2), end = date +
c(0, diff(date)/2) )
d$start.y = approx(d$date, d
G'day Gabor,
On Thu, 17 Mar 2011 20:38:21 -0400
Gabor Grothendieck wrote:
> > Or am I missing something?
O.k., because the residuals don't add to zero, there may be a non-zero
correlation between residuals and fitted values, which messes up the
equation at the variance level.
> Try it on an ex
Hi!
Not an elegant solution, but seems to work:
date <- c(1:300)
flow <- sin(2*pi/53*c(1:300))
levels <- factor(rep(c("high","med","low"),100))
data <- cbind.data.frame(date, flow, levels)
colours <- as.numeric(levels)+1
# interpolate
resolution <- 0.001
appres <- approx(date,flow,seq(min(date
On Thu, Mar 17, 2011 at 8:22 PM, Berwin A Turlach
wrote:
> G'day Gabor,
>
> On Thu, 17 Mar 2011 11:36:38 -0400
> Gabor Grothendieck wrote:
>
>> The idea is that if you have a positive quantity that can be broken
>> down into two nonnegative quantities: X = X1 + X2 then it makes sense
>> to ask wh
On Thu, Mar 17, 2011 at 10:06:21PM +, Khanvilkar, Shashank wrote:
> Hello,
> Thanks in advance for any help,
>
> I have read a CSV file in which there is a column for an IP addr as in:
>
> tmpInFile$V2
> [1] "74.125.224.38" "74.125.224.38" "129.46.71.19" "129.46.71.19"
> [5] "129.46.71.1
On Thu, Mar 17, 2011 at 07:52:29PM -0400, Gabor Grothendieck wrote:
> ...
> > I hope to learn what I'm misunderstanding.
> ...
>
> This line in the code above produces an invalid object:
> class(df$time) <- "POSIXct"
>
> It should be:
> class(df$time) <- c("POSIXct", "POSIXt")
Ah. Thank you, Si
G'day Gabor,
On Thu, 17 Mar 2011 11:36:38 -0400
Gabor Grothendieck wrote:
> The idea is that if you have a positive quantity that can be broken
> down into two nonnegative quantities: X = X1 + X2 then it makes sense
> to ask what proportion X1 is of X. For example: 10 = 6 + 4 and 6 is
> .6 of
On Thu, Mar 17, 2011 at 4:54 PM, David Wolfskill wrote:
> On Thu, Mar 17, 2011 at 10:23:33AM -0400, Gabor Grothendieck wrote:
>> On Thu, Mar 17, 2011 at 9:38 AM, David Wolfskill wrote:
>> ...
>> > But the X-axis labels show up as "large integers" -- the POSIXct values
>> > are apparently treated
use 'unlist'
> x <- list(1:10, 20:27, 30:50)
> x
[[1]]
[1] 1 2 3 4 5 6 7 8 9 10
[[2]]
[1] 20 21 22 23 24 25 26 27
[[3]]
[1] 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
> quantile(unlist(x))
0% 25% 50% 75% 100%
1.0 15.0 31.0 40.5 50.0
>
On Thu, Mar 17, 2011
try this:
> x
[1] "74.125.224.38" "74.125.224.38" "129.46.71.19" "129.46.71.19"
"129.46.71.19" "129.46.71.19"
[7] "129.46.71.19" "129.46.71.19" "129.46.71.19" "129.46.71.19"
"129.46.71.19" "129.46.71.19"
> table(x)
x
129.46.71.19 74.125.224.38
10 2
> which.max(table
?unlist
quantile(unlist(data))
> Hi dear all,
>
> It may be a simple question, i have a list output with different number of
> elements as following;
>
> [[1]]
> [1] 0.86801402 -0.82974691 0.3974 -0.98566707 -4.96576856
> -1.32056754
> [7] -5.54093319 -0.07600462 -1.34457280 -1.04080125
Thank you for our reply. It's a pity, that 2 variables defined by different
formula have same name. If the variables had been named differently, I
wouldn't have problem at all and it looks like it's done on purpose.
Because I test a quality of data (performance of collecting data) not a
model which
Dear List,
This is an embarrassing question, but I can seem to make this work
How do I
change the font size on the xlab and on the numbers shown in the x-axis on
the time series plot below. The arguments cex.lab and cex.axis do not seem
to be 'passing' to the plot function.
plot(ts(rnorm(100), s
Hi All,
I'm trying to plot data that is a time series of flows that are associated
with a specific level, and I would like each level to represent a colour
in a line plot. Here is some data that approximates what I'm using:
date=c(1:300)
flow=sin(2*pi/53*c(1:300))
levels=c(rep(c("
Hello,
Thanks in advance for any help,
I have read a CSV file in which there is a column for an IP addr as in:
tmpInFile$V2
[1] "74.125.224.38" "74.125.224.38" "129.46.71.19" "129.46.71.19"
[5] "129.46.71.19" "129.46.71.19" "129.46.71.19" "129.46.71.19"
[9] "129.46.71.19" "129.46.71.19
Hey all!
I am working on my master thesis and I am desperate with my model.
It looks as following:
Y(t) = β1*X1(t) + β2*X2(t) + δ*(β1*((1+c)/(δ+c))+β2)*IE(t) -
β2*α*((1+c)/(δ+c))*(δ+g)* IE(t-1)
note: c and g is a constant value
The problem I encounter is that between IE(t) and IE(t-1) there is
Hi dear all,
It may be a simple question, i have a list output with different number of
elements as following;
[[1]]
[1] 0.86801402 -0.82974691 0.3974 -0.98566707 -4.96576856 -1.32056754
[7] -5.54093319 -0.07600462 -1.34457280 -1.04080125 1.62843297 -0.20473912
[13] 0.30659907 2.669081
On Thu, Mar 17, 2011 at 10:23:33AM -0400, Gabor Grothendieck wrote:
> On Thu, Mar 17, 2011 at 9:38 AM, David Wolfskill wrote:
> ...
> > But the X-axis labels show up as "large integers" -- the POSIXct values
> > are apparently treated as numeric quantities for this purpose.
> ...
> Please cut this
I attach the data (csv format). There are the 3 coordinates, (but as there are
not so many points I wanted two do 3 analysis in each of them collapsing one
variable).There are two variables to study I have posted the data as a ratio
between both states and as a percentage state between both sta
On Thu, Mar 17, 2011 at 10:32 AM, Joshua Ulrich wrote:
> On Wed, Mar 16, 2011 at 6:58 PM, jctoll wrote:
>> Hi,
>>
>> I'm struggling to figure out the way to change the name of a column
>> from within a loop. The problem is I can't refer to the object by its
>> actual variable name, since that wi
Try this:
lapply(1:5,function(i){i;function()i})[[2]]()
or
lapply(1:5,function(i){i;function(j=i) j } )[[2]]()
## both give 2
Now try:
lapply(1:5,function(i){function(j=i) j } )[[2]]()
## gives 5 !
The problem is that if you do:
lapply(1:5,function(h){function(h)h)
what you get is a l
Thanks Jim,
It turns out that the problem was that all columns had been converted
to factors. Once I converted them back to numeric variables the code
worked fine. If anybody is wondering how, you can do this with the
following:
mynumber <- as.numeric(levels(myfactor))[myfactor]
There are plen
Hi,
I think it's a side effect of lazy evaluation, where you should
probably use the ?force like a jedi,
lapply(1:5,function(k){force(k) ; function(){k}})[[2]]()
HTH,
baptiste
On 18 March 2011 07:01, jamie.f.olson wrote:
> So, I've been confused by this for a while. If I want to create funct
So, I've been confused by this for a while. If I want to create functions in
an apply, it only uses the desired value for the variable if I create a new
local variable:
> lapply(1:5,function(h){k=h;function(){k}})[[1]]()
[1] 1
> lapply(1:5,function(k){function(){k}})[[1]]()
[1] 5
>
Normally, a
On Thu, Mar 17, 2011 at 02:54:49PM -0400, Jim Silverton wrote:
> I have a matrix say:
>
> 23 1
> 12 12
> 00
> 0 1
> 0 1
> 0 2
> 23 2
>
> I want to count of number of distinct rows and the number of disinct element
> in the second column and put these counts in a column. SO at the en
Hi all,
I would like to fit a gamm model of the form:
Y~X+X*f(z)
Where f is the smooth function and
With random effects on X and on the intercept.
So, I try to write it like this:
gam.lme<- gamm(Y~ s(z, by=X) +X, random=list(groups=pdDiag(~1+X)) )
but I get the error messag
Hi,
thank you for your elaborate answer. I downloaded Prof. Dayton's pdf and
will read it tomorrow.
A friend also told me that our professor said you can actually compare
AICs for different distributions. Apparently it's not correct strictly
speaking, because of the two different likelihoods,
Hey everyone,
Nevermind, I figured it out:
ggplot(data, aes(x1, y1))+geom_boxplot()+scale_y_log2(lim=c(...))
:)
Cheers,
Godwin
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-he
Hey everyone,
I'm having a little trouble with ggplot. I have two sets of y-values, one
whose range is contained in the other. Due to the nature of the y-values, I
wish to scale the y axis with a log base two transformation. Furthermore, I
wish to plot the two sets of y-values as boxplots in separ
Dear Jim,
17.03.2011 20:54, Jim Silverton wrote:
I have a matrix say:
23 1
12 12
00
0 1
0 1
0 2
23 2
I want to count of number of distinct rows and the number of disinct element
in the second column and put these counts in a column. SO at the end of the
day I should have:
c(1, 1
Did you post your data or hypothetical data?
Usually that helps make your problem more clear and more interesting
( likely to get a useful response to your post).
From: tintin...@hotmail.com
To: r-help@r-project.org
Date: Thu, 17 Mar 2011 17:38:14
I have a matrix say:
23 1
12 12
00
0 1
0 1
0 2
23 2
I want to count of number of distinct rows and the number of disinct element
in the second column and put these counts in a column. SO at the end of the
day I should have:
c(1, 1, 1, 2, 2, 1, 1) for the distinct rows and c(1, 1, 1
Is X a numeric variable or a factor? If it's numeric try
gam.lme<- gamm(Y~ s(z, by=X), random=list(groups=pdDiag(~1+X)) )
... since otherwise the separate X term is confounded with s(z, by=X).
(gam detects such confounding and copes with it, but gamm can't).
Simon
On 17/03/11 17:47, I
Scarlet,
If the mfrow is being overridden, perhaps the rimage package might be
able to piece the individual plots...
--
Muhammad Rahiz
Researcher & DPhil Candidate (Climate Systems & Policy)
School of Geography & the Environment
University of Oxford
On Thu, 17 Mar 2011, scarlet wrote:
Jim,
Hi all,
I would like to fit a gamm model of the form:
Y~X+X*f(z)
Where f is the smooth function and
With random effects on X and on the intercept.
So, I try to write it like this:
gam.lme<- gamm(Y~ s(z, by=X) +X, random=list(groups=pdDiag(~1+X)) )
but I get the error messag
It is all a matter of what you are comparing too, or what the null model is.
For most cases (standard regression) we compare a model with slope and
intercept to an intercept only model (looking at the effect of the slope), the
intercept only model fits a horizontal line through the mean of the
Jim,
Thanks for looking into this. The c without paste works. If the rq model
overrides the mfrow, I think I will have to piece together individual plots
using other software.
scarlet
--
View this message in context:
http://r.789695.n4.nabble.com/plotting-multiple-figures-on-one-page-tp338298
> paste(samp, ".pdf", sep="")
[1] "20110317.pdf"
> paste(samp, ".csv", sep="")
[1] "20110317.csv"
On Thursday, March 17, 2011 at 10:05 AM, pierz wrote:
I would like to use samp as a part of a filename that I can change. My source
&
I suggest that you post this on the R-sig-mixed-models list where you
are more likely to find those with bothe interest and expertise in
these matters.
-- Bert
On Thu, Mar 17, 2011 at 7:44 AM, Franssens, Samuel
wrote:
> Hi,
>
> I have the following type of data: 86 subjects in three independent
me, and I would like to be able
> to type in the date (later perhaps automate this using list.files) and then
> read the csv and write the pdf automatically. I have tried different
> combinations with "" and () around samp, but I keep getting the error
> "object 'sa
Hi,
I have the following type of data: 86 subjects in three independent groups
(high power vs low power vs control). Each subject solves 8 reasoning problems
of two kinds: conflict problems and noconflict problems. I measure accuracy in
solving the reasoning problems. To summarize: binary respo
Dear,
I'm trying to run diagnostics on MCMC analysis (fitting a log-linear
model to rates data). I'm getting an error message when trying
Gelman-Rubin shrink factor plot:
>gelman.plot(out)
Error in chol.default(W) :
the leading minor of order 2 is not positive definite
I take it that somewher
On Mar 17, 2011; 04:29pm Thierry Onkelinx wrote:
>> You cannot compare lm() with lme() because the likelihoods are not the
>> same. Use gls() instead of lm()
And perhaps I should have added the following:
First para on page 155 of Pinheiro & Bates (2000) states, "The anova method
can be used to
different
combinations with "" and () around samp, but I keep getting the error
"object 'samp.csv' not found".
samp <- "20110317"
read.csv(file=samp.csv,...)
#next R processes some code that works fine, and then should save the
figure:
pdf(file=samp.pdf,...)
de
Thats exactly what I would like to do. Any idea on good text? I've consulted
severel texts, but no one defined R^2 as R^2 = 1 - Sum(R[i]^2) /
Sum((y[i])^2-y*)) still less why to use different formulas for similar model
or why should be R^2 closer to 1 when y=a*x+0 than in general model y=a*x+b.
fr
On Mar 17, 2011; 04:29pm Thierry Onkelinx wrote:
>> You cannot compare lm() with lme() because the likelihoods are not the
>> same. Use gls() instead of lm()
Hi Thierry,
Of course, I stand subject to correction, but unless something dramatic has
changed, you can. gls() can be used if you need to
Yes they are. I had edited the reply, but It didn't help.
Correction:
2)I meant zero slope, no zero intercept.
--
View this message in context:
http://r.789695.n4.nabble.com/Strange-R-squared-possible-error-tp3382818p3384648.html
Sent from the R help mailing list archive at Nabble.com.
Thank you for your very comprehensible answer.
I a priori know that model y=a*x+0 is right and that I can't get x=constant
nor y=constant.
I'm comparing performance of data gathering in my data set to another data
sets in which performance gathering is characterized by R-squared . The data
in dat
I rewrite my previous comand, CVlm works now. It was related to having the
same names in the df than in the formula included at CVlm .
Now I'd like to know:
Q1: if the seed is of any special importance, or I can type just seed=29
like in the example?
Q2: I typed:
CVlm(df = mydf, form.lm = fo
On Thu, 17 Mar 2011, Henrik Bengtsson wrote:
On Thu, Mar 17, 2011 at 6:58 AM, cchace wrote:
I frequently get a segmentation fault error when using the "plot" command.
It happens about half the time.
We are running an old version of R (R version 2.8.0 (2008-10-20) on Linux.
I did a quick sea
Dear R Users, R Core
Team,
I have a two dimensional space where I measure a numerical value in two
situations at different points. I have measured the change and I would like to
test if there are areas in this 2D-space where there is a different amount of
change (no change, increase, decrease).
Hello Allan,
Thanks the response. Provides me hope. I appreciate [3], might even go with
that.
And for posterity, here's the code (assuming pastebin never expires)
[1] Test string : http://pastebin.com/FyAFzmTv
[2] Pattern (modified as per your suggestion) : http://pastebin.com/s7VT0r5K
pattern
Taby,
First, it is better to reply to the whole list (which I have included on
this reply); there is a better chance of someone helping you. Just
because I could help with one aspect does not mean I necessarily can (or
have the time to) help with more.
Further comments are inline below.
On
On Wed, Mar 16, 2011 at 3:49 PM, derek wrote:
> k=lm(y~x)
> summary(k)
> returns R^2=0.9994
>
> lm(y~x) is supposed to find coef. a anb b in y=a*x+b
>
> l=lm(y~x+0)
> summary(l)
> returns R^2=0.9998
> lm(y~x+0) is supposed to find coef. a in y=a*x+b while setting b=0
>
> The question is why do I g
On Wed, Mar 16, 2011 at 6:58 PM, jctoll wrote:
> Hi,
>
> I'm struggling to figure out the way to change the name of a column
> from within a loop. The problem is I can't refer to the object by its
> actual variable name, since that will change each time through the
> loop. My xts object is A.
>
1. There is a CRAN mirror at Bristol (roughly 100km from
Swansea?). They must have (or have had) some very active R users there.
2. Sundar Dorai-Raj and I developed a local R Archive Network
and subversion repository internal to a company where we used to work.
I believe t
Dear Andrew,
The reported df in lavaan is 0 which is correct. It is because this
path model is saturated. "28" is not the df, it is the no. of pieces
of information. The no. of parameter estimates is also 28. Thus, the
df is 0.
However, you are correct that there are only 13, not 28, free
paramet
Your model is saturated.
I think lavaan calculates the number of degrees of freedom this way:
DF = n*(n + 1)/2 - t - n.fix*(n.fix + 1)/2
n = number of variables
t = number of free parameters
n.fix = number of fixed exogenous variables
So, if you fix the exogenous variables, as in mimic = "Mplus
see inline.
On Thu, Mar 17, 2011 at 4:58 AM, Rubén Roa wrote:
> Hi Alexx,
>
> I don't see any problem in comparing models based on different distributions
> for the same data using the AIC, as long as they have a different number of
> parameters and all the constants are included.
> For example
Hi Erin,
On Thu, Mar 17, 2011 at 1:27 AM, Erin Hodgess wrote:
> Dear R People:
>
> Hello again!
>
> I found something unusual in the behavior of the "endpoints" function
> from the xts package:
>
>> x1 <- ts(1:24,start=2008,freq=12)
>> dat <- seq(as.Date("2008/01/01"),length=24,by="months")
>> li
On Thu, Mar 17, 2011 at 6:58 AM, cchace wrote:
>
> I frequently get a segmentation fault error when using the "plot" command.
> It happens about half the time.
>
> We are running an old version of R (R version 2.8.0 (2008-10-20) on Linux.
>
> I did a quick search for this problem and didn't find a
On 2011-03-17 02:08, derek wrote:
Exuse me, I don't claim R^2 can't be negative. What I say if I get R^2
negative then the data are useless.
I know, that what Thomas said is true in general case. But in my special
case of data, using nonzero intercept is nonsense, and to get R^2 less than
0.985 i
Dear Mark,
You cannot compare lm() with lme() because the likelihoods are not the same.
Use gls() instead of lm()
library(nlme)
data("sleepstudy", package = "lme4")
fm <- lm(Reaction ~ Days, sleepstudy)
fm0 <- gls(Reaction ~ Days, sleepstudy)
logLik(fm)
logLik(fm0)
fm1 <- lme(Reaction ~ Days, r
On Thu, Mar 17, 2011 at 9:38 AM, David Wolfskill wrote:
> I've used barplot(), including the anmes.arg parameter, on data frames
> successfully, but I'm even newer to using zoo than I am to R. :-}
>
> I am working on a functon that accepts a data frame ("df") as its
> primary argument, extracts i
On Thu, Mar 17, 2011 at 2:42 PM, Joanne Demmler wrote:
> Dear R users,
Hi
>
> we are currently trying to set up a R working group at Swansea University.
> I would be very grateful to get some information and feedback from R users
> that have done something similar, in particular regarding:
We s
I frequently get a segmentation fault error when using the "plot" command.
It happens about half the time.
We are running an old version of R (R version 2.8.0 (2008-10-20) on Linux.
I did a quick search for this problem and didn't find anything. Apologies if
I missed it.
*** Process received
I rewrite my previous comand, CVlm works now. It was related to having the
same names in the df than in the formula included at CVlm .
Now I'd like to know if the seed is of any special importance, or I can type
just seed=29 like in the example.
Thanks in advance, u...@host.com
--
View this me
> 2) I don't want to fit data with linear model of zero intercept.
> 3) I dont know if I understand correctly. Im 100% sure the model for my data
> should have zero intercept.
> The only coordinate which Im 100% sure is correct. If I had measured quality
> Y of a same sample X0 number of times I wo
LMM without Random effect:
I want to run an LMM both with and without the random factor (ID). And then
extract the log-lik values from the two models in order to generate a
p-value.
with random factor as:lmer(y~x+(1|ID),data)
Question: can I simply substitute a dummy var (e.g. populated by z
Exuse me, I don't claim R^2 can't be negative. What I say if I get R^2
negative then the data are useless.
I know, that what Thomas said is true in general case. But in my special
case of data, using nonzero intercept is nonsense, and to get R^2 less than
0.985 is considered poor job (standard R^2>
On Mar 17, 2011; 11:43am Baugh wrote:
>> Question: can I simply substitute a dummy var (e.g. populated by zeros)
>> for "ID" to run the model
>> without the random factor? when I try this R returns values that seem
>> reasonable, but I want to be sure
>> this is appropriate.
If you can fit the
> Date: Wed, 16 Mar 2011 13:50:37 -0700
> From: solomon.mess...@gmail.com
> To: r-help@r-project.org
> Subject: Re: [R] How to make sure R's g++ compiler uses certain C++ flags
> when making a package
>
> Looks like the problem may be that R is automatic
library(mgcv)
b <- gam(arcsine.success ~ s(date.num,clutch.size,k=50))
vis.gam(b,theta=30)
will fit a thin plate spline and plot it. k is an upper limit on the
number of degrees of freedom for the TPS, but the actual degrees of
freedom are chosen automatically (use the 'method' argument of gam
I've used barplot(), including the anmes.arg parameter, on data frames
successfully, but I'm even newer to using zoo than I am to R. :-}
I am working on a functon that accepts a data frame ("df") as its
primary argument, extracts information from it to create a zoo, then
generates a plot based on
Dear R users,
we are currently trying to set up a R working group at Swansea University.
I would be very grateful to get some information and feedback from R
users that have done something similar, in particular regarding:
- help with giving a general overview of what R is capable of
- help wi
On Wed, Mar 16, 2011 at 12:52 AM, wrote:
> The Lattice auto.key argument has a bug in R.12.2.
>
> R version 2.12.2 (2011-02-25)
> Platform: i386-pc-mingw32/i386 (32-bit)
>
> other attached packages:
> [1] lattice_0.19-17
>
> loaded via a namespace (and not attached):
> [1] grid_2.12.2
>
> If
Arni Magnusson hafro.is> writes:
>
> I have been reading about autocorrelation in linear models over the last
> couple of days, and I have to say the more I read, the more confused I
> get. Beyond confusion lies enlightenment, so I'm tempted to ask R-Help for
> guidance.
>
> Most authors are
I have been trying to use lavaan (version 0.4-7) for a simple path model,
but the program seems to be computing far less degrees of freedom for my
model then it should have. I have 7 variables, which should give (7)(8)/2 =
28 covariances, and hence 28 DF. The model seems to only think I have 13
D
matrix(a, ncol=3, nrow=4, byrow=TRUE)
or
> dim(a) <- c(3,4)
> a <- t(a)
> a
[,1] [,2] [,3]
[1,]123
[2,]456
[3,]789
[4,] 10 11 12
Depending on the context of the problem.
Sarah
On Thu, Mar 17, 2011 at 7:59 AM, Bodnar Laszlo EB_HU
wrote:
> Hi again
On Thu, Mar 17, 2011 at 3:54 AM, wrote:
> It doesn't work (in R) because it is not written in R. It's written in some
> other language that looks a bit like R.
It parses in R, so I would say it was written in R.
To paraphrase Obi-wan, it's just not the R you are looking for.
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