Hi all,
I have run a ridge regression on a data set 'final' as follows:
reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$u,
lambda=seq(0,10,0.01))
Then I enter :
select(reg) and it returns: modified HKB estimator is 19.3409
modified L-W
R-3.0.0 prints an error message at startup. I uninstalled a previous
version and installed this on Windows 8. Uninstall of previous version
deleted all previously installed packages and I forgot to keep a list
of them. How do I correct this error?
Error: requested primitive type is not
On 30/04/2013 07:44, Anupam Tyagi wrote:
R-3.0.0 prints an error message at startup. I uninstalled a previous
version and installed this on Windows 8. Uninstall of previous version
deleted all previously installed packages and I forgot to keep a list
of them. How do I correct this error?
BINGO!!!
Many thanks arun!
You are a genious!!!
montserrat
-Mensaje original-
De: arun [mailto:smartpink...@yahoo.com]
Enviado el: lunes, 29 de abril de 2013 18:33
Para: De Castro Pascual, Montserrat
CC: R help
Asunto: Re: [R] rbinding some elements from a list and obtain another list
Wow! It's true!!!
Many thanks!!!
montserrat
-Mensaje original-
De: MacQueen, Don [mailto:macque...@llnl.gov]
Enviado el: lunes, 29 de abril de 2013 22:39
Para: De Castro Pascual, Montserrat; r-help@r-project.org
Asunto: Re: [R] rbinding some elements from a list and obtain another list
Hi All,
I am trying to shift from running mixed models in SAS using PROC MIXED
to using lme4 package in R. In trying to match the coefficients of R
output to that of SAS output, I came across this problem.
The dataset I am using is this one:
Hi David,
Sorry if I have explain me wrong!
Mylist-list(A=data.frame, B=data.frame, C=data.frame, D=data.frame)
I know that this is a malformed command, only was a fast way to descrive
mylist.
Thank you!
-Mensaje original-
De: David Winsemius [mailto:dwinsem...@comcast.net]
Enviado el:
Hi Indrajit,
In your first SAS code, change to type=un. cs imposes the (somewhat
dubious) assumption that the variance of both the intercept and slope are
equal. If you are using lme4, all random effects in a single block (e.g.,
(1 + month | batch) the 1 = intercept and month = random slope)
Hi all,
I was running the adf test in R.
CODE 1:
adf.test(data$LOSS)
Augmented Dickey-Fuller Test
data: data$LOSS
Dickey-Fuller = -1.9864, Lag order = 2, p-value = 0.5775
alternative hypothesis: stationary
CODE 2:
adf.test(diff(diff(data$LOSS)))
Augmented Dickey-Fuller
Thanks a lot Joshua.
Regards,
Indrajit
On Tue, Apr 30, 2013 at 1:13 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
Hi Indrajit,
In your first SAS code, change to type=un. cs imposes the (somewhat
dubious) assumption that the variance of both the intercept and slope are
equal. If you are
Hi,
I am using Extrafont package to install more fonts for my graphs. My
primary graphic tool is ggplot2. I seem to have problem installing the
package, but could not pinpoint where it is.
I try to follow the instruction here:
https://github.com/wch/extrafont
I guess (but am not sure)
Dear All,
Is it possible to add a percentage completion bar to R? I find it
frustrating when running long calculations that I don't know at what
point the process is. It would be very helpful so I am not guessing if I
should wait a few minutes or hours.
On 30/04/2013 10:40, Fabio Berzaghi wrote:
Dear All,
Is it possible to add a percentage completion bar to R? I find it
frustrating when running long calculations that I don't know at what
point the process is. It would be very helpful so I am not guessing if I
should wait a few minutes or
On Tue, Apr 30, 2013 at 10:40 AM, Fabio Berzaghi f...@dmu.dk wrote:
Dear All,
Is it possible to add a percentage completion bar to R? I find it
frustrating when running long calculations that I don't know at what point
the process is. It would be very helpful so I am not guessing if I should
Hi,
May be this helps:
funcName- function(df1, x){
whatCol=df1[[x]]
print(Got it)
print(whatCol)
}
funcName(df,ColA)
#[1] Got it
#[1] 1 2 3 4 5
funcName(df,ColB)
#[1] Got it
#[1] A B C D E
#Levels: A B C D E
A.K.
I am trying to extract the 2nd column from a dataframe using a function
a) This looks like homework. The Posting Guide clearly indicates that this list
is not for homework help.
b) This is a statistics theory question that happens to use R, not an R
question that happens to be about statistics. Also off-topic per the Posting
Guide... there are other forums for
Hi,
Not sure if this is what you meant.
dat1- read.table(text=
id t scores scores2 scores3
2 0 1.2 1.4 1.9
2 2 2.3 2.5 2.2
2 3 3.6
Thanks to you all for your replies. I didn't realize bigmemory is only
available in Unix environments - when I saw
install.packages('bigmemory')
Installing package into C:/Users/BenC/Documents/R/win-library/3.0
(as lib is unspecified)
--- Please select a CRAN mirror for use in this session
This comes up regularly. Type ?print.survfit and look at the comments there under
value.
Terry T.
- begin included message
Hi,
I'm not sure if this is the proper way to ask questions, sorry if not. But
here's my problem:
I'm trying to do a bootstrap estimate of the
very helpful!! Thanks a lot.
From: Jim Lemon j...@bitwrit.com.au
Cc: r-help@r-project.org r-help@r-project.org
Sent: Monday, April 29, 2013 6:53 PM
Subject: Re: [R] biplot for principal componens analysis
On 04/30/2013 08:24 AM, capricy gao wrote:
I
-Original Message-
I have a dataset which for the sake of simplicity has two endpoints. We would like to test
if two different end-points have the same eventual meaning. To try and take an example
that people might understand better:
Lets assume we had a group of subjects who all
Hi there
I am a Masters student at the University of Stellenbosch. I have been using R
to analyze the data, using the GLS model, of one of my experiments.
The problem that I am having is that whenever I run my model using:
Question about package Coxme:
I develop a cox model that includes a variable treatment with 3 levels (A, B,
C):
model_alea_int - coxme(Surv(delai, status) ~ (1|trt)+ strata(center) , data)
I am surprised that the output given in R is 3 coefficients for random effects
whereas only 2 dummy
Dear R users,
For the purpose of validating a prediction model using validate() from the rms
package, I am running into some trouble with using the fastbw() function
breaking up natural groups of variables.
Is there any way I can specify to keep certain variable together? In
particular, if
Hi there,
I've been battling with an extension of this in my own data: getting
appropriate error bars once data is stacked in a bar graph.
(original question:
http://r.789695.n4.nabble.com/ggplot2-se-variable-in-geom-errorbar-s-limits-
td3311176.html). It wouldn't let me reply to that thread.
I am trying to find the percentage of the parameters explaining the bacterial
community composition. I have one data matrix with relative abundance of OTUs
and one with environmental parameters. I used varpart in vegan package but the
values in the venn diagram is bigger than 100% in total.How
Hi,
I was wondering if there is a package/function for Panel Granger
non-causality tests? I am interested in Toda-Yamamoto procedure in panel
data setting.
Thank you,
--
View this message in context:
http://r.789695.n4.nabble.com/Panel-Granger-Causality-Tests-tp4665834.html
Sent from the R
On Apr 30, 2013, at 6:00 AM, arun wrote:
Hi,
May be this helps:
funcName- function(df1, x){
whatCol=df1[[x]]
print(Got it)
print(whatCol)
}
funcName(df,ColA)
#[1] Got it
#[1] 1 2 3 4 5
funcName(df,ColB)
#[1] Got it
#[1] A B C D E
#Levels: A B C D E
To ST;
You should
Hello everyone,
I'm trying to perform some trend analysis using the package Kendall and for
some observations I get the following error WARNING Error exit, tauk2, IFAULT
= 12
Could anyone tell me how to create a logical vector of observations having its
element equal to TRUE if this error
That is unlikely in the absence of a reproducible example [1].
[1]
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
---
Jeff NewmillerThe . . Go
Hi Olvia,
Welcome to the R-help list.
Are you seriously proposing to add error bars to a stacked barplot? Why?
Do you mean a dodged barplot?
My first thought is that there probably are several better ways of presenting
your data than stacked bar charts--I sometimes think that anything other
I noticed that the points on the biplot are not exactly the same as the
predicted values.
Could any body give me a hint about why?
Thanks.
To: Jim Lemon j...@bitwrit.com.au
Cc: r-help@r-project.org r-help@r-project.org
Sent: Tuesday, April 30, 2013 9:51
Dear Rxperts,
Is there a simpler way to generate multipanel grouped individual profile
plots? All individuals of a group within a panel have the same color. As
of now I am using lattice::xyplot to get the desired effect. Please feel
free to suggest other ideas.
Also, I am trying to create a
Hi I have been trying for days now to read the contents of this xml file into
R. I tried the simpler solution xmlToDataFrame, which worked on another
file, but somehow it doesn't work for this file.
I have tried different other solutions, but I just can't seem to get it
quite right.
doc -
Actually, Jeff, it may not be.
The OP has likely overfit -- as a guess, Depth is essentially
continuous and as.factor() produces too many levels. I would recommend
that the OP find a local statistical consultant to help him, as he
seems out of his depth, statistically. Failing that, he should
Another alternative might be to post the the r-sig-mixed-models list
gls() is part of nlme.
-- Bert
On Tue, Apr 30, 2013 at 11:12 AM, Bert Gunter bgun...@gene.com wrote:
Actually, Jeff, it may not be.
The OP has likely overfit -- as a guess, Depth is essentially
continuous and as.factor()
Dear All,
I would like to store quote as part of an vector.
For instance, I would like to get an character object as
x = 12ab34
or
y = c(1, 2, , a, b, , 3, 4)
Is that possible? Thank you.
Best wishes,
Jie
[[alternative HTML version deleted]]
y = c(1, 2, '', a, b, '', 3, 4)
y
#[1] 1 2 \ a b \ 3 4
A.K.
- Original Message -
From: Jie jimmycl...@gmail.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Tuesday, April 30, 2013 2:46 PM
Subject: [R] Quote as element of a vector/list
Dear All,
I would like
Hi,
Try this:
dat1-read.table(text=
ID X0 X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15
1 5184 0 0 1 0 0 0 0 0 0 1 0 0 0
0 0 0
2 6884 0 0 1 0 0 1 0 0 0 0 0 0 0
0 0
Tena koe Jie
Try
x - ' 12ab34 '
plot(1:10, xlab=x)
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jie
Sent: Wednesday, 1 May 2013 6:46 a.m.
To: r-help@r-project.org
Subject: [R] Quote as element of a
Folks,
This is probably a help me google this properly, please-type
of question.
In TIBCO Spotfire, there is a procedure called line
similarity. I use this to determine which observations show a growing, stable
or declining pattern... sort of like a
Here is one way to, for each row in the data.frame v, regress the numbers in
columns 2 through 4 on the numbers 1 through 3, storing only the slopes, and
then creating a column saying if the slope is greater than zero or not.
v[,Beta] - vapply(seq_len(nrow(v)),
1. Read an Introduction to R or other R tutorial to learn how R works.
2. You apparently wish to apply a function, f, to each row of a data
frame or matrix to classify it as growing, declining, etc. Only you
know what that function should look like. Write it.
3. Apply it using ?apply or perhaps
Hi,
You could also do:
v- read.table(text=
Name Year_1_value Year_2_value Year_3_value
A 1 2 3
B 2 7 19
C 3 4 2
D 10 7 6
E 4 4 5
F NA 3 6
,sep=,header=TRUE,stringsAsFactors=FALSE)
names(v)[-1]-gsub((.*\\d+)_.*$,\\1,names(v)[-1])
v2- v
v1-reshape(v,direction=long,varying=2:4,sep=_)
Looks like my post might get some problems, so I re-wrote my question plus some
new one...
I noticed that the points on the biplot are not exactly the same as the
predicted values.
Another relevant question: should I expect that all the vector points have the
same length if chose
On Tue, Apr 30, 2013 at 5:12 PM, Dorte Klerke dkle...@hotmail.com wrote:
Hi I have been trying for days now to read the contents of this xml file into
R. I tried the simpler solution xmlToDataFrame, which worked on another
file, but somehow it doesn't work for this file.
How is this
Hi All,
i've problem to understand how to work with R to generate a hierarchical
clustering
my data are in a csv and looks like :
idcode,count,temp,sal,depth_m,subs
16001,136,4.308,32.828,63.46,47
16001,109,4.31,32.829,63.09,49
16001,107,4.302,32.822,62.54,47
16001,87,4.318,32.834,62.54,48
Hello,
I am work with a linear regression model:
y=ax+b with the function of lm.
y= observed migration distance of butterflies
x= predicted migration distance of butterflies
Usually the result will show
if the linear term a is significantly different from zero based on the
p-value.
Now
Greetings to r-help land.
I've run into some program crashes and I've traced them back to methods()
behavior
after the package gdata is loaded. I provide now a minimal re-producible
example. This seems bugish to me. How about you?
dat - data.frame(x = rnorm(100), y = rnorm(100))
lm1 - lm(y ~ x,
It is easy to construct your own test. I test against null of 0 first so I
can be sure I match the right result from summary.lm.
## get the standard error
seofb - sqrt(diag(vcov(lm1)))
## calculate t. Replace 0 by your null
myt - (coef(lm1) - 0)/seofb
mypval - 2*pt(abs(myt), lower.tail = FALSE,
Or use an offset
lm( y ~ x+offset(x), data = dat)
The offset gives x a coefficient of 1, so the coefficient of x in this
model is the difference between the coefficient of x in the model without
an offset and 1 -- the thing you want.
-thomas
On Wed, May 1, 2013 at 2:54 PM, Paul Johnson
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