Dear All,
How to calculate multiple correlation coefficient R, using R.
regards,
Sukumar
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Dear All,
How to calculate multiple correlation coefficient R, using R.
regards,
Sukumar
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Error in La.svd(x, nu, nv) : error code 1 from Lapack routine dgesdd
what resources are there to track down errors like this
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At 3:50 PM -0700 5/3/10, John Kane wrote:
I think that you are correct. R has the annoying habit of
converting character data to factors when you don't want it to while
it is importing data. This is because the in the option
"stringsAsFactors" is set to TRUE for some weird historical reasons
thanks.. worked
On Mon, May 3, 2010 at 3:24 PM, Ted Harding wrote:
> On 03-May-10 21:19:34, steven mosher wrote:
> > The situation arises where I open a file to write a data.frame
> > to it. with write.table.
> >
> > multiple lines are written to the file and the file is kept in
> > Append=TRUE
On Mon, 3 May 2010, Tim Clark wrote:
Dear List,
I am trying to do model averaging for a negative binomial model
using the package AICcmodavg. I need to use glm() since the package
does not accept glm.nb() models. I can get glm() to work if I first
run glm.nb and take theta from that model,
On Mon, 3 May 2010, Carl Witthoft wrote:
I wanted to ask Eric a question or two about the rtiff package, but his
listed email address bounces w/ 550 error. Does anyone know how to reach
him, or whether he's actively maintaining rtiff?
He is not. The latest version of rtiff was done by the
Hi Roscoe,
I couldn't make it though that excel worksheet without my eyes glazing
over... but do you want something like
data(mtcars)
tab <- xtabs(~cyl + gear, data=mtcars)
tab/sum(tab)
?
Best,
Isat
On Mon, May 3, 2010 at 10:08 PM, Roscoe wrote:
> Dear R-Help,
>
> First post, be gentle...
>
>
Hi Joseph,
How about this?
matplot(cbind(m0, m1, m3, m4), type = 'l', lty = 1)
legend('topright', paste('m', c(0, 1, 3, 4), sep = ""), lty = 1, col = 1:4)
See ?matplot and ?legend for details.
HTH,
Jorge
On Mon, May 3, 2010 at 6:42 PM, <> wrote:
> R-listers:
>
> I have searched the help file
On May 3, 2010, at 9:48 PM, Daniel Malter wrote:
Hi, on the one hand, you write "fairly large," on the other hand,
you write
"should be readable by anything." The warnings indicate that you are
plain
out of memory at some point. Not too surprising, given that your
dataset has
about 45
Dear R-Help,
First post, be gentle...
I am new to R: I can perform basic functions using R code - import and
export data, load packages and not much else at this stage.
Objective: to create a Joint Probability Table and save as a csv file
(with headers).
Input: A csv file (with headers) contai
Thank you. It works!
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R-listers:
I have searched the help files and everything I have related to R
graphics. I cannot find how to graph y against
several distributions on a single graph. Here is code for creating 4
Poisson distributions with different mean values, although I would
prefer having it in a loop: The to
Dear List,
I am trying to do model averaging for a negative binomial model using the
package AICcmodavg. I need to use glm() since the package does not accept
glm.nb() models. I can get glm() to work if I first run glm.nb and take theta
from that model, but is there a simpler way to estimate
This is not really the correct forum, but very briefly, and in rough order of
complexity, you can do a summary (which counts the occurences of the
categories in a factor), you could look at crosstabulations of questions
(e.g ?table, ?tabulate, ?chisq.test), and also measures of association in
tabl
Ok I will try that. I think I set it to False when I tried it the first
time, maybe that was my mistake
On Mon, May 3, 2010 at 3:28 PM, Ista Zahn wrote:
> Hi Steve,
> I think you just need to set col.names = FALSE (instead of col.names
> =NULL) on subsequent writes.
>
> -Ista
>
> On Mon, May 3,
Dear Tal,
There are several approaches in doing it (see Steiger, 2003). It
should not be difficult to implement them in R.
Steiger, J.H. (2003). Comparing correlations. In A. Maydeu-Olivares
(Ed.) Psychometrics. A festschrift to Roderick P. McDonald. Mahwah,
NJ: Lawrence Erlbaum Associates.
http
Something like this?
plot(0,0)
legend.text<-c("VAR 1","Higher","Average","Lower","VAR
2","Higher","Average","Lower")
legend.pch=1:8
legend.col=c(0,1,1,1,0,1,1,1)
legend("bottomright",legend=legend.text,pch=legend.pch,col=legend.col,ncol=2)
-tgs
On Mon, May 3, 2010 at 3:10 PM, Nish wrote:
>
> H
Hi, on the one hand, you write "fairly large," on the other hand, you write
"should be readable by anything." The warnings indicate that you are plain
out of memory at some point. Not too surprising, given that your dataset has
about 45 rows and 720 columns. You may search the r-help files firs
The stringAsFactors = False option did not work either.
False <> FALSE
in addition Vincent was getting an error since he used stringAsFactors, instead
of stringsAsFactors as the argument name.
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Check you input data. You have some non-numeric characters in columns where
you are expecting numerics.
The parameter is
stringsAsFactors=FALSE
You had it spelt wrong. After reading in your data, do the conversion to
numeric and then examine which locations contain NA; this will point you to
t
I wanted to ask Eric a question or two about the rtiff package, but his
listed email address bounces w/ 550 error. Does anyone know how to
reach him, or whether he's actively maintaining rtiff?
If anyone's interested, my primary desire is for rtiff (or other tool)
to provide me with the raw
On May 3, 2010, at 7:21 PM, vincent.deluard wrote:
Hi John,
Thanks for responding so quickly.
So I tried:
Test2[1:3,1:3])
'data.frame': 3 obs. of 3 variables:
$ Date : Factor w/ 2480 levels "A UN Equity",..: 1033 2396 613
$ X4.20.2010: Factor w/ 1994 levels "0.24","0.33",..: 953
Hi John,
Thanks for responding so quickly.
So I tried:
Test2[1:3,1:3])
'data.frame': 3 obs. of 3 variables:
$ Date : Factor w/ 2480 levels "A UN Equity",..: 1033 2396 613
$ X4.20.2010: Factor w/ 1994 levels "0.24","0.33",..: 953 497 1814
$ X4.19.2010: Factor w/ 1957 lev
On May 3, 2010, at 6:16 PM, Carson Baughman wrote:
All-
Thank you in advance for any help you might be able to lend.
Here is
my issue. I am trying to open a fairly large .dat file. The file
originally was downloaded as a GZ file but I unzipped it (with 7-
zip) into
it's current 1.8
Hi David,
Thanks for the prompt answer.
In this file, I have cut the data to the first 15 columns.
Here is the code:
t2 <- read.csv('Test2.csv', sep=",", header=TRUE)
> Test2[2,2]
[1] 18.72
1994 Levels: 0.24 0.33 0.49 0.53 0.65 0.67 0.72 0.76 0.88 0.9 0.94 ... 98.88
> Test2[2,2]+1
A small executable example would help here. We don't even know how you are
graphing this. There are several packages including lattice and ggplot as well
as the base graphs.
If you are doing this in base graphics have a look at ?text
--- On Mon, 5/3/10, Nish wrote:
> From: Nish
> Subject:
I think that you are correct. R has the annoying habit of converting
character data to factors when you don't want it to while it is importing data.
This is because the in the option "stringsAsFactors" is set to TRUE for some
weird historical reasons.
Try the command str(insert name of data)
Dear R gurus,
A re-post due to suggestions from moderators.
Thanks to tips from Gabor and Felix, I was able to make some progress.. How
do I control the position of panel.text? I would like the text appear at a
specific position (say, top right corner) inside a panel.
Below is the sample code
All-
Thank you in advance for any help you might be able to lend. Here is
my issue. I am trying to open a fairly large .dat file. The file
originally was downloaded as a GZ file but I unzipped it (with 7-zip) into
it's current 1.86 gig .dat format. I know that the data is "just a plain
A
In the ESL(2) (by F-H-T), they have labels on their Lasso graphs
(specifically for the South African Heart Disease data).
So my question is, how do I label each variable in a similar way on my
graph.
The following code should produce a plot with 9 enumerated variables:
library(glmnet)
x=mat
Hello,
I am new to R and need some help with the legend. How can I add a legend for
two variables (in two columns) each having multiple values to be explained
in the legend. For example:
Var 1 Var 2
- Higher - Higher
- Avg - Avg
- Lower
On May 3, 2010, at 6:22 PM, vincent.deluard wrote:
Hi there,
This will sound very stupid because I just started using R but I see
you had
similar problems.
I just loaded a very large dataset (2950*6602) from csv into R. The
format
is ticker=row, date=column.
Not a particularly precis
Hi Steve,
I think you just need to set col.names = FALSE (instead of col.names
=NULL) on subsequent writes.
-Ista
On Mon, May 3, 2010 at 5:19 PM, steven mosher wrote:
> The situation arises where I open a file to write a data.frame to it. with
> write.table.
>
> multiple lines are written to the
On 03-May-10 21:19:34, steven mosher wrote:
> The situation arises where I open a file to write a data.frame
> to it. with write.table.
>
> multiple lines are written to the file and the file is kept in
> Append=TRUE
> mode.
>
> If one sets the col.names to the names of the variables being
> wri
Hi there,
This will sound very stupid because I just started using R but I see you had
similar problems.
I just loaded a very large dataset (2950*6602) from csv into R. The format
is ticker=row, date=column.
Every time I want to compute basic operations, R returns "In Ops.factor: not
meaningful
Dear John,
The R documentation is of course not "perfect", but it is (in my
estimation) quite good. See more specific responses below, keeping in
mind that I'm actually pretty ignorant of this things myslef.
On Mon, May 3, 2010 at 4:44 PM, John Sorkin wrote:
> Many thanks to both David and Ista.
The situation arises where I open a file to write a data.frame to it. with
write.table.
multiple lines are written to the file and the file is kept in Append=TRUE
mode.
If one sets the col.names to the names of the variables being written, you
have output
that looks like this...
name1 name2 na
Dear all,
I'm looking for an implementation of the generalized extended Kalman filter
for survival data, presented in this article Fahrmeir (1994) - 'dynamic
modelling for discrete time survival data'. The same author also publish a
Bayesian version of the algorithm 'dynamic discrete-time duration
Many thanks to both David and Ista. I think that my question regarding
extracting items returned from lmer points out a weakness of R, viz. less than
perfect documentation. I know that R is written by volunteers, all of whom I
cherish, yet having to ask the listserve for a question as basic as
After having read through some literature on R, I am happy with collecting
it, and visualizing numbers. Unfortunately I do find it hard to compare and
visualize string data. The data is from a LimeSurvey with lots of YES/NO or
multi-choice questions (check-boxes)
Among the check-boxes there a coup
Thanks David, the text() worked for me. I wasnt able to correctly use plot
with points and the ecdf function.
David Winsemius wrote:
>
>
> Yes, ecdf() which returns a function
>
> points(c(50,55,60,65), ecdf(1:1000)( c(50,55,60,65) ), col=c("red",
> "green", "yellow", "blue") )
>
> Draws
I tried with this:
> sessionInfo()
R version 2.10.1 (2009-12-14)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
[
On May 3, 2010, at 3:16 PM, SamT wrote:
For simplicity we'll assume my code looks as follows
plot(ecdf(1:1000))
which produces a diagonal line.
Ideally, I'd like to highlight or label in a different color certain X
values (say 50,55,60,65)
Is there a function which will allow me to do this
For simplicity we'll assume my code looks as follows
plot(ecdf(1:1000))
which produces a diagonal line.
Ideally, I'd like to highlight or label in a different color certain X
values (say 50,55,60,65)
Is there a function which will allow me to do this?
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On 03/05/2010 2:43 PM, anderson nuel wrote:
I will sur that problem is coming from the first part of the test.
data[pa,k] is a vector because pa is a vector.
Coud you help me to solve this error.
If you want to test that all elements match, you should use
all(data[pa, k] == df[, j]) &&
On Mon, May 3, 2010 at 7:21 PM, wei x1 wrote:
>
> hell all:
>
> I have a vector as follows:
>
>> head(res)
> 1007_s_at.value 1053_at.value 117_at.value 121_at.value
> 1255_g_at.value
> 0.225801033 0.009747404 0.709517954 0.008825740 0.496859178
> 1294_at.value
> 0.0
?rev
On Mon, May 3, 2010 at 4:06 PM, Steve Hempell wrote:
> Can you reverse the color scheme order in rcolorbrewer. If so, how?
>
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>
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Can you reverse the color scheme order in rcolorbrewer. If so, how?
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I will sur that problem is coming from the first part of the test.
data[pa,k] is a vector because pa is a vector.
Coud you help me to solve this error.
Best Regards
2010/4/30 Duncan Murdoch
> On 30/04/2010 4:19 AM, anderson nuel wrote:
>
>> Dear r-help,
>>
>> Could you help me to find a
hell all:
I have a vector as follows:
> head(res)
1007_s_at.value 1053_at.value117_at.value121_at.value 1255_g_at.value
0.225801033 0.009747404 0.709517954 0.008825740 0.496859178
1294_at.value
0.005091231
after I convert the res into a data frame I got
Hello all,
I believe this can be done using bootstrap, but I am wondering if there is
some other way that might be used to tackle this.
#Let's say I have two pairs of samples:
set.seed(100)
s1 <- rnorm(100)
s2 <- s1 + rnorm(100)
x1 <- s1[1:99]
y1 <- s2[1:99]
x2 <- x1
y2 <- s2[2:100]
#And both
Or better:
tapply(x0, y0, mean)
On Mon, May 3, 2010 at 2:35 PM, Henrique Dallazuanna wrote:
> Try this:
>
> rowsum(x0, y0)[,1]/table(y0)
>
>
> On Mon, May 3, 2010 at 2:11 PM, song song wrote:
>
>> x0=rnorm(100)
>> y0=rpois(100,3)+1
>>
>> ind=as.data.frame(table(y0))
>>
>> ind1=ind[,1]
>> ind2=i
I believe you are trying to find the mean of x0 for each level (group) of y0.
try this:
by(x0, y0, mean)
or if you want a vector (e.g., to merge into a matrix)
c(by(x0, y0, mean))
Best regards,
Josh
On Mon, May 3, 2010 at 10:11 AM, song song wrote:
> x0=rnorm(100)
> y0=rpois(100,3)+1
>
> in
Try this:
rowsum(x0, y0)[,1]/table(y0)
On Mon, May 3, 2010 at 2:11 PM, song song wrote:
> x0=rnorm(100)
> y0=rpois(100,3)+1
>
> ind=as.data.frame(table(y0))
>
> ind1=ind[,1]
> ind2=ind[,2]
>
> phi=NULL
> for (i in 1:length(ind2)){
> phi[i]=sum(x0[y0==ind1[i]])/ind2[i]
> }
>
>[[alterna
On May 3, 2010, at 12:28 PM, SamT wrote:
Hi,
I have plotted a cdf using the ecdf function with plot().
Code would make this concrete.
I want to
highlight/identify points on the same plot. Also, because this is a
cdf I
am not sure of the y coords for the point, otherwise I thought of
as.vector(do.call(c, lapply(split(x0, y0), mean)))
Test with data generated according to your code:
> phi=NULL
> for (i in 1:length(ind2)){
+ phi[i]=sum(x0[y0==ind1[i]])/ind2[i]
+ }
> phi
[1] -0.18922774 0.36333115 -0.04295032 -0.13892563 -0.03968301 0.33326034
[7] 0.28649576 -0.03786830
> a
Here is a workaround:
library(zoo)
# test data
z <- zoo(cbind(1:5, NA, c(1:3, NA, 5), NA))
ix <- colSums(!is.na(z)) > 0
z[, ix] <- na.approx(z[, ix])
On Mon, May 3, 2010 at 12:41 PM, Abiel X Reinhart
wrote:
> I am trying to run na.approx on a zoo object in which some of the columns
> contain
Hi Marshall,
I'm not aware of any packages that implement these features as you
described them. But most of the tasks are already fairly easy in R --
see below.
On Mon, May 3, 2010 at 11:18 AM, Marshall Feldman wrote:
>
> Thanks for getting back so quickly Ista,
>
> I was actually casting about fo
On May 3, 2010, at 9:59 AM, someone wrote:
I dont want to apply the unique for all columns but just the ID
column.
dataset[ !duplicated(dataset$ID), ]
--
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West Hartford, CT
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Hi,
I have plotted a cdf using the ecdf function with plot(). I want to
highlight/identify points on the same plot. Also, because this is a cdf I
am not sure of the y coords for the point, otherwise I thought of using
highlight in the NCStats package.
Thanks
--
View this message in context:
On May 3, 2010, at 11:43 AM, serdal ozusaglam wrote:
Hi Serdal,
> There is a lot of confusion here (how much is yours and how much is
> mine remains to be seen). See specific comments in line.
Also inline comments.
>
> On Mon, May 3, 2010 at 9:19 AM, serdal ozusaglam
> wrote:
>>
>> Dear R u
I am trying to run na.approx on a zoo object in which some of the columns
contain nothing by NA values. When I do this I get the following error:
Error in approx(x[!na], y[!na], xout, ...) :
need at least two non-NA values to interpolate
Is there a way I can use na.approx with my dataset so t
x0=rnorm(100)
y0=rpois(100,3)+1
ind=as.data.frame(table(y0))
ind1=ind[,1]
ind2=ind[,2]
phi=NULL
for (i in 1:length(ind2)){
phi[i]=sum(x0[y0==ind1[i]])/ind2[i]
}
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How about
yourdata[ !duplicated(yourdata$ID), ]
?
See ?duplicated for more information.
HTH,
Jorge
On Mon, May 3, 2010 at 9:04 AM, someone <> wrote:
>
> as a r noob i am having another problem:
> i have a big dataframe where each row corresponds to one entry and each
> column is a field...
>
Hi:
Here are three solutions; since this question comes up fairly often, you can
find
other solutions in the R-help archives.
(1) Use functions from base R: split the data frame by ID, extract the first
record from each split and slurp them together with rbind():
> do.call(rbind, lapply(sp
Hi Serdal,
> There is a lot of confusion here (how much is yours and how much is
> mine remains to be seen). See specific comments in line.
Also inline comments.
>
> On Mon, May 3, 2010 at 9:19 AM, serdal ozusaglam
> wrote:
>>
>> Dear R users,
>>
>>
>> I think i have a simple question w
On 05/03/2010 08:47 AM, Chris Evans wrote:
Thanks Tal& Thomas, I am now experimenting with both SWord and R2wd and
both are certainly a huge step forward for me, tied as I am to Word and
the Windoze/M$ world for now.
Chris
Note that many of the general solutions offered produce documents (.do
Thank you for your advice, ill try to be more explicit now, i wasnt in
the first mail because i thought it is a simple question to answer,
so; i have a panel data which contains 48858 observations during 3 year
therefore, there are 146574 observations in total,
i have 22 different industri
Thank you for your advice, ill try to be more explicit now, i wasnt in the
first mail because i thought it is a simple question to answer,
so; i have a panel data which contains 48858 observations during 3 year
therefore, there are 146574 observations in total,
i have 22 different industries d
Thanks for getting back so quickly Ista,
I was actually casting about for any examples of R software that deals
with this kind of structure. But your question is a good one. Here are a
few things I'd like to be able to do:
* Store data in R at the finest level of detail but easily refer to
could you please elaborate a little more on that?
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names()
-
Lanna Jin
lanna...@gmail.com
510-898-8525
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I dont want to apply the unique for all columns but just the ID column.
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See
example(gam)
particularly
plot(gam.object,se=TRUE).
On Mon, May 3, 2010 at 5:20 AM, Oscar Saenz de Miera wrote:
>
> To whoever it may correspond,
>
> My name is Oscar Saenz and I am working on my thesis in Spain.
>
> I am using GAMs in "R" and, now that I have estimated my models, I need t
plus, it took long time to complete this analysis for only 24 records!
Although I get the error msg, the result seems to be correct.
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Sent from the R help mailing list arch
I am conducting a very simple t test for two genes using lapply (i try to
avoid loop since i will have thousands of genes later on). however, I got
strange error msg like the followings. It looks that R is complaining my
factor has only one level, which is not the case (I check many times).
> res
Does anyone know how to deal with DATES in MySQL database connections?
dbWriteTable converts dates in a data frame to a character.
Thanks!
--
Dr. Brett Ginsburg
Assistant Professor
Department of Psychiatry
The University of Texas Health Science Center at San Antonio
San Antonio, TX 78229
>
210-5
On May 3, 2010, at 10:38 AM, Ista Zahn wrote:
Hi Serdal,
There is a lot of confusion here (how much is yours and how much is
mine remains to be seen). See specific comments in line.
Also inline comments.
On Mon, May 3, 2010 at 9:19 AM, serdal ozusaglam
wrote:
Dear R users,
I think i h
Hi Serdal,
There is a lot of confusion here (how much is yours and how much is
mine remains to be seen). See specific comments in line.
On Mon, May 3, 2010 at 9:19 AM, serdal ozusaglam
wrote:
>
> Dear R users,
>
>
> I think i have a simple question which i want to explain by an example;
>
> i hav
Hi John,
David's advice is probably sufficient, but I thought I would just add
the following example, showing how to find out what extractor
functions have been defined for a particular s4 class:
class(fm2) #find out the class of fm2 ("mer")
showMethods(classes="mer") # find out what methods are d
On May 3, 2010, at 9:41 AM, John Sorkin wrote:
R 2.10.0
windows XP
I am trying to get the coefficients returned from lmer. Normally I
would use the names function to get the objects returned by the
function. This does not work in lmer, but it does for lm:
When you read the help page for
hello,
i did median based linear regression and computed conf.intervals for my
coefficients.
but something must have went wrong as the estimates are out of estimates
confidence bounds.
does someone know what is the problem here.
i get warning messages about wilcox.test is not able to do calculat
thanks for your effort.
to be more precise:
ID , OS, time and many more are the columns.
each entry is a row.
when I do:
x <- unique(dataset$ID)
It just gives me a list of all IDs (levels).
I want to get a dataframe where just one entry (row) for each ID is
included...
like:
userA , Win, 12:
if that doesn't work, maybe also try:
if x is your data frame with length a columns, "unique(x[,1:a])".
-
Lanna Jin
lanna...@gmail.com
510-898-8525
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Did you try: if x is the data frame, unique(x)?
-
Lanna Jin
lanna...@gmail.com
510-898-8525
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_
Hi Marshell,
What exactly do you mean by "handles this kind of data structure"?
What do you want R to do?
Best,
Ista
On Mon, May 3, 2010 at 9:44 AM, Marshall Feldman wrote:
> Hello,
>
> Hierarchical factors are a very common data structure. For instance, one
> might have municipalities within st
as a r noob i am having another problem:
i have a big dataframe where each row corresponds to one entry and each
column is a field...
for instance, i have the column ID and time and many more...
Id like to get a dataframe where all IDs are just included once (some users
with that ID might have sev
Thanks Tal & Thomas, I am now experimenting with both SWord and R2wd and
both are certainly a huge step forward for me, tied as I am to Word and
the Windoze/M$ world for now.
Chris
Tal Galili sent the following at 01/05/2010 09:44:
> Hi all,
> I forwarded this question to the r-com mailing li
Hello,
Hierarchical factors are a very common data structure. For instance, one
might have municipalities within states within countries within
continents. Other examples include occupational codes, biological
species, software types (R within statistical software within analytical
software),
R 2.10.0
windows XP
I am trying to get the coefficients returned from lmer. Normally I would use
the names function to get the objects returned by the function. This does not
work in lmer, but it does for lm:
<- lmer(Reaction ~ Days + (1|Subject) + (0+Days|Subject), sleepstudy)
> names(fm2)
NU
Try, unique(dataset[,1:a]), where a is the number of columns that you have.
1:a would apply the unique to all columns.
-
Lanna Jin
lanna...@gmail.com
510-898-8525
--
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Sent fr
Thanks Tal/Petr, I regularly use the clipboard option but for most of
what I do people would prefer to see things in proportional fonts and
I'd love to be able to reformat tabulations easily to get nice layouts.
Of course, I've got very adept at global search and replace of
successive spaces with
Dear R users,
I think i have a simple question which i want to explain by an example;
i have several 2-digit industry codes that i want to use for conducting
by-industry analysis but i think there is a problem with the degrees of freedom!
for example, when i do my analysis without any 2-digit
Hi,
The model specification formula language introduced in Chambers and
Hastie potentially handles rather complex models. Typically the user
specifies the model and in a separate argument specifies the method. For
example, one specifies a general linear model with glm(formula,family).
But wit
To whoever it may correspond,
My name is Oscar Saenz and I am working on my thesis in Spain.
I am using GAMs in "R" and, now that I have estimated my models, I need to plot
the predicted smooth functions against the dependent variable (just as in
Carlslaw et al. 2007, fig. 1*).
Otherwise,
On 05/03/2010 09:43 PM, Gabriele Esposito wrote:
Good afternoon,
I am looking for a way to do a scatterplot of 4 values summing to 1
inside a 3D symplex, i.e. an equilateral pyramid. With the function
triax.plot I can do that with 3 values summing to 1, but I can't find
an equivalent with an ext
On 05/03/2010 05:15 PM, John Poulsen wrote:
Hello,
I have used layout() to produce to 2 plots on a page, leaving a plotting space above them. I would
like "Legend", which will actually be a real legend, to be centered above the two graphs.
Right now I am only able to position "Legend" above
See the describe functions in these packages:
Hmisc
pysch
prettyR
On Mon, May 3, 2010 at 8:12 AM, adrien Penac wrote:
> Thank a lot for these answers.
>
> Some of you wondered why I needed
> to do that!
>
> In fact, I have not so big data.frame whith many
> columns (98) and many of them are simi
Thank a lot for these answers.
Some of you wondered why I needed
to do that!
In fact, I have not so big data.frame whith many
columns (98) and many of them are similar (many binary answers, some
factor data and a few quantitative datas).
As I am a lazy guy, I
wanted to do a sort of function
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