Got it. Thanks very much.
On 5/28/07, Prof Brian Ripley [EMAIL PROTECTED] wrote:
How about 'traceback'? (It does not necesssarily show all the frames, but
it does help and the exceptions are fairly esoteric.)
On Mon, 28 May 2007, ronggui wrote:
In that the meaning of parent.frame depends
Thanks Brian,
I have
Load glx
in the Module section of xorg.conf -
However, based on my latest probing the graphics card is not in the
opensuse xgl database, and the 3D acceleration is not enabled. So I will
have to wait till a new package comes available with the driver for the
intel GMA
dear useRs,
I have several datasets which are in repeated measurements format and I
would like to extract the first measurement for each individual. In some
cases measurements of certain characteristics are also stacked on top of
each other and thats where the problem lies. A prototype of the
Deepayan Sarkar a écrit :
On 5/27/07, Patrick Giraudoux [EMAIL PROTECTED] wrote:
Dear all,
After going through the Lattice doc and R-help list and google, I got
the feeling that there is no function in lattice or other package to
compute a pie chart object of class trellis. Although pie
Dear Gabor,
In order to perform your suggestion I needed to split my 'big' 720*5551
matrix into small ones of the type: 720*400 due to memory constraints.
But after performing the task I get less rows in the new matrix. For
example:
zz1-zz[,1:400]
dim(zz1)
[1] 720 400
zz1[,1]
1985-01-05
Adrian Dragulescu wrote:
Check the documentation link from
http://cm.bell-labs.com/cm/ms/departments/sia/project/trellis/software.writing.html
Adrian
On 5/26/07, Tyler Smith [EMAIL PROTECTED] wrote:
Hi,
I've just produced my first lattice plot - the graphic is very
impressive, but I
I've been using heatmap(), hclust(), and cutree(), and I'm trying to figure
out the column labels of the left (as opposed to the right) cluster.
Using cutree(x, k=2), I get two clusters labeled 1 and 2. How do you figure
out if cluster 2 is the one associated with the left cluster on the heatmap
Dear R-experts,
Sorry if I've overlooked a simple solution here. I have calculated a
proportion of the number of observations which meet a criteria, applied to
five years of data. How can I break down this proportion statistic for each
year?
For example (data in zoo format):
Thank you, Ravi.
You probably will have noticed, that the attachment didn't make
it to the mailing list.
The reason is that the we let the mailing list software strip
binary attachments which can easily be misused to spread
viruses; see -- http://www.r-project.org/mail.html (search attachment)
Duncan == Duncan Murdoch [EMAIL PROTECTED]
on Fri, 25 May 2007 15:38:54 -0400 writes:
Duncan On 5/25/2007 1:09 PM, Prof Brian Ripley wrote:
On Fri, 25 May 2007, Martin Maechler wrote:
path.expand(~)
[1] /home/maechler
Yes, but beware that may not do what
On Mon, 28 May 2007, Martin Maechler wrote:
Duncan == Duncan Murdoch [EMAIL PROTECTED]
on Fri, 25 May 2007 15:38:54 -0400 writes:
Duncan On 5/25/2007 1:09 PM, Prof Brian Ripley wrote:
On Fri, 25 May 2007, Martin Maechler wrote:
path.expand(~)
[1] /home/maechler
LuckeJF == Lucke, Joseph F [EMAIL PROTECTED]
on Fri, 25 May 2007 12:29:49 -0500 writes:
LuckeJF Most standard tests, such as t-tests and ANOVA,
LuckeJF are fairly resistant to non-normalilty for
LuckeJF significance testing. It's the sample means that
LuckeJF have to be
apply(dfr,1,FUN=function(x){
cat(c(x[1],
ifelse(x[2]==x[3],x[2],paste(x[2],x[3],sep=-)),\n),
file=filename.txt)
})
This code assumes the data frame with at least 3 columns, errors will occur
if there will be
On Mon, 2007-05-28 at 10:58 +0800, Ruixin ZHU wrote:
Hi everybody,
When I followed a practice example, I got an error as follows:
###
cc-read.table('example5_2.dat',header=TRUE)
Here is one way to break it down to years.
x -open high lowclose hc lc
+ 2004-12-29 4135 4135 4106 4116 8 -21
+ 2004-12-30 4120 4131 4115 4119 15 -1
+ 2004-12-31 4123 4124 4114 4117 5 -5
+ 2005-01-04 4106 4137 4103 4137 20 -14
+ 2005-01-06 4085 4110 4085
Without a small reproducible example there is not much to say.
Try cutting the columns down to half successively until you have
an object with 4 columns that exhibits the same behavior and then
do the same with rows until you get a 4x6 example.
Here is another, slightly shorter, solution:
This is really a query about MySQL which I am trying to use to set up a
database which I will then access with RODBC.
I have my data in a .csv file, and some of the fields are date/time
fields. I tried to create a table using mysql with the definition of the
date/time field given by
CallDate
Here are a couple of solutions:
1. using zoo package
First add Date to the header so there
are the same number of column headers as columns and
then read in using read.zoo. Then aggregate over years
using mean. For more on zoo try library(zoo); vignette(zoo)
and for more on dates see the R
I do this by having a POSIXct variable in R and using sqlSave in RODBC to
write the SQL table. Examples are part of my test suite, so I am sure
they do work.
It is not at all clear how you are trying to import the data into MySQL.
On Tue, 29 May 2007, David Scott wrote:
This is really a
Jim Lemon a écrit :
Patrick Giraudoux wrote:
Dear all,
After going through the Lattice doc and R-help list and google, I got
the feeling that there is no function in lattice or other package to
compute a pie chart object of class trellis. Although pie charts
are obviously not considered
P
Yes indeed. Thats' likely what I am going to do. Anyway, to plot axes,
labels of sophisticated graphs on maps may be interesting anyway. For
instance, we are monitoring fox and hare populations in tens of game
areas. Drawing observations (panel.xyplot) over time and representing
the
Without any code in your post (see last line of every post to r-help)
its impossible to really know what you are doing but here is an
example:
library(RODBC)
channel - odbcConnect(db01) # existing data base
# add a table with a datetime field
sqlQuery(channel, create table table02 (field01
Hi R-programmers !
I would like to perform a linear model regression month by month using the
'lm' function and i don't know how to do it.
The data is organised as below:
Month ExcessReturn Return STO
8 0.047595875 0.05274292 0.854352503
8 0.016134874 0.049226941 4.399372005
8
Hello,
I have been trying to compute the following:
for(t in 3:n){
h[t]-(omegac/(1-betac))+alphac*Y[t-1]
b[t]-Y[t-1]
bb[t]-0
for(j in 2:(t-1)){
h[t]-h[t]+alphac*((betac)^(j-1))*Y[t-j]
Hi R-programmers !
I would like to perform a linear model regressio using the 'lm' function and
i don't know how to do it.
The data is organised as below:
Month ExcessReturn Return STO
8 0.047595875 0.05274292 0.854352503
8 0.016134874 0.049226941 4.399372005
8 -0.000443869
Hi
I am trying to install the package pheno, but it needs the package
nprq by Roger Koenker et al. which I can I find this package? It does
not seem to be on CRAN and googling also doesn't give me an URL - is it
still somewhere available?
Thanks,
Rainer
--
NEW EMAIL ADDRESS AND ADDRESS:
(... sorry! the first post was in HTML format ...)
Dear R-users,
I want to perform a One-Sample parametric bootstrapped
Kolmogorov-Smirnov GoF test (note package Matching provides
ks.boot which is a 2-sample non-parametric bootstrapped K-S
version).
So I wrote this code:
---[R Code] ---
Hi,
I have 2 text files the first one is in the following format:
CC: Some statements
...
this is continued for 11-12 lines of variable length.
then a line containing the variable names used in the file and then the
variable fields with their
Roberto Brunelli wrote:
I'm using an S4 object with a slot of type 'call': I would like to be able to
initialize
it with something like NULL or NA (indicating that there is no information in
the slot)
but the value should comply with the fact that it must be of type call.
Is there any
Taka Matzmoto wrote:
Dear R-users,
I have a data.frame having NA (e.g., 2nd, 4th rows, etc).
Start End Length Variable_name
[1,] 1 1 1a
[2,] NANANA
[3,] 2 2 1b
[4,] NANANA
[5,] 3 6 4c
[6,] 7 10 4d
:
:
:
I like to
In case you haven't seen this, there is an example in Paul Murrell's book
that plots temperatures on a map using 'thermometer' charts. I would
imagine it should be relatively straight forward to combine the floating.pie
function with Paul's grid-base code (but I have not tried it myself).
On Mon, 28 May 2007, Uwe Ligges wrote:
Roberto Brunelli wrote:
I'm using an S4 object with a slot of type 'call': I would like to be able
to initialize
it with something like NULL or NA (indicating that there is no information
in the slot)
but the value should comply with the fact that
Here are three ways. The first uses a common
error term and the others separate. They all
give the same coefficient estimates.
Lines - Month ExcessReturn Return STO
8 0.047595875 0.05274292 0.854352503
8 0.016134874 0.049226941 4.399372005
8 -0.000443869 0.004357305 -1.04980297
9
Prof Brian Ripley wrote:
On Mon, 28 May 2007, Uwe Ligges wrote:
Roberto Brunelli wrote:
I'm using an S4 object with a slot of type 'call': I would like to be
able to initialize
it with something like NULL or NA (indicating that there is no
information in the slot)
but the value
Hi,
I was not able to find a way such that R CMD build run Sweave with eval=FALSE.
I have a package where I want the R code in the vignettes to run only during R
CMD check but not on R CMD build. Is there any way to achieve this ?
thank you.
On 5/28/2007 2:11 PM, johan Faux wrote:
Hi,
I was not able to find a way such that R CMD build run Sweave with
eval=FALSE. I have a package where I want the R code in the vignettes to run
only during R CMD check but not on R CMD build. Is there any way to achieve
this ?
You can use the
On Mon, 28 May 2007, Martin Maechler wrote:
LuckeJF == Lucke, Joseph F [EMAIL PROTECTED]
on Fri, 25 May 2007 12:29:49 -0500 writes:
LuckeJF Most standard tests, such as t-tests and ANOVA,
LuckeJF are fairly resistant to non-normalilty for
LuckeJF significance testing. It's the
Hello,
I have installed R2.5.0 from sources ( x86_64 )
and added the package RODBC
and now I am trying to connect to a mysql database
In windows R after installing the 3.51 driver
and creating the dsn by specifying server, user, and password
it is easy to connect with
channel - odbcConnect(dsn)
yOn Mon, 28 May 2007, Bill Szkotnicki wrote:
Hello,
I have installed R2.5.0 from sources ( x86_64 )
and added the package RODBC
and now I am trying to connect to a mysql database
In windows R after installing the 3.51 driver
and creating the dsn by specifying server, user, and password
it
This may sound like a very naive question, but...
give two lists of coordinate pairs (x,y - Cartesian space) is there any simple
way to compute the affine transformation matrix in R.
I have a set of data which is offset from where i know it should be. I have
coordinates of the current data,
Isn't this just a regression (hopefully with a near-zero error).
coef(lm(cbind(xnew, ynew) ~ xold + yold))
should do what I think you are asking for. (I am not clear which
direction you want the transformation, so choose 'old' and 'new'
accordingly.)
On Mon, 28 May 2007, Dylan Beaudette
Hi All
Had anyone of you seen if it is possible to split a large lme() job to be
processed by multiple cpus/computers?
I am just at the very beginning to understand related things, but does the
lme()
use solution finding functions like nlm() mle(), and I-dont-know-what-else of
the standard R
tmp - data.frame(y=rnorm(12), a=factor(rep(letters[1:4],3)))
tmp
y a
1 -0.60866099 a
2 0.55500538 b
3 0.12231693 c
4 -0.24613790 d
5 -0.09253593 a
6 -1.54652581 b
7 0.17204210 c
8 -1.22778942 d
9 1.22151194 a
10 -0.43982577 b
11 -1.25444287 c
12 -0.97251060 d
tmp.aov -
Don't know what it happening, but when I compared the output of 'str(tmp.aov)'
before and after the 'source', I saw at least the following differences:
before:
$ terms:Classes 'terms', 'formula' length 3 y ~ a
.. ..- attr(*, variables)= language list(y, a)
.. ..- attr(*, factors)=
dat -
+ Month ExcessReturn Return STO
+ 8 0.047595875 0.05274292 0.854352503
+ 8 0.016134874 0.049226941 4.399372005
+ 8 -0.000443869 0.004357305 -1.04980297
+ 9 0.002206554 -0.089068828 0.544809429
+ 9 0.021296551 0.003795071 0.226875834
+ 9 0.006741578 0.014104606
I have a matrix of characters (actually numbers that have been read in as
numbers), and I'd like to remove the NA.
I'm familiar with na.omit, but is there an equivalent of na.omit when the NA
are the actual characters NA?
Thanks,
Andrew
[[alternative HTML version deleted]]
Roberto Brunelli [EMAIL PROTECTED] writes:
I'm using an S4 object with a slot of type 'call': I would like to
be able to initialize it with something like NULL or NA (indicating
that there is no information in the slot) but the value should
comply with the fact that it must be of type call.
HiI have a class which would like to hold
a connection to ODBC data connection. The handle to the connection is
something like (according to RODBC manual):channel =
odbcConnect(DSN,uid=user,pwd=password)However, I would like to
make a slot of my foo class to hold this object. For
Hi all,
I am trying to figure out the formula used by R's Spearman rho (using
cor(method=spearman)) because I can't seem to get the same value as by
calculating by hand. Perhaps I'm using cor wrong, but I don't know
where. Basically, I am running these commands:
Since they are characters you can just compare for them. You did not show
what your data looks like, or what you want to do if there are NA. Do you
want the row removed? You can use 'apply' to test a row for NA:
x - matrix(0,5,5)
x[1,3] - x[4,4] - NA
x
[,1] [,2] [,3] [,4] [,5]
[1,] 0
Hi,
You can try with
cor.test(rank(y[1]),rank(y[2]))
On 5/29/07, Raymond Wan [EMAIL PROTECTED] wrote:
Hi all,
I am trying to figure out the formula used by R's Spearman rho (using
cor(method=spearman)) because I can't seem to get the same value as by
calculating by hand. Perhaps I'm
Surely all you need to do is change them to missing and use na.omit.
e.g.
x - matrix(letters, 13, 2)
x[4] - NA # put one in the middle
is.na(x[x == NA]) - TRUE
(x - na.omit(x))
Bill Venables
CSIRO Laboratories
PO Box 120, Cleveland, 4163
AUSTRALIA
Office Phone (email preferred): +61 7 3826
You can also use type.convert() if you did want to convert your characters to
numbers and NA's to NA's so that you can use na.omit().
x - matrix(0,5,5)
x[1,3] - x[4,4] - NA
newx - apply(x,2,type.convert)
newx
[,1] [,2] [,3] [,4] [,5]
[1,]00 NA00
[2,]0000
Hi,
Chung-hong Chan wrote:
Hi,
You can try with
cor.test(rank(y[1]),rank(y[2]))
Thanks for this! It didn't solve my problem, but it helped me realize
that the formula I was using by hand is invalid for the tie case. I
just realized that with R's cor function, the Pearson correlation
54 matches
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