Mike Williamson this.is.mvw at gmail.com writes:
I have a (relatively long) function script that generates a levelplot.
I don't want to include all of the code, so I have attached in file
miniDataSet.csv an example for the data set miniDataSet that is called
below.
Thanks for providing a
Luc, Eik, Jorge,
thanks to all of you!
Nice to see how many different solutions there are for the same
problem. :)
Best
Henning
Am 07.04.2009 um 22:25 schrieb Jorge Ivan Velez:
Dear Hans,
Try also:
x - structure(c(2, 1, 2, 1, 3, 2, 4, 1, 4, 3, 4, 3, 5, 2, 5, 1, 6,
1), .Dim = c(2L,
Thanks for reply David,
but the problem is that the place i get those files from saves them in the
old Excel format (Excel 5.0), not Excel 97-2003, so neither of the packages
(RODBC, xlsReadWrite) can read it.
Your question is a bit confusing. This is the current state of play
regarding
Dear R-mailing list,
I am using the fCopulae package.
Where can I find the relevant references for functions such as elliptical2d( )
? I am particularly interested in references pointing out how the correlation
coefficients can be incorporated in such density functions.
Thanks in advance
Hi,
I have problem. In the function below (test and test2) i want the function
test not to print the variable data but i want the function test2 to use the
variable test$data.
This is the creation of the variable data:
matrice=c(1:10)
matrice=matrix(matrice,nrow=5,ncol=2)
This is the
Hi R-experts,
I'm using the svm implentation available in the library 'e1071',
to train a classifier using cross validation
(...)
obj = tune.svm(truth ~ .,
data=examples,cost=2^(grid.cost),gamma=2^(grid.gamma),cross=k)
and to save the best model into a file as ?write.svm suggests:
Hello!
The following code is an implementation of a Poisson regression. It
generates some data-samples and computes the beta values with the negative
log likelihood function.
Now, my task is to compute the asymptotic convergence intervalls for the
values of beta but I dont know how to implement
Hi,
We are trying to write control messages (errors, warnings, ...) to a tktable
while the program is running. Until now we only managed to write all the
messages at once at the end of execution. How can we refresh the window in
order to show the tclArray content?
On Wed, 8 Apr 2009, Yuri Volchik wrote:
Thanks for reply David,
but the problem is that the place i get those files from saves them in the
old Excel format (Excel 5.0), not Excel 97-2003, so neither of the packages
(RODBC, xlsReadWrite) can read it.
RODBC supports any format for which an
On Tue, Apr 7, 2009 at 11:06 PM, jpearl01 joshea...@hotmail.com wrote:
There was an error in the file... an extraneous comma. That's taken care of.
however, my tree prints out an image that doesn't seem like a mst. Attached
is the csv file I used...
Well, it looks definitely a tree to me.
Thanks for a great package Duncan,
one of the most amazing things i saw for R platform.
In the meanwhile i can do with you suggestion: par3d(ignoreExtent=TRUE);
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Sent from the R help mailing
On Wed, Apr 08, 2009 at 10:02:10AM +0200, alberto cassese wrote:
Hi,
I have problem. In the function below (test and test2) i want the function
test not to print the variable data but i want the function test2 to use the
variable test$data.
This is the creation of the variable data:
Dear R and Tinn-R users,
i recently switched to Tinn-R and sending code to R works fine (R 2.8.1,
Tinn-R 2.2.0.2, OS Windows XP). However, i encountered a problem when
trying to send plots to pdf files like this:
library(lattice)
pdf(plot1.pdf)
PLOT-(xyplot, ...)
PLOT
dev.off()
The file
why
rm(list=ls(all=TRUE),envir=globalenv())
is ok but
ca-function() rm(list=ls(all=TRUE),envir=globalenv())
ca()
does not work?
Thank you!
v
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Sent from the R help
I must be missing something obvious here:
According to the help page for read.spss, the reencode option is only
active when R is run under a UTF-8 locale.
read.spss can only import the SPSS file when run under a iso88591(5)
locale, under a UTF-8 locale I get:
Error in read.spss(wo.sav) : error
Taraxacum88 wrote:
why
rm(list=ls(all=TRUE),envir=globalenv())
is ok but
ca-function() rm(list=ls(all=TRUE),envir=globalenv())
ca()
does not work?
ls(all=TRUE) from within your function lists all local variables, not
globals. You need envir=globalenv() there too.
Duncan Murdoch
Dear Henning,
You need to print() lattice plots when using a device:
library(lattice)
pdf(plot1.pdf)
PLOT-(xyplot, ...)
print(PLOT)
dev.off()
So this is not due to TINN-R.
HTH,
Thierry
ir. Thierry Onkelinx
On 8 Apr 2009, at 11:44, Duncan Murdoch wrote:
Mark Heckmann wrote:
Dear Duncan,
Thanks for the reply. This works, but unfortunately I need a
different
solution.
My script is supposed to run completely automated and the graphics
I produce
vary in size each time I run the script. But I
Dear Thierry,
thanks for your help, with solved the problem, although i don't why,
because when using the R editor accessible via the R console i created many
many lattice plots with the code i posted, i.e. without the print()
command.
Anyway, thanks a lot,
Henning
Dear All,
to my surprise as.factor does not accept a levels argument. Maybe I
did not read the documentation well enough. See the example below. I
wanted to use ch1 as factor in the newdata argument of survfit, so I
assumed that I could write as.factor(ch1, levels=ch1), since the
order
Hello,
I am a phd student in Bioinformatics and I am using the Random Forest
package in order to classify my data, but I have some questions.
Is there a function in order to visualize the trees, so as to get the rules?
Also, could you please provide me with the code of randomForest function,
as I
Thank you, Jim. I see, the fact that in the documentation you find only
as.factor(x) means that it does not accept more arguments.
Does as.factor have speed advantages over factor, or is there a
different cause for it's existence?
Heinz
At 13:50 08.04.2009, jim holtman wrote:
as.factor
Hi Melissa,
so what you want to calculate is something like
S-cumsum(lambs)-(1:length(lambs))*mean(lambs)
since (with x.bar=mean(x))
S_1=x_1-x.bar
S_2=S1+(x_2-x.bar)=x_1-x.bar+x_2-x_bar=x_1+x_2-2*x.bar
...
S_n=S_(n-1)+x_n-x.bar=sum_i (x_i)-n*x.bar
Eik.
m.mcquil...@lancaster.ac.uk schrieb:
Mark Heckmann wrote:
Dear Duncan,
Thanks for the reply. This works, but unfortunately I need a different
solution.
My script is supposed to run completely automated and the graphics I produce
vary in size each time I run the script. But I want the graphics to be
fitted to my .pdf output without
Hi Henning,
thanks for your help, with solved the problem, although i don't why,
because when using the R editor accessible via the R console i created many
many lattice plots with the code i posted, i.e. without the print()
command.
At the command line, R objects (including lattice plots
I read records using scan:
dat-data.frame(scan(file=KDA.csv,what=list(t=%m/%d/%y
%H:%M,f=0,p=0,d=0,o=0,s=0,a=0,l=0,c=0),skip=2,sep=,,nmax=np,flush=TRUE,na.strings=c(I/OTimeout,ArcOff-line)))
which results in:
dat[1:5,]
t fp d o sa l c
1 1/21/09 5:01 16151 8.2 76
Is there a way to do it in R?
Especially generating plot in EPS/PDF format.
By transparent I mean clear (not white) background.
I want to attached it to dark PPT slides.
- Gundala Viswanath
__
R-help@r-project.org mailing list
clr -
function ()
rm(list = ls(pos = .GlobalEnv), pos = .GlobalEnv)
this works for me
HTH,
baptiste
On 8 Apr 2009, at 10:50, Taraxacum88 wrote:
why
rm(list=ls(all=TRUE),envir=globalenv())
is ok but
ca-function() rm(list=ls(all=TRUE),envir=globalenv())
ca()
does not work?
Thank
Hello R users,
I've used the following help two compare two regression line slopes.
Wanted to test if they differ significantly:
Hi,
I've made a research about how to compare two regression line slopes
(of y versus x for 2 groups, group being a factor ) using R.
I knew the method based on the
On Tue, 2009-04-07 at 16:33 -0400, Donald Braman wrote:
I've been playing around with various table tools, trying to construct a
fairly simple cross-tab. It shouldn't be hard, but for some reason it
turning out to be (for me).
If I want to see how many men and how many women agree with a
as.factor does not accept levels as an argument. use the first form
that you have
factor(ch1, levels=ch1)
On Wed, Apr 8, 2009 at 7:36 AM, Heinz Tuechler tuech...@gmx.at wrote:
Dear All,
to my surprise as.factor does not accept a levels argument. Maybe I did not
read the documentation well
Dear Duncan,
Thanks for the reply. This works, but unfortunately I need a different
solution.
My script is supposed to run completely automated and the graphics I produce
vary in size each time I run the script. But I want the graphics to be
fitted to my .pdf output without specifying the height
I want to draw a grid rectangle without a border.
?gpar says:
Specifying the value NULL for a parameter is the same as not specifying any
value for that parameter, except for col and fill, where NULL indicates not
to draw a border or not to fill an area (respectively).
Hi Benedikt
in principle this is correct. A p-value less than a prespecified
alpha-level leads to rejection of the null hypothesis H0.
I'm a little bit puzzled about your given p-value, since actually your
slopes are the same, so H0 is valid but rejected. This might be a matter
of chance, but
with base graphics,
par(bg=NA)
see ?par
Hope this helps,
baptiste
On 8 Apr 2009, at 12:54, Gundala Viswanath wrote:
Is there a way to do it in R?
Especially generating plot in EPS/PDF format.
By transparent I mean clear (not white) background.
I want to attached it to dark PPT slides.
-
Rolf Turner-3 wrote:
On 8/04/2009, at 1:27 PM, Ben Bolker wrote:
snip
I agree that that the individual-level random effect is probably
the issue.
I played with this some today but didn't manage to resolve it --
tried JAGS/R2jags and glmer from lme4 but didn't manage to
Try varying the arguments to this to accommodate
the precise format of your data. See the three zoo
vignettes, ?read.zoo and R News 4/1 for dates
and times.
Lines - t,f,p,d,o,s,a,l,c
+ 1/21/09 5:01,16151,8.2,76,30,282,1060,53,7
+ 1/21/09 5:02,16256,8.3,76,23,282,1059,54,7
+ 1/21/09
Hello everybody,
I have a question. I would like to get a correlation between
constitutive and induced plant defence which I messured on 30 plant
species. So I have table with Species, Induced defence (ID), and
constitutive defence (CD). Since Induced and constitutive defence are
not
Hans Ekbrand wrote:
I must be missing something obvious here:
According to the help page for read.spss, the reencode option is only
active when R is run under a UTF-8 locale.
Not in my version:
reencode: logical: should character strings be re-encoded to the
current locale. The
if you not persist in using pdf, the emf format works very well with PPT
and has a transparent background.
Its the way you copypaste pdf-graphs form Acrobat to PPT which causes
the solid (white) background - unfortunately ppt has no builtt in
support for pdf-graphs.
By default pdfs have a
It is just a simple version of 'factor'. The only speed advantage it
might have is that it checks to see if it is a factor first. Here is
the definition:
) as.factor
function (x)
if (is.factor(x)) x else factor(x)
environment: namespace:base
You can always list out what the function does to
On Wed, Apr 08, 2009 at 03:03:06PM +0200, Peter Dalgaard wrote:
Hans Ekbrand wrote:
I must be missing something obvious here:
According to the help page for read.spss, the reencode option is only
active when R is run under a UTF-8 locale.
Not in my version:
reencode: logical: should
Jim - you are right, I should have looked before. So there is a
difference that should also effect the dropping of unused levels.
Thanks,
Heinz
At 15:31 08.04.2009, jim holtman wrote:
It is just a simple version of 'factor'. The only speed advantage it
might have is that it checks to see if
I have source a script running with no errors that ends in these lines:
==
for (i in 2:3) win.graph()
dev.set(2)
levelplot(avg ~ procs * size, rrt, drape=T,
colorkey=T,main=ops/sec,scales=list(x=list(tick.number=15),y=list(log=10)))
dev.set(3)
levelplot(avg ~ procs * size, ort,
Hi Jacy,
I'll take my chances answering your question.
I'd suggest using the paste() function and then transforming the string
into a formula object:
my.formula =
as.formula(paste(y~x1+x2+,paste(colnames(REG)[2:num],collapse=+),
sep= )) ;
aov(my.formula,data=REG) ;
The first line
Currently I am using the R write command to output results to a *.txt file
and then copying those results into an initialization file. In an attempt to
continue to automate the process I would like to have R write to the location
in the existing initialization file, instead of me copying the
On Wed, 2009-04-08 at 10:05 -0400, SHANE MILLER, BLOOMBERG/ 731 LEXIN
wrote:
I have source a script running with no errors that ends in these lines:
==
for (i in 2:3) win.graph()
dev.set(2)
levelplot(avg ~ procs * size, rrt, drape=T,
Numbers like ``1239178547.653775 is inserted into a vector. I print the vector:
route_9_80_end
[1] 1239178522 1239178526 1239178524 1239178524 1239178524 1239178523
1239178524 1239178522 1239178521 1239178565 1239178566 1239178566
[13] 1239178565 1239178566 1239178566 1239178565 1239178566
Hans Ekbrand wrote:
On Wed, Apr 08, 2009 at 03:03:06PM +0200, Peter Dalgaard wrote:
Hans Ekbrand wrote:
I must be missing something obvious here:
According to the help page for read.spss, the reencode option is only
active when R is run under a UTF-8 locale.
Not in my version:
reencode:
[Ooops, brain/fingers not well connected today, sent prematurely]
On Wed, 2009-04-08 at 10:05 -0400, SHANE MILLER, BLOOMBERG/ 731 LEXIN
wrote:
I have source a script running with no errors that ends in these lines:
==
for (i in 2:3) win.graph()
dev.set(2)
levelplot(avg ~ procs *
./configure works with no errors ?
I have compiled on mac os x and debian so my advice may be
circumspect, but the errors I don't know what they are.
On Tue, Apr 7, 2009 at 10:17 PM, tomkur2006-takeh...@yahoo.com wrote:
Hello,
Is there any binary version of R 2.8.x available for x86 Solaris
Seems like the place where i get those files from has an error somewhere when
they create the file which doesn't allow ODBC to see sheet with data, i.e
sqlTables returns:
[1] TABLE_CAT TABLE_SCHEM TABLE_NAME TABLE_TYPE REMARKS
0 rows (or 0-length row.names)
When i open this file in excel
Looks like that code implements a non-exhaustive variant of the
randomization test, sometimes called a permutation test. Assuming you
have a data frame with columns ID CD, this should do it:
n.obs = 100
ID=rnorm(n.obs)
CD=rnorm(n.obs)
obs.cor = cor(ID,CD)
num.permutations = 1e4
perm.cor =
Assuming you have a data frame with columns ID CD, this should do it
Oops, the code I sent doesn't assume this. It assumes that you have
two vectors, ID CD, as generated in the first 3 lines.
On Wed, Apr 8, 2009 at 11:31 AM, Mike Lawrence mike.lawre...@dal.ca wrote:
Looks like that code
*Hits head*
Of course, the approach taken by your Genstat code of only shuffling
one variable is sufficient and faster:
n.obs = 100
ID=rnorm(n.obs)
CD=rnorm(n.obs)
obs.cor = cor(ID,CD)
num.permutations = 1e4
perm.cor = rep(NA,num.permutations)
start.time=proc.time()[1]
index = 1:n.obs
for(i in
Numbers like ``1239178547.653775 is inserted into a vector. I print
the vector:
route_9_80_end
[1] 1239178522 1239178526 1239178524 1239178524 1239178524
1239178523 1239178524 1239178522 1239178521 1239178565 1239178566
1239178566
[13] 1239178565 1239178566 1239178566 1239178565
*sigh*
and the elapsed time line should be:
cat('Elapsed time:',proc.time()[1]-start.time)
On Wed, Apr 8, 2009 at 11:35 AM, Mike Lawrence mike.lawre...@dal.ca wrote:
*Hits head*
Of course, the approach taken by your Genstat code of only shuffling
one variable is sufficient and faster:
On Wed, Apr 08, 2009 at 04:17:51PM +0200, Peter Dalgaard wrote:
Hans Ekbrand wrote:
Someone running foreign 8.34 that is willing to test my SPSS-file?
Someone with an SPSS file problem willing to help test the prereleases? :-)
http://sociologi.cjb.net/temp/test.sav
--
Hans Ekbrand
On 4/8/2009 11:01 AM, Camarda, Carlo Giovanni wrote:
Dear R-users,
within the rgl-package, I would have a question about the usage of persp3d in combination of rgl.viewpoint.
I am not able to figure out how to let a 3D plot rotating around likewise the example in ?rgl.viewpoint. It seems
Hi,
I drew an ellipse with the package ellipse. Now I would like to know if
a point is inside the ellipse. Is any R functions to do it without
computing the equation of the ellipse manually? Thanks.
For example, if I do plot(ellipse(0.8), type = 'l'), I would like to
know if (0,1) belongs
At 00:48 04/04/2009, Duncan Murdoch wrote:
On 03/04/2009 5:37 PM, Emmanuel Charpentier wrote:
Le vendredi 03 avril 2009 à 14:17 -0400, stephen sefick a écrit :
I am starting to use R for almost any sort of calculation that I need.
I am a biologist that works in the states, and there is
OK, I needed to plot a set of vectors on top of a contour plot. I
figured out a way to do this. I create a panel function that calls
larrows() with arguments constructed from my vector data. Then, when
I go to do the contour plot, I call contourplot() with the panel
argument set to point to my
Eik Vettorazzi wrote, On 08.04.2009 15:08:
By default pdfs have a transparent background, see ?pdf and there the bg
part,
which can be proven by a minimal R+ LaTeX example
#R code
pdf(test.pdf)
plot(1,1)
dev.off()
#minimal tex example
\documentclass[a4paper,12pt]{article}
Dear Duncan,
thanks for your prompt reply.
I was tempted to use movie3d, but below I copied what I get on my PC. Then, I
have ImageMagick installed and, though I set convert equal to FALSE, I was not
able to find where the function writes the .png files.
Maybe I lack experience in
Converting dates is getting stranger still. I am coercing a data frame
into a ts as follows:
tst1-as.POSIXct(1/21/09 5:01,format=%m/%d/%y %H:%M)
tst2-as.POSIXct(1/28/09 3:40,format=%m/%d/%y %H:%M)
tsdat-as.ts(dat,start=tst1,end=tst2,frequency=1)
This generates a ts object. But strangely enough
On 4/8/2009 11:47 AM, Camarda, Carlo Giovanni wrote:
Dear Duncan,
thanks for your prompt reply.
I was tempted to use movie3d, but below I copied what I get on my PC. Then, I have ImageMagick installed and, though I set convert equal to FALSE, I was not able to find where the function writes
Hello,
I am working with S-plus TimeSeries Object. I wonder how can I change the
column names of the variable instead of using the one default?
i.e to change DEV.MSCI to other name
PositionsDEV.MSCI
04/30/1980 00:00:00.000 -0.0150328542
05/31/1980 00:00:00.000
Hi everyone,
I try to make a netcdf file which disposes a difference between 2 variables
of 2netcdf files the same dimension
When I programmed under R, everything is ok but when I put the code under
EOF of a sh script, an error occurs:
Error, passed variable has a dim that is NOT of class
Dear R Users,
I am using the reshape package to reformat gridded data into column format
using the code shown below. However, when I display the resulting object, a
single column is fomed (instead of three) and all the latitude values (which
should be in either column one or two) are
On 4/8/2009 11:17 AM, Alain Guillet wrote:
Hi,
I drew an ellipse with the package ellipse. Now I would like to know if
a point is inside the ellipse. Is any R functions to do it without
computing the equation of the ellipse manually? Thanks.
For example, if I do plot(ellipse(0.8), type =
Hi there,
I have been using the example provided bellow for a while, and It was
working without any problem. Nevertheless, just since 2-3 days is not
working, probably because I did update.packages(). I have tried to
re-install the older versions of the packages Hmisc() and xtable(), but
Is this what you are looking for:
x
Positions DEV.MSCI
04/30/1980 00:00:00.000 -0.0150328542
05/31/1980 00:00:00.000 0.0005087752
06/30/1980 00:00:00.000 0.0586794492
07/31/1980 00:00:00.000 0.0458505592
08/31/1980 00:00:00.000 0.0350926728
colnames(x)[2] - NewName
x
I am inserting a DNA sequence into a plot, and hope to colourize each
of the four nucleotide of the DNA sequence with a unique colour i.e.,
A (red), C (green), G (blue, and T (yellow). I use the
following codes, but the DNA sequence only shows as red
DNA - ACGT
plot(1, xlim = c(0,1), ylim =
Mike Lawrence wrote:
Looks like that code implements a non-exhaustive variant of the
randomization test, sometimes called a permutation test.
Isn't it the other way around? (Permutation tests can be exhaustive by
looking at all permutations, if a randomization test did that, then it
On 4/8/09, Tobias Verbeke tobias.verb...@openanalytics.be wrote:
Hi Henning,
thanks for your help, with solved the problem, although i don't why,
because when using the R editor accessible via the R console i created many
many lattice plots with the code i posted, i.e. without the print()
Use 'text' to write out each one:
plot(0, type='n', xlim=c(0,1), ylim=c(0,1))
text(seq(0,1,length=10), rep(0.5,10), LETTERS[1:10], col=1:10)
On Wed, Apr 8, 2009 at 12:15 PM, Daren Tan darenta...@gmail.com wrote:
I am inserting a DNA sequence into a plot, and hope to colourize each
of the
On 4/8/09, Cable, Samuel B Civ USAF AFMC AFRL/RVBXI
samuel.ca...@hanscom.af.mil wrote:
OK, I needed to plot a set of vectors on top of a contour plot. I
figured out a way to do this. I create a panel function that calls
larrows() with arguments constructed from my vector data. Then, when
Hi everyone,
I am trying to compare proportions among groups using the logistic
regression
approach as follows:
1) Fit the model log(p_i/(1-p_i)) = M + G_i, where p_i is the probability
of success in group i and G_i is the effect of group i, i=1,..,I.
2) Test the hypotheses:
Ho: G_1 = G_2
On 4/7/09, Don McKenzie d...@u.washington.edu wrote:
I'm plotting the following (stripped of inessentials)
xyplot(sd ~ distance |
wshed,data=sdvar.df,scales=list(x=list(log=TRUE),y=list(log=TRUE)))
sdvar.df is a data frame, sd and distance are numeric, wshed is an ordered
factor
This sounds very similar to what I've been working on, but I'm not sure
without an example.
My solution has been to use an optimization that normalizes inside the
objective function. The betas that are provided by optim are not
normalized, however since they were normalized inside the objective
Yes, ./configure works with no error.
I guess it's a problem of my environment setup. It's my first attempt to build
it myself, and I have downloaded the binaries and installed it previously.
--- On Wed, 4/8/09, stephen sefick ssef...@gmail.com wrote:
From: stephen sefick ssef...@gmail.com
Dear R users,
I have the following problem. Suppose I have the following toy data set:
data
m1 m2 m3 m4 m5 state
[1,] 1 0 1 13 23 2
[2,] 0 1 0 23 94 2
[3,] 1 0 0 45 56 1
[4,] 0 1 0 35 84 2
[5,] 1 1 0 98 37 1
[6,] 1 1 0 68 1 2
where the
Hello everyone,
I'm trying to use mle from package stats4 to fit a bi/multi-modal
distribution to some data, but I have some problems with it.
Here's what I'm doing (for a bimodal distribution):
# Build some fake binormally distributed data, the procedure fails also with
real data, so the
I am not sure what you mean. Of course you can plot it using different
layouts, e.g. with layout.reingold.tilford (after choosing the root
vertex in some way) and then it looks like a usual tree plot, but why
would that be any better?
I'd like to be able to distinguish between the nodes better.
I was taught that Fisher proposed the F-test as a computationally
simpler approximation to what he called a Randomization test,
consisting of exhaustive permutations. I never looked at the original
Fisher reference myself, so this may be false.
However, I haven't observed a consistent
Does solaris have a package managment system? And do you have all of
the packages for building? It looks like you do, but I am not sure.
On Wed, Apr 8, 2009 at 12:42 PM, tomkur2006-takeh...@yahoo.com wrote:
Yes, ./configure works with no error.
I guess it's a problem of my environment
have you tried using zoo and then using the function as.ts()
On Wed, Apr 8, 2009 at 11:56 AM, am...@xs4all.nl wrote:
Converting dates is getting stranger still. I am coercing a data frame
into a ts as follows:
tst1-as.POSIXct(1/21/09 5:01,format=%m/%d/%y %H:%M)
tst2-as.POSIXct(1/28/09
can you provide a reproducible example?
On Wed, Apr 8, 2009 at 11:59 AM, Steve Murray smurray...@hotmail.com wrote:
Dear R Users,
I am using the reshape package to reformat gridded data into column format
using the code shown below. However, when I display the resulting object, a
single
Hans Ekbrand wrote:
On Wed, Apr 08, 2009 at 04:17:51PM +0200, Peter Dalgaard wrote:
Hans Ekbrand wrote:
Someone running foreign 8.34 that is willing to test my SPSS-file?
Someone with an SPSS file problem willing to help test the prereleases? :-)
http://sociologi.cjb.net/temp/test.sav
No
Your problem is related to the theory underlying linear models (and is an
example as to why it is important to understand the theory, not just know how
to plug numbers into a computer).
The lm function is based on theory that assumes the y variable in normally
distributed with the mean of that
On Wed, Apr 8, 2009 at 6:12 PM, jpearl01 joshea...@hotmail.com wrote:
I am not sure what you mean. Of course you can plot it using different
layouts, e.g. with layout.reingold.tilford (after choosing the root
vertex in some way) and then it looks like a usual tree plot, but why
would that be any
I don't understand your first question, but, since no one else has
responded I can answer your second question. panel.bwplot, unlike
panel.xyplot doesn't use panel.superpose when groups is not NULL. In
order to get an analogous result you need to specify that you want to
use panel.superpose.
cols
On Wed, Apr 08, 2009 at 07:12:23PM +0200, Peter Dalgaard wrote:
Apparently, you can work around it like this
lc - Sys.setlocale(LC_CTYPE)
Sys.setlocale(LC_CTYPE, da_DK)
x - read.spss(~/Desktop/downloads/test.sav, reencode = latin1)
Sys.setlocale(LC_CTYPE, lc)
-- which doesn't strike me as
Hi Greg,
Thanks for your guidance.
In this case, the evidence is that the primary subpopulation of the data,
accounting for observes the standard statistical model in the sense that Rice
uses the term. It may by all accounts be normally distributed, and a Q-Q shows
a large portion of the
Hello R-Help,
I ran some analysis and were hit with some low Z-score. I tried to convert
it to a p-value, however, it seems like the ceiling is around 1e-16.
1-pnorm(8)
[1] 6.661338e-16
1-pnorm(9)
[1] 0
Do you have any suggestion how I can display a very low p-value in the form
of scientific
Hi All,
I have a very simple graph:
cars - c(1, 3, 6, 4, 9)
trucks - c(2, 5, 4, 5, 12)
year - c(2004, 2005, 2006, 2007, 2008)
df2-data.frame(cars,trucks,year)
xyplot(cars+trucks~year, data=df2, type=o)
I need to show the values of cars on the graph. How can I do this?
Thanks.
--
View this
The source code of the whole package is available on CRAN. All packages
are submitted to CRAN is source form.
There's no rule per se that gives the final prediction, as the final
prediction is the result of plural vote by all trees in the forest.
You may want to look at the varUsed() and
?options look at scipen
On Wed, Apr 8, 2009 at 2:38 PM, Bhoom Suktitipat suktiti...@gmail.com wrote:
Hello R-Help,
I ran some analysis and were hit with some low Z-score. I tried to convert
it to a p-value, however, it seems like the ceiling is around 1e-16.
1-pnorm(8)
[1] 6.661338e-16
in this case you need to use the 'lower.tail' argument, e.g.,
pnorm(8:15, lower.tail = FALSE)
I hope it helps.
Best,
Dimitris
Bhoom Suktitipat wrote:
Hello R-Help,
I ran some analysis and were hit with some low Z-score. I tried to convert
it to a p-value, however, it seems like the
with ggplot2,
d - melt(df2,id=year)
qplot(year,value,data=d,colour=variable,geom=c(line,point)) +
geom_text(data= subset(d, variable == cars), aes(label=value))
with lattice, my best guess would be to use grid.text in a custom
panel function.
Hope this helps,
baptiste
On 8 Apr 2009, at
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