Dear All
I want to generate variable with Bivariate Normal Distribution by
use mean1 = a, variance1 = b, mean2 = c, variance2 = d, rho = e.
How I can do this.
Many Thanks.
IRD
[[alternative HTML version deleted]]
Hi
r-help-boun...@r-project.org napsal dne 20.10.2010 07:58:29:
Hi Josh,
What I'm really trying to do is to refer to objects whose names I have
stored in a vector. This example was arbitrary.
I do a lot of looping through files in the working directory, or through
objects in the namespace,
On Tue, Oct 19, 2010 at 10:58 PM, Daniel Weitzenfeld
dweitzenf...@gmail.com wrote:
Hi Josh,
What I'm really trying to do is to refer to objects whose names I have
stored in a vector. This example was arbitrary.
I do a lot of looping through files in the working directory, or through
objects
Il giorno mar, 19/10/2010 alle 11.12 -0700, Dieter Menne ha scritto:
Thanks Dieter for your help, but unfortunately your suggestion results
only in changing the color of the *lines* and not the color of the
*area* of the polygon.
I also tried to call col from within the panel.superpose
Ottorino-Luca Pantani wrote:
Thanks Dieter for your help, but unfortunately your suggestion results
only in changing the color of the *lines* and not the color of the
*area* of the polygon.
The following complete code works for me. Do you have the current version
of lattice/R
You could e.g. use the package mvtnorm (not horribly fast, but handy):
(assuming that rho is the correlation)
library(mvtnorm)
m - c(mean1, mean2) #mean vector
cov - rho*sqrt(variance1*variance2)
sig - matrix(c(variance1, cov, cov, variance2), nrow=2) #covariance matrix
rmvnorm(100, mean=m,
Il giorno mer, 20/10/2010 alle 00.05 -0700, Dieter Menne ha scritto:
The following complete code works for me. Do you have the current
version
of lattice/R installed?
So the problem could be somewhere else. I suspected some of this kind.
RShowDoc(NEWS, package = lattice)
tell me that is
You could also use lattice + Hmisc:
# Fake data (taken from Dennis Murphy's reply):
mortrate - round(runif(100), 3)
moe = 1.96 * sqrt(mortrate * (1 - mortrate))/10
dd - data.frame(rate = mortrate, CI.lower = mortrate - moe, CI.upper =
mortrate + moe, hosp = factor(paste('H', 1:100, sep = '')))
#
Hi:
scale_manual() is a little tricky when you build the legend from within the
plot. I used shorter labels than you, but this worked for me:
ggplot(mydata, aes(y=score2, x=score1)) +
geom_point() +
stat_quantile(quantiles=c(0.50), aes(colour='red'), size = 1) +
Hi,
I am trying to calculate confidence intervals using ci.numeric from epicalc
package. If I generate a normal set of data and find the 99% and 95% CI, they
seem too narrow to me. Am I doing something wrong?? The IQR goes from -0.62 to
0.62, so I thought the CI limits should be more extreme
Hi all,
We currently faced a problem about how to pass a matrix with different
types to a COM function call thru the RDCOMClient asCOMArray().
As we all know that matrix in R can't hold of elements with different
types. So we managed to create a matrix of lists:
rates - matrix(
sachinthaka.abeywardana wrote:
I get the error:
A-read.table(P:/temp.csv,header=TRUE, sep=,);
Error in scan(file, what, nmax, sep, dec, quote, skip, nlines,
na.strings, :
line 11018 did not have 85 elements
but when i open it in excel everything is fine.
Better open it with
That works - Thanks Ravi.
Muhammad Rahiz
On Tue, 19 Oct 2010, Ravi Varadhan wrote:
You can do this.
dsm = c(800,600,NA,525,NA,450,400,NA,NA,NA,0)
s3 = seq(0.05,1.05,0.1)
plot(s3,dsm,col=blue,las=1,xlab=fraction,ylab=distance (km))
fc - function(x,a,b){a*exp(-b*x)}
fm -
Hi Dieter:
I think the OP wanted both lines and shading; from your code I could get the
shading but not the lines. This is what it took for me to get the lines
(note the type and col.line changes in xyplot() ):
panel.bands - function(x, y, upper, lower,
subscripts, ..., font,
djmuseR wrote:
Hi Dieter:
I think the OP wanted both lines and shading; from your code I could get
the
shading but not the lines. This is what it took for me to get the lines
(note the type and col.line changes in xyplot() ):
xyplot(est ~ x | cond, group = grp, data = data, type =
Hi,
I use a mboost model to predict my dependent variable on new data. I get the
following warning message:
In bs(mf[[i]], knots = args$knots[[i]]$knots, degree = args$degree, :
some 'x' values beyond boundary knots may cause ill-conditioned bases
The new predicted values are partly
On Wed, Oct 13, 2010 at 3:46 AM, Mikhail Titov m...@gmx.us wrote:
Hello!
I posted a feature request at lattice page on r-forge at
https://r-forge.r-project.org/tracker/index.php?func=detailaid=1127group_id=638atid=2570
,
but I decided to duplicate it here to make sure that I understand
Hello,
I am trying to achieve something which I *think* is possible using rowsum,
but a little help should be useful:
Consider the following dataframe DF0:
A B C
89 1 140
89 06 20
89 29 137
89 52 13
89 57 10
89 97 23
89
Dear List,
I am relatively new to R and am trying to create more attractive plots than
excel can manage!
I have looked through the various programmes ggplot, lattice, hmisc etc but my
case seems to be not metnioned, maybe it is but i have not noticed - if this is
the case i apologise.
hi rob, i encountered the same problem on a students laptop and have no idea
anymore after trying updating, re-installing and just about everything. like
you say, the jri.dll file is exactly at the same place as on your machine.
let me know if and how you solved it! thanks, kat
--
View this
Works for me! Thanks, Dieter!
Regards,
Dennis
On Wed, Oct 20, 2010 at 2:07 AM, Dieter Menne
dieter.me...@menne-biomed.dewrote:
djmuseR wrote:
Hi Dieter:
I think the OP wanted both lines and shading; from your code I could get
the
shading but not the lines. This is what it took
Hi:
Here's one way, although it can be improved a bit.
d1 - aggregate(C ~ A, data = subset(DF0, B == 1), FUN = sum)
d2 - subset(DF0, B != 1)
# B not in d1, so need to replace it
d1
A C
1 52 124
2 57 64
3 89 192
d1$B - rep(1, nrow(d1))
d1 - d1[, c(1, 3, 2)] # reorder columns to permit
Senior Quantitative Research Analyst
Based in Irish Life Investment Managers, Lower Beresford Place, Dublin 1
Irish Life Investment Managers (ILIM), the investment management company within
the Irish Life Permanent Group manages assets of circa EUR30bn and provides
fund management
Or even better (doh!)...
library(plyr)
ddply(DF0, .(A, B), summarise, C = sum(C))
A B C
1 52 1 124
2 52 59 38
3 52 97 75
4 57 1 64
5 57 6 26
6 57 114 12
7 89 1 192
8 89 6 20
9 89 29 137
10 89 52 13
11 89 57 10
12 89 97 23
which means that aggregate(),
Am 19.10.2010 23:23, schrieb Greg Snow:
Look at my.symbols and ms.image in the TeachingDemos package.
Thank you, that's I was looking for.
Knut
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Proof of last assertion - sorry, I just had to follow through...
library(data.table)
DT0 - data.table(DF0, key = 'A, B')
DT0[, sum(C), by = 'A, B']
A B V1
[1,] 52 1 124
[2,] 52 59 38
[3,] 52 97 75
[4,] 57 1 64
[5,] 57 6 26
[6,] 57 114 12
[7,] 89 1 192
[8,] 89 6
Hi:
Your confidence intervals are so short that the size of the point in the
graphics region covers the endpoints! You also have a wide range of
simulated means (0 - 52) and actual values (0 - 54). Here are some measures
of your CIs:
with(simvsact, max(simCI.upper - simCI.lower))# maximum
This summaryBy worked wonders! Thanks a lot djmuseR, I own you a beer!
library(doBy)
summaryBy(C ~ A + B, data = DF0, FUN = sum)
A B C.sum
1 52 1 124
2 52 5938
3 52 9775
4 57 164
5 57 626
6 57 11412
7 89 1 192
8 89 620
9 89 29 137
10 89
update.packages() updates all packages in all libraries listed in
.libPaths() unless you specify an explicit library.
It may happen that the version number has not changed and you just want
to reinstall for your upgraded R. In that case use:
update.packages(checkBuilt=TRUE)
Best,
Uwe Ligges
Thanks Uwe,
It may operate like that on most peoples machines, but either its not
operating like that on mine. Or I have another problem :-(
As u can see from my code below I've run update.packages(checkBuilt=TRUE)
and my 'private' library is in my LibPaths()...
However when I try to load the
Another option to consider:
x
A B C
1 89 1 140
2 89 6 20
3 89 29 137
4 89 52 13
5 89 57 10
6 89 97 23
7 89 1 37
8 89 1 12
9 89 1 3
10 52 1 11
11 52 1 31
12 52 1 16
13 52 1 6
14 52 1 10
15 52 1 13
16 52 1 10
17 52 1 25
18 52 1 2
On 20.10.2010 13:59, Chris Howden wrote:
Thanks Uwe,
It may operate like that on most peoples machines, but either its not
operating like that on mine. Or I have another problem :-(
As u can see from my code below I've run update.packages(checkBuilt=TRUE)
and my 'private' library is in my
I have two vectors
min
0.2, 0.3, 0.6, 0.1
max
0.4,0.5,0.7,0.4
Is there a way to combine these two vector so that the values will be taken
from the vectors alternating so that I will get a new
minmaxminmaxminmax-vector?
minmax
0.2,0.4,0.3,0.5,0.6,0.7,0.1,0.4
Any help is much appreciated
--
Thanks Uwe,
I was wondering if it was something like that.
I'll delete the base packages from my personal library.
And just as a comment...although I'm a rather new user to R (as U may have
guessed). I gather that every now and then popular and necessary packages
are added to base R.
So I'm
one way is the following:
min.x - c(0.2, 0.3, 0.6, 0.1)
max.x - c(0.4, 0.5, 0.7, 0.4)
c(rbind(min.x, max.x))
I hope it helps.
Best,
Dimitris
On 10/20/2010 2:13 PM, fugelpitch wrote:
I have two vectors
min
0.2, 0.3, 0.6, 0.1
max
0.4,0.5,0.7,0.4
Is there a way to combine these two
Try this:
c(matrix(c(Min, Max), 2, byrow = T))
On Wed, Oct 20, 2010 at 10:13 AM, fugelpitch jo...@runtimerecords.netwrote:
I have two vectors
min
0.2, 0.3, 0.6, 0.1
max
0.4,0.5,0.7,0.4
Is there a way to combine these two vector so that the values will be taken
from the vectors
fugelpitch jonas at runtimerecords.net writes:
I have two vectors
min
0.2, 0.3, 0.6, 0.1
max
0.4,0.5,0.7,0.4
Is there a way to combine these two vector so that the values will be taken
from the vectors alternating so that I will get a new
minmaxminmaxminmax-vector?
minmax
Hi,
Can anyone give me a pointer on howto make a package execute a function at
loading?
Following an older post (http://bit.ly/cS1Go4), I'd like to do something
along the lines of
.localstuff - new.env()
.localstuff$OftenUsedData - read.csv(...)
upon loading the package ...
Thanks, Joh
On Wed, 20 Oct 2010, Uwe Ligges wrote:
On 20.10.2010 13:59, Chris Howden wrote:
Thanks Uwe,
It may operate like that on most peoples machines, but either its not
operating like that on mine. Or I have another problem :-(
As u can see from my code below I've run
Dear R users,
I want to model some time series data with the help of dynamic factor
analysis, as described by A.F.Zuur in Dynamic factor analysis to estimate
common trends in fisheries time series.
Unfortunately the only implementation of the method I could find sofar is in
the non free software
Hi:
Even more fun with Deducer in 2.12.0:
I downloaded the JGR 64-bit executable from R-forge and installed the latest
binary of the package from there a half hour ago. When I try the JGR
executable, I get the same error that Rob and Kat reported. When I try to
call Deducer from within R-2.12.0
Hi.
I want to use the text() command to produce graphics text in the (Windows)
font Tahoma, but I don't know how to do this. I appreciate any
suggestions/ideas you may have.
All Best,
Ethan
[[alternative HTML version deleted]]
__
Hello
I've used read.table to read a file that contains numbers such as 0.1
when I write them back with write.table those numbers appear as 1e-5
How can I keep the old format?
thanks
--
View this message in context:
Hello
How can I select several not continuous rows ?
If I wanted to select rows 1 to 7 I'll write
mydata[,1:7]
But what if I need to select rows 1 to 5 and 10 to 15?
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Sent
2010/10/20 David A. dasol...@hotmail.com
I am trying to calculate confidence intervals using ci.numeric from
epicalc package. If I generate a normal set of data and find the 99% and 95%
CI, they seem too narrow to me. Am I doing something wrong??
Yes.
set.seed(123)
x- rnorm(200,0,1)
Hello
(1) How can I compare two Pearson correlation coefficients for significant
differences without the use of the raw data?
(2) How can I compare two linear regression coefficients for significant
differences without the use of the raw data?
(3) How can I compare two multiple correlation
Does anyone know the why for this error when doing a logistic regression with
binomial family?
Thank you
Kind regards
ana
Error in family$linkfun(mustart) : Value -1717986918 out of range (0, 1)
in attachment is my data set and the comands I used were
for adults
1) See the rw-FAQ and read and edit the file etc/Rdevga
2) Set up your own font family: see ?windowsFont.
Both are mentioned in the obvious place: the help file for ?windows
(see the posting guide).
On Wed, 20 Oct 2010, Ethan Arenson wrote:
Hi.
I want to use the text() command to produce
Dear R experts,
I'm beginner.
My question about ANOVA for unequal sample sized data should be obsolete but
I can not clarify it.
I have a dataset from 23 males and 18 females.
I measured one condition('cond') with 4 levels.
So I'd like to see main effect of gender, cond and gender by cond
Hi,
On Wed, Oct 20, 2010 at 9:25 AM, skan juanp...@gmail.com wrote:
Hello
How can I select several not continuous rows ?
If I wanted to select rows 1 to 7 I'll write
mydata[,1:7]
But what if I need to select rows 1 to 5 and 10 to 15?
mydata[, c(1:5, 10:15)]
-Ista
--
View this
Dear List,
I am relatively new to R and am trying to create more attractive plots than
excel can manage!
I have looked through the various programmes ggplot, lattice, hmisc etc but my
case seems to be not metnioned, maybe it is but i have not noticed - if this is
the case i apologise.
Try this:
format(0.1, scientific = F)
On Wed, Oct 20, 2010 at 11:21 AM, skan juanp...@gmail.com wrote:
Hello
I've used read.table to read a file that contains numbers such as 0.1
when I write them back with write.table those numbers appear as 1e-5
How can I keep the old
mydata[c(1:5, 10:15), ]
HTH,
Dennis
On Wed, Oct 20, 2010 at 6:25 AM, skan juanp...@gmail.com wrote:
Hello
How can I select several not continuous rows ?
If I wanted to select rows 1 to 7 I'll write
mydata[,1:7]
But what if I need to select rows 1 to 5 and 10 to 15?
--
View this
mydata[c(1:5, 10:15),]
On Wed, Oct 20, 2010 at 11:25 AM, skan juanp...@gmail.com wrote:
Hello
How can I select several not continuous rows ?
If I wanted to select rows 1 to 7 I'll write
mydata[,1:7]
But what if I need to select rows 1 to 5 and 10 to 15?
--
View this message in
Hi,
You are doing the right thing by being concerned about the warning messages.
The warning message of `NaN' usually arises due to illegal arithmetic
operations (in the real number field) such as logarithm or square root of a
negative number. This could happen during the intermediate stages of
Hi Jeong,
On Wed, Oct 20, 2010 at 5:25 AM, BumSeok Jeong bumseok.je...@gmail.com wrote:
Dear R experts,
I'm beginner.
My question about ANOVA for unequal sample sized data should be obsolete but
I can not clarify it.
I have a dataset from 23 males and 18 females.
I measured one
I think that your first problem is that you have a very large range of
values and the CIs are small in comparison, so you won't see any
difference on the plots. Do you want to plot each of the 35 values
showing the complete range and then where the actual value lies either
inside/outside the
On Oct 20, 2010, at 9:25 AM, skan wrote:
Hello
How can I select several not continuous rows ?
If I wanted to select rows 1 to 7 I'll write
mydata[,1:7]
But what if I need to select rows 1 to 5 and 10 to 15?
mydata[ , c(1:5, 10:15) ]
--
David Winsemius, MD
West Hartford, CT
Hi,
I am estimating a (fixed-effects) model with plm, for which I would like
to get the fitted values. If I call fitted() on my estimated model, it
returns NULL.
How do I get the fitted values out of the plm object?
Thanks.
Max
__
On Oct 20, 2010, at 9:21 AM, skan wrote:
Hello
I've used read.table to read a file that contains numbers such as
0.1
when I write them back with write.table those numbers appear as 1e-5
How can I keep the old format?
To change globally for the session:
?options
options(scipen
On Oct 20, 2010, at 4:10 AM, David A. wrote:
Hi,
I am trying to calculate confidence intervals using ci.numeric from
epicalc package. If I generate a normal set of data and find the 99%
and 95% CI, they seem too narrow to me. Am I doing something wrong??
The next sentence suggests that
Hi Tim,
you have two problems at the same time:
1.) The warning you get means that you predictor (e.g. predictor1) has
another range in the training set than in the test set. In this case you
have data in you test set that lies outside of the range of the training
set (for predictor1). This
On Oct 20, 2010, at 5:03 AM, Gouveia, Ana wrote:
Does anyone know the why for this error when doing a logistic
regression with binomial family?
Thank you
Kind regards
ana
Error in family$linkfun(mustart) : Value -1717986918 out of range
(0, 1)
in attachment is my data set and the comands
On Oct 20, 2010, at 10:01 AM, David Winsemius wrote:
On Oct 20, 2010, at 9:25 AM, skan wrote:
Hello
How can I select several not continuous rows ?
If I wanted to select rows 1 to 7 I'll write
mydata[,1:7]
But what if I need to select rows 1 to 5 and 10 to 15?
mydata[ , c(1:5, 10:15) ]
I have a factor which is species and the variables are a few sites where this
species live.
When I plot:
plot(species.factor, minlatitudemaxlatitude)
the plot produced is automatically a boxplot which shows the range of the
species (which is what I want).
However, in the thin box which is
I can not understand why this fails
faicoutput2 - list(stuff21 = as.numeric(faicout$coefficients[2]),
+ stuff31=as.numeric(faicout$coefficients[3]),
+ stuff41=as.numeric(faicout$coefficients[4]),
+ stuff32=(stuff21-stuff31),
+
I want the reshaped data to look like this.
plate.id HYBwell.id rlt1.control1
well.id rlt1.control1 well.idrlt1.disease1 well.id
rlt1.disease2
1 P1 SKOV3hyb A10.190
A2 0.210 B10.217
HI! I'm student of university and i need an example (an application) of
Multivariate GARCH model for replicate it in my thesis. It's better with model
CCC e DCC.
Best Regards
Giuseppe
for answer: bepperozz...@hotmail.com
[[alternative HTML version deleted]]
Il giorno mer, 20/10/2010 alle 03.05 -0700, Dennis Murphy ha scritto:
Works for me! Thanks, Dieter!
Hi all.
Thanks again to you both for the help.
It was a problem due to a lack in updating ubuntu.
After the upgrade to Lucid I entirely forgot to
update /etc/apt/sources.list with the new lines.
Hello!
Here is my data:
x-data.frame(y=rnorm(100,0,1),a=rnorm(100,1,1),b=rnorm(100,2,1),weights=runif(100))
data.for.regression-x[1:3]
names(data.for.regression)
myweights-x$weights
I run simple weighted regression and everything is fine:
reg1-lm(y~., data.for.regression, weights=myweights)
I do not see that it has been created yet. You my have defined it
earlier in the 'list' expression, but as an object it is not available
yet. You might have to do something like uwing 'within'
x - list()
x - within(x, {
+ a = 1:10
+ b = 11:20
+ c = a + b
+ d = a * b})
x
$d
[1] 11 24 39
Sorry R community, I am still stucked wit this. How can avoid the x-limit
error?
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Sent from the R help mailing list archive at Nabble.com.
On Oct 20, 2010, at 10:45 AM, fugelpitch wrote:
I have a factor which is species and the variables are a few sites
where this
species live.
When I plot:
plot(species.factor, minlatitudemaxlatitude)
the plot produced is automatically a boxplot which shows the range
of the
species (which
On Wed, 20 Oct 2010, max.e.br...@gmail.com wrote:
Hi,
I am estimating a (fixed-effects) model with plm, for which I would like
to get the fitted values. If I call fitted() on my estimated model, it
returns NULL.
How do I get the fitted values out of the plm object?
I think there is
Hello!
Sometimes I have to produce several graphs at a time, but need to be
able to see them all one by one in the RGraphics window.
I do it manually like this:
I create some plot:
plot(1:5)
It opens the RGraphics window. I click on the window, go (in the
menue) to History-Recording, and then
I suspect that Dr. Jim is conflating the behavior of R w.r.t.
arguments in a function call (where assignment to one argument can be
seen by another argument) with the behavior of naming items in a
list expression.
--
David.
On Oct 20, 2010, at 11:01 AM, jim holtman wrote:
I do not see
?windows
On Wed, Oct 20, 2010 at 11:19 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Hello!
Sometimes I have to produce several graphs at a time, but need to be
able to see them all one by one in the RGraphics window.
I do it manually like this:
I create some plot:
Dear all
What is the preferred spreadsheet-like x-delimited data format for use
with R? Should I prefer tab-, comma-, space- or
some_other_delimitor-seprated data?
I'm asking this because I've been once bitten by CSV data containing `
' ' (apostrophe) symbols that R couldn't easily digest.
On Wed, Oct 20, 2010 at 11:25 AM, Liviu Andronic landronim...@gmail.com wrote:
Dear all
What is the preferred spreadsheet-like x-delimited data format for use
with R? Should I prefer tab-, comma-, space- or
some_other_delimitor-seprated data?
I'm asking this because I've been once bitten by
Your first example is all one statement. Within that statement you are defining
elements of a list. Unfortunately you are defining some of those elements in
terms of others, but they don't exist until the statement has completed. If you
break the statement up into an initial list formation
On Wed, Oct 20, 2010 at 7:53 AM, ottorino
ottorino-luca.pant...@unifi.it wrote:
Il giorno mer, 20/10/2010 alle 03.05 -0700, Dennis Murphy ha scritto:
Works for me! Thanks, Dieter!
Hi all.
Thanks again to you both for the help.
It was a problem due to a lack in updating ubuntu.
After the
It seems David and Phil have given you all the pieces and you are not
listening. Try:
h = sprintf('%06d', seq(07,18, by=1) ) # Note that h is text
myseries=sample(1:12, 12) #fake data to plot
tim=strptime(h,'%H%M%S')
library(zoo)
test=zoo(myseries, tim)
plot(test)
Rob
str(species.factor)
Factor w/ 81 levels Acer_platanoides_Bl,..: 12 12 55 55 76 76 52 52
67 67 ...
str(minmax)
int [1:162] 6163779 7262127 6163779 7422941 6163779 6644861 6163779
6644861 6163779 7262127 ...
That is, my factor is the species listed two times each
Acer_platanoides_Bl
Thanks a lot for the helpful reference, Jim.
So, the solution is very simple:
windows(record=TRUE)
for(i in 10:12){
plot(1:i,main=paste(Plot ,i,sep=))
}
Dimitri
On Wed, Oct 20, 2010 at 11:24 AM, jim holtman jholt...@gmail.com wrote:
?windows
On Wed, Oct 20, 2010 at 11:19 AM, Dimitri
Dear R-users,
Do you know if we can use the function lme in R for log-normal
distribution of parameters as used in Nonmem ?
theta=theta0*exp(eta)
In our model, the parameters follow the log-normal distribution so it's
not reasonable to deal with normal distribution which gives us negative
Well, I was listening, but the error was due to the extra function 'format'.
Without that, it works perfectly.
Thanks for the help,
Marco
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On Wed, Oct 20, 2010 at 11:08 AM, Manta mantin...@libero.it wrote:
Sorry R community, I am still stucked wit this. How can avoid the x-limit
error?
If the problem is that you wish to create and plot an hourly series at
12 successive hours starting at 7am then here are three ways depending
on
On Oct 20, 2010, at 9:14 AM, Dennis Murphy wrote:
Hi:
Even more fun with Deducer in 2.12.0:
I downloaded the JGR 64-bit executable from R-forge and installed the latest
binary of the package from there a half hour ago.
The old 1.61 loader only supports R 2.12. You'll need the new
My question is about the write.foreign() command in the foreign
package. I use a command like the following to try and output data
and a code file to read my data into SAS.
write.foreign(data.frame.object, datafile=filepath,
codefile=filepath, package=SAS, dataname=myData)
With my data
I need to write a table that can be opened in Excel or OpenOffice such that
there are newlines embedded within cells.
After much Googling and futzing, I can't figure out how to do this. The way
to do this within Excel is alt-Enter and I've tried '/n', '/n/r', '/r/n' per
some web suggestions
When I use grid.rect to print a multi-coloured grid, it is incredibly slow
compared to a single colour grid, or even a two colour grid.
I've set out some simplified examples below. This is something I run literally
thousands of times a day, so I would greatly appreciate any hints on how I
On Mon, Oct 11, 2010 at 12:55 PM, casperyc caspe...@hotmail.co.uk wrote:
And now I just wonder why the ' bty='n' ' won't work?
I did
dotplot(BATCH~RESPONSE,data=d,subset=Type=='SHORT',bty='n')
and tried other bty parameters, none is working
As David says, par settings don't work in
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Andrew Miles
Sent: Wednesday, October 20, 2010 10:10 AM
To: r-help@r-project.org
Subject: [R] Problem exporting data using write.foreign
My question is about the
On 20/10/2010 1:04 PM, Mark Kimpel wrote:
I need to write a table that can be opened in Excel or OpenOffice such that
there are newlines embedded within cells.
After much Googling and futzing, I can't figure out how to do this. The way
to do this within Excel is alt-Enter and I've tried '/n',
Dear list
I am interested n developing a graph using plot() and text() commands in which
several text strings will be placed within the plot area I would like some way
to get the size (width and height), either in pixels or using the plot's
coordinate system, of text strings, based on the
Hello.
Correct, IMHO: Achim's suggestion is the best way to get in-sample fitted
values, w.r.t. e.g. direct calculation as yhat=X%*%coef, as it implicitly takes
care of fixed effects etc..
A method for out-of-sample use is not implemented yet. It will eventually be,
but there is no schedule for
See ?strwidth
On Wed, Oct 20, 2010 at 3:36 PM, Barth B. Riley bbri...@chestnut.orgwrote:
Dear list
I am interested n developing a graph using plot() and text() commands in
which several text strings will be placed within the plot area I would like
some way to get the size (width and
I think Excel wants a \n for newlines
in a text cell entry but \r\n to separate
rows of a csv file. You may have to open
the file in binary mode and put in the \r\n
at line ends by hand to achieve this from R,
as it tranlates all \ns to \r\ns when
writing them to a file.
(\n is not the same as
Thanks.
It was 'bty=n' on a text graph, I was just having fun with it.
It does not really matter that much.
CasperYC
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I was able to figure this one out. It turns out that truncation was
not the problem, as I had no variables with names longer than 32
characters (that is quite a long name!). I document my process here
so that future users with the same problem can benefit.
First, I examined the code for
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