Thanks for the reply, Dieter.
I'm sorry I should have made it clear in my original post - the number
(output of which.max()) IS dependent on the grouping..
Thanks,
-Girish
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helloR wrote:
>
>
> this is the problem:
> load this R data frame over the internet and save it to your hard drive.
> http://rss.acs.unt.edu/Rdoc/library/twang/data/racepro ling.RData
> ...
>
> please show how to save a dataset of males only (the variable male=1) to a
> new dataframe. Then do
Girish A.R. wrote:
>
> Sorry, just realized that there was a typo in the following code of my
> original post. The correct code is as shown below (corrected
> 'data=swtop16' to 'data=DF'):
>
> sales <- xyplot(pct_inv_left ~ week_num|sku_num, data=DF,type =
> "l",lwd=2,panel = function(...) {
>
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing
nitish wrote:
>
> I have a dataset that goes like: dataset =
> t |x
> 0 |x1
> 1 |x2
> 2 |0
> 3 |0
> 4 |0
> 5 |0
> 6 |x3
> 7 |0
> 8 |0
> 9 |0
> 10 |x4
>
> and so on. I wish to detect the periodicity of occurrences. t is in
> seconds and x are arbitrary, whose magnitude i a
Is there a bug in tiff compression output in Mac version?
My program is correct when I run on windows version; However, in Mac, the
compression fail~
[[alternative HTML version deleted]]
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https://stat.ethz.ch/
I have a dataset that goes like: dataset =
t___|x___
0|x1
1|x2
2|0
3|0
4|0
5|0
6|x3
7|0
8|0
9|0
10 |x4
and so on. I wish to detect the periodicity of occurrences. t is in
seconds and x are arbitrary, whose magnitude i am not interested in. I
just wish to ge
Sorry, just realized that there was a typo in the following code of my
original post. The correct code is as shown below (corrected 'data=swtop16'
to 'data=DF'):
sales <- xyplot(pct_inv_left ~ week_num|sku_num, data=DF,type =
"l",lwd=2,panel = function(...) {
panel.abline(h = 75, lty
Hi folks,
I need some help in customizing the abline() function to be used in a
lattice plot. I have attached a reproducible example below.
I need help in the following snippet:
disc <- xyplot(cnt_gt50pct_disc ~ week_num|sku_num, data=DF,type =
"h",lwd=2,panel = function(...) {
panel
Try this
male.df <- subset(raceprofiling, (male==1))
female.df <- subset(raceprofiling, (male==0))
people.df <- rbind(male.df,female.df)
works?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of tanzia chaudhury
Sent: 09 December 2010
Hi,
Just as a note, it is preferred that you use plain text rather than
rich text or html emails for this list.
On Wed, Dec 8, 2010 at 7:32 PM, tanzia chaudhury
wrote:
>
> this is the problem:
> load this R data frame over the internet and save it to your hard drive.
> http://rss.acs.unt.edu/Rdo
Hello,
I have uploaded a csv file that looks like this:
> gc
alpha_id beta_id
1 142053 1
2 9454 1
3 295618 2
442691 2
5 389224 3
6 9455 3
The alpha_id contains 310660 unique values and the beta_id contain
When I make check the R 2.12
* DONE (mgcv)
make[2]: Leaving directory `/usr/local/R-2.12.0/src/library/Recommended'
make[1]: Leaving directory `/usr/local/R-2.12.0/src/library/Recommended'
r...@ip-192-168-1-66:/usr/local/R-2.12.0# make check-all
make[1]: Entering directory `/usr/local/R-2.12.0/tes
this is the problem:
load this R data frame over the internet and save it to your hard drive.
http://rss.acs.unt.edu/Rdoc/library/twang/data/racepro ling.RData
please show how to save a dataset of males only (the variable male=1) to a new
dataframe. Then do the same thing for females (male=0). T
Thank you Phil, your method worked great!
Again, I was about to write back and add some facts -- all of the columns
need to be 'character'... when I tried my and Jonathan's solutions with
columns that were factors, I got errors.
Here's what happened when I tried to use the 'dcast' method on my ac
On 2010-12-08 17:07, Anup Nandialath wrote:
Dear R-helpers,
My question is related to how to impose constraints when when sampling from a
distribution.
For example, suppose I'm sampling a vector from a multivariate normal
distribution
vbeta<- 100*diag(2)
mbeta<- c(1,1)
ans<- beta<- c(rmvno
Another way, using just base R is
reshape(df,idvar='y',timevar='x',v.names='z',direction='wide')
y z.A z.B z.C
1 D a b c
4 E d e f
7 F g h i
- Phil Spector
Statistical Computing Facility
Hi Jonathan,
Thank you very much! I was about to recall this after I found this
discussion (https://stat.ethz.ch/pipermail/r-help/2008-June/164440.html),
but I think your solution is more tidy.
Thank you very much!
Matt
On Wed, Dec 8, 2010 at 8:33 PM, Jonathan Christensen wrote:
> Matt,
>
> li
Matt,
library(reshape2)
wide.df <- dcast(df, y ~ x)
Works great for me.
Jonathan
On Wed, Dec 8, 2010 at 7:26 PM, Matthew Pettis wrote:
> Hi,
>
> I was wondering if there is an easy way that I am missing for turning a long
> dataframe into a wide one. Below is sample code that will make what
Hi,
I was wondering if there is an easy way that I am missing for turning a long
dataframe into a wide one. Below is sample code that will make what I have
and, in comments, the form of what I want:
# Have: dataframe like 'df'
df <- expand.grid( x=LETTERS[1:3], y=LETTERS[4:6])
df$z <- lett
Why 2.8? You should perhaps go for 2.12 ... I like 2.11 right now
because most of my work has been tested on it already and it is quite
stable with other packages.
But, the ideally you should go for 2.12 which is the most recent
release and has substantial enhancements.
Then you should post the e
I am a beginner with R. After I installed R 2.8.0, I tried to add a package.
R CMD INSTALL RHIPE.tar.gz
I got following error message:
cannot extract package from RHIPE.tar.gz
Can anyone tell me what's the problem?
--
Best Regards,
Zihang Yin
919-225-1717
Computer Science
Duke University
_
Thanks Peter and David,
Peter, that is exactly what I was looking for. Sadly I have even used is.na()
in the past but forgot about it. David, thanks for the tip. I am
embarrassed that I exposed the fact that I use a lot of loops.
findInterval() seems very handy.
Wade
On Wed, Dec 8, 2010 at 4:
Dear R-helpers,
My question is related to how to impose constraints when when sampling from a
distribution.
For example, suppose I'm sampling a vector from a multivariate normal
distribution
vbeta <- 100*diag(2)
mbeta <- c(1,1)
ans <- beta <- c(rmvnorm(1,mbeta,vbeta))
ans will thus be a vect
I think the authors of lars could do a better job of
the 'Value' section of the help page. Nevertheless, most
of the list elements are pretty obvious if you
understand the LARS procedure. Your questions indicate
that you would benefit from studying the references
listed on the help page before att
I am beginning to work with the 'ape' package in R, and have run into some
trouble. I generated a UPGMA tree based on DNA sequence distance in Paup*
and read it into R, where it became an object of class "phylo". However, I
need it to be classified as a "dendrogram" for my purposes (to use it to
or
thanks!
problem solved!
casper
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Hi Jim,
Thanks for your insight. I used Linux split to split my large file into
smaller partitions. On the server I work on, multipath I/O access is enabled
and we use RAID for storage; thus, I don't think I can put each partition on
a spindle. I'm able to open multiple files at a time into stdin
This is very helpful, Phil. Being new to R and trying to get a handle on
*apply, by, aggregate... this is great. Thank you.
Jim
-Original Message-
From: Phil Spector [mailto:spec...@stat.berkeley.edu]
Sent: Wednesday, December 08, 2010 2:36 PM
To: Jim Moon
Cc: r-help@r-project.org
Good Evening R-Help Community:
I have attached a file that contains the output from sessionInfo() and a
summary
of my Win XP system. I am running R 2.12.0 and using Tinn-R 2.3.6.2 as my
interface. When I attempt to call either the doBy or coin packages R generates
an error that I do not unders
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Julian TszKin Chan
> Sent: Wednesday, December 08, 2010 2:39 PM
> To: r-help@r-project.org
> Subject: [R] How can i select a set of element in a list ?
>
> How can i select a set
How can i select a set of element in a list ?
x<-list(1,2,3,4,5)
How can I create a new list which only have the last 4 elements of x.
I can do it with a loop, but I want to avoid that. Thanks!
example :
y<-vector("list",4)
for ( i in 1:length(y) )
y[[i]] <- x[[ i + length(x) - length(y) ]]
Jim -
Suppose the data frame is named "df". Then
df$std_dev = apply(df,1,sd,na.rm=TRUE)
will do what you want, although it will generate
some warning messages due to the first column.
df$sd_dev = apply(df[,sapply(df,is.numeric)],1,sd)
does the same, but won't produce warnings.
How might one calculate standard deviation, row-wise, for the numeric values in
a data frame such as this one
V1 V2 V3 V4 V5
1 rs11089130 0.4565 0.4574 0.4569 0.4572
2 rs738829 0.6548 0.6519 0.6448 0.6549
3 rs915674 0.7503 0.7500 0.7517 0.7502
and place the standard d
On 2010-12-08 12:24, casperyc wrote:
Thomas Stewart wrote:
It isn't clear to me what you want to do. Do you want the axes to show?
Do
you want labels for the lines? Do you want a legend? What is your
desired
output?
-tgs
On Wed, Dec 8, 2010 at 2:42 PM, casperyc wrote:
Hi All,
How do
On 2010-12-08 12:03, akch...@bidmc.harvard.edu wrote:
Hi All,
I am new to the R program (only my 2nd day trying it out!)
and this may seem like a really stupid question but I was
wondering if someone could help. I have managed to calculate
the pearson correlation coefficient for my data and now I
On Dec 8, 2010, at 3:10 PM, Wade Wall wrote:
Hi all,
How can one evaluate NAs in a numeric dataframe column? For
example, I have
a dataframe (demo) with a column of numbers and several NAs. If I
write
demo.df >= 10, numerals will return TRUE or FALSE, but if the value is
"NA", "NA" is re
On Tue, 7 Dec 2010 21:24:27 -0700, Greg Snow
wrote:
>> Sharing LaTeX documents with people using word processors only is
no more difficult than giving driving directions to someone who is
blindfolded . . . . If you just need basic input or approval then
give them a paper version or pdf file and
Dear all,
I'm new to R and time series analysis. I'd appreciate if you could shed
light on my problem.
Here is what I have been trying to do:
1. I fit the model ARIMA(1,0,0) with the training dataset xdata[1:100]
fitit = arima(xdata, order=c(1,0,0)
2. I have some current observations in the buff
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Wade Wall
> Sent: Wednesday, December 08, 2010 12:11 PM
> To: r-h...@stat.math.ethz.ch
> Subject: [R] evaluating NAs in a dataframe
>
> Hi all,
>
> How can one evaluate NAs in a
Hi!
> How can one evaluate NAs in a numeric dataframe column? For example, I have
> a dataframe (demo) with a column of numbers and several NAs. If I write
> demo.df >= 10, numerals will return TRUE or FALSE, but if the value is
> "NA", "NA" is returned. But if I write demo.df == "NA", i
> Date: Wed, 8 Dec 2010 19:25:09 +
> Subject: Re: [R] GIS Help: distance calculation based on ZIP Code
> From: b.rowling...@lancaster.ac.uk
> To: lihaw...@gmail.com
> CC: marchy...@hotmail.com; r-help@r-project.org
>
> On Wed, Dec 8, 2010 at 6:18
On 2010-12-08 12:10, Wade Wall wrote:
Hi all,
How can one evaluate NAs in a numeric dataframe column? For example, I have
a dataframe (demo) with a column of numbers and several NAs. If I write
demo.df>= 10, numerals will return TRUE or FALSE, but if the value is
"NA", "NA" is returned. But if
Thomas Stewart wrote:
>
> It isn't clear to me what you want to do. Do you want the axes to show?
> Do
> you want labels for the lines? Do you want a legend? What is your
> desired
> output?
>
> -tgs
>
> On Wed, Dec 8, 2010 at 2:42 PM, casperyc wrote:
>
>>
>> Hi All,
>>
>> How do I add t
Hi All,
I am new to the R program (only my 2nd day trying it out!) and this may seem
like a really stupid question but I was wondering if someone could help. I have
managed to calculate the pearson correlation coefficient for my data and now I
have a correlation coefficient matrix with a whole s
Hi all,
How can one evaluate NAs in a numeric dataframe column? For example, I have
a dataframe (demo) with a column of numbers and several NAs. If I write
demo.df >= 10, numerals will return TRUE or FALSE, but if the value is
"NA", "NA" is returned. But if I write demo.df == "NA", it returns as
It isn't clear to me what you want to do. Do you want the axes to show? Do
you want labels for the lines? Do you want a legend? What is your desired
output?
-tgs
On Wed, Dec 8, 2010 at 2:42 PM, casperyc wrote:
>
> Hi All,
>
> How do I add these axis labels?
>
> #
On Dec 8, 2010, at 2:40 PM, Erin Hodgess wrote:
Dear R People:
Suppose I have the following in a file:
1 1.171504 1.010070
2 -0.9317064 1.860900
3 -0.06522837 0.6561147
4 -1.817026 0.02619137
5 1.426983 0.5995691
6 -0.2844911 1.155561
7 -0.6920972 0.7633124
8 0.3129615 5.121108
I want to use
Yes, just set the colClasses argument to read.table (this will also tend to
speed up the reading, though only noticeable for really big files).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From:
Try this modification of your code.
-tgs
ynames <-
base.dat.sel2$base.dat.Covariate[order(base.dat.sel2$base.dat.US.Num.Obs.to.Achieve.Starting.Residual,decreasing=F)]
ynames <- as.character(ynames)
ynames[12]<-expression(theta[r])
ynames[13]<-expression(EC[gw])
ttt<-barplot(base.dat.sel2$base.d
Hi All,
How do I add these axis labels?
###
p=seq(0,1,length.out=500)
p=p[-c(1,length(p))]
g1=log(p/(1-p))
g2=qnorm(p)
g3=log(-log(1-p))
g4=-log(-log(p))
plot(p,g1,
'n',ylim=c(-5,5),las=1,
bty='n',
xaxt='n',yaxt='n',
xla
Dear R People:
Suppose I have the following in a file:
1 1.171504 1.010070
2 -0.9317064 1.860900
3 -0.06522837 0.6561147
4 -1.817026 0.02619137
5 1.426983 0.5995691
6 -0.2844911 1.155561
7 -0.6920972 0.7633124
8 0.3129615 5.121108
I want to use read.table to bring it in and I want the first colu
On 2010-12-08 09:16, Dick Knox wrote:
I am trying to understand the function acf
stats:::acf shows me the function
I am having trouble understanding the usage "$acf" in the following
acf<- array(.C(R_acf, as.double(x), as.integer(sampleT),
as.integer(nser), as.integer(lag.max), as
On Wed, Dec 8, 2010 at 6:18 PM, XINLI LI wrote:
> Thank you very much, I will look into it.
If it really is only 200 then I'd start typing them into Google's
direction finder and typing in the distance/time given. Even if you
only do one per minute they'll all be done in half a day easily. Job
d
Hi!
I have a large dataset where I need to recompute a value in each row based
on z-sores from another look-up table. The look-up table is arranged by raw
score in the first column and age in the first row, eg:
20 30 40
1 .3 .5 .7
2 .2 .3 .4
3 .1 .4 .7
the main matrix will have ages
Hello,
I've been looking through ?phantom and ?expression and this forum for
examples of how I might be able to manipulate some of the names that appear
on the y-axis of the barplot below. For example, the "gw" in "ECgw" would
appear as a subscript...or "qr" would be the theta symbol followed by
I have simple scatterplot of temperature data taken daily from many years so
at particular date there are usually more than one measurement, but not all
data is complete.
Now I want to draw line that will show me probability of finding data
between defined bands of temperatures, for example +5 de
On Wed, 8 Dec 2010, S.M. Raghavan wrote:
Hi all,
I am trying to fit a logistic regression for a bivariate response using five
independent variables in a stepwise procedure. My outputs look okay but does
any one know (or is there any literature on) how the confidence intervals
are calculated for
HI:
Is there a way to display a dataset on a dialog window? I am creating an
application
with Visual Basic and R, and I want the user to be able to see the dataset used
print out
on a dialog window. Not sure if there is a better way to do this, but basically
when the user
click a button on a VB
Thank you very much, I will look into it.
Best,
xing
On 12/8/10, Mike Marchywka wrote:
>
>
>
>
>
>
>
>
>
>
>
>
>
>> From: ggrothendi...@gmail.com
>> Date: Wed, 8 Dec 2010 10:50:40 -0500
>> To: lihaw...@gmail.com
>> CC: r-help@r-project.org
>> Subject: Re
Matthew Dowle wrote:
Might Wayland fix it in Narwhal ?
I hope those names mean something to Rainer, because they mean nothing
to me.
Duncan Murdoch
"Duncan Murdoch" wrote in message
news:4cff7177.7030...@gmail.com...
On 08/12/2010 6:07 AM, Rainer M Krug wrote:
-BEGIN PGP SIGNED ME
On Wed, 2010-12-08 at 08:48 -0800, anna_m wrote:
> Hi all,
> I try to interpolate a data set in the form:
>
> time Erg
> 0.00 48.65
> 1.50 56.08
> 3.00 38.33
> 4.50 49.65
> 6.00 61.39
> 7.50 51.25
> 9.00 50.450
Dear community,
I have now taken my R-file from lectures and intend to use it at home, but
have a problem
reading the Data from the file. I have installed and loaded the Package
xlsReadWrite so far.
I have also "Changed directory".
This is what I have entered
daten=read.xls("Daten A2")
This is
You might be better off partitioning the file before processing with
R. If you are planning on using "skip = n" to skip over records
before processing, then the last thread you would start would have to
read through 7/8 of the file before starting. The actual I/O, plus
the looking for the line fe
Thanks for the three great answers! For those who are curious, I timed the
three approaches:
nr <- 15812
nc <- 64636
mymat <- matrix(nrow=nr, ncol=nc)
mymat[1,1] <- 1 # see note below
# mydf is created elsewhere
dim(mydf)
# 109102633
colnames(mydf)
# "x" "y" "a"
# approach 1:
# mymat[
Hi here is the code as example
lars is in package lars
> x<-matrix(rnorm(20*5,0,1),20,5)
> bs<-matrix(sample(seq(1:10),5),5,1)
> er<-rnorm(20,0,1)
> y<-x%*%bs+er
> lobj<-lars(x,y,type="lasso")
> names(lobj)
[1] "call" "type" "df" "lambda" "R2"
[6] "RSS""Cp
Hi all,
I want to estimate parameters from a VARMA(p,q)-Modell.
The equations of the model or the model structures is given by:
Xt=beta1+beta2*Xt-1+beta3*Yt-1+epsilon1
Yt=beta4+beta5*Yt-1+espilon2
epsilon1 and espilon2 are white noise.
Xt is given by a vector of n elements e.g. (2,
Is anyone aware of a way to seasonally adjust time series data using X-12 ARIMA
and TRAMO/SEATS from within R? I know that that one can seasonally adjust data
with gretl, which I understand offers some level of R integration. However, all
the examples I've seen of gretl/R integration involve wor
> From: ggrothendi...@gmail.com
> Date: Wed, 8 Dec 2010 10:50:40 -0500
> To: lihaw...@gmail.com
> CC: r-help@r-project.org
> Subject: Re: [R] GIS Help: distance calculation based on ZIP Code
>
> On Wed, Dec 8, 2010 at 10:41 AM, XINLI LI wrote:
>
Provide at least a subset of the data you want to plot (e.g., 10 data
points). I assume that the data provides enough information to determine
how long each of the branches/cross ties are for the elements that you want
to plot. I don't use UniCox, nor do I know that the list of items that it
retu
On Dec 8, 2010, at 11:59 AM, S.M. Raghavan wrote:
Hi all,
I am trying to fit a logistic regression for a bivariate response
using five
independent variables in a stepwise procedure. My outputs look okay
but does
any one know (or is there any literature on) how the confidence
intervals
a
See McCullagh and Nelder's GLM book for details -- and also probably
V&R's MASS for a concise summary, although I don't have it at hand and
can't be sure it's there. Really, practically any book on GLM should
have details.
**HOWEVER** You should realize that all these references are "wrong"
in the
I am trying to understand the function acf
stats:::acf shows me the function
I am having trouble understanding the usage "$acf" in the following
acf <- array(.C(R_acf, as.double(x), as.integer(sampleT),
as.integer(nser), as.integer(lag.max), as.integer(type ==
"correlation"
Hi all,
I am trying to fit a logistic regression for a bivariate response using five
independent variables in a stepwise procedure. My outputs look okay but does
any one know (or is there any literature on) how the confidence intervals
are calculated for the reported odds ratios..?
Thanks!
I think
?table
is what you want.
-- Bert
On Wed, Dec 8, 2010 at 1:19 AM, Ivan Calandra
wrote:
> Hi,
>
> Still another way would be to count the levels of the factors (if it is
> indeed a factor):
> length(levels(example$V1))
>
> Ivan
>
> Le 12/8/2010 06:08, Jorge Ivan Velez a écrit :
>>
>> Hi,
Hi all,
I try to interpolate a data set in the form:
timeErg
0.0048.65
1.5056.08
3.0038.33
4.5049.65
6.0061.39
7.5051.25
9.0050.45
10.50 55.11
12.00 61.12000
Upcoming R Programming Techniques Courses
http://www.xlsolutions-corp.com/Rcourses
*** R Fundamentals and Programming Techniques
Washington DC, Dec 16-17, 2010
San Francisco, Dec 16-17
More on website
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Ask for group discount and reserve you
My apologies to Martin.
I should have known how prompt he would be and I should have
been more on the ball in checking incoming mail.
The new version of robustbase on R-Forge does indeed fix this.
Thanks to the robustbase team.
Peter
On 2010-12-08 07:57, Peter Ehlers wrote:
On 2010-12-08 06:3
try tapply() if you want the values for all levels of x, or calculate your
mean after a subset()
Check the documentation on both of these.
Mike
On Wed, Dec 8, 2010 at 9:54 AM, madr wrote:
>
> for example I have matrix with two columns
>
> x,y
> 1,0.56
> 2,9.55
> 2,7.56
> 5,2.55
> 5,0.56
> 3,0.
On 2010-12-08 06:39, kv wrote:
Thanks Peters, what i suspected (i.e. ties). For what it's worth i would add
that the problem happends when length(data)>100 (i.e. the second condition
in the mc.default() function).
Best,
I don't think that the length of 100 is special (other
than in the 'doRef
Thanks Jim,
"Ave" does what I wanted.
It is simpler and probably more efficient
than unlisting Sn as I tried.
Still I remain puzzled with the structure
of the by() or tapply() output and how
to access the individual cumsums.
Yes the split command is useful for checking
the result.
Gerrit.
Op 1
for example I have matrix with two columns
x,y
1,0.56
2,9.55
2,7.56
5,2.55
5,0.56
3,0.55
2,0.56
2,1.56
so I need to take average from y values placed where x==2
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Thank you Jorge, this is exactly what I needed!
Another solution that works too, suggested by Steve:group.med <- tapply( rt,
list(group, session), median)
From: jorgeivanve...@gmail.com
Date: Wed, 8 Dec 2010 09:43:47 -0500
Subject: Re: [R] How can I calculate the median for each factor combinatio
On Wed, Dec 8, 2010 at 10:41 AM, XINLI LI wrote:
> Dear R Group Users:
>
> If there is a resource or easy way to calculate the distance
> between zipcodes, for example, if I have the zipcode for 200 patients,
> and the zipcode of a hospital, how to calculate the travel distance of
> these indiv
Prof. Maechler , Koller,
Thank you very much, problem solved.
Best,
Seasons greetings.
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Sent from the R help mailing list archive at Nabble.com.
__
Dear R Group Users:
If there is a resource or easy way to calculate the distance
between zipcodes, for example, if I have the zipcode for 200 patients,
and the zipcode of a hospital, how to calculate the travel distance of
these individual patient to the hospital based on the zipcode. Your
inp
Thank you, Peter.
(see inline)
On Wed, Dec 8, 2010 at 14:58, Peter Ehlers wrote:
> On 2010-12-07 12:01, kv wrote:
>>
>> hello list,
>>
>> i'm a bit puzzled by the error message i get when i copy past this in R:
[.]
>> in my computer adjbox says:
>>
>> "maximal number of iterations (10
Thank you.
I tried this but not sure if I have implemented this correctly ... basically
if function1 hangs .. I need the timeout to be triggered and then the
process moves to the next function call.
I have this:
function1 <- function(x){
setTimeLimit(elapsed = 5*60, transient = FALSE)
step 1
ste
Might Wayland fix it in Narwhal ?
"Duncan Murdoch" wrote in message
news:4cff7177.7030...@gmail.com...
> On 08/12/2010 6:07 AM, Rainer M Krug wrote:
>> -BEGIN PGP SIGNED MESSAGE-
>> Hash: SHA1
>>
>> On 12/08/2010 12:05 PM, Duncan Murdoch wrote:
>>> Rainer M Krug wrote:
>>> Hi
>>>
>>> rgl
Hi Marianne,
Please consider the following:
with(table, aggregate(rt, list(group, session), FUN = median))
HTH,
Jorge
On Wed, Dec 8, 2010 at 9:29 AM, Marianne Stephan <> wrote:
>
> Hello everybody,
> I would like to calculate the median for each factor combination, with only
> one value per fa
Thanks Peters, what i suspected (i.e. ties). For what it's worth i would add
that the problem happends when length(data)>100 (i.e. the second condition
in the mc.default() function).
Best,
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Hello everybody,
I would like to calculate the median for each factor combination, with only one
value per factor combination as an output.
Could anybody help me?
For example:
# make table
g<-1:2
group<-rep(g, each=5)
session<-c(1,1,2,2,2,1,1,1,2,2)
rt<-seq(length=10,300, 800)
rt<-round(rt, dig
On 2010-12-07 12:01, kv wrote:
hello list,
i'm a bit puzzled by the error message i get when i copy past this in R:
data<-c(16,13,17,4,15,24,59,18,33,8,42,19,20,4,11,9,3,7,10,3,3,67,4,4,13,16,6,3,3,6,3,4,35,10,16,11,24,7,47,8,8,2,12,3,8,4,3,6,6,10,2,9,3,15,21,13,8,16,2,5,14,9,21,4,9,11,36,4,8,4
On 12/7/2010 9:35 PM, Yihui Xie wrote:
shell(paste("yap", "C:/WINDOWS/TEMP/Rtmpz0QkT8/file311f289a.dvi"))
I can confirm that using shell() directly on the .dvi file generated by
latex() works, while system() does not -- it hangs
as before.
However, Yihui's patch, in this form still hangs, so m
You cannot, plotMap() decides things for you. For more flexibility than:
polys <- data.frame(PID=rep(1,4), POS=1:4, X=c(0,1,1,0), Y=c(0,0,1,1))
plotMap(polys, xlim=c(-.5,1.5), ylim=c(-.5,1.5), projection="LL")
from the help page, consider trying:
library(sp)
library(maptools)
sp_polys <- PolySe
Thank you for your reply. My question was how to create Fig. 2? I am using
UniCox; the aa=uniCoxCV gives list of aa$ devcvm aa$ncallcvm aa$se.devcvm,
aa$devcv, aa$ ncallcv and aa$ folds, which data I should plot here to create
the graph in Fig. 2?
Many thanks,
Sam
--- On Wed, 8/12/10, jim h
Rosario Garcia Gil slu.se> writes:
>
> Hello
>
> I hope my question makes sense. It is possible to specify the shape
paramenters in a glm model with family Gamma?
>
It doesn't look that way, but you can do a Gamma model with
a specified shape parameter (albeit much less efficiently) via
li
Do a little reading on how to use the graphics commands. I would look at
plot
lines
segments
a combination of those will let you easily create the output. It
would seem you need some data object that has the dimensions of the
bar lengths you want to create, but given that, it is not much of a
p
For (at least) boxplot() and hist(), you can set plot=FALSE
But since we don't know what type of plot you want, difficult to be sure.
Ivan
Le 12/8/2010 14:14, Joel a écrit :
Hi is there any command or setting that allows you to do the plot command but
it dose not print the plot on screen?
So wh
> Date: Tue, 7 Dec 2010 17:22:57 -0800
> From: ryan.steven.gar...@gmail.com
> To: r-help@r-project.org
> Subject: [R] Parallel Scan of Large File
>
>
> Is it possible to parallel scan a large file into a character vector in 1M
> chunks using scan() wit
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