Displaying 20 results from an estimated 10000 matches similar to: "accessing current factor in tapply"
2001 Sep 25
1
blues in c
G'Day,
I'm a little confused why the c function has the followng effect on
classes - is this a feature ? My workround [ class(cc) <- c("POSIXt",
"POSIXct") ] seems to do the job.
Many thanks
Bernie McConnell
"R version 1.3.1, 2001-08-31" on NT
> aa <- as.POSIXct("2001-09-23")
> bb <- as.POSIXct("2001-09-24")
> cc
2004 Nov 11
2
RODBC & POSIX & Daylight Saving blues
Dear All,
The recent improvement in RODBC to recognize datetimes in tables has
exposed my ongoing confusion.
All my data are obtained from a satellite system (Argos) which tags events
in the GMT time zone. Daylight saving is ignored. To my way of thinking
this means that
1. twelve-o-clock means halfway through the day regardless of season, and
2. the difftime of any two dates where
2001 Apr 20
5
map projections
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2001 May 10
2
memory blues
G'Day again,
I am attempting to read a large MSAccess file into R, but get memory
problems. With the first 100 rows of the table ("Macca99") things, as
shown below, are fine and the resulting object is 33,780 bytes. But when I
read the entire table ("MaccaDiv99") which is 218,000 rows R falls over
with the message:
Rgui.exe - Application error
The instruction at
2001 Apr 25
2
POSIX revisited
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2002 Oct 29
2
RODBC blues
Good Day,
Perhaps I have missed a posting but ... I cannot find the RODBC package on
cran at www.stats.bris.ac.uk/R/.
Many thanks
Bernie McConnell
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2001 Jan 02
1
chron blues
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2011 Jun 30
4
aggregating data
Hi,
I am interested in using the cast function in R to perform some aggregation. I did once manage to get it working, but have now forgotten how I did this. So here is my dilemma. I have several thousands of probes (about 180,000) corresponding to each gene; what I'd like to do is obtain is a frequency count of the various occurrences of each probes for each gene.
The data would look
2001 Feb 07
1
removing trailing spaces
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2009 Dec 10
3
An error message I don't recognize
I have recently been told I will have to maintain some CentOS servers at
work. Since I have only been using Slackware for the last 16 years, I
decided to install CentOS on one of my servers at home to get an idea of
the differences. I installed CentOS 5.4 from CD with no problems, did a
yum update, set up a couple of samba shares and started to copy over
some files from one of my other
2007 Oct 26
2
Implementation of a Speex based hardware VOCODER
Hi everyone,
I?m a graduate student in a Brazilian Intitute of Technology, and I?m
doing some academic research regarding secure voice transmission over phone
lines. One of our reserach goals is to implement a hardware vocoder, with low
bit rates, and a preferably free algorithm, to be used in this secure voice
system.
Actually, there is a functional system using a proprietary AMBE
2010 Jan 14
1
Problem with checkinstall
I installed checkinstall 1.6.2 on CentOS 5.4 VM (VMWare Server on
WinXP). After getting the dependencies installed it compiled with no
errors. But when I run it in its own source directory, I keep getting an
error that I can't track down. The message is:
install: cannot change ownership of '/usr/local/lib/installwatch.so': No
such file or directory.
But not only does the file
2009 Dec 20
1
Removable drive configuration
I have a box running CentOS 5.3 with two Dataport removable drive bays
installed on the second IDE interface (/dev/hdc and hdd). I want to
configure it so I can plug in and mount various drives at different
times, including different size drives. So far it will only recognize
the first drive I plug in, and only if I boot the box after inserting
the drive.
Is there any way to set it up so I
2008 Sep 28
2
using tapply on a data frame in a function
Hello,
I'm trying to use tapply to find group means in a function. It works
outside of a function, but I get the error message from the following code:
"Error in tapply(index, cluster, mean) : arguments must have same length."
Any suggestions? Thanks.
eric
d <- data.frame(cbind(cluster=1:2, value1=1:10, value2=11:20))
d
FindClusterTraits <- function(framename, index){
2004 May 13
2
tapply & hist
I'm learning how to use tapply.
Now I'm having a go at the following code in which dati contains almost 600
lines, Pot - numeric - are the capacities of power plants and SGruppo - text
- the corresponding six technologies ("CCC", "CIC","TGC", "CSC","CPC", "TE").
.....................................................
2010 Feb 02
3
tapply for function taking of >1 argument?
I'm sure I can put this together from the various 'apply's and split, but I
wonder if anyone has a quick incantation:
E.g. I can do tapply( data, groups, mean)
but how can I do something like: tapply( list(data,weights), groups,
weighted.mean ) ?
(or: mapply is to sapply as ? is to tapply )
Thanks for your help.
--
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2008 Aug 07
6
multiple tapply
Hi folk,
I tried this and it works just perfectly
tapply(iris[,1],iris[5],mean)
but, how to obtain a single table from multiple variables?
In tapply x is an atomic object so this code doesn't work
tapply(iris[,1:4],iris[5],mean)
Thanx and great summer holidays
Gianandrea
--
View this message in context: http://www.nabble.com/multiple-tapply-tp18868063p18868063.html
Sent from the R help
2008 Nov 14
1
# values used in a function in tapply
Hello,
I am using tapply to pull out data by the day of week and then perform
functions (e.g. mean). I would like to have the number of values used for
the calcuation for the functions, sorted by each day of week. A number of
entries in any given column are NAs.
I have tried the following code and simple variants with no luck.
for (i in 1:length(a[1,])){
x<-tapply(a[,i],a[,1],mean,
2009 Apr 13
3
tapply output as a dataframe
i use tapply and by often, but i always end up banging my head against
the wall with the output.
is there a simpler way to convert the output of the following tapply to
a dataframe or matrix than what i have here:
# setup data for tapply
dt = data.frame(bucket=rep(1:4,25),val=rnorm(100))
fn = function(x) {
ret =
c(unname(quantile(x,probs=seq(.25,.75,.25),na.rm=T)),mean(x,na.rm=T))
}
a =
2009 Oct 15
1
tapply() and using factor() on a factor
Dear List,
Shouldn't result1 and result2 be equal in the following case?
Note that log$RequestID is a factor. That is, is.factor(log$RequestID)
yields TRUE.
result1 <- tapply(log$Flag,factor(log$RequestID),sum)
result2 <- tapply(log$Flag,log$RequestID,sum)
Yet, when I summarize the output, I get the following:
summary(result1)
Min. 1st Qu. Median Mean 3rd Qu.