Cron still works, but launchd/launchctl seems to be preferred by some
if you're on Mac OS X.
J.
On Fri, Nov 19, 2010 at 7:09 PM, Steve Lianoglou
wrote:
> Hi,
>
> On Fri, Nov 19, 2010 at 10:09 AM, amit jain wrote:
>> Hi All,
>>
>> Can anyone point to any package/resouce to schedule a job in R wh
e.
>
> Wade
>
> On Fri, Nov 5, 2010 at 4:12 PM, Jeffrey Spies wrote:
>>
>> Perhaps this will help:
>>
>> > test1 <- test2 <- data.frame(col1=factor(c(1:3), labels=c("a", "b",
>> > "c")))
>> > test3 <-
Perhaps this will help:
> test1 <- test2 <- data.frame(col1=factor(c(1:3), labels=c("a", "b", "c")))
> test3 <- data.frame(col1 = 1:3)
Now:
> test2[2,1] <- test1$col1[1]
> test2$col1
[1] a a c
Levels: a b c
vs
> test3[2,1] <- test1$col1[1]
> test3$col1
[1] 1 1 3
Because test3's first column,
Take a look at the parallel computing section of:
http://cran.r-project.org/web/views/HighPerformanceComputing.html
specifically the line concerning the foreach package.
Jeff.
On Sun, Oct 31, 2010 at 6:38 PM,
wrote:
>
> Hi all,
>
> Just following on from a previous thread (for loop). Is there
Hi, all,
Let's say I have some time series data--10 subjects measured 20
times--that I plot as follows:
library(ggplot2)
dat <- data.frame(subject=as.factor(rep(1:10, each=20)),
time=rep(1:20, 10), measure=as.vector(replicate(10, rnorm(20,
mean=runif(1, 0, 15), sd=runif(1, 1, 3)
p <- qplot(ti
I would demonstrate one of the many LaTeX table functions. Off hand,
packages xtable, hmisc, and quantreg all have functions that convert R
objects to LaTeX tables.
If they're unwilling to work in LaTeX, you can use something like
LaTeXiT or Laeqed to create PDFs or PNGs of the tables for inserti
To get just the list you wanted, Gabor's solution is more elegant, but
here's another using the apply family. First, your data:
dat <- scan(file="/g/bork8/waller/test_COGtoPath.txt",what="character",sep="\n")
I expect dat to be a vector of strings where each string is a line of
values separated
ying,
Jeff.
On Sat, Oct 9, 2010 at 1:04 PM, David Winsemius wrote:
>
> On Oct 9, 2010, at 12:46 PM, Jeffrey Spies wrote:
>
>> Jim's solution is the ideal way to read in the data: using the sep=";"
>> argument in read.table.
>>
>> However, if y
If you just want a list of matrices and their counts, you can use
Peter's list of matrices, L, and then:
With plyr:
require(plyr)
count(unlist(lapply(L, toString)))
Without plyr:
as.data.frame(table(unlist(lapply(L, toString
Cheers,
Jeff.
On Sat, Oct 9, 2010 at 12:44 PM, Peter Ehlers wr
Jim's solution is the ideal way to read in the data: using the sep=";"
argument in read.table.
However, if you do for some reason have a vector of strings like the
following (maybe someone gives you an Rdata file instead of the raw
data file):
MF_Data <- c("106506;AIG India Liquid Fund-Institutio
Here's another method without using any external regular expression libraries:
dat <- read.table(tc <- textConnection(
'0,1 1,3 40,10 0,0
20,5 4,2 10,40 10,0
0,11 1,2 120,10 0,0'), sep="")
mat <- apply(dat, c(1,2), function(x){
temp <- as.numeric(unlist(strsplit(x, ',')))
min(temp
Hi, Michael,
When I teach/preach on R, I emphasize the language's focus on data,
both in its objects and operations. It might seems basic, but it's
fundamental to most of the features you and others have mentioned. As
a statistical programming language, what we intend to do with R is
often very na
This should do it:
df <- data.frame(a=character(0), b=character(0), stringsAsFactors=F)
because:
typeof(factor(0))
is "integer"
while:
typeof(character(0))
is "character".
Cheers,
Jeff.
On Wed, Oct 6, 2010 at 1:00 PM, N David Brown wrote:
> Does anyone know why a data frame created with
An alternative to Peter's solution:
createVariable <- function(name) {assign(name, NULL, envir=.GlobalEnv)}
Jeff.
On Wed, Oct 6, 2010 at 12:32 PM, Ralf B wrote:
> Can one create a variable through a function by name
>
> createVariable <- function(name) {
> outputVariable = name
>
sample(1:1,100),])".
> Now it include all 2 variable.
> Thank you for your answer to inspire.
> Jumlong
>
> 2010/10/5 Jeffrey Spies
>>
>> We'll probably need much more info, but this should get you started:
>>
>> nameOfDataSet[sample(1:1,
We'll probably need much more info, but this should get you started:
nameOfDataSet[sample(1:1, 100),]
You can replace the 1 with dim(nameOfDataSet)[1] to make it more dynamic.
Jeff.
On Tue, Oct 5, 2010 at 3:07 AM, Jumlong Vongprasert
wrote:
> Dear all.
> I have data with 2 variable x,y
As Joshua said, in your example sim isn't be declared anywhere
(neither in the environment nor as an argument in a function), but you
might try something more R-ish:
prop.doubles <- function(sim){
sum(sample(1:6, sim,replace=T)==sample(1:6,sim,replace=T))/sim
}
prop.doubles(1000)
Cheers,
Jef
read.table('path.to.file.tar.gz')
or any of its derivatives (read.csv, read.csv2, etc.). Of course
you'll need to set the arguments appropriately. For example:
read.table('path.to.comma.delimited.tar.gz', sep=",", header=T)
Cheers,
Jeff.
On Mon, Oct 4, 2010 at 6:48 PM, wenyue sun wrote:
> Hey
The first argument in download.packages should be of type character or
a vector of characters.
This worked for me:
install.packages('reshape2')
as did:
download.packages('reshape2', '~/Downloads/')
Cheers,
Jeff.
On Sun, Oct 3, 2010 at 8:57 PM, Chris Howden
wrote:
> Hi everyone,
>
>
>
> I’m
This is certainly not my area of expertise, but like Peter mentioned,
Jeff Terpstra published this:
http://www.jstatsoft.org/v14/i07
which has R code listed as supplements. Joe McKean seems to keep an
updated version of that code here:
http://www.stat.wmich.edu/red5328/WWest/
And Brent Johnson
Firstly, `*` is the multiplication operator in R. Secondly, you'll
need to convert your factors to numerics:
L<-0.559*as.numeric(T.Grade)-0.896*as.numeric(Smoking)+0.92*as.numeric(Sex)-1.338
Cheers,
Jeff.
2010/10/3 笑啸 :
> dear professor:
> I am a doctor of urinary,and I am developing a nomogra
You should examine what is being looped over when you use a for loop
with the "i in dataframe" syntax:
j<-1; for(i in ex){ cat('step', j, i, sep=" ", fill=T); j<-j+1}
As you can see, each column in ex is being set to i for each step of
the for loop. Instead, it seems that you want to step over e
If I understand your problem correctly, I think you need to be doing:
summary(data.name)
The functions read.dta and read.spss both return things (data frames,
if you use the to.data.frame=T argument with read.spss). So whatever
variable you set is what you should be doing a summary on. In this
Correcting Karena's solution:
for(i in 1:22) {
chrn <- paste("chr",i,sep="")
chrn=MEDIPS.readAlignedSeqences(BSgenome="hg19", file=chrn,numrows=
?) # no quotes around chrn
chrn=MEDIPS.genomeVector(data=chrn, bin_size=50,extend=250)
frames=? # I don't know how you get this variable
...
ou
You can use na.omit on x after it is passed into your apply function
if na.rm == T, or simply pass your na.rm to functions that use it,
such as sum.
Hope that helps,
Jeff.
On Fri, Oct 1, 2010 at 11:07 AM, Ochsner, Scott A wrote:
> Hi,
>
> Take a matrix with missing values:
>
>> X = matrix(rnorm
Hi all,
Is there an LR decomposition function in R and, if not, how can we get the
non-compact representation of Q from QR decomposition?
Thanks,
Jeff.
--
View this message in context:
http://www.nabble.com/LR-Decomposition--tp18072588p18072588.html
Sent from the R help mailing list archive a
niqueCount<-"))
>
> setReplaceMethod("uniqueCount",
> signature=c(x="MyMatrix", value="numeric"),
> function(x, ..., value) {
> [EMAIL PROTECTED] <- value
> x
> })
>
> uniqueCount(x) <- uniqueCount(x)
Howdy all,
I have a problem that I'd like some advice/help in solving---it has to do
with R's pass-by-value system. I understand the issue, but am wondering if
anyone has found a working solution in dealing with it for cases when one
wants to modify an object inside of a method, specifically whe
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