Hi,
I was wondering if there is there a way you can schedule an R script to
run automatically through a scheduled task in windows or similar?..
Would R have to be open on the user's PC or could it be closed providing
we pointed it correctly at R?...
Thank you
Alex
Alex Johnson
Hello
I don't arrive to change the title of the legend
My code is :
library(ggplot2)
ma - max(General$AgeChangCat) ; mi - min(General$AgeChangCat)
Test$Recrutement - factor(Test$CadNonCadRecrut)
p -
ggplot(Test, aes(x=factor(Cat1), y=AgeChangCat )) +
ylim(mi,ma) +
geom_point() +
Le jeudi 10 octobre 2013 à 21:45 -0700, Ira Sharenow a écrit :
Thanks for the suggestion. From R version 3.0.2, I tried
testDF7 = iconv(x = test07 , from = UCS-2, to = )
Encoding(testDF7)
[1] unknown
testDF7[1:6]
[1] NA NA NA NA NA NA
So using UCS-2 produced
if you want to use R itself, you could try --
# check your time zone's abbreviation
Sys.time()
# subtract the time you want the program to run from the current time,
# including your time zone..mine is EDT
Sys.sleep( as.POSIXct( 2013-10-11 06:30:00 EDT ) - Sys.time() )
-- at the very top of
Hi R wizards,
Can any one tell me how to code to remove the title panels in a lattice
graph and to make my confidence intervals black instead of coloured?
I have the following code:
library(effects)
library(lme4)
data(mao1, package=lme4)
fm1 - lmer(frat ~ flandusenumb + ground.cover_lo +
(I'm afraid this post didn't reach the list on last Wednesday, here it is again
)
hi R-list,
And sorry for my frenglish !
I am running R Good Sport release ( i386-w64-mingw32/i386 (32-bit) ) )
under Windows 7 Professional, Service Pack 1.
My perl executable is ActivePerl build 817 [257965]
I have recently installed R and am trying to do some work on it.
Congratulations; you are learning one of (possibly the) world's most powerful
statistics environments.
To be honest I'm finding it a PAIN to [use?].
You can have cheap, powerful, easy to use statistics software. Pick any two.
On 10/10/2013 11:33 PM, Rebecca Stirnemann wrote:
Hi Michael,
Thanks! That worked. Which is so brilliant!
A couple of questions. In regards to display.
Do you know how to add labels on to the graph? The code below doesn't
work.
Not surprising, since your data, mao1, is not in the lme4
Hi
I usually use scale
something like
scale_fill_discrete(name = Fancy Title)
shall do the trick
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Arnaud Michel
Sent: Friday, October 11, 2013 11:02 AM
To: R
Hi
not sure if it is the most efficient and clever solution
ifelse(b7, NA, cumsum(c(0,diff(!(b7)))==1))
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Benjamin Gillespie
Sent: Thursday, October 10, 2013 12:39
Hello all -
I have an example column in a dataFrame
id.name
123.45
123.45
123.45
123.45
234.56
234.56
234.56
234.56
234.56
234.56
234.56
345.67
345.67
345.67
456.78
456.78
456.78
456.78
456.78
456.78
456.78
456.78
456.78
...
[truncated]
And I'd like to create a second vector of sequential
OK It is right
Thank you Petr
Michel
Le 11/10/2013 14:58, PIKAL Petr a écrit :
Hi
I usually use scale
something like
scale_fill_discrete(name = Fancy Title)
shall do the trick
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
Also,
it might be faster to use ?data.table()
library(data.table)
dt1- data.table(dat1,key='id.name')
dt1[,x:=seq(.N),by='id.name']
A.K.
On , arun smartpink...@yahoo.com wrote:
Hi,
Try:
dat1-
structure(list(id.name = c(123.45, 123.45, 123.45, 123.45, 234.56,
234.56, 234.56, 234.56,
Hello everybody,
I have count data and with these data, I would like to build a mixed
model by using the function glmer(). In a first time, I calculated the c-hat of
a simple model with glm() to verify overdispersion and I found a c-hat = 18. I
also verified overdispersion in the mixed model
To close the loop on this, I can report that I solved the convergence issue by
updating the original semBoot code. I thought I might be using something out of
date, so I ran sem:::bootSem.sem and grabbed the function code.
I updated the gist where I create bootSem2() to stop R from crashing on
Hi
I named your data test
test$x-1
test$x-unlist(lapply(split(test$x, test$id.name), cumsum))
test
id.name x
1 123.45 1
2 123.45 2
3 123.45 3
4 123.45 4
5 234.56 1
6 234.56 2
7 234.56 3
8 234.56 4
9 234.56 5
10 234.56 6
11 234.56 7
12 345.67 1
13 345.67 2
14 345.67 3
Steps:
1. write your code in R command line format
2. save to a .sh file
3. Add to cron of linux machine
Regards,
Vivek
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Johnson, Alex
Sent: Friday, October 11, 2013 4:05 PM
To:
On 11-10-2013, at 15:26, Steven Ranney steven.ran...@gmail.com wrote:
Hello all -
I have an example column in a dataFrame
id.name
123.45
123.45
123.45
123.45
234.56
234.56
234.56
234.56
234.56
234.56
234.56
345.67
345.67
345.67
456.78
456.78
456.78
456.78
456.78
456.78
Success!
Thanks to everyone who helped. I needed to have the right file encoding
parameter when using read.table().
test08 = read.table(test.csv, sep = ,, header = TRUE,
stringsAsFactors = FALSE, fileEncoding = UCS-2)
Upon further research:
Also,
df1 - do.call(rbind,lapply(split(df,df$group),function(x)
data.frame(dt2=seq(x[1,1],x[nrow(x),1],by=15 min),group=x[1,2])))
id1- df1[,2][match(as.POSIXct(dt),df1[,1])]
id1[is.na(id1)]- 0
identical(id1,id)
#[1] TRUE
A.K,
On Thursday, October 10, 2013 9:29 PM, arun smartpink...@yahoo.com
I'd wager that the problem lies in the expectations. Karl: Read up on what
order() actually does; it is not what I think you think it does.
Perhaps
str.dat[match(gen.names, str.dat$ind.names),]
or
rownames(str.dat) - str.dat$ind.names
str.dat[gen.names,]
was intended? (Both untested, some
And if you need some extra digits:
require(Rmpfr)
testfn-function(x){2^x+3^x-13}
myint-c(mpfr(-5,precBits=1000),mpfr(5,precBits=1000))
myroot-unirootR(testfn, myint, tol=1e-30)
myroot
John Nash
On 13-10-11 06:00 AM, r-help-requ...@r-project.org wrote:
Message: 33
Date: Thu, 10 Oct 2013
-Original Message-
I'm using R version 3.0.0 on a mac. I'm having trouble getting order to
behave as I expect it should. I'm trying to sort a data.frame according
to a character vector. I'm able to sort the data.frame, but it retruns an
unexpected result. I have no idea where the
On Oct 10, 2013, at 8:16 AM, Zev Ross wrote:
Hi All,
I'm trying to edit a file in place using system2 and sed from within R. I can
get my command to work unless there is a backslash in the command in which
case I'm warned about an unrecognized escape. So, for example:
system2(sed -i
Is it easy or difficult to label the abscissa of a scatter graph as
1/trueScaleValue at that point?
--
View this message in context:
http://r.789695.n4.nabble.com/labeling-abscissa-using-a-function-of-the-plotted-scale-tp4678075.html
Sent from the R help mailing list archive at Nabble.com.
Hi,
In the example you showed:
m1- matrix(0,length(vec),max(vec))
1*!upper.tri(m1)
#or
m1[!upper.tri(m1)] - rep(rep(1,length(vec)),vec)
#But, in a case like below, perhaps:
vec1- c(3,4,5)
m2- matrix(0,length(vec1),max(vec1))
indx -
At this point 3 functions have been suggested and I'll add a 4th:
f1 - function(x)unlist(lapply(unname(split(rep.int(1L,length(x)), x)),
cumsum))
f2 - function(x)unlist(sapply(rle(x)$lengths, function(k) 1:k ))
f3 - function(x)ave(x,x,FUN=seq)
f4 - function(x)ave(seq_along(x), x,
I think all of the above call lapply(split()) at some point and that can use
a lot of memory when there are lots of unique values in x. You can use
a sort-based algorithm to avoid that problem.
E.g.,
Sequence -
function(nvec) {
# like base::sequence, but faster for long nvec. If
system2(sed, args=c(-i, s/oldword\\s/newword/g, d:/junk/x/test.tex))
/Henrik
On Fri, Oct 11, 2013 at 8:58 AM, David Winsemius dwinsem...@comcast.net wrote:
On Oct 10, 2013, at 8:16 AM, Zev Ross wrote:
Hi All,
I'm trying to edit a file in place using system2 and sed from within R. I
can
Are you familiar with the sos package? Consider the following:
library(sos)
op - findFn('orthogonal polynomial') # 165 links in 35 pkgs
ops - findFn('orthogonal polynomials')#158 links in 35 pkgs
op. - op |ops# 194 links in 43 pkgs
save(op., file='orthopoly.rda')
summary(op.)
Thanks to Henrik and David for responses. Both were right. A small edit
to my description of the problem (which has now been solved). I said
that the first version of system2 ( system2(sed -i s/oldword/newword/g
d:/junk/x/test.tex)) worked fine but in fact it was not working, it
just wasn't
Look at the ?Startup help page in R. It shows a couple of ways to have
code run automatically when R starts (and can depend on which folder R
starts from). So you could have the windows task scheduler run R and use
the above to set the script to run.
Also look at ?Rscript for a way to run a
On 10.10.2013 21:40, Sheri wrote:
Hi everyone,
I am hoping someone can help with my attempted use of the expression
function. I have a long series of text and variable to paste together
including a degree symbol. The text is to be placed on my scatter plot
using the mtext function.
Using
Hi everybody,
I thought I was using the get() fn correctly here to loop over multiple
data frame names in an R for() loop. Can someone advise?
miss-c(#NULL!,999)
d-c(d1,d2,d3,d4)
for(i in 1:4){
+
+ miss1-ifelse(i=2,miss[1],miss[2])
+ miss1
+
+
I think you want 'assign' at that point. Would suggest using a 'list'
to store the input instead of unique named objects. 'list's are
easier to manage.
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On
Dear Ryan,
On 11 October 2013 16:50, Ryan Morrison ryan.r.morri...@me.com wrote:
The cpquery function is definitely easier to use given that my evidence is
contained in lists.
For the record, this is not a general solution to the bug. It's true
that likelihood weighting (method = lw) is easier
Dear Ryan,
On 11 October 2013 20:44, Ryan Morrison ryan.r.morri...@me.com wrote:
Thanks for the clarification. For my immediate purposes I think using the
likelihood
weighting method will suffice. I'm calculating conditional probabilities
based on a
single instantiation each time I use the
Attempting to follow the OP's conditions and assuming I understood
them correctly, here is one way to wrap this up into a function:
makeMat - function(x)
{
stopifnot(is.integer(x))
nr - length(x)
nc - max(x)
# Initialize a matrix of zeros
m - matrix(0, nr, nc)
#
On Oct 11, 2013, at 9:48 AM, Hurr wrote:
Is it easy or difficult to label the abscissa of a scatter graph as
1/trueScaleValue at that point?
It's easy.
?axis
--
View this message in context:
simpler (and sloppier) but with **no looping or apply's **
**IFF* the matrix is structured as in the OP's example, then lower.tri
(or upper.tri) should be used:
n - 4 ## number of columns in matrix -- note that I changed it from
the example; does not have to be square
x - 1:3 ## the number of
Thanks Dennis. I noticed I didn't take the 0 value into consideration and
also didn't check the unsorted vector.vec1- c(2,4,1)
is.numeric(vec1)
#[1] TRUE
makeMat(as.integer(vec1))
makeMatrix2- function(x){
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
m1-
Hi Arun,
Many thanks for that solution; it works well :)
I also have some data where times and dates don't follow a 15 min intervals,
so, for example:
dates=rep(01/02/13,times=20)
Nice, thanks Petr :)
Ben Gillespie, Research Postgraduate
o---o
School of Geography, University of Leeds, Leeds, LS2 9JT
o---o
Tel: +44(0)113 34 33345
Mob: +44(0)770
Seems like a bug in the code:
x- c(3,4,1)
n- 3
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
n- 4
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid
Hi
I have a question regarding plots in R. I have data from the SP 500 in the
format:
date close change
1980-01-07 109.92 3.4
I plotted the data with plot(spdata$date, log(spdata$close), type=p)
Now I want to ad the colors green and red to the data frame. if the change
is
Dear all,
I am using the DEoptim package. My optimization syntax has three decesion
variables. If I increase the upper bound of the first or third variables,I
get an error related to singularity of a matrix. However, I need to
increase the upper bound of the first and third variables. The true
Thanks Marco,
The cpquery function is definitely easier to use given that my evidence is
contained in lists.
Thanks for your help!
---
Ryan
Ryan Morrison, PE
PhD Candidate
University of New Mexico
Department of Civil Engineering
Centennial Engineering Center, Room 3057
Phone: 505-633-5506
Hello,
Working on an ubuntu 13.04 with a sessionInfo() specified below, I try to
update my packages, and RSQLite update consistently fails like this:
update.packages(checkBuilt=TRUE, ask=FALSE)
Warning: package 'RUnit' in library '/usr/lib/R/site-library' will not be
updated
trying URL
Hi Marco,
Thanks for the clarification. For my immediate purposes I think using the
likelihood weighting method will suffice. I'm calculating conditional
probabilities based on a single instantiation each time I use the function. The
evidence I use for instantiation changes for each i+1 in my
Hi Ben,
No problem.
Try:
df1 - do.call(rbind,lapply(split(df,df$group),function(x) {
data.frame(dt2=dt[dt=x[1,1] dtx[length(x),1]],group=x[1,2]) }))
id1 - df1[,2][match(as.POSIXct(dt),df1[,1])]
id1[is.na(id1)]- 0
identical(id,id1)
#[1] TRUE
A.K.
On Friday, October 11, 2013 5:43 PM,
I have a java class with routines (and their tests) that I would like to
use in R so I don't have to have two copies of important subroutines.
I have looked at rjava, but can't grasp it all and don't know what are
the important items to observe first so I don't get into too much
trouble later. I
Thanks.
--
View this message in context:
http://r.789695.n4.nabble.com/labeling-abscissa-using-a-function-of-the-plotted-scale-tp4678075p4678103.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
On 10/11/2013 11:45 PM, Mubar wrote:
Hi
I have a question regarding plots in R. I have data from the SP 500 in the
format:
date close change
1980-01-07 109.92 3.4
I plotted the data with plot(spdata$date, log(spdata$close), type=p)
Now I want to ad the colors green and red
Hello,
Iâm unsuccessfully trying to apply piecewise linear regression over each of
22 groups. The data structure of the reproducible toy dataset is below. Iâm
using the âsegmentedâ package, it worked fine with a data set that
containing only one group (âLot.Runâ).
$ Cycle :
I am trying to get the calculated mean and the symbol of x-bar to show in
the title of multiple histograms in R. Here is the code I have for one of
the histograms:
hist(outcome[,11], main= Heart Attack (expression(bar(x))) =
(mean(outcome[,11])), xlab =30-day Death Rate, xlim = c(min(hist_min),
Hello,
I'm trying to find a solver that will work for the mixed complementarity
problem (MCP). I've searched the CRAN task view page on optimization and
mathematical programming as well as many google searches to no avail. Does
anyone know if there is an MCP solver available for R?
Thanks
On 10/12/13 11:39, Emily Jean Fales wrote:
I am trying to get the calculated mean and the symbol of x-bar to show in
the title of multiple histograms in R. Here is the code I have for one of
the histograms:
hist(outcome[,11], main= Heart Attack (expression(bar(x))) =
(mean(outcome[,11])), xlab
Your examples are the problem:
On Fri, Oct 11, 2013 at 2:43 PM, arun smartpink...@yahoo.com wrote:
Seems like a bug in the code:
x- c(3,4,1)
n- 3
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
## This can't
You have posted neither code nor your data, contrary to what the
posting guide asks.
How do you expect anyone to help?
Cheers,
Bert
On Fri, Oct 11, 2013 at 4:54 AM, Aya Anas aa...@feps.edu.eg wrote:
Dear all,
I am using the DEoptim package. My optimization syntax has three decesion
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