See
http://cran.r-project.org/bin/macosx/RMacOSX-FAQ.html#I-see-no-text-in-a-Quartz-plot_0021
And the default font is not serif ... that FAQ says it is Arial, but I
do not know if that is current.
Mac-specific questions to R-sig-mac please. (This must be Mac-specific
as the default device,
Hi Mitchell,
It's a good suggestion, and I had thought of this, but one concern is that
the model fit at the end of the terminal nodes would need to be a
generalized linear model (the predicted outcomes are factors). Given that
the models need to be fit automatically without human guidance, I'm n
I don't know this particular package well, but I believe "party" contains
something called "mob" which creates a regression tree terminating in
different models at each node. Could that be adapted to your project?
On Thursday, August 7, 2014, Craig Aumann wrote:
> I'm struggling with the best w
I'm struggling with the best way to apply different predictive models over
different geographical areas of a raster stack.
The context of the problem is that different predictive models are
developed within different polygonal regions of the overall study area.
Each model needs to be used to pred
Are you aware you have so-called smart quotes in your R code? That can
cause all sorts of interesting errors, though I don't get the exact
one you report.
Sarah
On Thu, Aug 7, 2014 at 6:37 PM, Fisher Dennis wrote:
> R 3.1.1
> OS X (and Windows)
>
> Colleagues
>
> I have some code that manages fi
R 3.1.1
OS X (and Windows)
Colleagues
I have some code that manages files. Previously (as late as 3.1.0), the
command:
file.info(FILENAME)$mtime == “”
yielded T/F
Now, it triggers an error:
Error in as.POSIXlt.character(x, tz, ...) :
character string is not in a stan
On Aug 7, 2014, at 12:59 PM, Tim Blass wrote:
> Hello,
>
> I am using R 3.1.1 on a (four year old) MacBook, running OSX 10.9.4.
>
> I just tried making and labeling a plot as follows:
>
>> x<-rnorm(10)
>> y<-rnorm(10)
>> plot(x,y)
>> title(main="random points")
>
> which produces a scatter pl
The problem is that you are not actually 'mapping' any variables to
the fill and colour aestethics so ggplot wont produce legends for
those. I'm not sure ggplots are appropiate for what you're trying to
do here but you can sure hack around it a bit, for instance try:
ggplot(tabu, aes(x=weeks, y=T)
Hello,
I am using R 3.1.1 on a (four year old) MacBook, running OSX 10.9.4.
I just tried making and labeling a plot as follows:
> x<-rnorm(10)
> y<-rnorm(10)
> plot(x,y)
> title(main="random points")
which produces a scatter plot of the random points, but without the title
and without any numbe
Hi,
I am having certain issues in using GP_fit() of GPfit package. Basically
each time I call this function, it gets hanged. What I have is a data
frame(matrix) which has 2 columns and some 2000 rows and first I normalize
the values in the range [0,1]. I also have a output status vector which has
I am trying to do a dose response in my dataset, but nothing go a head.
I am adapting a script shared on the web, but I unable to make it
useful for my dataset. I would like to got the LC50 for each Isolado
and if there are differences between then.
My data is https://dl.dropboxusercontent.com/u/
correcting a typo (400 MB, not GB. Thanks to David Winsemius for
reporting it). Spencer
###
Thanks to all who replied. For the record, I will summarize here
what I tried and what I learned:
Mike Harwood suggested the ff package. David Winsemius suggested
data.
Thanks to all who replied. For the record, I will summarize here
what I tried and what I learned:
Mike Harwood suggested the ff package. David Winsemius suggested
data.table and colbycol. Peter Langfelder suggested sqldf.
sqldf::read.csv.sql allowed me to create an SQL
Dear R-users,
I am looking for a weighted knn-search function, but i cannot manage to find
one. There are several options of weighted knn classifiers, but i would rather
use a simple 'search function' (such as get.knnx). Anyone knows a search
function with "weight" option ?
Thanks
[[al
I prefer the idiom
c(TRUE, a[-1] != a[-length(x)])
because it works for character and other data types as well.
I also find that thinking in terms of runs instead of subscripting
tricks is easier.
__
R-help@r-project.org mailing list
https://stat.ethz
Better:
b <- c(a[1]-1,a[-length(a)])
On 07 Aug 2014, at 17:28, Bart Kastermans wrote:
> For readability I like:
>
>> b <- c(0,a[-length(a)])
>> which(a != b & a == 0)
> [1] 4 12 18
>> which(a != b & a == 1)
> [1] 1 6 16 23
>
>
> On 07 Aug 2014, at 17:23, William Dunlap wrote:
>
>> My sol
For readability I like:
> b <- c(0,a[-length(a)])
> which(a != b & a == 0)
[1] 4 12 18
> which(a != b & a == 1)
[1] 1 6 16 23
On 07 Aug 2014, at 17:23, William Dunlap wrote:
> My solution may be a bit clearer if you define the function isFirstInRun
> isFirstInRun <- function(x) {
> if (le
My solution may be a bit clearer if you define the function isFirstInRun
isFirstInRun <- function(x) {
if (length(x) == 0) {
logical(0)
} else {
c(TRUE, x[-1] != x[-length(x)])
}
}
Then that solution is equivalent to
which(isFirstInRun(a) & a==1)
If 'a' contains NA's then
On 07 Aug 2014, at 11:16 , jim holtman wrote:
> rle
...with a little tinkering, like
> m <- c(1,cumsum(rle(a)$lengths)+1)
> m
[1] 1 4 6 12 16 18 23 34
then look at every 2nd element, discarding the last.
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solb
> a<-c(1,1,1,0,0,1,1,1,1,1,1,0,0,0,0,1,1,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1)
> which( a==1 & c(TRUE, a[-length(a)]!=1) )
[1] 1 6 16 23
> which( a==0 & c(TRUE, a[-length(a)]!=0) )
[1] 4 12 18
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Aug 6, 2014 at 7:12 PM, Johnnycz wrote:
> Hello,eve
Hi All
Following is my dataset.
dput(tabu)
structure(list(weeks = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30), values = c(9.45, 7.99, 9.29, 11.66, 12.16, 10.18, 8.04,
11.46, 9.2, 10.34, 9.03, 11.47, 10.51, 9.4, 10.08, 9.37, 10
Hello,
runSum calculates a running sum looking back a fixed distance n, e.g. 20.
How do I calculate a dynamic runSum function for an xts object?
In
otherwords, I want to calculate a running sum at each point in time
looking back a variable distance. In this example, values governed by
the vect
Dear R people,
I’ve been doing some hpc using R and openmpi. Unfortunately I’ve encoutred a
major problem and it’s nature is hard to pin down:
Essentially I call mpirun Rscipt … as soon as the script reaches a
foreach()%dopar% it halts indefinitely. I’ve attached the qsub script:
#!/bin/bash
#
Hi Jenny,
Have you tried igraph before? See, http://igraph.org/r/doc/
There are couple of centrality measures there.
Best,
-m
On 6 August 2014 02:50, Jenny Jiang wrote:
> Dear R-help,
>
> My name is Jenny Jiang and I am a Finance Honours research
> student from the University of New South W
Hi Scott,
The trick is to postpone plotting (plot = FALSE) and then do plotdev() with the
required limits:
library(plot3D)
x <- z <- seq(-4, 4, by=0.2)
y <- seq(-6, 6, by=0.2)
M <- mesh(x,y,z)
R <- with(M, sqrt(x^2 + y^2 +z^2))
p <- sin(2*R)/(R+1e-3)
x.limits <- c(-2, 2)
y.limits <- c(-2, 2)
sli
Josh,
Thank you for your detailed answer.
Best,
Ryan
On 7 Aug 2014, at 16:21, Joshua Wiley wrote:
> Hi Ryan,
>
> It does work, but the *apply family of functions always pass to the first
> argument, so you can specify e2 = , but not e1 =. For example:
>
>> sapply(1:3, `>`, e2 = 2)
> [1] FALSE
rle
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 Wed, Aug 6, 2014 at 10:12 PM, Johnnycz wrote:
> Hello,everybody,
> I have a sequence,like
> a<-c(1,1,1,0,0,1,1,1,1,1,1,0,0,0,0,1,1,0,0,0,0,0,1
> Igor Sosa Mayor
> on Wed, 6 Aug 2014 18:13:56 +0200 writes:
> Sverre Stausland writes:
>>> citation()
>> Error: $ operator is invalid for atomic vectors In
>> addition: Warning message: In packageDescription(pkg =
>> package, lib.loc = dirname(dir)) : no package
28 matches
Mail list logo
