On Tue, Nov 26, 2013 at 1:02 AM, C. Alina Cansler acans...@uw.edu wrote:
Don,
This looks helpful:
https://stat.ethz.ch/pipermail/r-help/2011-March/272361.html
Yes, he's a helpful chap.
The fundamental problem here is the colour palette. When I was a boy
all we had was a pen plotter with
On 26/11/2013 08:25, Barry Rowlingson wrote:
On Tue, Nov 26, 2013 at 1:02 AM, C. Alina Cansler acans...@uw.edu wrote:
Don,
This looks helpful:
https://stat.ethz.ch/pipermail/r-help/2011-March/272361.html
Yes, he's a helpful chap.
The fundamental problem here is the colour palette. When I
On Tue, Nov 26, 2013 at 9:58 AM, Prof Brian Ripley
rip...@stats.ox.ac.uk wrote:
But the image function (and probably levelplot) doesn't allow that so
Mis-information alert! The help says
col: a list of colors such as that generated by ‘rainbow’,
‘heat.colors’,
Hi
It is work for split/lapply or sapply approach.
ff-function(data) {ss-lm(y~x, data); c(coef(ss), summary(ss)$adj.r.squared)}
lapply(split(data[,1:2], data$city), ff)
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On
The functions 'age.window' and 'cal.window' in the package 'eha' are
designed to perform 'rectangular cuts' in the Lexis diagram. So, in your
case,
require(eha)
dat - data.frame(enter = rep(0, length(Survival_days), exit =
Survival_days, event = Outcome)
dat.1 - age.window(dat, c(0, 2190))
Dear all,
Thank you so much for all the replies.
The following worked exactly as needed:
library(HSAUR2, lib.loc=/home/liv/R/i686-pc-linux-gnu-library/3.0)
library(reshape2, lib.loc=/usr/local/lib/R/site-library)
data(womensrole)
zzz - melt(womensrole,id.var=c(education,gender))
-
The freetype people fixed the 2nd set of issues with system fonts shipped with
Mac OS X, and released 2.5.1 almost immediately after that. So there are
new bundles under http://sourceforge.net/projects/outmodedbonsai/files/R/ .
Just a reminder that the official R binaries for windows/mac OS X
Dear Forumites,
Hi, I'm a long time eavesdropper, first time poster, but I simply couldn't
find any answer to this perhaps rather naive question:
I am trying to see if my data is significantly different from a null
hypothesis using GLMMs.
I would like to run a GLMM because I have random
Hi there, I am learning to use and understand R but have come up against a
problem. I'm sure it is a pretty simple thing, but I can't find a solution.
I wonder if anyone could help?
I have been using the The R Book by Crawley and working through some of
the examples before trying to apply them to
Hi all,
I'm sorry, maybe the solution in my problem is easy, but i am still new to R,
and any help
here is appreciated!
I want to create a data frame, or a matrix in which the output of each loop
will be stored.
Here is the program
n-c(10,8,7,5)
n_goal-c(8,9,9,4)
w-c(0.1,0.1,0.1,0.1)
Fantastic. Thanks very much! Is there an easy way to plot the points and
the 4 areas?
Best,
Ioanna
-Original Message-
From: David Carlson [mailto:dcarl...@tamu.edu]
Sent: 25 November 2013 15:21
To: 'IOANNA'; r-help@r-project.org
Subject: RE: [R] Aggregating spatial data
Something
-Original Message-
I am attempting to bin a vector of numbers between 0 and 1 into
intervals of 0.001 but many values at the endpoints of the intervals
are getting binned into the wrong interval. For example, the first 3
rows are binned incorrectly here:
From: Jim Holtman
FAQ
You need to look at the full accuracy of the number representation:
df
x cutR
1 0.30800 [0.308,0.309)
2 0.42200 [0.422,0.423)
3 0.17400 [0.173,0.174)
4 0.04709 [0.047,0.048)
Vuew(print(df$x, digits = 20))
Error: could not find function Vuew
No suitable frames for recover()
AFAICT, your code should not be expected to run unless you have an
attach(lizards) prior to the glm() call. Either that or you need data=lizards
in the call itself.
-pd
On 26 Nov 2013, at 12:43 , philthe1st philip.br...@bristol.ac.uk wrote:
Hi there, I am learning to use and understand R but
Hello all,
I have to export the graph to SVG format. I was able to do that.
The problem is that I'm using this SVG file in a XML-FO processor that
doesn't support percentages for the rgb color, only absolute numbers.
Is there any way I can control it from the R code?
Thanks!
--
Igor.
And in fact
attach(lizards)
is included on page 560. You must have left it out of your
example code. Without it, your first example would not have run
either.
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
Jonas Josefsson jonas.josefsson at slu.se writes:
Hi!
(I was initially going to say that this question would probably be
better on r-sig-mixed-mod...@r-project.org, but now that I've been
through it I've changed my mind -- there aren't really any issues here
that are specific to mixed
The points are easy:
plot(y~x, w, pch=., asp=1)
The four polygons are just four polygon() calls:
polygon(c(.5, .5, 50.5, 50.5), c(.5, 50.5, 50.5, .5))
polygon(c(.5, .5, 50.5, 50.5), c(50.5, 100.5, 100.5, 50.5))
polygon(c(50.5, 50.5, 100.5, 100.5), c(.5, 50.5, 50.5, .5))
polygon(c(50.5, 50.5,
I am using R3.0.2 on a windows 64 bit machine. I building a .rnw file
from an .r file. I am using MikTex as the engine for latex. The
function dtaStata - read.dta(core2.dta, convert.dates=TRUE) works
well in .R but; When I try reading a dta (stata) file I get an error
subscript out of bounds. The
Hi,
Try:
Lines1 - readLines(textConnection(contig number 11
tttgctcggaatc
contig number 23
gcacttccttattatacaggtaaaccgtatttggat
contig number 3
aaagctcggaatt))
seq1 - ncattccattcattaattaattaatgaatgaatgn
concatenated_contig -
Hello,
I solved my problem with new version :
- R version 3.0.2
- pbdMPI version 0.2-1
- Intel compiler version 13.1.3
- OpenMPI version 1.6.5
Options for compiling R :
export CC=icc
export CXX=icpc
export F77=ifort
export FC=ifort
export AR=xiar
export LD=xild
export IFLAGS=-O3 -fp-model
Hi,
My problem is as follows:
INPUT:
Frequency from one column and value of Piglets from another one
OUTPUT:
Repeat this Piglet value as per the Frequency
i.e.
Piglet 1, Frequency 3, implies 1,1,1
Piglet 7, Frequency 2, implies 7,7
SOLUTION:
This is what I have tried so far:
1. A helper
Thank you for your replies,
A massive oversight on my part. I did include it the first time but not
later on. Sorry to waste your time with this simple question.
Thanks,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/error-in-eval-tp4681170p4681187.html
Sent from the R
I would like to try to run the quantstrat package (located here:
https://r-forge.r-project.org/R/?group_id=316). However, if I try to load
quantstrat, I get a warning that it is not available for R v3.0.1, so perhaps
that is the end of it.
Does anyone know if it is possible to run quantstrat
On Tue, Nov 26, 2013 at 11:34 AM, Ira Fuchs irafu...@gmail.com wrote:
I would like to try to run the quantstrat package (located here:
https://r-forge.r-project.org/R/?group_id=316). However, if I try to load
quantstrat, I get a warning that it is not available for R v3.0.1, so perhaps
that
Hi,
If I understand the question don't you simply want:
with(df.1, rep(Piglets, times=Frequency))
[1] 5 7 7 8 8 8 9 9 9 10 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11
[26] 11 12 12 12 12 12 13 13 13 14 14
Sarah
On Tue, Nov 26, 2013 at 9:59 AM, Burhan ul haq ulh...@gmail.com wrote:
Thanks to everyone who weighed in on this. I found a naive solution that was
good enough for my needs, and it may take me a bit to get the subtleties
of your comments.
On Nov 26, 2013, at 2:12 AM, Barry Rowlingson b.rowling...@lancaster.ac.uk
wrote:
On Tue, Nov 26, 2013 at 9:58 AM, Prof
Thanks. Since quantstrat showed a Last change date of only 6 days ago and a
Build status of Current, I blithely assumed that the command that is shown
for installing it would work:
install.packages(quantstrat, repos=http://R-Forge.R-project.org;)
Installing package into
On Tue, Nov 26, 2013 at 11:55 AM, Ira Fuchs irafu...@gmail.com wrote:
Thanks. Since quantstrat showed a Last change date of only 6 days ago and a
Build status of Current, I blithely assumed that the command that is shown
for installing it would work:
install.packages(quantstrat,
Hello;
I have the following table
m
V2 V1 V3
1 C/L 0 732179
3 C/S 0 803926
19 D/F 0 724924
17 D/I 0 755841
13 D/L 0 731904
15 D/S 0 798289
11 I/F 0 871670
9 I/I 0 897718
5 I/L 0 2628113
7 I/S 0 2628113
2 C/L 1 1107269
4 C/S 1 1395714
20 D/F 1 1181282
18 D/I 1
Hi,
Not sure whether this is what you wanted.
df.1[rep(1:nrow(df.1),df.1[,2]),]
A.K.
On Tuesday, November 26, 2013 12:31 PM, Burhan ul haq ulh...@gmail.com wrote:
Hi,
My problem is as follows:
INPUT:
Frequency from one column and value of Piglets from another one
OUTPUT:
Repeat this
On 26-11-2013, at 15:59, Burhan ul haq ulh...@gmail.com wrote:
Hi,
My problem is as follows:
INPUT:
Frequency from one column and value of Piglets from another one
OUTPUT:
Repeat this Piglet value as per the Frequency
i.e.
Piglet 1, Frequency 3, implies 1,1,1
Piglet 7, Frequency
This is the kind of question where including the output of sessionInfo() is a
really good idea.
This would, for example, answer the question of whether you are in fact using
the x64 version of R.
---
Jeff Newmiller
Hello;
I have the following table
m
V2 V1 V3
1 C/L 0 732179
3 C/S 0 803926
19 D/F 0 724924
17 D/I 0 755841
13 D/L 0 731904
15 D/S 0 798289
11 I/F 0 871670
9 I/I 0 897718
5 I/L 0 2628113
7 I/S 0 2628113
2 C/L 1 1107269
4 C/S 1 1395714
20 D/F 1 1181282
18 D/I 1
Thanks for your help! I got quantstrat working after building it along with the
FinancialInstrument and blotter packages (as well as foreach).
Ira
On Nov 26, 2013, at 12:59 PM, Joshua Ulrich wrote:
On Tue, Nov 26, 2013 at 11:55 AM, Ira Fuchs irafu...@gmail.com wrote:
Thanks. Since
You reused 'i' in the inner 'for' loop. This looks like it give the
correct answer:
n-c(10,8,7,5)
n_goal-c(8,9,9,4)
w-c(0.1,0.1,0.1,0.1)
matrix-mat.or.vec(6,4)
FI-function(n_t) {
+
(((n_goal[1]-n[1]+W[1]-f[1]+n_t[1])^2)+((n_goal[2]-n[2]+W[2]-f[2]+n_t[2]-n_t[1])^2)+
+
All interesting suggestions.
I guess a better example of the code would have been a good idea. So,
I'll put a relevant snippet here.
Rows are cases. There are multiple cases for each ID, marked with a
date. I'm trying to calculate a time recency weighted score for a
covariate, added as a new
Hi R Experts,
I need your help with two question regarding randomForest.
1. When I run a Random Forest model how do I extract the formula I used
so that I can store it in a character vector in a dataframe?
For example the dataframe might look like this if I am running models using the
Hi,
On Tue, Nov 26, 2013 at 11:41 AM, Noah Silverman noahsilver...@ucla.edu wrote:
All interesting suggestions.
I guess a better example of the code would have been a good idea. So,
I'll put a relevant snippet here.
Rows are cases. There are multiple cases for each ID, marked with a
I could use a little help writing a panel function to append text to each
panel of a lattice::barchart(). Below is a modified version of the barley
dataset to illustrate.
data(barley)
# add a new variable called samp.size
barley$samp.size-round(runif(n=nrow(barley), min=0, max=50),0)
# Below is
Hello,
I am using R, 2.15.2, on a 64-bit Linux box. I run R through Emacs' ESS.
R runs in a French, Canadian-French, locale and lately I got surprising
results
from functions making factor variables from character variables. Many of
the
variables in input data.frames are character variables
Hi,
May be:
barplot(m$V3,names.arg=paste(m$V2,m$V1,sep=\n),col=rainbow(10))
#or
barplot(m$V3,names.arg=m$V2,col=rainbow(20),legend=m$V1, args.legend = list(x =
topright,box.lwd=0,border=FALSE))
A.K.
On Tuesday, November 26, 2013 1:18 PM, Adel ESSAFI adeless...@gmail.com wrote:
Hello;
I have
On 2013-11-27 10:07, Tim Sippel wrote:
I could use a little help writing a panel function to append text to
each
panel of a lattice::barchart(). Below is a modified version of the
barley
dataset to illustrate.
data(barley)
# add a new variable called samp.size
Hi
Try
barchart(yield ~ variety | site, data = barley,
groups = year, layout = c(1,6), stack = TRUE,
auto.key = list(space = right),
ylab = Barley Yield (bushels/acre),
scales = list(x = list(rot = 45)),
panel = function(x,y,...){
Hi
If you want to track the individual fish try turning the graph around. not
all is displayed - just a concept graph
windows(10,8)
xyplot(gspd_mps ~ as.POSIXct(x)|TagID, dat,
as.table = TRUE,
layout = c(1,14),
groups = Station,
strip = FALSE,
type
Thanks Duncan, although that helps me learn something new about lattice
plots (I never had played with the as.table argument before), I need
something a bit more elegant. I need to generate about 10 plots, each with
up to 20 panels. So the panel-by-panel approach is a bit cludgier than I am
Tomer Czaczkes Tomer.Czaczkes at biologie.uni-regensburg.de writes:
Dear Forumites,
Hi, I'm a long time eavesdropper, first time poster, but I simply couldn't
find any answer to this perhaps rather naive question:
I am trying to see if my data is significantly different from a null
Hi,
For the first part, you could do:
fmi2 - fmi
attributes(fmi2$terms) - NULL
capture.output(fmi2$terms)
#[1] Species ~ Sepal.Length + Sepal.Width + Petal.Length + Petal.Width
A.k.
On Tuesday, November 26, 2013 3:55 PM, Lopez, Dan lopez...@llnl.gov wrote:
Hi R Experts,
I need your help
Hi Arun,
Thanks for your help. Sorry for my late response. Take care and stay fine.
Regards,
Halim
On Sun, 24 Nov 2013 07:45:24 -0800 (PST), arun wrote
Hi Halim,
I guess this works for you. Modifying Jeff's solution:
volinp-c(0,0.000467,0.002762,0.008621,0.020014,0.038907,0.067094)
Dear All,
In the following simple case I can't seem to get an improved fit,
despite trying all of the control possibilities. As there seem to be
no examples anywhere which show use of functions such as dnorm
within a formula, and as I am not confident at all that my formula is
correctly
PLEASE REMOVE THIS MESSAGE AND PREVIOUS COPIES ! - Many thanks
Dear All,
In the following simple case I can't seem to get an improved fit,
despite trying all of the control possibilities. As there seem to be
no examples anywhere which show use of functions such as dnorm
within a formula, and as
Bonjour, j'ai une série de 18 groupes de données de 20 réplicats chaque à
comparer (dont plusieurs controles). Est-ce que je peux faire un test de Tukey
pour analyser autant de données? si oui, comment? sinon, quel test je peux
faire?
J'ai fait un bartlett.test pour vérifier l'égalité des
Hi all,
I am attempting to create a weights object and perform a Moran I test as
well. I have a very large spatial weights matrix (roughly 22,000x22,000)
that was created in Excel and read into R, and I'm now trying to implement:
library(spdep)
SW=mat2listw(matrix)
I am getting the following
Hi,
I am a novice of R and I have attended the coursea courses for computing
for R analysis.
Recently I am using both the lattice and ggplots package to create
scatterplot in R.
when using lattice package, I can contrl the whole graph size(including
plot area and title/footnote) with below codes:
I was trying to find root for two non-linear equations. Â My equation is
similar as follows:
 120 â 5* exp(b0+b1*4)
/ (1+ exp(b0+b1*4) ) = 0
 690 â 31*
exp(b0+b1*4) / (1+ exp(b0+b1*4) ) =0
Â
How could find value of bo and b1 root values?
Thank you in advance.
Â
Saymi
On Nov 26, 2013, at 5:03 PM, Béatrice Perron beatrice_per...@hotmail.com
wrote:
Bonjour, j'ai une série de 18 groupes de données de 20 réplicats chaque à
comparer (dont plusieurs controles). Est-ce que je peux faire un test de
Tukey pour analyser autant de données? si oui, comment? sinon,
Bonjour,
Il serait préférable de poster votre question en anglais. Mais à mon
sens, ceci est plutôt une question de statistique générale, sans
rapport direct avec R.
Please post your question in English, as everybody would be able to
understand it. But I would ask a local statistician for
I was wondering if anyone knew of a package that contained a function
of Buse's (http://www.jstor.org/stable/2683631 ) GLS R2 equation? If
not, I would greatly appreciate any pointers about how I would
implement Buse's equation using the results from nlme::gls function!
Thanks very much for your
Hi,
A big thanks to everyone who replied. But special ones to Berend for
pointing out my mistakes, that will really help me in future.
Cheers !
On Tue, Nov 26, 2013 at 11:19 PM, Berend Hasselman b...@xs4all.nl wrote:
On 26-11-2013, at 15:59, Burhan ul haq ulh...@gmail.com wrote:
Hi,
Hi
I do not know what in.data is and I am not familiar with ggplot2
and I am not sure what you mean
I think you need to keep things simple and explain what you want and dput a
dataset
Try this
xyplot(test2 ~ test1, data = in.data,
groups = SEX,
par.settings =
Dear Jeff;
Below is the output for sessioninfo()
R version 3.0.2 (2013-09-25)
Platform: x86_64-w64-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=English_United Kingdom.1252 LC_CTYPE=English_United
Kingdom.1252
[3] LC_MONETARY=English_United Kingdom.1252 LC_NUMERIC=C
[5] LC_TIME=English_United
Hello,
Using 'sos':
library(sos)
findFn('buse')
Hope this helps,
Pascal
On 27 November 2013 14:55, Sheri O'Connor socon...@lakeheadu.ca wrote:
I was wondering if anyone knew of a package that contained a function
of Buse's (http://www.jstor.org/stable/2683631 ) GLS R2 equation? If
not, I
Here is the error am getting
Error: chunk 1
Error in tt[[ll[v]]] : subscript out of bounds
Error in rle(filenames) : 'x' must be an atomic vector
Calls: Anonymous - Anonymous - RweaveTryStop
Execution halted
On Wed, Nov 27, 2013 at 9:25 AM, Keniajin Wambui kiang...@gmail.com wrote:
Dear Jeff;
On 27-11-2013, at 05:16, Binod Manandhar gibi...@yahoo.com wrote:
I was trying to find root for two non-linear equations. My equation is
similar as follows:
120 – 5* exp(b0+b1*4)
/ (1+ exp(b0+b1*4) ) = 0
690 – 31*
exp(b0+b1*4) / (1+ exp(b0+b1*4) ) =0
How could find value of bo and
Could it be that your r-script is saved in a different encoding than
the one used by R (which will probably be UTF8 since you're working on
linux)?
--
Jan
gerald.j...@dgag.ca schreef:
Hello,
I am using R, 2.15.2, on a 64-bit Linux box. I run R through Emacs' ESS.
R runs in a
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