Hello
My colleagues asked me to install R with module shiny.
R version 3.0.1 compiled fine on Solaris 10 and is running well.
I tried to install shiny. With the dependencies Rcpp should be installed
before. But the compile step did fail. See below.
The initial error message is Error in
Hi Greg
On 21/06/13 08:02, Greg Dropkin wrote:
hi
I'm trying to understand (a little) the code behind summary.gam, and have
the Biometrika article referred to in the Help. But am stuck early on.
In the code, starting at line 167:
if (est.disp)
rdf - residual.df
else rdf - -1
res -
On Jun 20, 2013, at 16:21 , Annette Timm wrote:
Hi All
I am having some trouble making a test for linearity of my passing bablok
regression.As far as I am concerned i need to make cusum test for linearity
to get a p-value. I can make a cusum plot, but I can not figure out how to
test for
Hey,
I don't really get the output of the performed Levene's test.
So in short.
two data sets, one with 75 elements, one with 283.
And I want to check whether they have the same variance.
I did a LeveneTest
And this is the output:
Levene's Test for Homogeneity of Variance (center = median)
hi
I'm trying to understand (a little) the code behind summary.gam, and have
the Biometrika article referred to in the Help. But am stuck early on.
In the code, starting at line 167:
if (est.disp)
rdf - residual.df
else rdf - -1
res - testStat(p, Xt, V, df[i], type = p.type,
res.df = rdf)
I received the same error message. With me, the problem was that I had the
cells in my CSV file in Excel formatted as percentages. R read them as
factors and not as numeric values. After changing the cell format from
percentage to general in Excel, the problem was solved.
--
View this message
Hello everybody
I have this problem: I need to match an addresses database F1 with the
information contained in a toponymic database F2.
The format of F1 is given by three columns and 800 rows, with the
columns being:
A1. Street/Road/Avenue
A2. Name
A3. Number
Consider for instance Avenue J.
Once you have this matrix:
mx - cbind(x1, x2, x3, x4)
x1 x2 x3 x4
[1,] 22 14 24 10
[2,] 27 12 1 2
[3,] 23 13 6 2
[4,] 27 6 21 2
[5,] 14 29 12 19
[6,] 5 14 11 29
[7,] 10 10 27 25
[8,] 16 29 28 10
[9,] 22 22 19 7
[10,] 30 20 5 22
[11,] 2 12 24 15
[12,] 5 21
I don't really get the output of the performed Levene's test.
One answer would be to read John Fox's rather good book, for which car is the
corresponding R package.
Another is to look up leven'es test; a reasonably authoritative web reference
for the calculation is
Dear list,
I have already posted to sci.stat.consult, however, I was hoping that
there might be some experience in the R community on the following question:
I am wondering whether anyone has experience with variants of (parametric) Gini
coefficients (or alternatives) which allow for
Hello,
The Pr(F) is a regular p-value, with the usual meaning.
Rui Barradas
Em 21-06-2013 11:05, pieter escreveu:
k.
Thanks.
Sorry i'm not the biggest statician, but I still find the output write
out not the clearest of all. Anyway, that is probably a personal
problem. :-)
Just one question.
Hello,
Just to add to the last post, you can try to get that p-value using what
that output gives you:
df1 - 1
df2 - 356
F_stat - 10.948
pf(F_stat, df1, df2, lower.tail = FALSE) # same value
Hope this helps,
Rui Barradas
Em 21-06-2013 12:11, Rui Barradas escreveu:
Hello,
The Pr(F) is a
Hi all,
So this code works for three of the six seamounts. The problem I am
encountering is the fact that some of the seamounts doesn't have all the
depth categories (i.e missing the deep category) and thus gives me this
error Error in plot.window(...) : need finite 'ylim' values.
How can I fix
Dear R users,
I have a seasonal time series of period 4 (my_ts).
I would like to apply to my_ts a Seasonal ARMA(2,0) process (only the seasonal
part), with Seasonal AR coefficients respectively 0.54 and 0.5.
I tried to use the arima command from the stats package, but this code is not
correct:
Hi,
Try:
dat1- read.csv(mydata.csv,sep=\t,row.names=1) #sep could be different in
your case.
dat1
# A B C D E
#Jhon 1 3 6 9 8
#Kelly 2 3 4 6 9
#Audrey 3 5 6 9 7
A.K.
Hi,
I am calling my csv data with 37 columns and 16 rows into R. However I really
need to keep coloumns names as
I think you have to assume there could be other coding errors as
well, such as misspellings or abbreviating Street as St. You
probably will need to use sub() to correct A2 and possibly A1 before
trying to merge. To figure where the problems are, you might try
something like this. The last command
Hello,
I am trying to superpose on a single panel two QQ plots with the lattice
qqmath function.
Here is a reproducible example of the problem I am facing:
# Generate data
shape = 8
rate = c(rep(1/4,100),rep(1/3,100))
x = rgamma(200,shape,rate)
groups = gl(2,100,200,labels=LETTERS[1:2])
#
Dear All,
I have the following code:
a -rnorm(5000,10,2)
hdr.den(a,prob = c(25, 50, 75,95))
is there a way to change the colors of the horizontal bars representing the
given highest density regions from the current white, green, red and black to
colors of the shades of grey going from lighter
Hello R experts,
I want to define desired numbers to a vector based on the present
numbers. Can anybody please help me?
Obviously I found worst ways to do it, but I believe there must be any
better way.
I have vector as
X
[1] 43 43 43 43 0 39 13 39 50 39 39 23 23 32 0 13 32 23 32 23 0 13
Hello,
I'm not sure I understand. You want to attribute a color number 1:7 to
each element of your vector? Maybe the following will do.
x - scan(text =
43 43 43 43 0 39 13 39 50 39 39 23 23 32 0 13 32 23 32 23 0 13 13 0
)
cols - rep(1:7, rle(x[order(x)])$lengths)[order(order(x))]
Your rate variable is not a single value, but a vector of 200!.
I'm surprised you get anything at all. You have to pick one rate
value to use to create the values for the x-axis. For example run
rate - 1/4
And then run the qqmath() function.
Or, change the rate variable in the ggamma()
Thanks a lot Rui,
It helps :)
I can modify this according to my need.. I didn't know about rle function.
Thanks again,
Mitra
On 21 June 2013 23:32, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
I'm not sure I understand. You want to attribute a color number 1:7 to
each element of your
I suspect the OP may want
rep(1:length(unique(x)), rle(x[order(x)])$lengths)[order(order(x))]
to allow for variable numbers of unique values.
Clint
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department
Yes I have already changed that as this was only a small vector, my real
vector is quite big..
thanks a ton.
Bets wishes,
Mitra
On 21 June 2013 23:43, Clint Bowman cl...@ecy.wa.gov wrote:
I suspect the OP may want
rep(1:length(unique(x)), rle(x[order(x)])$lengths)[**order(order(x))]
to
Hello,
It's not your first post, you should know to say from which package is
the function you're using.
As for the question the answer is no, those values are hard coded in the
source. What you can do is to download the source and change file hdr.den.R
Hope this helps,
Rui Barradas
Em
Rui,
package is hdrcde, and thanks for the input,
Andras
--- On Fri, 6/21/13, Rui Barradas ruipbarra...@sapo.pt wrote:
From: Rui Barradas ruipbarra...@sapo.pt
Subject: Re: [R] hdr.den plot colors help
To: Andras Farkas motyoc...@yahoo.com
Cc: r-help@r-project.org
Date: Friday, June 21,
Hi,*
*I would like to set the entries of an array that appear as NA to a
certain integer, but did not find a way to do so. The related code is given
below:
A - array(-1000 , dim=c(100) );
for(i in 1: 100 ){
A[i] =B[i,57] - B[i,5];
}
A [ which( A == NA) ] - 900;
The NAs still appear as
You should tell us where hdr.den() comes from as it is not one of
the default packages and there are thousands of packages. Some
searching suggests you are using package hdrcde.
Looking at the source code for hdr.den() it appears that the package
author used the default palette in selecting
There exists an option corr.bias for making regression analyses with the
R package randomForest. The manual warns “Experimental. Use at your own
risk.” What does corr.bias precisely do and why can it be dangerous to
use it? As far as I know the bias correction is based on a linear
regression
Error in `[-` (`* tmp *`, i, j, value = -1.37244): index out of bounds
what kind of error? how to solve it?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
You really really need to do your homework.
?NA
-- Bert
On Fri, Jun 21, 2013 at 8:59 AM, Yasin Gocgun entropy...@gmail.com wrote:
Hi,*
*I would like to set the entries of an array that appear as NA to a
certain integer, but did not find a way to do so. The related code is given
below:
A
Does it work if you ignore NAs when calculating the max?
ylim=c(0,max(sm.mean$Total_Ch + 0.1, na.rm=TRUE))
Jean
On Fri, Jun 21, 2013 at 7:30 AM, Tinus Sonnekus tsonne...@gmail.com wrote:
Hi all,
So this code works for three of the six seamounts. The problem I am
encountering is the fact
Read 'An Introduction to R', found on the R web site. It is short and worth
reading in its entirety. Section 2.5 on missing values answers your question:
2.5 Missing values
In some cases the components of a vector may not be completely known. When an
element or value is not available or
HI,
May be this also works:
plot(x,col=as.numeric(factor(x)))
A.K.
- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: Suparna Mitra suparna.mitra...@gmail.com
Cc: r-help@r-project.org
Sent: Friday, June 21, 2013 11:32 AM
Subject: Re: [R] How to define desired numbers to a
Hi, Christian.
It sounds like to the maintainer feels that the package needs improvements,
i.e., it may not function as expected. However, you may contact him
directly using maintainer(randomForest).
Best,
Benjamin
On Fri, Jun 21, 2013 at 9:36 AM, Christian Kampichler
Hi
What about delivering further information about data, code, R and OS version.
Most probably you are tryind to index some object in cycle above its dimension.
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of
Ah, I see.
You could try something like this ... create a list of your data at the
outset, then just use the for() loop to create the plots.
Jean
# fake data with two Seamounts ... one with a missing depth
SChla - structure(list(Seamount = c(Atlantis, Atlantis, Atlantis,
Atlantis, Atlantis,
On Jun 21, 2013, at 3:39 AM, Stefan Sobernig wrote:
Dear list,
I have already posted to sci.stat.consult, however, I was hoping that there
might be some experience in the R community on the following question:
I am wondering whether anyone has experience with variants of (parametric)
Dear All,
wonder if you would provide your insights on the following: the code:
library(lattice)
y -c(1:58)
x -runif(58,5,10)
z -runif(58,8,12)
dataset -data.frame(y,x)
dotplot(y ~ x, data = dataset)
dataset -data.frame(y,z)
dotplot(y~z,data = dataset,col=red)
I would like to overlay the two
Hello,
Consider the following transcript
x=NULL
x$date=10
x
$date
[1] 10
or
x=NULL
x[10]=10
x
[1] NA NA NA NA NA NA NA NA NA 10
Wouldn't one expect that x remain NULL despite the further additions (i.e
the x$date, x[10]) etc? Is coercing appropriate here?
Cheers
Saptarshi
dotplot is a lattice function. add is an argument to some base
graphics. Never the twain shall meet. There is no add argument to
dotplot -- did you read the lattice Help for dotplot (found under
xyplot)??
I don't understand what you mean by overlay dotplots, but lattice
does this via conditioning
Well, feel free to make up any semantics that you like for a language
you write, but R already has its own and tells you what to expect:
From ?NULL:
NULL can be indexed (see Extract) in just about any syntactically
legal way: whether is makes sense or not, the result is always NULL.
Objects with
Maybe something like this?
dataset - data.frame(x=c(y, y), y=c(x, z), g=rep(1:2, each=58))
dotplot(x~y, groups=g, data=dataset)
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original
On Jun 21, 2013, at 1:21 PM, Saptarshi Guha wrote:
Hello,
Consider the following transcript
x=NULL
x$date=10
x
$date
[1] 10
or
x=NULL
x[10]=10
x
[1] NA NA NA NA NA NA NA NA NA 10
Wouldn't one expect that x remain NULL despite the further additions (i.e
the x$date,
In a similiar manner with additions
dat - data.frame(x=x,y=y,z=z)
dotplot(y~z+x,data = dat, pch = 16, type= c(p,g))
HTH
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
Email: home: mac...@northnet.com.au
At 07:00 22/06/2013, you
On 22/06/13 01:48, Stefano Sofia wrote:
Dear R users,
I have a seasonal time series of period 4 (my_ts).
I would like to apply to my_ts a Seasonal ARMA(2,0) process (only the seasonal
part), with Seasonal AR coefficients respectively 0.54 and 0.5.
I tried to use the arima command from the stats
Hi,
May be this helps:
###Added more lines of fake data
fn_hp- read.table(text=
id date wl_m wet cuml_day
585 fn 2012-03-03 0.1527048 1 1
586 fn 2012-03-04 0.2121408 1 2
587 fn 2012-03-05 0.1877568 1 3
Dear pieter,
On Fri, 21 Jun 2013 10:12:26 +0200
pieter dencous...@gmail.com wrote:
Hey,
I don't really get the output of the performed Levene's test.
So in short.
two data sets, one with 75 elements, one with 283.
And I want to check whether they have the same variance.
I did a
Vallejo, Roger roger.vall...@ars.usda.gov asked:
I would like to know if we can estimate Rg between two binary traits
(disease status: alive vs. dead) with the R package.
My data: we have 100 full-sib (FS) families,
and two random samples (each with n= 200 FS fish) from each FS family
were
Hello,
I've recently upgraded to R 2.11.1 on Mac OS X 10.8.4. Now when I
start R up I get the following message:
Error in identical(call[[1L]], quote(doTryCatch)) :
7 arguments passed to .Internal(identical) which requires 5
Error in normalizePath(dirname(pkgpath), /, TRUE) :
3
I have a complicated, multi-part question. My apologies if I do not make
myself clear. I am also a fairly novice R user, so forgive me if this seems
rudimentary. I want to calculate a index of colocation for whale dive data
and prey distribution data. This entails:
Calculating a frequency
Hi
Following on David's rate argument
try (with modifications of pch and grid)
rate - 1/4
shape = 8
rate = c(rep(1/4,100),rep(1/3,100))
x = rgamma(200,shape,rate)
groups = gl(2,100,200,labels=LETTERS[1:2])
dat = data.frame(x=x, gp=groups)
qqmath(~ x, data = dat,
groups = gp,
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