Yep, but I figured that out quite fast ;0)
Thanks for giving me a hand ... you want believe a many times I skimmed the
cbind help without actually seeing this ... well, it was 0:30 ...
Thanks again, Joh
Tony Plate wrote:
Whoops, it looks like there's a typo in ?cbind (R version 2.6.0 Patched
Dear all,
I'm analysing a survey, and creating a barchart of the different responses
for each question. The questions are grouped according to a number of
categories, so I'm using lattice to create a plot with each question in a
category on it. The problem is that the response set for
On Dec 18, 2007 9:39 PM, David Winsemius [EMAIL PROTECTED] wrote:
Armin Goralczyk [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
It's not the spaces, the problem is the tag (sorry that I didn't
specify this), or maybe the string []. I am working on a Mac OS X 10.4
with R version 2.6.
Martin Tomko wrote:
Dear list,
I have a vector (array, table row, whatever is best) of frequency values
for categories (or bins), and I need to find the median category.
Trivial to do by hand, but I was wondering if there is a means to do it
in R in an elegant way.
The obvious
Hi,
I was just wondering if prop.trend.test() is equivalent to the
Cochran-Armitage Trend test?
Thank you!
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PLEASE do read
Mark Leeds schrieb:
That's because the first factor is being used as the baseline. This is
explained in any intro to GLMs text.
Thank you Mark, I didn't found that hint.
Knut
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R-help@r-project.org mailing list
Hi
suppose I have two arrays x1,x2 of dimensions a1,b1,c1 and
a2,b2,c2 respectively.
I want x = x1 + x2 with dimensions c(max(a1,a2), max(b1,b2),max
(c1,c2))
with
x[a,b,c] = x1[a1,b1,c1] + x2[a2,b2,c2] ifa =min(a1,a2) , b=min
(b1,b2), c=min(c1,c2)
and the other bits either x1 or x2
On Tue, 2007-12-18 at 16:27 -0600, Naiara Pinto wrote:
Dear all,
I would like to use a tree regression method to analyze my dataset. I
am interested in the fact that random forests creates in-bag and
out-of-bag datasets, but I also need an estimate of support for each
split. That seems hard
On Wed, 2007-12-19 at 14:59 +0800, gallon li wrote:
I have the following list of observations of calendar time:
[1] 03-Nov-1997 09-Oct-1991 27-Aug-1992 01-Jul-1994 19-Jan-1990 12-Nov-1993
[7] 08-Oct-1993 10-Nov-1982 08-Dec-1986 23-Dec-1987 02-Aug-1995 20-Oct-1998
[13] 29-Apr-1991
livia wrote:
Hello everyone,
I have got a question about a simple calculation. If I would like to
calculate 50/12 and return the result as 4 and the remainer 2. Is there a
function of doing this?
Many thanks.
?%% to see how to get the remainder. You might put the result and
remainder
Hello R Users,
I am interested in using R to generate quantitative structure-activity
relationships (QSARs) for small molecules given a set of molecular
descriptors and biological data tab-delimited or excel file. In which fist
value colum of each row is biological value and rest all are its
Try this:
%add% - function(x1, x2)
{
dim1 - dim(x1)
dim2 - dim(x2)
seq1 - list();for(i in 1:length(dim1)) seq1[[i]]=seq(dim1[i])
filter1 - paste(seq1, collapse=,)
cmd1 - paste(out[, filter1, ], sep=)
seq2 - list();for(i in 1:length(dim2)) seq2[[i]]=seq(dim2[i])
Hi all,
I have observed that when using the randomForest package to do regression, the
predicted values of the dependent variable given by a trained forest are not
centred and have the wrong slope when plotted against the true values.
This means that the R^2 value obtained by squaring the
I have a sample of observations:
yy
[1] 0. 2.3972 4.3500 -4.1972 0.6361
[6] 1.0806 5.9056 -1.8722 2.1333 -1.1806
[11] 3.6167 0.8778 8.3389 3.8417 1.
[16] -3.7611 -11.6778 -2.0306
Dear all,
I am trying to wrap a *nix shell script around R for a particular
purpose, for which I need to get R to execute predetermined commands
but retain interactivity and allow user input during their execution.
A straight redirection of standard input is therefore not appropriate,
and
I am not sure I really understand what you want but
will this work?
tt-c(03-Nov-1997,09-Oct-1991,27-Aug-1992,01-Jul-1994,19-Jan-1990,
12-Nov-1993,08-Oct-1993,10-Nov-1982,08-Dec-1986,23-Dec-1987,02-Aug-1995,
20-Oct-1998,29-Apr-1991,16-Mar-1994,20-May-1991,28-Dec-1987,14-Jul-1999,
I get unexpected behavior from readLines() and
scan() depending on how the file is opened with
gzfile or unz. More specifically:
file - gzfile(file.gz)
readLines(file,1)
[1] a\tb\tc
readLines(file,1)
[1] a\tb\tc
close(file)
It seems that the stream is rewound between calls to
readLines.
Dear R Users,
I am working for the United Nations to construct a complete life table
from an abridged table.
I want to use the code of Hydman Filter by Rob J Hydman but an error
sentence always appears and it simply doesn't run--
source(C:/R/Jamie/HymanFilter.R)
Error in
Hi all
I can't find what I am looking for so I am asking here. I have a
dataset that looks something like this.
Year season percent_below
2000 Winter 6.9179870
2000 Spring 1.6829436
2000 Summer 1.8463501
2000 Autumn 3.8184993
2001 Winter 2.8832806
2001
Dear R Users,
I am working for the United Nations to construct a complete life table
from an abridged table.
I want to use the code of Hydman Filter by Rob J Hydman but an error
sentence always appears and it simply doesn't run--
source(C:/R/Jamie/HymanFilter.R)
Error in
R does this sort of thing easily without any parse/eval acrobatics needed.
E.g., you can do:
stu - function(x) {return( 1 + (2*x*x) - (3*x) )}
(x - 0:3)
[1] 0 1 2 3
stu(x)
[1] 1 0 3 10
metafun - function(FUN, data) FUN(data)
metafun(stu, x)
[1] 1 0 3 10
# if you want to be
If all your entries are double precision then you are
using 8 bytes per entry, so 20,000*n entries are just
160,000*n bytes, i.e. less than 160*n Kb. If your n is
100 you get 16 Mb which is not that much (especially
if you pre-allocate it only once). So just use the
matrix and don't worry!
---
I used the command below, but R gives me the error message--syntax error.
can anyone see the mistakes I made?
optimize(function(x,y)
+ ((327.727-(1-0.114^10)*y*(1-x)/x/(1-x^y))+(9517.336-327.727
*(1+(1-x)*(1+y)/x-327.727)))^2
+ interval=c(0,1))
At the same time, I use nlm() but R gives me the
Alternatively
levels(df$binname)[which(df$freq =
0.5*cumsum(df$freq)[nrow(df)])[1]]
--- Chuck Cleland [EMAIL PROTECTED] wrote:
Martin Tomko wrote:
Dear list,
I have a vector (array, table row, whatever is
best) of frequency values
for categories (or bins), and I need to find the
median
System:
Linux kernel 2.6.22-14
Ubuntu 7.10 gutsy
ESS 5.3.0 on Emacs 22.1.1
R version 2.6.0
Colleagues
I would like to use the user contributed package clim.pact. I'm having
trouble with the dependency of clim.pact on ncdf. Like others I am
finding that R CMD INSTALL does not find the netcdf.h
Hi Everyone,
I've got a question about data representation. I have some psychometric
data with 5 scores for 15 different groups. I've been asked to show
some kind of mean plots. The data below is the mean and SD for a given
group, unfortunately my employer doesn't want me posting full
On Wednesday 19 December 2007, Max wrote:
Hi Everyone,
I've got a question about data representation. I have some psychometric
data with 5 scores for 15 different groups. I've been asked to show
some kind of mean plots. The data below is the mean and SD for a given
group, unfortunately my
R-users
E-mail: r-help@r-project.org
I found the answer myself.
'.Fortran(baklo,' in lo.wam() and .Fortran(bakfit,in
s.wam() may carry out backfitting. But I cannot
create an R code which gives the same results as those of
bakfit. If someone knows the detail of bakfit algorithm,
please let
On 12/19/07, Dylan Beaudette [EMAIL PROTECTED] wrote:
On Wednesday 19 December 2007, Max wrote:
Hi Everyone,
I've got a question about data representation. I have some psychometric
data with 5 scores for 15 different groups. I've been asked to show
some kind of mean plots. The data
50 %% 12
[1] 2
50 %/% 12
[1] 4
?Arithmetic
--- livia [EMAIL PROTECTED] wrote:
Hello everyone,
I have got a question about a simple calculation. If
I would like to
calculate 50/12 and return the result as 4 and the
remainer 2. Is there a
function of doing this?
Many thanks.
--
Dear all,
I'm trying to estimate the parameters of a special case of a poisson
model, where the specified equation has an integral and several fixed
parameters.
I think that the MLE command in STATS4 package could be a good choice,
but it's a little complicated. I've got some problems with the
Won't it be simpler to do:
for (i in 1:12){
data - my.fun(my.list[i]))
save(data,file = paste(data,i,.RData, sep=)) }
--- Marie Pierre Sylvestre
[EMAIL PROTECTED] wrote:
Dear R users,
I am analysing a very large data set and I need to
perform several data
manipulations. The dataset is
Max wrote:
Hi Everyone,
I've got a question about data representation. I have some psychometric
data with 5 scores for 15 different groups. I've been asked to show
some kind of mean plots. The data below is the mean and SD for a given
group, unfortunately my employer doesn't want me
Hello All,
I have had considerable bad luck with attempting the following with for
loops. Here is the problem:
# Suppose we have a data.frame with the following data, which can be
considered a type of edgelist (for those with networks backgrounds):
#
# V1 V2
# 1 A
#
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