Hi,Dear all R experts,
I am trying to do the 2-way contingency table analysis by fitting
the loglinear models. However, I found my table has several empty
cells which are theoretically missing values.I have no idea of how
to solve them coz we cannot compute the simulated p-value with
Hi,
I have 3 vectors which I want to plot as in one plot.
I was wondering why this code of mine only show the last vector:
__BEGIN__
library(lattice)
out_fname - paste(MyPlot.png,sep=)
trellis.device(png, color=TRUE)
png(out_fname)
plot(vect1,type=l,col=red,xlab=Nof Genes, ylab=RMSD)
Stationarity is a statement about a stochastic process, not about a single
realization. It is a statement about what might have happened, not what
did happen.
A sine wave with a random (uniform) wave is stationary, and indeed the
superposition of such waves is the spectral decomposition. A
Dear R community
I have a problem to access particular list. I have a code given below where
there is recursive process. It is not possible to run it because there are few
other functions involved inside like sv, LN, RN etc.
k=0; n=0; variable=c(); vr-list()
func - function(data,testdata)
Dear R-users,
Attach with is my data..what i want to do is finding a suitable distribution
for my data..I want to run a few test like the poisson and the exponential
distribution. Please help me on how to find the p-value for poisson as well as
the exponential distribution without knowing
Dear R-users,
Attach with is my data..what i want to do is finding a suitable distribution
for my data..I want to run a few test like the poisson and the exponential
distribution. Please help me on how to find the p-value for poisson as well as
the exponential distribution without
Hi All,
I am a new member to R programming.
Am generating some visuals by using Cairo library. But Cairo is not
compatible with all compilers(Box plot,histogram and RNA degradation
plots-I would prefer to use some libraries rather than R native
functions).Can anyone suggest an alternative for
Stephan:
the Naive Bayes model consists of several tables, one for each
(categorical) predictor. Using
m = naiveBayes( ... )
m$tables
you will get (as the help page says):
tables: A list of tables, one for each predictor variable. For each
categorical variable a table giving, for
Thanks to everyone who responded to my email.
Moshe's email explains clearly why my matrices were not positive
definite for certain negative correlations.
I now have better understanding of the problem.
Thanks
Mizan
2008/6/27 Moshe Olshansky [EMAIL PROTECTED]:
If the main diagonal element of
Dear everyone:
I am now doing one exercise with hclust and i do not know how to deal with
the
reslut as produced by it!
My aim is to find out the number of clusters and what are the members for
each
cluster?
So I am writing to here to get your help
thanks in advance!
chunping
Sorry for the wrong placing of my last email.
Here a solution to my problem and incomprehension respectively:
# Example 95%
x - rnorm(1000, mean = 0, sd = 1)
y - rnorm(1000, mean = 1, sd = 1.3)
kerneld - kde2d(x, y, n = 100, lims = c(-5.0, 5.0, -5.0, 5.0))
pp - array()
for (i in 1:1000){
A solution with ggplot2 would be
dataset - data.frame(cat = factor(rbinom(1000, 1, prob = .33), labels =
LETTERS[1:2]), startyear = floor(runif(1000, 2001, 2009)), studentid =
1:1000)
library(ggplot2)
ggplot(dataset, aes(x = startyear, colour = cat)) + stat_bin(aes(y =
..count..), geom=line,
Hello every body,
I am quite a new user of R so I beg your pardon for this naive question and
the lake of syntax with wich I ask it:
I have a data frames like this:
colacolb
1 c
1 i
1 i
1 c
2 i
2 c
2 c
2 i
2 i
...
1
I would like
Hi
Is there a way to remove blank characters from the end of strings in a
vector? Something like the =TRIM functions of the OpenOffice
spreadsheet. E.g.,
a - c(hola, Yes , hello )# I'd like to get:
c(hola, Yes, hello)
Thanks
Juli
--
http://www.ceam.es/pausas
Is this what you want:
x - c(hola, Yes , hello )
gsub( *$, , x)
[1] hola Yes hello
On Fri, Jun 27, 2008 at 4:34 AM, juli pausas [EMAIL PROTECTED] wrote:
Hi
Is there a way to remove blank characters from the end of strings in a
vector? Something like the =TRIM functions
Thank you both.
Katharine approach seemed to me easier to implement so I used it
successfully. However I still wonder why in linear model using lm I can
use
for (i in ) lm( data[,i]~data[,1])
but in nls the same construction fails. I understand that it is sometimes
difficult to set
Hi , Ping,
First you should report more detail about your problem and make sure
it is a R specific one.
To find out more on using hclust, I suggest you try help(hclust). I
am not sure about what result
you want to obtain since the result of hclust seems straight forward.
By the way,
There is also the trim command in the gdata package. Removes blanks
from the front of the string as well which may not be what you want.
Regards
JS
---
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of jim holtman
Sent: 27 June 2008 10:27
To: juli
Thank you Berwin.
Ok, I take your point. I normally do nls modelling interactively but this
time I was given a set of data so I tried to use an approach which I use
quite often in lm or in plotting to pdf file. I obviously was not
successful and there is nothing about it in documentation or
Hi,
I want to plot a series of numbers ( which lie in the range 1-20, there are
many such numbers) as a barplot. However instead of representing the values
in terms of height, i want to assign one unique color to each number. How
can this be done in R?
Thanks,
archana,
II yr PhD student,
Dept of
Hi
you probably want to look to interp from arima package
Petr Pikal
[EMAIL PROTECTED]
724008364, 581252140, 581252257
[EMAIL PROTECTED] napsal dne 26.06.2008 20:00:48:
Hello list,
I'm new to R and I have a problem :-) Below is what my data file that
looks
like. I tried to import and
There are well-informed answers given as examples on the sub() help page.
Hint 1: there is no need to globally substitute patterns anchored at the
end: they can only match in one place. There is also no need to
substitute for .
Hint 2: 'blank characters' and 'spaces' are not the same
As has already been pointed out, the syntax of formulae in nls() is not
the same as in linear models. So things which are valid for linear models
are not necessarily valid for others.
Since nls() is sparsely documented on the help page, you need to look at
the references (and you can also
I'm working with Windows XP and R 2.6.0
R.Version()
$platform
[1] i386-pc-mingw32
-Lauri
2008/6/27, Lauri Nikkinen [EMAIL PROTECTED]:
Hello,
Is there a way to control pointsize of pdf:s produced by Sweave? I
would like to have the same pointsize from (not a working example)
Hi
you mix base and lattice graphics.
[EMAIL PROTECTED] napsal dne 27.06.2008 08:25:55:
Hi,
I have 3 vectors which I want to plot as in one plot.
I was wondering why this code of mine only show the last vector:
__BEGIN__
library(lattice)
out_fname - paste(MyPlot.png,sep=)
why not
Hi R gurus
I have a matching problem that I cant solve. I have tried multiple solutions
and searched varius help-sites but I cant get it to work.
This is the problem
myexstrings = c(*AAA.AA,BBB BB,*.CCC.,**dd- d)
what I want do do is to remove any non-characters in the beginning and
everything
Please discuss this with the package maintainer (see the posting guide).
He will need lots of details you have omitted, including the precise OS
and the C++ compiler used.
BTW, this illustrates a common problem with packages using C++, which
people often test only under one compiler, and
Hello,
Is there a way to control pointsize of pdf:s produced by Sweave? I
would like to have the same pointsize from (not a working example)
pdf(file=C:/temp/example.pdf, width=7, height=7, bg=white, pointsize=10)
plot(1:10)
etc..
dev.off()
as
\documentclass[a4paper]{article}
On 27/06/2008 6:23 AM, Lauri Nikkinen wrote:
I'm working with Windows XP and R 2.6.0
R.Version()
$platform
[1] i386-pc-mingw32
-Lauri
2008/6/27, Lauri Nikkinen [EMAIL PROTECTED]:
Hello,
Is there a way to control pointsize of pdf:s produced by Sweave? I
would like to have the same pointsize
Wanding,
I'm the maintainer of igraph, but missed your previous email.
Yes, currently the released version of igraph fails to compile
with gcc 4.3.x. I made the required modifications to fix this,
but these are still in the igraph development tree, as there has been
no release since that.
pdf.options() seems to be a new function (from 2.7.0), so I quess I'll
have to upgrade or write my own hook function for Sweave. Thanks.
Best
Lauri
2008/6/27, Duncan Murdoch [EMAIL PROTECTED]:
On 27/06/2008 6:23 AM, Lauri Nikkinen wrote:
I'm working with Windows XP and R 2.6.0
Dear R Users,
Is anyone aware of a package which calculates the Yule Kendall resistant
(to errors,outliers) measure of skewness ? An easy calculation to perform,
but was just wondering if a package exists (as the contents of that
package would probably include other cool things I would also be
On 27/06/2008 7:12 AM, Lauri Nikkinen wrote:
pdf.options() seems to be a new function (from 2.7.0), so I quess I'll
have to upgrade or write my own hook function for Sweave. Thanks.
I'd recommend upgrading. I think it would be difficult to do this with
a hook function: you'd basically need
Yes, I think so too. I already tried with
options(SweaveHooks=list(fig=function() pdf(pointsize=10)))
but as you said it tries to open pdf device and Sweaving fails...
Best
Lauri
2008/6/27, Duncan Murdoch [EMAIL PROTECTED]:
On 27/06/2008 7:12 AM, Lauri Nikkinen wrote:
pdf.options() seems to
On 27 Jun 2008, at 12:23, Tom.O wrote:
Hi R gurus
I have a matching problem that I cant solve. I have tried multiple
solutions
and searched varius help-sites but I cant get it to work.
This is the problem
myexstrings = c(*AAA.AA,BBB BB,*.CCC.,**dd- d)
what I want do do is to remove any
David Afshartous [EMAIL PROTECTED] wrote:
All,
I have data across 5 time points that I am graphing via xyplot, along with
error bars. For one of the variables I have missing data for two of the
time points. The code below is okay but I can't seem to get the lines to
connect
On Fri, 2008-06-27 at 15:17 +0530, Archana Vagish wrote:
Hi,
I want to plot a series of numbers ( which lie in the range 1-20, there are
many such numbers) as a barplot. However instead of representing the values
in terms of height, i want to assign one unique color to each number. How
can
Hello list!
I have a set of data like this:
alldata[1:5,]
breaks numbers disttype moltypetype
1 0.0006598 Gapped Distances 5S Between species
2 0.407 0 Gapped Distances 5S Between species
3 0.8135228 Gapped Distances 5S Between
Hello,
It's just a strange coincidence that someone posted just very recently a
question about matching. I know there are several match function in the base
package (such as match, pmatch, charmatch, and the gsub etc) but I can't
seem to use them wisely to be able to get what I need.
suppose I
Well I have tried that and it's unfortuanally not the solution.
This return all the characters in the string, but I dont want the characters
after the ending non-character symbol. Only the starting characters ore of
interest.
gsub(\\W*,, myexstrings,perl=T)
[1] A B CCC ddd
Regards
this should do what you want:
myexstrings = c(*AAA.AA,BBB BB,*.CCC.,**dd- d)
a = gsub(^\\W*,, myexstrings,perl=T)
b = gsub(\\W.*, , a, perl=T)
b
[1] AAA BBB CCC dd
first one, removes any non-word characters from the beginning (as you
already figured out)
second one, removes any remaining
Dear R-users,
The following provides a basis to illustrate some questions I have about
xyplot() customization.
x - rep(1:2, times = 6)
y - numeric()
for (i in 1:3) {
y - c(y, 1:4*10^(i-1))
}
g - rep(1:6, each = 2)
ax - rep(1:3, each = 4)
tmp - data.frame(x,y,g)
xyplot(y ~ x | g,
All,
I have a long vector that contains an even number of entries. I'd like to
switch the 1st and 2nd entry, the 3rd and 4th, and so on, without writing a
loop.
This code works:
X = c(8, 10, 6, 3, 20, 1)
index = c(2,1,4,3,6,5)
X[index]
But for a long list is there a way to generate the index?
On 27 Jun 2008, at 13:56, Tom.O wrote:
Well I have tried that and it's unfortuanally not the solution.
This return all the characters in the string, but I dont want the
characters
after the ending non-character symbol. Only the starting characters
ore of
interest.
gsub(\\W*,,
You can do it without and index like this:
c(matrix(X, 2)[2:1,])
or if you need the index for some purpose apart from this:
c(matrix(seq_along(X), 2)[2:1,])
On Fri, Jun 27, 2008 at 10:11 AM, David Afshartous
[EMAIL PROTECTED] wrote:
All,
I have a long vector that contains an even number of
on 06/27/2008 09:11 AM David Afshartous wrote:
All,
I have a long vector that contains an even number of entries. I'd like to
switch the 1st and 2nd entry, the 3rd and 4th, and so on, without writing a
loop.
This code works:
X = c(8, 10, 6, 3, 20, 1)
index = c(2,1,4,3,6,5)
X[index]
But for a
Try this:
ave(X, rep(1:(length(X)/2), each = 2), FUN=rev)
On Fri, Jun 27, 2008 at 11:11 AM, David Afshartous
[EMAIL PROTECTED] wrote:
All,
I have a long vector that contains an even number of entries. I'd like to
switch the 1st and 2nd entry, the 3rd and 4th, and so on, without writing a
On 27 Jun 2008, at 14:30, francogrex wrote:
Hello,
It's just a strange coincidence that someone posted just very
recently a
question about matching. I know there are several match function in
the base
package (such as match, pmatch, charmatch, and the gsub etc) but I
can't
seem to use
Thanks Henrique, Marc, and Gabor!
On 6/27/08 10:17 AM, Henrique Dallazuanna [EMAIL PROTECTED] wrote:
Try this:
ave(X, rep(1:(length(X)/2), each = 2), FUN=rev)
On Fri, Jun 27, 2008 at 11:11 AM, David Afshartous [EMAIL PROTECTED]
wrote:
All,
I have a long vector that contains an
Henrique Dallazuanna wrote:
Try this:
ave(X, rep(1:(length(X)/2), each = 2), FUN=rev)
Also,
ix - 2*rep(1:(length(X)/2-1), each = 2) + 2:1
X[ix]
or
ix - seq_along(X) + c(1,-1)
X[ix]
On Fri, Jun 27, 2008 at 11:11 AM, David Afshartous
[EMAIL PROTECTED] wrote:
All,
I have a long
Hi, xiechao
i don't think that is a R specific problem. you mean u got two
random variables X,Y and both
of them binomial distributed and you want to find the distribution of
a new variable Z = X/Y.
That is a basic transformation problem. u can start with introducing a
new r.v. namely W,
Thanks guys, all of you. You have just made this weekend a much more happier
weekend.
Regards Tom
Hans-Jörg Bibiko wrote:
On 27 Jun 2008, at 13:56, Tom.O wrote:
Well I have tried that and it's unfortuanally not the solution.
This return all the characters in the string, but I dont
I'm trying to replace NA with 0 value...
I've write a loop, but don't work...
Where's the problem?
cimfasy_rwl
1991 0.92 0.72 0.50 1.29 0.54 1.22
1992 2.15 1.28 1.23 2.26 1.22 3.17
1993 1.50 0.87 1.68 1.97 0.83 2.55
1994
Hi Karin,
Try this
xyplot(numbers~sqrt(breaks)|moltype+disttype,
groups = type,
data = alldata,
panel = function(x, y, ...){
panel.abline(h = 0, col.line = 1)
panel.xyplot(x, y, ...,
pch = 1,
col.symbol = 1)})
Look at panel.abline in the lattice help for more details. Here, h = 0
will plot a
Hi
I am looking to generate permutations for data tested using the
friedman.test. Just read a paper by Joachim Rohmel, The permutation
distribution of the Friedman test (Computational Statistics Data Analysis
1997, 26: 83-99). He offers APL code for carrying out these procedures;
wondering if
Don't use a loop for this. Do this.
X - matrix(c(NA, 2, NA, 4),ncol=2) # Sample matrix
X[is.na(X)] - 0 # Do this instead
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Alfredo
Alessandrini
Sent: Friday, June 27, 2008 10:37 AM
To:
Hi,
I have a problem in assessing the list element.
i have list called geneset it contains the following elements
geneset
[[1]]
V1 V2 V3 ...V200
Genenamegene1 gene2gene200
[[2]]
V1V2 V3 V4...V[240]
Hi Ramya,
Try something like this:
as.character(unlist(lapply(geneset,function(x) x[1])))
HTH,
Jorge
On Fri, Jun 27, 2008 at 10:33 AM, Rajasekaramya [EMAIL PROTECTED]
wrote:
Hi,
I have a problem in assessing the list element.
i have list called geneset it contains the following
Here is a solution using strapply from the gsubfn package:
library(gsubfn)
strapply(myexstrings, (\\w+).*, backref = -1, simplify = c)
It matches the first string of word characters following by
anything else and then returns the first backreference in
each match, i.e. the portion within
Don't use a loop for this. Do this.
I need to use a loop...
I've many data.
Alfredo
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and
Hi,
cimfasy_rwl[ is.na(cimfasy_rwl) ] -0
Or did I understood Your right?
HTH,
Kimmo
Alfredo Alessandrini wrote:
I'm trying to replace NA with 0 value...
I've write a loop, but don't work...
Where's the problem?
cimfasy_rwl
1991 0.92 0.72 0.50 1.29 0.54 1.22
Dear Petr,
Thanks so much for your detailed guidance.
I'll have a look at your suggestions.
- Gundala Viswanath
Jakarta - Indonesia
On Fri, Jun 27, 2008 at 7:09 PM, Petr PIKAL [EMAIL PROTECTED] wrote:
Hi
you mix base and lattice graphics.
[EMAIL PROTECTED] napsal dne 27.06.2008 08:25:55:
Peng Jiang wrote:
Hi, xiechao
i don't think that is a R specific problem. you mean u got two random
variables X,Y and both
of them binomial distributed and you want to find the distribution of
a new variable Z = X/Y.
That is a basic transformation problem. u can start with introducing a
try this:
firstgenes = lapply(geneset, function(x){return(x[1,1])})
firstgenes = do.call(rbind(firstgenes))
on 06/27/2008 10:33 AM Rajasekaramya said the following:
Hi,
I have a problem in assessing the list element.
i have list called geneset it contains the following elements
oh, unlist - very nice function, thanks :)
on 06/27/2008 11:23 AM Jorge Ivan Velez said the following:
Hi Ramya,
Try something like this:
as.character(unlist(lapply(geneset,function(x) x[1])))
HTH,
Jorge
On Fri, Jun 27, 2008 at 10:33 AM, Rajasekaramya [EMAIL PROTECTED]
wrote:
Hi,
I
pdf( yourfile.pdf, height=22, width=17)
#yourcode
dev.off()
then use the gimp (free) to transform it to .png or whatever else (pdf makes
good graph)
On Thu, Jun 26, 2008 at 8:59 PM, Daniel Folkinshteyn [EMAIL PROTECTED]
wrote:
not sure why it doesn't work, but try the following:
first, plot
Like this:
for (i in files_rwl) {
thisfile - get(i)
thisfile[is.na(thisfile)] - 0
assign(i, thisfile)
}
It's likely that you could condense it, but using the long form makes it
clear what's happening.
Sarah
--
Sarah Goslee
http://www.functionaldiversity.org
Please consider the following PCA example;
my.df -
data.frame(A=(x - rnorm(100,mean=100, sd=10)),
B=(y - x + rnorm(100,mean=10, sd=10)))
plot(my.df)
my.pc -
prcomp(my.df, center=TRUE, scale=TRUE)
biplot(my.pc)
my.x - (my.pc$x)[,1]
my.y - (my.pc$x)[,2]
plot(my.x, my.y,
Dear K. Elo,
You're right!
cimfasy_rwl=read.table(textConnection(1991 0.92 0.72 0.50
1.29 0.54 1.22
1992 2.15 1.28 1.23 2.26 1.22 3.17
1993 1.50 0.87 1.68 1.97 0.83 2.55
1994 0.69 0.00 0.76 1.89 0.60 0.87
Hi Rajasekaramya
sapply(geneset[1:5431], [[, V1)
or
Bioconductor::subListExtract(geneset[1:5431], V1)
Martin
Rajasekaramya wrote:
Hi,
I have a problem in assessing the list element.
i have list called geneset it contains the following elements
geneset
[[1]]
for (i in files_rwl) {
thisfile - get(i)
thisfile[is.na(thisfile)] - 0
assign(i, thisfile)
}
It's likely that you could condense it, but using the long form makes it
clear what's happening.
..it's work
Thanks...
Alfredo
hi, all-
i wrote a function that accept multiple arguments, but don't know how to
assign names automatically. run the following code:
foo - function (...) {
x = list(...)
names(x) - deparse(substitute(...))
x
}
a = 1; b = 2; c = 3
y - foo( a, b, c)
names(y)
as you can see, only the first
On Fri, 27 Jun 2008, stephen sefick wrote:
pdf( yourfile.pdf, height=22, width=17)
#yourcode
dev.off()
then use the gimp (free) to transform it to .png or whatever else (pdf makes
good graph)
So do R's graphics devices
If you do want to convert PDF to PNG there are much better ways
On Fri, 27 Jun 2008, Dan Bolser wrote:
Please consider the following PCA example;
my.df -
data.frame(A=(x - rnorm(100,mean=100, sd=10)),
B=(y - x + rnorm(100,mean=10, sd=10)))
plot(my.df)
my.pc -
prcomp(my.df, center=TRUE, scale=TRUE)
biplot(my.pc)
my.x - (my.pc$x)[,1]
my.y -
Hi,
I get the following error when I try to read in a CSV file:
Path-read.table('MetaCycSample2.csv',sep=',', header=FALSE)
Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, :
line 17 did not have 5 elements
Some of the rows have more columns than others. Also when
On Fri, 27 Jun 2008, Prof Brian Ripley wrote:
On Fri, 27 Jun 2008, Dan Bolser wrote:
Please consider the following PCA example;
my.df -
data.frame(A=(x - rnorm(100,mean=100, sd=10)),
B=(y - x + rnorm(100,mean=10, sd=10)))
plot(my.df)
my.pc -
prcomp(my.df, center=TRUE,
Hello all,
I am doing SAM and the median of positive genes in the permutated sets is =
false positives, and the parent set gives true positives.
FDR = FP/TP * 100.
My FDR comes to greater than 100.
Is that possible?
Please help!
Thanks,
-D.
[[alternative HTML version deleted]]
Dear Tolga,
See equation 2.6
herehttp://secamlocal.ex.ac.uk/people/staff/dbs202/cag/courses/MT37C/course/node14.html.
Also try this:
# Yule Kendall resistant measure of skewness
YK=function(x){
qs=quantile(x,probs=c(0.25,0.5,.75))
res=(qs[1]-2*qs[2]+qs[3])/(qs[2]-qs[1])
names(res)=NULL
res
}
#
Hi
Thank you very much for taking time to answer.
The solution of using hist(data) for the main dataset and adding
lines(density(refdata)) for the reference data seem to work great. I
forgot to mention one thing however, I need to have frequency on the y
azis instead of density as now.
I know
Try
read.table(,fill=TRUE)
On Fri, Jun 27, 2008 at 12:00 PM, [EMAIL PROTECTED] wrote:
Hi,
I get the following error when I try to read in a CSV file:
Path-read.table('MetaCycSample2.csv',sep=',', header=FALSE)
Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings,
Martin Morgan wrote:
Hi Rajasekaramya
sapply(geneset[1:5431], [[, V1)
or
Bioconductor::subListExtract(geneset[1:5431], V1)
oops, I meant using the 'Biobase' package from Bioconductor, e.g.,
library(Biobase)
subListExtract(geneset[1:5431], V1)
Sorry for the confusion,
Martin
Martin
This is solved for me. Upgrading R to 2.7.1 sorted out the issue somehow.
Maybe there is a problem with Ubuntu 8.04 and R 2.6.2. Or maybe some other
package was installed that fixed it. Whatever the case. Thanks!
Prof Brian Ripley wrote:
On Fri, 27 Jun 2008, stephen sefick wrote:
pdf(
DeaR list,
Pardon the stupidity of this question but I've been trying this for a
while now without success.
I've followed the example given in the green book programming with
data, and I now have a working example of a S4 class with a few
methods (plot, summary, as.data.frame). It's all
You evidence a good deal of confusion. VR's S PROGRAMMING, in particular
Chapter 3 (especially section 3.5 on Computing on the Language) would be
helpful to you. See also their example on p. 46.
However in brief:
foo - function (...) {
x -3
list(...)
}
returns the evaluated ... arguments as a
Dear Jorge,
Many thanks, this is great.
Tolga
Jorge Ivan Velez [EMAIL PROTECTED]
27/06/2008 18:11
To
[EMAIL PROTECTED]
cc
r-help@r-project.org
Subject
Re: [R] Yule Kendall resistant measure of skewness
Dear Tolga,
See equation 2.6 here. Also try this:
# Yule Kendall resistant
Try:
RSiteSearch(names of objects passed as ...)
On Fri, Jun 27, 2008 at 11:48 AM, whizvast [EMAIL PROTECTED] wrote:
hi, all-
i wrote a function that accept multiple arguments, but don't know how to
assign names automatically. run the following code:
foo - function (...) {
x =
Bogdan Tanasa [EMAIL PROTECTED] writes:
Hi everyone.
Actually, you don't want to say hi to _everyone_; you sent your mail
to three different news groups, but should have sent it to the single
news group that will be most helpful to you. You should have sent your
email to the Bioconductor
Hi!
Given a vector (or a factor within a df),i.e. v1 - c(1,1,1,2,3,4,1,10,3)
and a dictionary
cbind(c(1,2,3),c(1001,1002,1003))
is there a function (on the same line than recode() in car)
to get v2 as c(1001,1001,1001,1002,1003,4,1001,10,1003) ?
I'm using myself a function based on match()
I'll surely look into it.
Stephen
On Fri, Jun 27, 2008 at 1:43 PM, hippie dream [EMAIL PROTECTED]
wrote:
This is solved for me. Upgrading R to 2.7.1 sorted out the issue somehow.
Maybe there is a problem with Ubuntu 8.04 and R 2.6.2. Or maybe some other
package was installed that fixed it.
Dear Agustin,
Perhaps
v1 - c(1,1,1,2,3,4,1,10,3)
dput(as.numeric(ifelse(v1%in%c(1,2,3),paste(100,v1,sep=),v1)))
HTH,
Jorge
On Fri, Jun 27, 2008 at 2:41 PM, Agustin Lobo [EMAIL PROTECTED] wrote:
Hi!
Given a vector (or a factor within a df),i.e. v1 - c(1,1,1,2,3,4,1,10,3)
and a dictionary
if there's nothing specific for it, you could probably do it with merge?
on 06/27/2008 02:41 PM Agustin Lobo said the following:
Hi!
Given a vector (or a factor within a df),i.e. v1 - c(1,1,1,2,3,4,1,10,3)
and a dictionary
cbind(c(1,2,3),c(1001,1002,1003))
is there a function (on the same
on 06/27/2008 01:41 PM Agustin Lobo wrote:
Hi!
Given a vector (or a factor within a df),i.e. v1 - c(1,1,1,2,3,4,1,10,3)
and a dictionary
cbind(c(1,2,3),c(1001,1002,1003))
is there a function (on the same line than recode() in car)
to get v2 as c(1001,1001,1001,1002,1003,4,1001,10,1003) ?
I'm
HiI need help about anderson-darlin test for a weibull distribution in R.Thanks.
René
_
[[elided Hotmail spam]]
e ready.
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On Fri, Jun 27, 2008 at 12:05 AM, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
Hi All,
I am a new member to R programming.
Am generating some visuals by using Cairo library. But Cairo is not
compatible with all compilers
May I ask which compiler you are using? There are other vector
2008/6/27 Prof Brian Ripley [EMAIL PROTECTED]:
On Fri, 27 Jun 2008, Prof Brian Ripley wrote:
On Fri, 27 Jun 2008, Dan Bolser wrote:
Please consider the following PCA example;
my.df -
data.frame(A=(x - rnorm(100,mean=100, sd=10)),
B=(y - x + rnorm(100,mean=10, sd=10)))
if you want the frequency scale rather than density scale, then leave
hist as is (by default it uses the frequency scale), and rescale the
density by multiplying it by the appropriate NOBS.
on 06/27/2008 01:16 PM Thomas Frööjd said the following:
Hi
Thank you very much for taking time to
This one should be easy but it's giving me a hard time mostly because tapply
puts the results in a list. I want to calculate the cumulative sum of a
variable in a dataframe, but with the accumulation only within each level of
a factor. For a very simple example, take:
df -
Just use ?unlist
df$willdo - unlist(tapply(df$x, df$fac, cumsum))
df$ideal - df$willdo - df$x
Levi Waldron wrote:
This one should be easy but it's giving me a hard time mostly because tapply
puts the results in a list. I want to calculate the cumulative sum of a
variable in a dataframe, but
On Fri, 2008-06-27 at 16:52 -0400, Levi Waldron wrote:
This one should be easy but it's giving me a hard time mostly because tapply
puts the results in a list. I want to calculate the cumulative sum of a
variable in a dataframe, but with the accumulation only within each level of
a factor.
Try this:
df$wildo - ave(df$x, df$fac, FUN = cumsum)
On Fri, Jun 27, 2008 at 4:52 PM, Levi Waldron [EMAIL PROTECTED] wrote:
This one should be easy but it's giving me a hard time mostly because tapply
puts the results in a list. I want to calculate the cumulative sum of a
variable in a
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