michael wrote:
>
> Hi,
>
> Is there available package on the optimization function using
> Newton-Raphson method (iterative quadratic approximation)? I have been
> using
> the 'optim' function in R and found it really unstable (it depends heavily
> on the initial values and functional forms
Thanks Corey,
Ive looked into them before and I dont think they can help me with this
problem. The Big functions are great for handling and analysing data sets
that are too big for R to store in memory.
However I believe my problem goes 1 step beyond that. In that my distance
matrix has too
On 11/08/11 13:19, Jesse Brown wrote:
I've run across what I think is a small bug in the plotrix package.
I've tried to contact the maintainer (Jim Lemon) directly but email is
returned 'undeliverable' at the provided address.
What is the best method to push a patch to a CRAN package in this case
this may help others. among the set of font resources are the following:
http://yihui.name/en/2010/03/font-families-for-the-r-pdf-device/
http://stat.ethz.ch/R-manual/R-devel/library/grDevices/html/postscriptFonts.html
http://stat.ethz.ch/R-manual/R-patched/library/grDevices/html/embedFonts.html
R-helpers:
Is there an easy way to transfer a data frame to a list, where each
list element is a dataframe containing a single row from the original
data frame?
--j
--
Jonathan A. Greenberg, PhD
Assistant Project Scientist
Center for Spatial Technologies and Remote Sensing (CSTARS)
Department o
I've run across what I think is a small bug in the plotrix package.
I've tried to contact the maintainer (Jim Lemon) directly but email is
returned 'undeliverable' at the provided address.
What is the best method to push a patch to a CRAN package in this case?
Thank you,
Jesse
__
Answered my own question. ?split
Never mind!
--j
On Wed, Aug 10, 2011 at 7:35 PM, Jonathan Greenberg
wrote:
> R-helpers:
>
> Is there an easy way to transfer a data frame to a list, where each
> list element is a dataframe containing a single row from the original
> data frame?
>
> --j
>
> --
On Aug 10, 2011, at 10:03 PM, Rolf Turner wrote:
On 11/08/11 13:27, David Winsemius wrote:
On Aug 10, 2011, at 9:22 PM, Rolf Turner wrote:
On 11/08/11 13:11, David Winsemius wrote:
On Aug 10, 2011, at 8:23 PM, Andra Isan wrote:
Hi All,
I would like to create a matrix in R but I dont kno
On 11/08/11 13:27, David Winsemius wrote:
On Aug 10, 2011, at 9:22 PM, Rolf Turner wrote:
On 11/08/11 13:11, David Winsemius wrote:
On Aug 10, 2011, at 8:23 PM, Andra Isan wrote:
Hi All,
I would like to create a matrix in R but I dont know the size of my
matrix. I only know the size of th
On Aug 10, 2011, at 7:34 PM, Ward, Michael Patrick wrote:
I used this service several months ago and was very pleased with the
response.
I have a dataframe with several thousand lines and to each line I
need to apply a series of "if else" statements. For each row I need
either a value
Mike,
One thing I noticed that '&&' should be replaced by '&'. The former is
applied to the first element of the vector while the latter applicable
to the whole vector.
HTH
Weidong Gu
On Wed, Aug 10, 2011 at 7:34 PM, Ward, Michael Patrick
wrote:
>
> I used this service several months ago and
On Aug 10, 2011, at 9:22 PM, Rolf Turner wrote:
On 11/08/11 13:11, David Winsemius wrote:
On Aug 10, 2011, at 8:23 PM, Andra Isan wrote:
Hi All,
I would like to create a matrix in R but I dont know the size of
my matrix. I only know the size of the columns but not the size of
the rows.
On 11/08/11 13:11, David Winsemius wrote:
On Aug 10, 2011, at 8:23 PM, Andra Isan wrote:
Hi All,
I would like to create a matrix in R but I dont know the size of my
matrix. I only know the size of the columns but not the size of the
rows. So, is there any way to create a dynamic matrix of si
On Aug 10, 2011, at 8:23 PM, Andra Isan wrote:
Hi All,
I would like to create a matrix in R but I dont know the size of my
matrix. I only know the size of the columns but not the size of the
rows. So, is there any way to create a dynamic matrix of size NULL
by n_cols? and then add to that
I used this service several months ago and was very pleased with the response.
I have a dataframe with several thousand lines and to each line I need to apply
a series of "if else" statements. For each row I need either a value or a
blank/NA. Below is the series of if else statements I have bee
Hi All,
I would like to create a matrix in R but I dont know the size of my matrix. I
only know the size of the columns but not the size of the rows. So, is there
any way to create a dynamic matrix of size NULL by n_cols? and then add to that
matrix?
I know for a vector, I can do this: x= NULL
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of michael
> Sent: Wednesday, August 10, 2011 4:37 PM
> To: r-help@r-project.org
> Subject: [R] Need help on Newton-Raphson optimization
>
> Hi,
>
> Is there available package
On 11/08/11 11:36, michael wrote:
Hi,
Is there available package on the optimization function using
Newton-Raphson method (iterative quadratic approximation)? I have been using
the 'optim' function in R and found it really unstable (it depends heavily
on the initial values and functional f
Hi,
Is there available package on the optimization function using
Newton-Raphson method (iterative quadratic approximation)? I have been using
the 'optim' function in R and found it really unstable (it depends heavily
on the initial values and functional forms). If I have to code it by myself
PS to my previous posting: Also have a look at kmeansruns in fpc. This
runs kmeans for several numbers of clusters and decides the number of
clusters by either Calinski&Harabasz or Average Silhouette Width.
Christian
On Wed, 10 Aug 2011, Ken Hutchison wrote:
Hello all,
I am using the clust
There is a number of methods in the literature to decide the number of
clusters for k-means. Probably the most popular one is the Calinski and
Harabasz index, implemented as calinhara in package fpc. A distance
based version (and several other indexes to do this) is in function
cluster.stats in
Hello,
I'm attempting to plot out a series of bar charts of recency, frequency and
monetary (RFM) scores using recency as a conditioning variable so we can see
the details of the FM components. My dataset looks like this:
rfmCellPop{rfm_score, orecency, ofrequency, omonetary, cellpop}
Where r
Hi,
Does anyone have feed back on using flot (javascript plotting library) for R
data (e.g. ggplot2) visualisation on the web?
Thanks for your comments,
--
View this message in context:
http://r.789695.n4.nabble.com/convert-R-plots-into-annotated-web-graphics-tp1471984p3734407.html
Sent fro
On Aug 10, 2011, at 23:39 , David Winsemius wrote:
> Hopefully this expected invitation to grilling will not result in severe
> burns. I note that one of the *apply family would have been successful in
> returning the desired object in conjunction with c() :
>
> > rapply(x, c)
> [1] "a" "b" "c
Thanks Ista,
In my real code that is exactly what I'm doing, but I want to prepend the
names with a sequential number for easier reference once the pngs are made.
My initial thought was to add the sequential number to the data before
sending it to plyr and drawing it out there, but that seems lik
I'm looking for the best way to do the following:
run a set of GAM models, and then make predictions with new data.
My problem is the size of the gam model object, I would like to strip it
down to the bare minimum of information needed to apply the model to new
data. For example, if this wer
Hi Justin,
On Wed, Aug 10, 2011 at 5:04 PM, Justin Haynes wrote:
> If I have data:
>
> dat<-data.frame(a=rnorm(20),b=rnorm(20),c=rnorm(20),d=rnorm(20),site=rep(letters[5:8],each=5))
>
> And want to plot like this:
>
> ctr<-1
> for(i in c('a','b','c','d')){
> png(file=paste('/tmp/plot_number_',
On Aug 10, 2011, at 3:58 PM, Liviu Andronic wrote:
On Wed, Aug 10, 2011 at 9:32 PM, David Winsemius > wrote:
Thanks all! This is perfect, and very R-ish: never where a novice
would expect it to be.
Well, since `unlist` is linked in the See Also on the help page for
`list`,
I can only hope
Hi Dennis
That those files are in a directory/folder suggests that they were extracted
from their
zip (.xlsx) file. The following are the basic contents of the .xlsx file
1484 02-28-11 12:48 [Content_Types].xml
733 02-28-11 12:48 _rels/.rels
972 02-28-11 12:48 xl/_rels
On Wed, Aug 10, 2011 at 12:07 PM, Ken Hutchison wrote:
> Hello all,
> I am using the clustering functions in R in order to work with large
> masses of binary time series data, however the clustering functions do not
> seem able to fit this size of practical problem. Library 'hclust' is good
> (t
If I have data:
dat<-data.frame(a=rnorm(20),b=rnorm(20),c=rnorm(20),d=rnorm(20),site=rep(letters[5:8],each=5))
And want to plot like this:
ctr<-1
for(i in c('a','b','c','d')){
png(file=paste('/tmp/plot_number_',ctr,'.png',sep=''),height=8.5,
width=11,units='in',pointsize=9,res=300)
print
On Wed, Aug 10, 2011 at 11:25 PM, ONKELINX, Thierry
wrote:
> Just type glm at the prompt.
>
Typing glm.fit at the prompt will probably be more informative.
-thomas
--
Thomas Lumley
Professor of Biostatistics
University of Auckland
__
R-help@r-pr
Hi:
Your R terminology needs a little work, but I think this is what you're after:
# read your example data from e-mail into an R session
dd <- read.table(textConnection("
Ind1 11 00 12 15 28
Ind2 21 33 22 67 52
Ind3 22 45 21 22 56
Ind4 11 25 74 77 42
Ind5 41 32 67 45 22"), header
Try the flow cytometry clustering functions in Bioconductor.
-thomas
On Thu, Aug 11, 2011 at 7:07 AM, Ken Hutchison wrote:
> Hello all,
> I am using the clustering functions in R in order to work with large
> masses of binary time series data, however the clustering functions do not
> see
dear R-experts---can someone please refer me to the latest
installation instructions for graphics fonts in R (the pdf device)?
(I would like to install the Charter font from the texlive 2011
distribution under OSX.)
sincerely,
/iaw
Ivo Welch (ivo.we...@gmail.com)
___
On Wed, Aug 10, 2011 at 9:32 PM, David Winsemius wrote:
>> Thanks all! This is perfect, and very R-ish: never where a novice
>> would expect it to be.
>
> Well, since `unlist` is linked in the See Also on the help page for `list`,
> I can only hope you meant that in complete jest.
>
More or less.
Here is code to transform the matrix that by() or array(split())
produces, along with an example of the speed of the various
approaches. Using split(), either directly or via by() or tapply(),
saves a lot of time.
f0 <- function(df) {
# original code with typos fixed.
list_structure <- la
On Aug 10, 2011, at 3:10 PM, Liviu Andronic wrote:
On Wed, Aug 10, 2011 at 9:02 PM, R. Michael Weylandt
wrote:
unlist()
Thanks all! This is perfect, and very R-ish: never where a novice
would expect it to be.
Well, since `unlist` is linked in the See Also on the help page for
`list`, I
On Aug 10, 2011, at 2:02 PM, Zev Ross wrote:
> Hi,
>
> For some reason I'm finding that my table caption is disappearing if I print
> xtable output with the floating argument set to FALSE. Below is a very simple
> Sweave file that produces two tables the first has no caption and the second
> h
Sorry, that second line of code won't work: do it in 2.
f <- function() {return(sample(1:5,3,replace=T/F))}
replicate(100,f())
Michael
On Wed, Aug 10, 2011 at 3:06 PM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> To pick random elements to sample, you can just use the sample func
On Wed, Aug 10, 2011 at 9:02 PM, R. Michael Weylandt
wrote:
> unlist()
>
Thanks all! This is perfect, and very R-ish: never where a novice
would expect it to be.
Cheers
Liviu
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-
Hello all,
I am using the clustering functions in R in order to work with large
masses of binary time series data, however the clustering functions do not
seem able to fit this size of practical problem. Library 'hclust' is good
(though it may be sub par for this size of problem, thus doubly poo
To pick random elements to sample, you can just use the sample function
sample(1:5,3,replace=T/F) # pick true or false as needed for your data.
If you replicate this, you should have no problem.
replicate(100,function() return(sample(1:5,3,replace=T/F)))
This will be plenty fast, but if you get
unlist(x)
r-help-boun...@r-project.org wrote on 08/10/2011 01:58:57 PM:
> [image removed]
>
> [R] convert 'list' to 'vector'?
>
> Liviu Andronic
>
> to:
>
> r-help@r-project.org Help
>
> 08/10/2011 02:02 PM
>
> Sent by:
>
> r-help-boun...@r-project.org
>
> Dear all
> How does one conver
Check function unlist().
Best,
Dimitris
On 8/10/2011 8:58 PM, Liviu Andronic wrote:
Dear all
How does one convert a "non-symmetric" list to a vector? See below:
x<- list()
x[[1]]<- letters[1:5]
x[[2]]<- letters[6:10]
x[[3]]<- letters[11:12]
x
[[1]]
[1] "a" "b" "c" "d" "e"
[[2]]
[1] "f" "g
Hi,
For some reason I'm finding that my table caption is disappearing if I
print xtable output with the floating argument set to FALSE. Below is a
very simple Sweave file that produces two tables the first has no
caption and the second has a caption (if you want to see it
http://www.zevross.c
unlist()
Michael Weylandt
On Wed, Aug 10, 2011 at 2:58 PM, Liviu Andronic wrote:
> Dear all
> How does one convert a "non-symmetric" list to a vector? See below:
>
> > x <- list()
> > x[[1]] <- letters[1:5]
> > x[[2]] <- letters[6:10]
> > x[[3]] <- letters[11:12]
> > x
> [[1]]
> [1] "a" "b" "c"
Hello,
I am a R beginner and hoping to obtain some hints or suggestions about
using permutations to sort a data set I have.
Here is an example dataset:
Ind1 11 00 12 15 28
Ind2 21 33 22 67 52
Ind3 22 45 21 22 56
Ind4 11 25 74 77 42
Ind5 41 32 67 45 22
This will
Dear all
How does one convert a "non-symmetric" list to a vector? See below:
> x <- list()
> x[[1]] <- letters[1:5]
> x[[2]] <- letters[6:10]
> x[[3]] <- letters[11:12]
> x
[[1]]
[1] "a" "b" "c" "d" "e"
[[2]]
[1] "f" "g" "h" "i" "j"
[[3]]
[1] "k" "l"
> paste(x)
[1] "c(\"a\", \"b\", \"c\", \"d\"
Duncan:
Yup, you're right. Can't assign, just print.
-- Bert
On Wed, Aug 10, 2011 at 11:02 AM, Duncan Murdoch
wrote:
> On 10/08/2011 1:16 PM, Bert Gunter wrote:
>>
>> Duncan et. al:
>>
>> Inline below.
>>
>> On Wed, Aug 10, 2011 at 9:48 AM, Duncan Murdoch
>> wrote:
>> > On 10/08/2011 9:40 AM
The see also potion of paste gives you the functions you can use for this
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Soyeon Kim
> Sent: Wednesday, August 10, 2011 2:22 PM
> To: r-help@r-project.org
> Subject: [R] Opposite
or,
gsub('V','',vn)
On 8/10/2011 2:23 PM, Peter Langfelder wrote:
> On Wed, Aug 10, 2011 at 11:22 AM, Soyeon Kim wrote:
>> Dear All,
>>
>> I have vn variable
>>> vn
>> [1] "V300" "V376"
>> What I want to get is
>> 300 376
> as.numeric(substring(vn, 2))
>
> HTH
>
> Peter
>
> _
On Wed, Aug 10, 2011 at 11:22 AM, Soyeon Kim wrote:
> Dear All,
>
> I have vn variable
>> vn
> [1] "V300" "V376"
> What I want to get is
> 300 376
as.numeric(substring(vn, 2))
HTH
Peter
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailma
Dear All,
I have vn variable
> vn
[1] "V300" "V376"
What I want to get is
300 376
without V and "" from vn variable.
Could you help me about this issue?
Thank you,
Soyeon
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing l
Assuming your data is in a data.frame called df, try this:
attach(df)
TR.groups <- cut(TvarRatio, seq(0.07, 0.11, 0.01))
m <- table(Mutation_Status, TR.groups)
mut.no <- dim(m)[1]
barplot(m, col=seq(mut.no), xlab="TvarRatio", ylab="Frequency")
legend("topleft", dimnames(m)[[1]], fill=seq(mut.no),
On 10/08/2011 1:16 PM, Bert Gunter wrote:
Duncan et. al:
Inline below.
On Wed, Aug 10, 2011 at 9:48 AM, Duncan Murdoch
wrote:
> On 10/08/2011 9:40 AM, Johannes Radinger wrote:
>>
>> Jean,
>>
>> thank you for your answer.
>> especially the line X<- numeric(length(lT)) helped me a lot.
>>
>
Perhaps you could shade the bars as appropriate?
I'm not going to use your data because it's not an easily paste-able but how
about this:
x = rnorm(100)
y = sample(c("A","B"),100,replace=T,prob=c(0.7,0.3))
d = data.frame(level = x, status = y)
n = 10 # Number of bins
breaks = quantile(d$level, (
HI everyone,
I'm plotting a histogram in R and within that histogram i need to
demonstrate the percentage of another variable (Percentage of MutStatus)
within the bins plotted inthe histogramI don't know how to do that!
Data:Validation_Status Mutation_Status TvarRatio
WildtypeNone
Duncan et. al:
Inline below.
On Wed, Aug 10, 2011 at 9:48 AM, Duncan Murdoch
wrote:
> On 10/08/2011 9:40 AM, Johannes Radinger wrote:
>>
>> Jean,
>>
>> thank you for your answer.
>> especially the line X<- numeric(length(lT)) helped me a lot.
>>
>> Anyway, in my case I'd like to get a "dynamic"
I was going to suggest
> AB <- df[c("A","B")]
> ls2 <- array(split(df$C, AB), dim=sapply(AB, nlevels), dimnames=sapply(AB,
levels))
which produces a matrix very similar to what Duncan's by() call produces
> ls1 <- by(df$C, df[,1:2], identity)
E.g.,
> ls2[["a","X"]]
[1] 1 2
> ls1[["a","
Hi Frederic,
shouldn't there be an result for the 3rd row as well, eg ls1$b$Y?
Maybe this will do what you want?
dtf<-within(dtf,index<-factor(A:B))
tapply(dtf$C,dtf$index,list)
Hth.
Am 10.08.2011 16:30, schrieb Frederic F:
> Hello Duncan,
>
> Here is a small example to illustrate what I am
You might want to look into the packages bigmemory and biganalytics.
Corey
On Tue, Aug 9, 2011 at 8:38 PM, Chris Howden
wrote:
> Hi,
>
> Im trying to do a hierarchical cluster analysis in R with a Big Data set.
> Im running into problems using the dist() function.
>
> Ive been looking at a fe
On 10/08/2011 9:40 AM, Johannes Radinger wrote:
Jean,
thank you for your answer.
especially the line X<- numeric(length(lT)) helped me a lot.
Anyway, in my case I'd like to get a "dynamic" variable or better a
function for X. I mean if i try to call X I'd like that this drawing of random
numbe
On 10/08/2011 11:59 AM, Lippel, Anna wrote:
Hi Duncan,
I have tried to install a tar.gz package following your instructions
(https://stat.ethz.ch/pipermail/r-help/2008-August/169599.html) but without
success. Here are the steps I followed:
I installed the last version of Rtools and ran Rcmd I
On 10/08/2011 10:30 AM, Frederic F wrote:
Hello Duncan,
Here is a small example to illustrate what I am trying to do.
# Example data.frame
df=data.frame(A=c("a","a","b","b"), B=c("X","X","Y","Z"), C=c(1,2,3,4))
# A B C
# 1 a X 1
# 2 a X 2
# 3 b Y 3
# 4 b Z 4
### First way of getting the list
Don't they have their own support mailing list? You should review
their documentation for specifics.
--
David.
On Aug 10, 2011, at 12:24 PM, Dr. Alireza Zolfaghari wrote:
Hi list,
I used to work with RExcel in excel 2003. Now in 2007, I tried the
same
RExcel, but it did not work. I got R ve
Hi,
this command gives you all possible encoding options on your platform:
iconvlist()
hope it answers your question,
T
--
View this message in context:
http://r.789695.n4.nabble.com/scan-file-encoding-tp840838p3733327.html
Sent from the R help mailing list archive at Nabble.com.
___
Hello Denis,
> To borrow shamelessly from one of the prominent helpers on this list:
> "What is the problem you're trying to solve?" Â Â (attribution: Jim Holtman)
I'm trying to connect two sets of legacy R tools: the output of the
first one can be transformed in a data.frame without loss of
inf
Original-Nachricht
> Datum: Wed, 10 Aug 2011 09:38:38 -0400
> Von: Duncan Murdoch
> An: Johannes Radinger
> CC: r-help@r-project.org
> Betreff: Re: [R] function runif in for loop
> On 10/08/2011 7:28 AM, Johannes Radinger wrote:
> > Hello,
> >
> > I'd like to perform a regres
Jean,
thank you for your answer.
especially the line X <- numeric(length(lT)) helped me a lot.
Anyway, in my case I'd like to get a "dynamic" variable or better a
function for X. I mean if i try to call X I'd like that this drawing of random
number is performed. In the case now if I call X sever
Hi Duncan,
I have tried to install a tar.gz package following your instructions
(https://stat.ethz.ch/pipermail/r-help/2008-August/169599.html) but without
success. Here are the steps I followed:
I installed the last version of Rtools and ran Rcmd INSTALL rJava_0.8-8.tar.gz
and got the error m
Hello Duncan,
Here is a small example to illustrate what I am trying to do.
# Example data.frame
df=data.frame(A=c("a","a","b","b"), B=c("X","X","Y","Z"), C=c(1,2,3,4))
# A B C
# 1 a X 1
# 2 a X 2
# 3 b Y 3
# 4 b Z 4
### First way of getting the list structure (ls1) using imbricated lapply
Hello
I am trying to learn the spatial panel data analysis (newbie). I have the R
version 2.13.1 and I did download the spml package required for the spatial
panel data analysis. However, when I tried the analysis, I get the following
error message. could not find function "spfeml". Can somebo
Hi list,
I used to work with RExcel in excel 2003. Now in 2007, I tried the same
RExcel, but it did not work. I got R version 12. I downloaded/installed the
latest version of RExcel 3.2.0 from http://sunsite.univie.ac.at/rcom/. It
has added the RExcel add-ins, but when I click on starting R in add-
As it says in
?format
the digits argument specifies "... how many significant digits are to be
used ... enough decimal places will be used so that the smallest (in
magnitude) number has this many significant digits ..."
In your example, the last value in column "a" is 0.06348058 which is
writ
Try this:
merge(q1, q2, all = TRUE)
On Wed, Aug 10, 2011 at 12:04 PM, Anthony Ching Ho Ng
wrote:
> Dear R-help,
>
> I wonder if you could give me some suggestions in how to do a union
> join of two data frames as follow:
> -> union join the common column, and insert a 0 if one is missing.
>
> I
Dear Jen,
Actually you can check out what R does by looking at the source.
# first type the name of the function
> rstandard
function (model, ...)
UseMethod("rstandard")
# ?methods will list you the corresponding functions
> methods("rstandard")
[1] rstandard.glm rstandard.lm
# choose rstanda
We have released to CRAN a new version of the (recommended)
package Matrix, and of package lme4 yesterday.
Anyone who gets the new version of Matrix *MUST* re-install lme4
-- if (s)he is using lme4 at all. Technical details about that further below.
The fact that yesterday's version number of Ma
Hello
On Wed, Aug 10, 2011 at 2:31 PM, Jean V Adams wrote:
> The function format() might serve your needs.
>
This looks very promising, but yields some strange results. See below:
> x <- data.frame(a=rnorm(10), b=rnorm(10), c=rnorm(10), d=letters[1:10])
> x
a b
January,
It looks like you will need an interaction effect, perhaps
g <- lm( response ~ subject + group*time)
Please see the ancova function in the HH package.
install.packages("HH") ## if necessary
library(HH)
?ancova
Rich
On Wed, Aug 10, 2011 at 5:15 AM, January Weiner wrote:
> Hello,
>
On Aug 10, 2011, at 11:04 AM, Anthony Ching Ho Ng wrote:
Dear R-help,
I wonder if you could give me some suggestions in how to do a union
join of two data frames as follow:
-> union join the common column, and insert a 0 if one is missing.
I made a function to perform the following, and I kno
Dear R-help,
I wonder if you could give me some suggestions in how to do a union
join of two data frames as follow:
-> union join the common column, and insert a 0 if one is missing.
I made a function to perform the following, and I know it may not that
quite welly written, but it works.
Any sug
Thanks Patrick - at least I know I wasn't being too silly :-)
Jen
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Following the suggestion by Duncan Murdoch, this should work for you.
X <- runif(length(lT), lT, uT)
Jean
From:
"Johannes Radinger"
To:
Jean V Adams
Cc:
r-help@r-project.org
Date:
08/10/2011 08:40 AM
Subject:
Re: [R] function runif in for loop
Jean,
thank you for your answer.
especially
On 08/10/2011 10:03 AM, Jen wrote:
Hi,
I must be doing something silly here, because I can't get the studentised
and standardised residuals from r output of a linear model to agree with
what I think they should be from equation form.
x = seq(1,10)
y = x + rnorm(10)
mod = lm(y~x)
rstandard(mod)
r
R version 2.13.1
OS X (or Windows)
Colleagues,
I received a number of files with a .xls extension. These files open in XL
and, by all appearances, are XL files. However, it appears to me that the
files are actually XML:
> readLines(dir()[16])[1:10]
[1] ""
On 10/08/2011 10:16 AM, Barry Rowlingson wrote:
On Wed, Aug 10, 2011 at 2:34 PM, Duncan Murdoch
wrote:
> On 10/08/2011 5:58 AM, taraxacum wrote:
>>
>> Hi all,
>> I need to convert a floating point value from Microsoft Basic format to
>> IEEE
>> format.
>> There's a simple way to achieve t
So far we have received over 70 responses to our survey. If you have NOT
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On Wed, Aug 10, 2011 at 2:34 PM, Duncan Murdoch
wrote:
> On 10/08/2011 5:58 AM, taraxacum wrote:
>>
>> Hi all,
>> I need to convert a floating point value from Microsoft Basic format to
>> IEEE
>> format.
>> There's a simple way to achieve this in R or I have to write my own
>> function?
>> (e.g.
Hi,
I must be doing something silly here, because I can't get the studentised
and standardised residuals from r output of a linear model to agree with
what I think they should be from equation form.
Thanks in advance,
Jennifer
x = seq(1,10)
y = x + rnorm(10)
mod = lm(y~x)
rstandard(mod)
residuals
Can't you use sapply?
sapply(seq_len(10), function(i){SomeExpression(i)})
Best regards,
Thierry
> -Oorspronkelijk bericht-
> Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> Namens David Winsemius
> Verzonden: woensdag 10 augustus 2011 15:50
> Aan: Anthony Ching
On Aug 10, 2011, at 9:08 AM, Anthony Ching Ho Ng wrote:
Dear list,
I wonder if there a better way to have rbind/cbind/append to create
the first element (if it is empty) instead of doing the following in a
loop?
for (i in 1:10) {
if (i == 1) {
aRow = SomeExpression(i)
} else {
aRow =
On 10/08/2011 7:28 AM, Johannes Radinger wrote:
Hello,
I'd like to perform a regression using MCMCregress (MCMCpack).
One variable therefore should be a function rather than a variable:
I want to use X as an input and X should be defined as a random number between
to values. Therefore I want t
On 10/08/2011 5:58 AM, taraxacum wrote:
Hi all,
I need to convert a floating point value from Microsoft Basic format to IEEE
format.
There's a simple way to achieve this in R or I have to write my own
function?
(e.g. convert the C code below)
You'll need to write your own function. It can be v
To borrow shamelessly from one of the prominent helpers on this list:
"What is the problem you're trying to solve?"(attribution: Jim Holtman)
I have the sense you want to do something over many subsets of your
data frame. If so, breaking things up into lists of lists of lists is
not necessari
Anthony,
See ?make.names for a description of valid names. Here's an excerpt:
"A syntactically valid name consists of letters, numbers and the dot or
underline characters and starts with a letter or the dot not followed by a
number. ... The character "X" is prepended if necessary."
There is a
Johannes,
You have the loop set up right, you just need to add indexing to refer to
the looping variable, i.
lT <- sample(1:10)
uT <- sample(21:30)
X <- numeric(length(lT))
for (i in 1:length(lT)) X[i] <- runif(1, lT[i], uT[i])
X
Note that I changed the name of the result from T to X, because
Dear List,
I wonder why when using read.csv(), if the column name contains a
numeric i.e. a stock symbols-"0001.HK", it will automatically insert
an "X" character to the column names - "X0001.HK".
Now I have to manually do a loop and use substring() to remove the "X"
character. Any advice? Thanks
Hi:
Try this:
## Function that takes a data frame as input and outputs a data frame:
chrSumm <- function(d) { # d is a data frame
colnames(d) <- c("chr","start","end","base1","base2",
"totalreads","methylation","strand")
TR <- nrow(d)
RG1 <- sum(d['totalreads']
Gabe,
Since you didn't provide a small example of your data, I can't test out
your code. However, I used an example from the ?ordiellipse help page to
draw different colored ellipses (using the show.groups= argument) with
labels (using the label= argument).
Hope this helps.
library(vegan)
d
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