Hello R User,
I was trying to display r.squared and p value in table from regression, but I
could not display these parameters in the table (matrix)
for example
individual - c(1,1,6,8,8,9,9,9,12,12)
day - c(4,17,12,12,17,3,9,22,13,20)
condition - c(0.72, 0.72, 0.67, 0.73, 0.76, 0.65, 0.68,
Hi,
I am new to R and have been trying to utilize the IMA package for analyzing
Illumina methylation data.
The IMA.methy450R function is used to load the Illumina data set and the
manual indicates that this should create a QC.pdf file with basic quality
control information (Basic Quality Control
On Fri, 27 Apr 2012, Vale Fara wrote:
I am working with lotteries and I need to generate two sets of uniform
random numbers.
Requirements:
1) each set has 60 random numbers
random integers?
2) random numbers in the first set are taken from an interval (0-10),
whereas numbers in the second
Hello,
Why not == ?
x == y
which(x == y)
Hope this helps,
Rui Barradas
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Hello,
This solution is not very pretty but it works.
nms - unlist(d[1, ])
nm - unique(nms)
dd - na.exclude(sapply(nm, function(jj){
inx - nms %in% jj
do.call(rbind, as.list(d[, inx]))
}))
dd - dd[ dd[ , nm[1]] != nm[1], ]
dd - data.frame(apply(dd, 2,
Hello,
Hello again,
I am still having the same problem with the main loop.it succeeded to
finish reading and writing from the first file in my
listfile but failed when it moves to the second file giving this
error:Error: subscript out of bounds
Don't subscript then.
The only place where
Hello, again.
Yes ,It worked. So how can I insert it in my code? Please
Easily. Just replace your beginning of loop
for(n in 1:length(listfile))
by the for(...) above, and comment out the line
h=listfile[n]
since it's no longer needed.
Rui Barradas
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Hello,
Yes ,It worked. So how can I insert it in my code? Please
Simply like this:
#for (n in 1:length(listfile))
#{
# h=listfile[n]
#
for(h in listfile)
{
Rui Barradas
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I signed up. I'm doing a talk on real time text classification using
node.js and R
Cory
On Saturday, April 28, 2012 2:20:08 PM UTC-5, Barry Rowlingson wrote:
There's now a page on lanyrd (the social conference directory) for
useR! 2012 in Nashville:
http://lanyrd.com/2012/useR/
its
Hi, I am new to R. I was trying to use RDCOMClient package to access com
object based applications within R. This has been working fine for a while.
I have identical set up on two PCs (R-2.1.10 running on XP both on network
so same profile). Recently while installing a new package, I had to
Hi, I am very new to R so please excuse me if I am asking very obvious
questions.
I am trying to call a blackbox api function implemented in as a COM object
from R. The function definition says that
1. if calling from VBA, the first parameter should be set to Nothing
2. if calling from matlab,
I finally managed to get this done. I went to the contributed packages for
version 2.14 Windows 32 bit and downloaded the zip file. This got installed
smoothly in R 2.15. Don't know why I did not try this before.
Regards,
Indrajit
From: Prof Brian Ripley
You don't need quotes in the library statement, you can just use
library(parallel).
Regards,
Indrajit
From: Uwe Ligges lig...@statistik.tu-dortmund.de
To: ya xinxi...@163.com
Cc: r-help@r-project.org
Sent: Saturday, April 28, 2012 11:57 PM
Subject: Re: [R]
In the book, Regression Modeling Strategies, Frank Harrell uses the function
rcs() often.
The current version or R and Hmisc library don't appear to have this function.
What is an appropriate substitute?
Thank You
--
Noah Silverman
UCLA Department of Statistics
8208 Math Sciences Building
On 29.04.2012 09:28, Indrajit Sengupta wrote:
You don't need quotes in the library statement, you can just use
library(parallel).
Yes, a special (mis-)feature of library(). Since we are trying to teach
R here, we should provide clean R code. In an ideal world, we would be
able to say
On Apr 29, 2012, at 09:34 , Noah Silverman wrote:
In the book, Regression Modeling Strategies, Frank Harrell uses the
function rcs() often.
The current version or R and Hmisc library don't appear to have this
function.
What is an appropriate substitute?
The rms package, I
You came close.
Here is how it might be done:
individual - rep(c(1,1,6,8,8,9,9,9,12,12),2)
day - rep(c(4,17,12,12,17,3,9,22,13,20),2)
condition - rep(c(0.72, 0.72, 0.67, 0.73, 0.76, 0.65, 0.68, 0.78, 0.73,
0.71),2)
test - data.frame(individual, day, condition)
#ind.id - unique(test$individual)
Hi everyone
I am stuck on specifying my own maximum likelihood function for a
special poisson model.
My poisson model is as follow: O ~ Pois(b*N + b*RR*E)
With
O = observed cases
b = constant (known)
N = number of unexposed persons (known)
E = number exposed persons (known)
RR = relative risk
Ciao Roberto,
welcome to hyperSpec!
1. reproducible example
You can use the example data sets coming with hyperSpec to create examples that
everyone can easily reproduce.
2. there's a hyperSpec-help mailing list which you may want to join.
You can subscribe at
Hi, ur reply is much appreciated and I think that will work. But i don't know
how to incorporate that code into map.market function:
function (id, area, group, color, scale = NULL, lab = c(TRUE,
FALSE), main = Map of the Market, print = TRUE)
{
if (any(length(id) != length(area),
I am trying to make a function permute that takes two arguments: 1)
and alphabet consisting of a set of symbols and 2) a length being a
single integer N.With a safety check check in permute() that stops
execution (use the stop() function) with a warning when the number of
permutations would exceed
No its not perfectly working see below
j-c(1,1,2,2,3)
k-c(2,1,1,1,1)
j%%k
[1] 1 0 0 0 0
problem comes when one of my sequence have 1
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Sent from the R help
Interesting set of question.. I am completely new to R but let me try my
luck.
Random number in R
x - runif(60 , 0 , 10) # 60 numbers from 0 to 10
y- runif(60, 15 , 25) # same as above , from 15 to 25
The second part though. Do you mean ,
for( i in 1:length(x)) {
z = x[i] + y[i]
return z
}
I just tried it.
library - quantmod
library
[1] quantmod
typeof(library)
[1] character
So how do I , or anyone , assign a string without ?
library - quantmod
Error: object 'quantmod' not found
What about this?
library - c(quantmod)
Does it do the job? Sorry if it has been asked. I'm
Hi Uwe and Indrajit,
Thank you very much for the response.
Since we are talking about this package, can I ask more about how to use
it to deal with statistical issues please?
I have this data with categorical missing values in it. Now I am trying
to impute them using a function called mice
You seem confused the library Uwe is talking about is a function
not a string
These work
library(quantmod)
library(quantmod)
or
pkg - quantmod
library(pkg) # Doesn't work!
library(pkg, character.only = TRUE) # Works
Michael
On Sun, Apr 29, 2012 at 3:59 AM, ad...@use-r.com wrote:
I think I misread the question. Is this what you want
j - c(1,1,2,2,3)
k - c(2,1,1,1,1)
xx - j == k
which(xx == TRUE)
John Kane
Kingston ON Canada
-Original Message-
From: sagarnikam...@gmail.com
Sent: Sat, 28 Apr 2012 23:48:43 -0700 (PDT)
To: r-help@r-project.org
Subject: Re:
Ah .. another thing I learnt. Thanks
You seem confused the library Uwe is talking about is a function
not a string
These work
library(quantmod)
library(quantmod)
or
pkg - quantmod
library(pkg) # Doesn't work!
library(pkg, character.only = TRUE) # Works
Michael
On Sun,
On 29-04-2012, at 08:48, sagarnikam123 wrote:
No its not perfectly working see below
j-c(1,1,2,2,3)
k-c(2,1,1,1,1)
j%%k
[1] 1 0 0 0 0
problem comes when one of my sequence have 1
Did you try the other solutions that were suggested?
Berend
On 29-04-2012, at 13:09, Melinda Harwood wrote:
I am trying to make a function permute that takes two arguments: 1)
and alphabet consisting of a set of symbols and 2) a length being a
single integer N.With a safety check check in permute() that stops
execution (use the stop() function) with
Thanks for all the suggestions. This is my final code that seems to be
working:
alphabet = c(a,b,c,d)
holder = c()
permute = function(alphabet,n){
while((length(unique(holder))/n)(length(alphabet)^n)){
perm = sample(alphabet, replace=T, size=n)
holder = rbind(holder, perm,
Hi Tal,
Thank you so much for the help. I am afraid that I could not make you
understand of my problem. The code you wrote is for whole data set, but I
wanted to do regression based on Individual and put coefficient, r2 and p
value. Individual is group variable.
Here I again run the script
On Apr 28, 2012, at 10:46 AM, Daniel Malter wrote:
Hi,
I have a data frame whose first row (not the header) contains the true
column names. The same column name can occur multiple times in the
dataset.
Columns with equal names are not adjacent, and for each observation
only one
of the
Unfortunately, I'm not really making any progresses, despite a lot of effort.
I've compiled R on Mac OS X for myself using MacPorts and the error is now
state 28000, code 201 which is failed password authentification.
__
R-help@r-project.org mailing
Denis.Aydin at unibas.ch writes:
Hi everyone
I am stuck on specifying my own maximum likelihood function for a
special poisson model.
My poisson model is as follow: O ~ Pois(b*N + b*RR*E)
With
O = observed cases
b = constant (known)
N = number of unexposed persons (known)
E =
Perhaps NULL? Or maybe a vector of length 0you can also just leave
arguments missing in R thanks to lazy evaluation so there's not a single
equivalent to nothing
Hth,
Michael
On Apr 28, 2012, at 6:04 PM, lm shinilku...@hotmail.com wrote:
Hi, I am very new to R so please excuse me if I am
I believe the regularity of the problem allows a (to me, anyway)
simpler procedure.
td - t(apply(d,2, na.omit))
data.frame(split(as.numeric(td[,-1]),td[,1]))
-- Bert
On Sat, Apr 28, 2012 at 9:33 AM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
This solution is not very pretty but it
These sorts of issues are often machine dependent and not R package or R
software dependent, especially when it comes to processes like com. I a,m
not a RDCOMClient user, but it's been on my list to look at. You don't
provide much info on your various component versions (see posting guide) so
On Fri, Apr 27, 2012 at 02:08:09PM -0700, Hans Thompson wrote:
I meant (1,1) as an (x,y) coordinate. I am trying to find the function to
return the coefficient of the line running through:
B
x y
a 1 1
and all the points in:
A
x y
a 1 3
b 2 2
c 3 1
Hi.
Hi Uwe and Indrajit,
Thank you very much for the response.
Since we are talking about this package, can I ask more about how to
use it to deal with statistical issues please?
I have this data with categorical missing values in it. Now I am
trying to impute them using a function
I am not sure why you would want it, but here it is:
individual - c(1,1,6,8,8,9,9,9,12,12)
day - c(4,17,12,12,17,3,9,22,13,20)
condition - c(0.72, 0.72, 0.67, 0.73, 0.76, 0.65, 0.68, 0.78, 0.73, 0.71)
test1 - data.frame(individual, day, condition)
test1
library(plyr)
result=ddply(test1,
A) Yes, there is something wrong. Code not reproducible. Behavdat not found.
B) Last time I used it, Yahoo Mail could send in plain text. Admittedly that
has been awhile, but Google sez read [1] and/or [2].
[1] http://help.yahoo.com/tutorials/mail/mail_persmsg1.html
[2]
Hi,
I have a large data set that I input as a matrix, where I have 1:x rows
with names AX, then x+1: y rows named AY, etc. The idea is that I have to
count how many rows exactly I have with name AX and how many I have with
name AY (or find which row numbers have names AX). Is there any way in
On Apr 29, 2012, at 2:25 AM, billy am wickedpu...@gmail.com wrote:
Interesting set of question.. I am completely new to R but let me try my
luck.
Random number in R
x - runif(60 , 0 , 10) # 60 numbers from 0 to 10
y- runif(60, 15 , 25) # same as above , from 15 to 25
The second
Hi,
This is my code (my data is attached):
library(languageR)
library(rms)
library(party)
OLDDATA - read.csv(/Users/Abigail/Documents/OldData250412.csv)
OLDDATA$YD - factor(OLDDATA$YD, label=c(Yes, No))
OLDDATA$ND - factor(OLDDATA$ND, label=c(Yes, No))
attach(OLDDATA)
defaults - cbind(YD, ND)
Hi,
i am trying to run an ANCOVA and a bootstrapped ANCOVA analysis on a specific
data set. I am using the ancova and ancboot functions as in the following code:
setwd(C:/Users/User/Desktop/Rdatabilingualstudy2012)
bilingualismdata-read.spss(bilingualdataforconferences2012.sav,
On Sun, 29 Apr 2012 11:27:06 -0700 (PDT)
arun smartpink...@yahoo.com wrote:
I tried to get xyplot shrink fit for my mixed model._ But, the mixed model
line is not seen in the graph.
I would like to know if there is anything wrong with my code.
Typo? Try panel.abline(fixef(fm1),... or
Dear R group,
Tried to turn off the html in yahoo. So far not successful. Apologies!
I tried to get xyplot shrink fit for my mixed model. But, the mixed model line
is not seen in the graph.
I would like to know if there is anything wrong with my code.
Thanks,
A.K.
(fm1 - lmer(Response3
Hi,
thank you both for your replies, I really appreciate it!
To Mike: yes, random integers. Can I use the function round() as in
the example with 5 random numbers below?
To Billy: for the second part I got an error, but it may be that I
didn't properly set i...?
Here is the R output:
x -
I'd use a combination of rownames(), grepl() and sum().
Get the names with the first, test with the second and count the positives (by
coercing TRUE - 1) with the last
Michael
On Apr 29, 2012, at 3:46 PM, katarv katiasm...@gmail.com wrote:
Hi,
I have a large data set that I input as a
CRAN (and crantastic) updates this week
New packages
* anoint (1.0)
Maintainer: S. A. Kovalchik
Author(s): Ravi Varadhan rvarad...@jhmi.edu and Stephanie Kovalchik
License: GPL (= 2)
http://crantastic.org/packages/anoint
Tools for assessing multiple treatment-covariate
Hello,
katarv wrote
Hi,
I have a large data set that I input as a matrix, where I have 1:x rows
with names AX, then x+1: y rows named AY, etc. The idea is that I have
to count how many rows exactly I have with name AX and how many I have
with name AY (or find which row numbers have
I would assume that you would use 'sample' to draw the numbers:
sample(0:10,60,TRUE)
[1] 2 3 1 2 9 2 2 0 3
[10] 0 4 2 3 9 7 3 10 9
[19] 8 5 8 7 6 3 10 0 6
[28] 8 10 6 3 3 2 7 0 0
[37] 1 4 8 2 10 2 0 7 9
[46] 9 9 7 9 6 10 1 1 6
[55] 1 8 3 8 2
Hello R users,
I am trying to obtain a direct adjusted survival curve. I am sending my whole
code (see below). It's basically the larynx cancer data with Stage 1-4. I am
using the cox model using coxph option, see the fit3 coxph. When I use the
avg.surv option on fit3, I get the following
Right.
Michael Weylandt wrote
I'd use a combination of rownames(), grepl() and sum().
Get the names with the first, test with the second and count the positives
(by coercing TRUE - 1) with the last
Michael
On Apr 29, 2012, at 3:46 PM, katarv lt;katiasmirn@gt; wrote:
Hi,
I
On Sun, Apr 29, 2012 at 4:38 PM, Vale Fara vale...@gmail.com wrote:
Hi,
thank you both for your replies, I really appreciate it!
To Mike: yes, random integers. Can I use the function round() as in
the example with 5 random numbers below?
To Billy: for the second part I got an error, but it
On Mon, 30 Apr 2012, Vale Fara wrote:
ok, what to do is to generate two sets (x,y) of integer uniform random
numbers so that the following condition is satisfied: the sum of the
numbers obtained in x,y matched two by two (first number obtained in x
with first number obtained in y and so on)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Mike Miller
Sent: Sunday, April 29, 2012 5:21 PM
To: Vale Fara
Cc: r-help@r-project.org; billy am
Subject: Re: [R] generate random numbers for lotteries
On Mon, 30 Apr 2012,
Thanks Michael. I tried with NULL - It throws Type mismatch error and same
error while passing c() or defining a- c() and then passing a as parameter
Also tried with not passing any parameter, then it says 'Parameter not
optional'. I am afarid no joy. Thanks Regards,lmDate: Sun, 29 Apr 2012
Hello,
There are other types of empty objects in R (zero length or dimension).
Maybe some of these
x - list()
length(x)
x - matrix(list())
dim(x) - c(0, 0)
x
Matlab's [] is the empty matrix so maybe the second works.
Other possibilities could be x - numeric(0) (or integer(0) or
character(0)).
On Sun, 29 Apr 2012, Daniel Nordlund wrote:
I don't know what the OP is really trying to accomplish yet, and I am
not motivated (yet) to try to figure it out. However, all this
flooring and ceiling) and rounding is not necessary for generating
uniform random integers. For N integers in the
Frank, sorry if I seem impatient, but do you know approximately when
this will be fixed?
In the meantime, if I have
library(nlme)
library(rms)
d - data.frame(x = rnorm(50), y = rnorm(50))
and I'm interested in the 3 df test for x that I would get from
anova(Gls(y ~ rcs(x, 4), data=d,
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