We would like to announce the following statistics course;
Course: Introduction to Linear mixed effects models, GLMM and MCMC with R
When: 10-14 February, 2014
Where: Pousada de juventude parque das nacoes. Rua de Moscavide, Lt 47 –
101, 1998- 011. Lisbon, Portugal
Info:
On Thu, Nov 21, 2013 at 8:35 AM, Trevor Walker
trevordaviswal...@gmail.comwrote:
I often work with tree data that is sampled with probability proportional
to size, which presents a special challenge when describing the frequency
distribution.
The survey package does lots of calculations
On 13-11-20 10:45 PM, Earl Brown wrote:
R-helpers,
I'm using system() to run a shell script that uses a library written in C++ to
analyze natural language (FreeLing: http://nlp.lsi.upc.edu/freeling). When I
run the following code in RStudio (0.97.248) and R.app (1.62) on Max OSX
(10.7.5):
Hi,
I have 100 observations X1,X2,..,X100 and the confidence interval
limits for the mean at 5% level of significance.
I wanted to repeat this procedure say 50 times and see how many times the
hypothetical mean is included in the confidence intervals.
Analytically I have done this, but I am
you need to show the statement in context with the rest of the script. you
need to tell us what you want to do, not how you want to do it.
Sent from my iPad
On Nov 20, 2013, at 15:16, Noah Silverman noahsilver...@g.ucla.edu wrote:
Hello,
I have a fairly large data.frame. (About 150,000
Thanks, got it.
**
Bander Alzahrani, r
*
Date: Thu, 21 Nov 2013 09:14:11 -0600
Subject: Re: [R] frequency of numbers
From: deter...@umn.edu
To: cs_2...@hotmail.com
CC: r-help@r-project.org
If
If you just need a count of how many of each number you can just use
table().
tmp - c(111,106,117,108,120,108,108,116,113)
table(tmp)
tmp
106 108 111 113 116 117 120
1 3 1 1 1 1 1
On Thu, Nov 21, 2013 at 9:10 AM, b. alzahrani cs_2...@hotmail.com wrote:
hi guys
Assume I
R 3.0.1
Windows 7
Rstudio 0.97.551
Colleagues,
The last time I thought about using lmer to run an unbalanced repeated measures
ANOVA, I found that the package did not return p values or SEs. I believe this
was because Professor Bates had important questions about the theory behind the
Many thanks to Wolfgang Viechtbauer for the prompt and clear answer.
However, I'm unable to understand what .cmicalc mathematically does in the
code line:
cmi - .cmicalc(mi)
I look in the metafor documentation and in R (?.cmical and ??.cmicalc) but
I have no result. Please, can I have further
I have data in a list of the form: list(list(dataframe())). It have
attached some random data with this structure called l.Rdata. The
structures in l.Rdata are as follows: a list of 4 groups: a list of 12
statistics: (and for each statistic, a dataframe contains (1row,3cols)
of values named
Hi everyone,
If the function I am trying to optimize have the ... argument, how to pass
it to optim()?
For example, we want to minimize this very simple function:
foo - function(a, ...) {
var=list(...)
(a-1)^2+do.call(sum,var)
}
Although the ... in this function doesn't make any
Think about what you are actually getting from pnorm().
For each scalar s you want to get
pnorm(s,mu[2],sigma[2])pnorm(s,mu[3],sigma[3])
pnorm(s,mu[4],sigma[4])
(and then take products of things).
But when s is a vector you get
pnorm(s[1],mu[2],sigma[2])
Here is the simplest answer that I know.
outcome6 - ifelse(Survival_days 2190, 0, Outcome)
survfit(Surv(Survival_days, outcome6) ~ Gender)
This assumes that the variable Outcome is coded as 0/1. If it were instead FALSE/TRUE
then change the 0 to FALSE in the first line.
The logrank
Dear John,
The Anova() function in the car package will perform a traditional
unbalanced repeated-measures ANOVA (or MANOVA). See the R Journal article at
http://journal.r-project.org/archive/2013-1/fox-friendly-weisberg.pdf.
I hope this helps,
John
-Original Message-
From:
Hi,
The new lme4 package has a bootMer function, so it is fairly easy to
compute bootstrap statistics.
The following packages could be what you are looking for:
1) lmerTest (see ?lmerTest:::lmer)
2) car (see Anova)
3) afex (see mixed)
4) LMERConvenienceFunctions (see pamer.fnc)
HTH,
Denes
Well ...
1. Try the latest version of lmer/lme4 and see for yourself. I think
the answer is still no.
2. If 1 is true, then I think you need to ask whether you **want** to
use a package that returns p values and se's, whether it exists or
not.
Cheers,
Bert
On Thu, Nov 21, 2013 at 2:07 PM,
Noah,
If N is # of rows, k is # of unique IDs
Using which() is O(N), using which() in a loop is going to be O(Nk);
sorting the entire data is O(N ln N) and then you can process it in
contiguous blocks, no which required.
-Neal
On Thu, Nov 21, 2013 at 8:48 AM, William Dunlap
Hello,
?optim has a dots argument, so just pass whatever you want after 'gr'.
Something like the following.
optim(-3, foo, NULL, 1, 2)
Hope this helps,
Rui Barradas
Em 21-11-2013 22:05, Nan Wang escreveu:
Hi everyone,
If the function I am trying to optimize have the ... argument, how to
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Preetam Pal
Sent: Thursday, November 21, 2013 12:10 PM
To: r-help@r-project.org
Subject: [R] Plotting multiple confidence intervals in the same graph
Hi,
I have 100
Hi all,
I am using lattice::wireframe to create a surface. I need to change the
coloring so that it depends on the x or y variable (instead of on z). How
should this be done? The documentation says that the color is automatically
chosen to depend on the height (e.g. z).
Thanks!
Georgi
How about this?
require(FNN)
#FOR DEMONSTRATION PURPOSES, GENERATE 2D DATA
set.seed(3242)
X - matrix(runif(50),ncol=2)
plot(X,pch=16,cex=1.5, asp=1)
#PLOT GRID
grid - as.matrix(expand.grid(seq(0,1,by=.1),seq(0,1,by=.1)) )
abline(v=unique(grid[,1]),col=#FF30)
Hi
I missed the original posting
I think to get the full picture we need to a reproducible example eg by dput
of the data.
To keep all the information about a plot you can use the par.settings
argument which is an alternative to trellis.par.set().
Try names(trellis.par.get()) to get a list of
That works.
Thanks a lot!
On Wed, Nov 20, 2013 at 11:52 PM, David Winsemius dwinsem...@comcast.netwrote:
On Nov 20, 2013, at 8:39 PM, David Winsemius wrote:
On Nov 20, 2013, at 7:12 PM, dan wang wrote:
Hi all,
Can anyone help me with below integrate function?
Basically, I
Hi
Without knowing anything about the data and the warning messages, however
the p.vec value seems a little strange.
try
p.vec= seq(1.35, 1.75, by = 0.1)
It could be that you have run out of memory to do 50 sets of calculations
If ok with the result then repeat with a narrower focus. Most
there would be an
over-estimate of risk in proportion to the ratio of such cases to the entire
population.
--
David.
Chris
library(survival)
set.seed(20131121)
ngroup - 100
xxx - rep(0:1, e=ngroup)
ttt - rexp(length(xxx), rate=xxx+.5)
plot(survfit(Surv(ttt) ~ xxx))
survdiff(Surv
Hi R Experts,
About the data:
My data consists of people (ID) with years of service (Yos) for each year. An
ID can appear multiple times.
The data is sorted by ID then by Year.
Problem:
I need to extract ID data with non-sequential YoS rows. For example below that
would be all rows for ID 33
Dear All,
In the following simple case I can't seem to get an improved fit,
despite trying all of the control possibilities. As there seem to be
no examples anywhere which show use of functions such as dnorm
within a formula, and as I am not confident at all that my formula is
correctly
The line with the slow process (According to Rprof) is:
j - which( d$id == person )
(I then process all the records indexed by j, which seems fast enough.)
Using split() once (and using its output in a loop) instead of == applied to
a long vector many times, as in
for(j in
Hi,
I'm trying to fit regression model, but there is something wrong with it.
The dataset contains 85 observations for 85 students.Those observations are
counts of several actions, and dependent variable is final score. More
precisely, I have 5 IV and one DV. I'm trying to build regression model
Dan,
Does this do it?
## where dt is the data
tmp - split(dt, dt$ID)
foo - lapply(tmp, function(x) any(diff(x$YoS) 1))
foo - data.frame( ID=names(foo), gap=unlist(foo))
Note that I ignored dept.
Little hard to see how YoS can increase by more than one when the year
increases by only one ...
Hi,
You may try:
with(testSeq,ave(YoS,ID,FUN=function(x) c(0,diff(x
# [1] 0.0 1.0 0.0 1.0 1.0 0.0 2.1 0.0 1.0 3.0 1.0
testSeq[!!(with(testSeq,ave(YoS,ID,FUN=function(x) any(c(0,diff(x))1,]
A.K.
On Thursday, November 21, 2013 6:55 PM, Lopez, Dan lopez...@llnl.gov wrote:
Hi R
Gonçalo,
Interesting question. You can use the optim() function to optimize a
function over two dimensions. Here's an example.
# some clumpy cloud data
myx - c(0.3 + rnorm(50, 0, 0.05), 0.7 + rnorm(50, 0, 0.05))
myy - c(0.45 + rnorm(50, 0, 0.05), 0.65 + rnorm(50, 0, 0.05))
# size of the fixed
This certainly looks like a bug, and there are many ways of inducing bugs that
only show up with large datasets - buffer overruns, fields that are too small
to hold the number of rows, etc. Remember that there is NO official
documentation of the .sas7bdat format, everything has been reverse
Hi,
Try:
1st part:
res3 - vector()
for(i in 1:length(D)){
res3 - rbind(res3, read.table(paste0(getwd(),/IR/,D[i]),header=TRUE)[,2])
res3
}
dim(res3)
#[1] 15 1686
identical(res2,res3)
#[1] TRUE
2nd part:
vec1 - unlist(lapply(D,function(x) {x1 -
or use tapply(seq_len(nrow(d),d$id,...) or a wrapper version
thereof (by, aggregate,...)
However, it would not surprise me if this does not help. I suspect
that the problem is not what you think but in the code and context you
omitted, as others have already noted.
-- Bert
On Thu, Nov 21,
(1) Is this homework? (This list doesn't do homework for people!)
(Animals maybe, but not people! :-) )
(2) Your question isn't really an R question but rather a
statistics/linear modelling
question. It is possible that you might get some insight from Frank
Harrel's book
Regression
No, it's not homework, it's just some initial analysis, but still...
and thanks for recommendation.
On Thu, Nov 21, 2013 at 4:42 PM, Rolf Turner r.tur...@auckland.ac.nzwrote:
(1) Is this homework? (This list doesn't do homework for people!)
(Animals maybe, but not people! :-) )
(2) Your
Hi All,
I'm trying to figure out how in my data set to add a column including a
count of unique events based on date. Here is a part of my data set:
trialno event date time
3 11301pm_intake 2010-11-24
Hi, I have a cloud of randomly distributed points in 2-dimensional space and
want to set up a grid, with a given grid-cell size, that minimizes the distance
between my points and the grid nodes. Does anyone know of an R function or
toolbox that somehow addresses this problem? This is a problem
you didn't show us the code you used to generate the legend.
I'm guessing you want to add to the legend list something like lty=0:7 .
KB wrote
I recently started using R, so I'm not really experienced with it. My
question is on adjusting xyplots to get lty lines instead of coloured
lines.
Thank you, Peter. I'll generate a self-contained example and contact the
maintainer.
Xiaochun
-Original Message-
From: peter dalgaard [mailto:pda...@gmail.com]
Sent: Thursday, November 21, 2013 9:19 AM
To: Li, Xiaochun
Cc: r-help@R-project.org
Subject: Re: [R] Does function
Hi all,
I tried below two methods to calculate the integrate of a same function.
Two different results are given.
The first one should be the right answer.
Can any one help to explain why?
Another issue is, the first one is not convenient as I have to update the
mu and sigma outside the function.
Hi,
From the dscription, looks like you need ?rle()
vec1 - c(111, 106, 117, 108, 120, 108, 108, 116, 116, 113)
res - rle(vec1)$lengths
names(res) - rle(vec1)$values
res[res1]
#108 116
# 2 2
length(res[res1])
A.K.
On Thursday, November 21, 2013 10:12 AM, b. alzahrani
Hi Dawn,
unix is kind vague, what exactly is the OS you are using? Do you really
need to build from source?
Best,
Ista
On Wed, Nov 20, 2013 at 4:13 PM, Dawn dawn1...@gmail.com wrote:
Hi,
I am trying to install R on the Unix system. When I type './configure', it
seems many not installed on
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 11/21/13, 12:34 , Jim Holtman wrote:
you need to show the statement in context with the rest of the
script. you need to tell us what you want to do, not how you want
to do it.
Agreed - a few details will result in guesses (see my guess
That subset will give you right truncation, not right censoring. See code
below. Use Thomas's solution.
Chris
library(survival)
set.seed(20131121)
ngroup - 100
xxx - rep(0:1, e=ngroup)
ttt - rexp(length(xxx), rate=xxx+.5)
plot(survfit(Surv(ttt) ~ xxx))
survdiff(Surv(ttt) ~ xxx)
# impose
Neal Fultz nfultz at gmail.com writes:
Noah,
If N is # of rows, k is # of unique IDs
Using which() is O(N), using which() in a loop is going to be O(Nk);
sorting the entire data is O(N ln N) and then you can process it in
contiguous blocks, no which required.
-Neal
You might
Hi,
May be you can try:
###Use dput()
dat1 - structure(list(trialno = c(11301L, 11301L, 11301L, 11301L, 11301L,
11301L, 11301L, 11301L, 11301L, 11301L, 11302L, 11302L, 11302L,
11302L, 11302L, 11302L, 11302L, 11302L, 11302L, 11302L), event = c(pm_intake,
am_intake, pk1, pm_intake, am_intake,
.cmicalc is a non-exported function. You can see the code with:
getAnywhere(.cmicalc)
Best,
Wolfgang
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of petre...@unina.it
Sent: Thursday, November 21, 2013 13:39
To:
Hi,
On Nov 21, 2013, at 10:42 AM, MacQueen, Don macque...@llnl.gov wrote:
I have some processes where I do the same thing, iterate over subsets of a
data frame.
My data frame has ~250,000 rows, 30 variables, and the subsets are such
that there are about 6000 of them.
Performing a which()
I have some processes where I do the same thing, iterate over subsets of a
data frame.
My data frame has ~250,000 rows, 30 variables, and the subsets are such
that there are about 6000 of them.
Performing a which() statement like yours seems quite fast.
For example, wrapping unix.time() around
Hi,
May be you can try:
###Use dput()
dat1 - structure(list(trialno = c(11301L, 11301L, 11301L, 11301L, 11301L,
11301L, 11301L, 11301L, 11301L, 11301L, 11302L, 11302L, 11302L,
11302L, 11302L, 11302L, 11302L, 11302L, 11302L, 11302L), event = c(pm_intake,
am_intake, pk1, pm_intake, am_intake,
Dear all,
This question was asked a few years ago. Back then, the answer was NO. Just
wonder if the package has been updated to make it possible.
An example (Theoph is a dataset coming with R)
ggplot(data=Theoph, aes(x=Time, y=conc)) + geom_point() +
facet_wrap(~Subject)
gives me all 12
What the Data Munger Guru said.
Plus: this is almost certainly a job for ddply or data.table.
Noah Silverman-2 wrote
Hello,
I have a fairly large data.frame. (About 150,000 rows of 100
variables.) There are case IDs, and multiple entries for each ID, with a
date stamp. (i.e. records of
Hi Don,
Yes, I am error checking a dataset produced by a query. Most likely a problem
with the query but wanted to assess the problem first.
BTW Arun provided another solution which is similar to yours but uses the
function ave instead:
testSeq[!!(with(testSeq,ave(YoS,ID,FUN=function(x)
Hi,
Try:
dat1 - read.table(text=a b c d e
1 2 3 4 5
10 9 8 7 6,sep=,header=TRUE)
Names1- read.table(text=Original New
e ee
b bb
a aa
c cc
d dd,sep=,header=TRUE,stringsAsFactors=FALSE)
It is better to dput()
I'd recommend something along these lines:
set.seed(32438786)
plot.new()
plot.window(xlim=c(-.5,.5),ylim=c(1,50))
for(i in 1:50){
X - rnorm(100)
CI - confint(lm(X~1))
ifelse( sum(0 CI) == 2 | sum(0CI)==2 , line.col - red, line.col -
black )
lines(CI,c(i,i),lwd=2, col=line.col)
}
box()
Hi,
Try:
#Generating the files
fileList1 - paste0(rep(LETTERS[1:5],each=3),1:3,.txt)
set.seed(48)
lapply(fileList1,function(x) {m1 -
matrix(sample(1:20,1686*2,replace=TRUE),nrow=1686,ncol=2);
write.table(m1,paste0(/home/arunksa111/Trial6/IR/,x),row.names=FALSE,quote=FALSE)})
dir()
#[1]
Not sure this helps but...
##
# data frame with 30,000 ID's, each with 5 dates, plus some random data...
df - data.frame(id=rep(1:3, each=5),
date=rep(1:5, each=3),
x=rnorm(15), y=rnorm(15,
Maybe something like this, assuming mean=0:
samsize - 100
replicates - 50
pval - .05
samples - replicate(replicates, rnorm(samsize))
confint - t(apply(samples, 2, function(x)
c(mean(x)-qt(1-pval/2,
df=samsize-1)*sd(x)/sqrt(samsize),
mean(x)+qt(1-pval/2,
hi guys
Assume I have this dataframe:
v3$number_of_ones
[1] 111 106 117 108 120 108 108 116 116 113
Is there any command in r that gives me the frequency of these numbers (how
many each number is repeated e.g. the number 108 repeated 2 and 111 repeated
one an so on)
I have around 10^6
Dear R users: before I try to undertake my own rewrite, has anyone already
written a contour function that does not require a regularly spaced grid,
preferably plot.contour(x,y,z,...)?
(it would be nice if it were based on loess() and plot()? further
annotations, changes, etc., would then be
When I use run this expression
a - 2; b - 3; xyplot(1:10 ~ a*(1:10), sub = c(bquote(a == .(a) ~ b==.(b
the subtitle contains three copies of the a = 2 b = 3 phrase.
Why does it do that? How do I tell it to give me only one copy?
Rich
__
a - 2; b - 3; xyplot(1:10 ~ a*(1:10), sub = c(bquote(a == .(a) ~ b==.(b
the subtitle contains three copies of the a = 2 b = 3 phrase.
Why does it do that? How do I tell it to give me only one copy?
To avoid it don't wrap bquote() with c(). The following does what you asked
for:
a
On 11/22/13 18:47, William Dunlap wrote:
a - 2; b - 3; xyplot(1:10 ~ a*(1:10), sub = c(bquote(a == .(a) ~ b==.(b
the subtitle contains three copies of the a = 2 b = 3 phrase.
Why does it do that? How do I tell it to give me only one copy?
To avoid it don't wrap bquote() with c(). The
I don't think anything has changed in the fundamental design of ggplot2, so the
answer is still no.
But you can use some kind of loop (for, or *apply) to generate them
sequentially. What do you plan to do with them? Save as jpeg? Append into a
pdf? Embed in an Sweave or knitr file?
66 matches
Mail list logo
