Hello,
I tried to use mle to fit a distribution(zero-inflated negbin for
count data). My call is very simple:
mle(ll)
ll() takes the three parameters, I'd like to be estimated (size, mu
and prob). But within the ll() function I have to judge if the current
parameter-set gives a nice fit or
1. You need R version = R-2.12.0 on Windows in order to use a package
that is compiled under R = 2.12.0.
2. winbuilder gave some reports. That may have said if and why your
package could not be compiled under 64-bit R for some reason.
3. If you have a 32-bit-only package, you can only load it
On Mon, Jan 31, 2011 at 01:35:14PM -0800, Shishir Vor wrote:
Hi,
I'm a beginner with R.
I have two different? tables with
the variable name dmr1 and tp2 a given following. v1 is the common? column?
field of
the both tables. In the first table column v3 is always v2+1 while in
the
On Mon, Jan 31, 2011 at 09:42:27PM +0100, moleps wrote:
Dear all,
Given
rr-data.frame(r1-rnorm(1000,10,5),r2-rnorm(1000,220,5))
Hello.
There is already an answer to your question. However, i think
that the above command works in a different way than you
expect. The embedded assignments
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi
I want to increase the size of the text for xlab in the plot command:
plot(1,1, xlab=Label)
I tried cex, cex.caption and others, but none worked. Is there a way of
using cex=2 for the xlab?
Thanks,
Rainer
- --
Rainer M. Krug, PhD
Hello list.
For some reason, the makers of glmnet do not accept a dataframe as input.
They expect the input to be a matrix, where the dummies are already
precoded.
Now I have created a sample dataset with
. 11 factor columns with two levels
. 4 factor columns with three levels
. 135 continuous
Good morning,
I have an excel spreadsheet with similarities among objects. The format of
the file is the following:
1st row: empty cell,object-1-name,object-name-2,...,object-N
2nd row: object-name-1,0,s1,2,s1,3,...,s1,N
3rd row: object-name-2,s2,1,0,s2,3,...,s2,N
The table is symmetrical and
Hello,
I have some data where a number of events (the total amount varies)
occur at cumulating times, I would like to create a scatterplot
(easily achieved using plot etc) of these events (the events can
either be times using poxist or I can convert them into just seconds
which is probably
Hi all, suppose I have following user defined function:
fn1 = function(x, y, ...) {
z - x+y; u=y^2
# if something with name add exists in the
function argument then do some calculation and return that calculated value,
NOT z
Hi list,
I would like to estimate and forecast the seasonal component of a series. My
model which uses daily data would be something y t = alpha + beta x SeasComp
t + gamma x OtherRegressors t.
One approach to this would be use quarterly dummies, another to use a sine
function. The first would
Try cex.lab:
plot(1,1, xlab=Label, cex.lab = 2)
On Tue, Feb 1, 2011 at 7:31 AM, Rainer M Krug r.m.k...@gmail.com wrote:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi
I want to increase the size of the text for xlab in the plot command:
plot(1,1, xlab=Label)
I tried cex,
Try this:
sum(a * c(100, 10, 1)) + sum(b * c(10, 1))
On Tue, Feb 1, 2011 at 2:22 AM, ADias diasan...@gmail.com wrote:
Hi
I am trying to create a function that is able to calculate this sum:
a-c(2,3,5)
b-(8,7)
with a meaning 235 and b 87. So the result of this sum would be 235 +
87
=
Hi Jeff,
I am not sure what you want to do, but the 'graph' and 'igraph'
packages are for network, graph theory, i.e.
http://en.wikipedia.org/wiki/Graph_theory
If you think that this is the kind of tool you need, look over the
manual of the packages, and the documentation at the igraph homepage
NS == Nick Sabbe nick.sa...@ugent.be
on Tue, 1 Feb 2011 10:46:01 +0100 writes:
NS Hello list.
NS For some reason, the makers of glmnet do not accept a dataframe as
input.
NS They expect the input to be a matrix, where the dummies are already
NS precoded.
NS Now I have
On 11-01-31 6:51 AM, Sascha Vieweg wrote:
When Sweave outputs function code that spreads across many lines,
the default indent of inner lines is 4 spaces (plus the prompt).
How can I change that default to 2 spaces? I tried to adjust my
Sweave.sty with the option tabsize but that doesn't show an
Hi
Im just wondering if anyone knows if Rserve is parallel programed? and if so
how, do it send every new connection to a new node or?
Thx for the help
Joel
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On 31.01.2011 23:23, muchad...@gmail.com wrote:
Is there a limit to how many nested function calls R will tolerate? I mean,
if I have a function that calls another function that calls another
function, etc., is there a limit to how deep I can go?
I am getting an error that says a variable is
Can add that if it dossent split up the processes by default is there anyway
I can do it by myself?
//Joel
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Hello,
I have a data frame with one column set as the date with function as.Date.
In example:
data.frame( Date = c(2009-09-01,2009-09-02,2009-09-03,2009-09-04,
2009-09-05), Data = c( 10:14 ) )
Date Data
2009-09-01 10
2009-09-02 11
2009-09-03 12
2009-09-04 13
Is there a way to
Try this:
DF - data.frame( Date =
c(2009-09-01,2009-09-02,2009-09-03,2009-09-04,
2009-09-05), Data = c( 10:14 ) )
subset(DF, as.Date(Date) = '2009-09-02' as.Date(Date) = '2009-09-04')
On Tue, Feb 1, 2011 at 11:09 AM, Usuario R r.user.sp...@gmail.com wrote:
Hello,
I have a data frame with
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 02/01/2011 12:12 PM, Henrique Dallazuanna wrote:
Try cex.lab:
plot(1,1, xlab=Label, cex.lab = 2)
Thanks - that is what I was looking for.
Cheers,
Rainer
On Tue, Feb 1, 2011 at 7:31 AM, Rainer M Krug r.m.k...@gmail.com
Thanks to all for your answers!
The ppt file should be here: http://bit.ly/gi15gh
I also uploaded some model specifications and the results I want to replicate +
the random dataset (long format) I'm using.
The zip file is here:
Hello everybody.
I have this object
procedure property sensor_data sensor_date
1S_10 nord626821.0 2002-09-30T00:00:00+0200
2 S_10 nord626821.0 2002-12-05T00:00:00+0100
3S_10 nord626821.1 2008-07-31T00:00:00+0200
4 S_1000 nord
On Tue, Feb 01, 2011 at 10:51:12PM +1300, Surrey Jackson wrote:
Hello,
I have some data where a number of events (the total amount varies)
occur at cumulating times, I would like to create a scatterplot
(easily achieved using plot etc) of these events (the events can
either be times using
Hi,
I am doing program that takes samples y times from listMV and saves the
result to list MVdata. The problem is that I need sample mean or standard
deviation for each sample (times y) and for all samples together. How can I
do that? Mean() and sd() won´t work.
Thanks allready,
Titta
Rainer,
Take a look at this thread:
http://www.mail-archive.com/r-help@r-project.org/msg64936.html
Jeremy
Jeremy Hetzel
Boston University
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Hi Christofer,
just have a look at how other functions deal with this, e.g.
plot
function (x, y, ...)
{
#*snip*
hasylab - function(...) !all(is.na(pmatch(names(list(...)),
ylab)))
#*snip*
got it?
cheers
Am 01.02.2011 11:29, schrieb Bogaso Christofer:
Hi all, suppose I have
Try this:
vapply(replicate(y, sample(listMV, 5), simplify = FALSE),
function(x)c(mean(x), sd(x)), c(Mean = 0, Sd = 0))
On Tue, Feb 1, 2011 at 7:56 AM, Titta peltok...@gmail.com wrote:
Hi,
I am doing program that takes samples y times from listMV and saves the
result to list MVdata. The
Hi Im trying to make a package in R just to learn how it works.
One thing I dont manage to do is to make a hidden environment or variable so
to say.
I want a hash list that I create my useing:
TestEnv-new.env()
assign(Key1,Ans1,envir=TestEnv)
assign(Key2,Ans2,envir=TestEnv)
Then I just have
During R CMD check I get the following note:
* checking data for non-ASCII characters ... NOTE
Note: found 9 marked UTF-8 string(s)
How can I search my code files for non ASCII chars using R?
Thanks,
Mark
--
Mark Heckmann (Dipl. Wirt.-Ing.)
phone +49 (0) 421/1614618
Empfehlen Sie GMX DSL
Hello,
I'm trying to modify my r-script to use RODBC instead of DBI/RMySQL (no more
ready-to-use package for windows).
I would like to copy a data.frame of 44 columns to a table of 45 columns
(the 45th is an autoincremental column).
With the following commands,
colnames(df)- a vector with
Hi,
I have a seemingly easy question that has been keeping be busy for quite a
while. The problem is the following:
0.1 + 0.1 + 0.1 == 0.3
[1] FALSE
Why is this false? Another example is
0.2 + 0.1 == 0.3
[1] FALSE
or
0.25 + 0.05 == 0.20 + 0.10
[1] FALSE
However, I do get TRUE if I use
I have a factor vector of subject races (Asian, Black, Hispanic, White; n=30)
that I want to plot with a Cleveland dotplot or dotchart.
I tried the following in R2.12.1 :
dotchart(table(school$Race))
Error in plot.xy(xy.coords(x, y), type = type, ...) : invalid plot type
Using the same data
I'm simulating a Markov process using a vector of proportions. Each
value in the vector represents the proportion of the population who are
in a particular state (so the vector sums to 1). I have a square matrix
of transition probabilities, and for each tick of the Markov clock the
vector is
Hi,
Its a bit messy, but here is a quick work around:
dotchart(as.matrix(table(infert$education))[,1])
I wonder if this has something to do with the addition of table
methods for points() and lines(). I'll post back if I find anything
more out.
Cheers,
Josh
On Tue, Feb 1, 2011 at 6:52 AM,
My question, buried in this rant, is is there a mail list
or other means for identifying sites with information likely
to be important to many R users but the data is difficult to obtain
due to the site's choice of technology?
Quite often, people here ask questions about scraping html
to get
Hello,
I'm doing the following:
library(ncdf)
library(fields)
library(animation)
saline - open.ncdf(salinity_1990.nc)
salt = get.var.ncdf(nc=saline, varid=Salinity)
# create an animation of the complete temporal domain in the ncdf file.
saveHTML({
for (i in 1:364) {
Hi Michael,
this is R-FAQ 7.31
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
cheers.
Am 01.02.2011 14:49, schrieb mlancee:
Hi,
I have a seemingly easy question that has been keeping be busy for quite a
while. The problem is the
Hi,
If you only want the final matrix, i.e. in this case the pm at 10
months, then you might be better off looking at something like the
square-and-multiply algorithm
(http://en.wikipedia.org/wiki/Exponentiation_by_squaring) rather than a
brute force multiplication.
Martyn
-Original
Hi:
Try this:
d - read.table(textConnection(
+ procedure property sensor_data sensor_date
+ 1S_10 nord626821.0 2002-09-30T00:00:00+0200
+ 2 S_10 nord626821.0 2002-12-05T00:00:00+0100
+ 3S_10 nord626821.1 2008-07-31T00:00:00+0200
+ 4
See 'Writing R Extensions' ...
'Function showNonASCII in package tools can help in finding non-ASCII
bytes in files.'
On Tue, 1 Feb 2011, Mark Heckmann wrote:
During R CMD check I get the following note:
* checking data for non-ASCII characters ... NOTE
Note: found 9 marked UTF-8
Hi All,
I'm trying to parallelize some code using Rmpi and I've started with a
sample 'hello world' program that's available at
http://math.acadiau.ca/ACMMaC/Rmpi/sample.html. The code is as
follows;
# Load the R MPI package if it is not already loaded.
if (!is.loaded(mpi_initialize)) {
if you have a homogeneous mc (= a constant transition matrix), your
state at time 10 is given by (chapman-kolmogorov)
p10=p0 %*% tm^(10)
so you need a matrix power function.
You can use the eigendecomposition and some linear algebra
A^n=(VDV^{-1})^n)=VD^nV^{-1}
dd-eigen(tm,symmetric=F)
Thanks Eik for your help. So far I am able to do following things:
fn1 = function(x, y, ...) {
+ x1 - x+y
+ if(is.null(names(list(...
+ res - x1
+ else
+ res - get(names(list(...)))
+ return(res)
+
+ } ;
fn1(-2,2)
[1] 0
fn1(-2,2, zz=3:5)
Error in
Hi:
Not sure, but perhaps you had something like this in mind:
y - runif(100)
w - matrix(y, nrow = 10) # fills in column-wise
v - apply(w, 2, cumsum)
plot(1:100, y, type = 'l', col = 'blue')
lines(1:100, as.vector(v), type = 'l', col = 'red')
Reshaping into a matrix and using apply() to
Eik Vettorazzi E.Vettorazzi at uke.uni-hamburg.de writes:
if you have a homogeneous mc (= a constant transition matrix), your
state at time 10 is given by (chapman-kolmogorov)
p10=p0 %*% tm^(10)
so you need a matrix power function.
There are matrix exponential functions in the Matrix and
almost there
fn1 - function(x, y, ...) {
x1 - x+y
if(is.null(names(list(...
res - x1
else {
res - list(...)
#further operations on res$aa res$zz, e.g.
#res-res$zz+res$aa
#or
#for (a in res) {
#do something
#}
}
res
}
fn1(-2,2)
fn1(-2,2, zz=3:5,aa=6:8)
see the
Dear R-users,
I'm running a lmer model using the lme4 package. My dependent variable is
dichotomous and I'm using the binomial family. The results
are slightly different from the HLM results based on a Bernoulli
distribution. I read that a Bernoulli distribution is an extension of a
binomial
Hello thank you for so much input. I am afraid that I am fairly new to this
and some of the material is above my understanding for the time being and I
am not looking at anything very complex as you will see from the program I
will include.
I have been talking to a fellow colleague I am working
I wanna to do a Regression type 2 or Regression with X measured with
erroranybody knows how can i make it in R??
thanks!
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Thanks a lot!
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__
R-help@r-project.org mailing list
yes there are. But Christofer doesn't need exp(A), but A^n.
But there is a matpow-function %^% in this package, which is a little
bit slower, I think:
library(expm)
states-1000
tm - matrix(runif(states^2),nrow=states) # random transition matrix
for illustration
tm - t(apply(tm,1,function (x)
I need some help in defining a print method for my new S4 class
definition. So fer I have worked like this:
setClass(MyClass, sealed=F, representation(slot1 = list,#create a
new class
slot2= vector,
slot3 = vector,
slot4 = vector))
On Tue, Feb 01, 2011 at 05:49:10AM -0800, mlancee wrote:
Hi,
I have a seemingly easy question that has been keeping be busy for quite a
while. The problem is the following:
0.1 + 0.1 + 0.1 == 0.3
[1] FALSE
Why is this false? Another example is
0.2 + 0.1 == 0.3
[1] FALSE
or
Hi,
this is R FAQ 7.31.
http://cran.r-project.org/doc/FAQ/R-FAQ.html
HTH,
Stephan
Am 01.02.2011 14:49, schrieb mlancee:
Hi,
I have a seemingly easy question that has been keeping be busy for quite a
while. The problem is the following:
0.1 + 0.1 + 0.1 == 0.3
[1] FALSE
Why is this false?
Hello,
I am trying to fit my Elisa results (absorbance readings) to a standard
curve. To create the standard curve model, I performed a 4-parameter
logistic fit using the 'drc' package (ExpectedConc~Absorbance). This gave me
the following:
FourP
A 'drc' model.
Call:
drm(formula = Response ~
On 2011-02-01 07:09, Joshua Wiley wrote:
Hi,
Its a bit messy, but here is a quick work around:
dotchart(as.matrix(table(infert$education))[,1])
I wonder if this has something to do with the addition of table
methods for points() and lines(). I'll post back if I find anything
more out.
Will
EV == Eik Vettorazzi e.vettora...@uke.uni-hamburg.de
on Tue, 1 Feb 2011 18:02:42 +0100 writes:
EV yes there are. But Christofer doesn't need exp(A), but
EV A^n.
EV But there is a matpow-function %^% in this package, which is a little
EV bit slower, I think:
EV
On Feb 1, 2011, at 10:41 AM, misil wrote:
I wanna to do a Regression type 2 or Regression with X measured with
erroranybody knows how can i make it in R??
thanks!
There are several prior posts in the archives ... although perhaps not
the Nabble archives since they start throwing away
Hi:
Here are a few ways to do this.
A useful approach is to use replicate() to generate the samples, flatten the
resulting matrix into a data frame and call one or more packages that are
well capable of handling multiple outputs per data subset.
Step 1: Generate the samples and rearrange into a
On Feb 1, 2011, at 11:08 AM, Christopher Anderson wrote:
Hello,
I am trying to fit my Elisa results (absorbance readings) to a
standard
curve. To create the standard curve model, I performed a 4-parameter
logistic fit using the 'drc' package (ExpectedConc~Absorbance). This
gave me
the
Dear R guru:
If I got a variable
aaa- up.6.11(16)
how can I extract 16 out of the bracket?
I could use substr, e.g.
substr(aaa, start=1, stop=2)
[1] up
But it needs start and stop, what if my start or stop is not fixed, I
just want the number inside the bracket, how can I achieve
On 02/01/2011 07:31 AM, Megh Dal wrote:
I need some help in defining a print method for my new S4 class
definition. So fer I have worked like this:
setClass(MyClass, sealed=F, representation(slot1 = list,#create a
new class
slot2= vector,
slot3 =
Try this:
gsub(.*\\((\\d+)\\).*, \\1, aaa)
On Tue, Feb 1, 2011 at 3:42 PM, Yan Jiao y.j...@ucl.ac.uk wrote:
Dear R guru:
If I got a variable
aaa- up.6.11(16)
how can I extract 16 out of the bracket?
I could use substr, e.g.
substr(aaa, start=1, stop=2)
[1] up
But it needs
Yan -
Here's one way. It assumes there's exactly one set of
brackets in the string, and they can be anywhere:
aaa- up.6.11(16)
sub('^.*?\\((.*)\\).*$','\\1',aaa)
[1] 16
- Phil Spector
Statistical Computing
Date: Tue, 1 Feb 2011 15:41:01 +
From: alex.sm...@gmail.com
To: spencer.gra...@structuremonitoring.com
CC: r-help@r-project.org; maech...@stat.math.ethz.ch
Subject: Re: [R] Positive Definite Matrix
Hello thank you for so much input. I am afraid that I am fairly new to this
and some of
On Tue, Feb 1, 2011 at 12:42 PM, Yan Jiao y.j...@ucl.ac.uk wrote:
Dear R guru:
If I got a variable
aaa- up.6.11(16)
how can I extract 16 out of the bracket?
I could use substr, e.g.
substr(aaa, start=1, stop=2)
[1] up
But it needs start and stop, what if my start or stop is
I am new to R. So I posted the program I wrote so I can get help on where it
is going wrong and if I could approach it using the Sylvesters criterion and
why it comes up with an error when I use the eigenvalues. The matrices I
provided are generated randomly the purpose of my question is not any
see inline below.
On Tue, Feb 1, 2011 at 2:25 PM, Mike Marchywka marchy...@hotmail.com wrote:
Date: Tue, 1 Feb 2011 15:41:01 +
From: alex.sm...@gmail.com
To: spencer.gra...@structuremonitoring.com
CC: r-help@r-project.org; maech...@stat.math.ethz.ch
Subject: Re: [R] Positive Definite
On Tue, Feb 1, 2011 at 10:51 AM, Luana Marotta lucsmaro...@gmail.com wrote:
Dear R-users,
I'm running a lmer model using the lme4 package. My dependent variable is
dichotomous and I'm using the binomial family. The results
are slightly different from the HLM results based on a Bernoulli
R-help readers:
A new e-book entitled 25 Recipes for Getting Started with R is now
available from O'Reilly Media:
http://oreilly.com/catalog/9781449303235
This collection of how-to recipes will get new R users up-and-running
quickly with basic statistics, graphics, and regression. The
May I ask a clinical question? For a trial, we have a treatment group of small
size, say 30 patients. We want to selectmatching control patients from a bigger
group (100 patients) in terms of several clinical variables, such as age, tumor
stage etc. This practice is to select the closest
Dear R,
Can't I cluster a dataset into k clusters where k is exactly the number of
observations? I have version 12.2 installed. See this example
a - matrix(1:100, 20)
kmeans(a, 20)
Error: number of cluster centres must lie between 1 and nrow(x)
This is a bit ad-hoc but I known R from version
On Tue, Feb 1, 2011 at 9:23 AM, Peter Ehlers ehl...@ucalgary.ca wrote:
Will this suffice:
dotchart(c(table(infert$education)))
That is probably the cleanest answer for now (I tried something like
that with as.vector(), but lost the names). Another reasonable option
would be:
My terminology is probably way off. I´ll try again in plain english.
I´d like to generate a scatter plot of r1 r2 and color code each pair
according to the probability of observing the pair given that the two samples
(r1 r2) are drawn from two independent normal distributions.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of moleps
Sent: Tuesday, February 01, 2011 11:32 AM
To: Peter Ehlers
Cc: r-help@r-project.org
Subject: Re: [R] p value for joint probability
My terminology is probably way
Date: Tue, 1 Feb 2011 12:10:31 -0800
From: nord...@dshs.wa.gov
To: r-help@r-project.org
Subject: Re: [R] p value for joint probability
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of moleps
Sent: Tuesday,
Look at the optmatch package.
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
email: rvarad...@jhmi.edu
- Original
On Feb 1, 2011, at 2:31 PM, moleps wrote:
My terminology is probably way off. I´ll try again in plain english.
I´d like to generate a scatter plot of r1 r2 and color code each
pair according to the probability of observing the pair given that
the two samples (r1 r2) are drawn from two
On Tue, Feb 01, 2011 at 08:31:57PM +0100, moleps wrote:
My terminology is probably way off. I´ll try again in plain english.
I´d like to generate a scatter plot of r1 r2 and color code each pair
according to the probability of observing the pair given that the two samples
(r1 r2) are
Another search term is geometric mean regression.
For simple models you can try the lmodel2 package.
Michael
On 2 February 2011 04:31, David Winsemius dwinsem...@comcast.net wrote:
On Feb 1, 2011, at 10:41 AM, misil wrote:
I wanna to do a Regression type 2 or Regression with X measured
Allright.. Appreciate the input on non-zero terminology (:-). What I wanted was:
rr-data.frame(r1=rnorm(1000,10,5),r2=rnorm(1000,220,5))
with(rr,plot(r1,r2))
r3-kde2d(r1,r2,lims=c(2,18,200,240))
filled.contour(r3)
//M
On 1. feb. 2011, at 21.26, David Winsemius wrote:
On Feb 1, 2011, at
Hello
I am trying to find a way to find the max value, for only a subset of a
dataframe, depending on how the data is grouped for example,
How would I find the maxmium responce, for all the GPR119a condition below:
responce,mouce,condition
0.105902,KO,con
0.232018561,KO,con
Hello Chris,
You may also use the R-package calib.
Hugo
On Tuesday 01 February 2011 17:08:13 Christopher Anderson wrote:
Hello,
I am trying to fit my Elisa results (absorbance readings) to a standard
curve. To create the standard curve model, I performed a 4-parameter
logistic fit
Christopher Anderson recsa at channing.harvard.edu writes:
Hello,
I am trying to fit my Elisa results (absorbance readings) to a standard
curve. To create the standard curve model, I performed a 4-parameter
logistic fit using the 'drc' package (ExpectedConc~Absorbance). This gave me
the
Would the following work?
legend(210, 110, bquote(r2 ==
.(format(summary(regression)$adj.r.squared,digits=3
See ?plotmath and ?bquote
Jeremy
Jeremy Hetzel
Boston University
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Correction, I forgot the caret:
legend(210, 110, bquote(r^2 ==
.(format(summary(regression)$adj.r.squared,digits=3
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Hello everybody,
is there a way to tell Sweave to skip the R code and process only the
LaTeX part?
I know it sounds a little bit weird, but the R code of my document is
complete and _slow_ and I need to add the LaTeX part.
Thanks,
franZ
__
Hi there,
How can I superscript the 2 of r2 =... in the legend below?
legend(210, 110, paste(r2 = ,
format(summary(regression)$adj.r.squared,digits=3), sep=))
I usually use expression(paste(..., but it won't work this time because
format(summary(... needs to be evaluated.
Thanks in
PLEASE HELP
I actually want to do the following:
a[j] = (1/(j!))*Î (i-1-d), j = 500, Î means product i = 1 to
j
Â
Yet, j! will stop at 170 and Î (i-1-d) at 172; so, a[j] will
not exceed 170.
I would like to have at least 200 a[j].
Â
WHAT SHOULD I DO?
Â
PLEASE SEE MY CODE FOR
Dear all, I have having a strange problem with upper.tri()/lower.tri()
functions which I used to use to make a matrix symmetric. However for a
specific large matrix, they seem not working. Initial I had following matrix:
MyMatrix
[,1] [,2] [,3] [,4]
I have this function and want to run it parallel with different sets of data.
Using SNOW and clusterApplyLB.
system.time(out - mclapply(cData, plotGraph)) #each cData contains 100X6000
doubles
system.time(out - mclapply(cData2, plotGraph))
system.time(out - mclapply(cData3, plotGraph))
Thanks, jthetzel. That works. Can you plot a legend with 2 lines using
bquote? e.g.
r^2 = x
rmse = y
, and \n don't seem to work.
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?tapply would be one way to get the max for all
conditions at once.
Your example is not reproducible, so I cannot give you a
reproducible answer, but adapt the following:
tapply(df$responce, df$condition, max)
A. Ramzan wrote:
Hello
I am trying to find a way to find the max value, for only
max(subset(dataName, condition==GPR119a)$responce)
У Аўт, 01/02/2011 у 19:05 +, A. Ramzan піша:
Hello
I am trying to find a way to find the max value, for only a subset of a
dataframe, depending on how the data is grouped for example,
How would I find the maxmium responce, for all the
I have a vector of unique elements that I want to replicate a variable number
of times depending on the element (lengths all 800). However I noticed
that the resulting length was not the sum of the each argument. The
following example demonstrates this.
I am confused as to why this works:
If your data frame is called myDF,
max(myDF[myDF[, condition] == GPR119a, responce])
Original message
Date: 01 Feb 2011 19:05:46 +
From: r-help-boun...@r-project.org (on behalf of A. Ramzan ar...@cam.ac.uk)
Subject: [R] (no subject)
To: r-help@r-project.org
Hello
I am trying to
max(dat[dat$condition==GPR119a, responce])
aggregate(dat$responce, by=list(dat$condition), max)
see ?[ and ?aggregate
-Ista
On Tue, Feb 1, 2011 at 2:05 PM, A. Ramzan ar...@cam.ac.uk wrote:
Hello
I am trying to find a way to find the max value, for only a subset of a
dataframe, depending on
Here is a paper that gives an review of matching
Elizabeth A. Stuart. (2010) Matching Methods for Causal Inference: A
Review and a Look Forward. Statistical Science; 25(1):1-21.
In the paper, there is a list of R packages that does matching.
hope this helps.
On Tue, Feb 1, 2011 at 2:01 PM,
On Feb 1, 2011, at 3:51 PM, Kiogou Lydie wrote:
I actually want to do the following:
a[j] = (1/(j!))*Π (i-1-d), j = 500, Π means product i = 1 to j
What is the purpose of this effort?
Yet, j! will stop at 170 and Π (i-1-d) at 172; so, a[j] will
not exceed 170.
I would like to have at
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