I tried sprintf and paste, which solves one of my problem i.e.
replacing the variable name by its value. However, >= is not renderend
correctly as the inequality symbol.
On Tue, Mar 24, 2009 at 12:24 PM, Gabor Grothendieck
wrote:
> For the legend argument try this:
>
> leg = as.expression(c(sprin
I'm not sure what you meant by "a topic on newton's method"
(algorithm? demo?), but the demonstration in the package 'animation'
might help:
install.packages('animation')
par(pch = 20)
ani.options(nmax = 50)
newton.method(function(x) 5 * x^3 - 7 * x^2 - 40 *
x + 100, 7.15, c(-6.2, 7.1))
Regar
For the legend argument try this:
leg = as.expression(c(sprintf("Up (>= %d)", threshold),
"Normal", "Down", "NA"))
or use paste in place of sprintf.
On Mon, Mar 23, 2009 at 11:53 PM, Daren Tan wrote:
> I need to have the maths symbol for >= in the legend, and to
> substitute threshold v
the plotmath help page should give the answer
http://www.stat.ucl.ac.be/ISdidactique/Rhelp/library/base/html/plotmath.html
On Mar 24, 2:53 pm, Daren Tan wrote:
> I need to have the maths symbol for >= in the legend, and to
> substitute threshold variable with its value. Somehow, various
> attemp
Try
Wood S.N. (2006) Generalized Additive Models: An Introduction
with R. Chapman and Hall/CRC Press.
listed in the references in the help file of the function.
It's a great read.
Andrew
On Mon, Mar 23, 2009 at 07:36:44PM -0700, oliviax wrote:
>
> I am writing my thesis with the fun
I need to have the maths symbol for >= in the legend, and to
substitute threshold variable with its value. Somehow, various
attempts weren't successful. Please help.
threshold <- 0.5
plot(NA, xlab="", ylab="", main="", axes=F, xlim=c(0,1), ylim=c(0,1),
xaxs="i", yaxs="i")
legend(x=0, y=1, fill=c(
I am writing my thesis with the function gam(), with the package {mgcv}.
My command is: gam(y~s(x1,bs="cr")+s(x2, bs="cr")).
I need help to know what are the default basis funcitons for gam. I have not
found any detailed reference for this.
Can anyone help me with this??
--
View this messa
Try this:
library(plotrix)
plot(x_vals, y_vals)
abline(h=min(y_vals), col="darkred")
boxed.labels(max(x_vals) - strwidth(bottom_label)/2, min(y_vals) +
strheight(bottom_label), bottom_label)
See ?boxed.labels for more.
On Mon, Mar 23, 2009 at 11:01 PM, Jason Rupert wrote:
>
> The result of the
Hello all,
This is something that I am sure has a really suave solution in R, but I can't
quite figure out the best (or even a basic) way to do it.
I have a simple linear regression that is fit with lm for which I would like to
estimate the x intercept with some measure of error around it (conf
?rect
--
David Winsemius
On Mar 23, 2009, at 11:01 PM, Jason Rupert wrote:
The result of the code shown below is posted at the following URL:
http://n2.nabble.com/Trying-to-properly-label-abline-td2524629.html
I would like to figure out a better way to label the horizontal
abline.
I t
The result of the code shown below is posted at the following URL:
http://n2.nabble.com/Trying-to-properly-label-abline-td2524629.html
I would like to figure out a better way to label the horizontal abline.
I tried multiplying the text y position by a scale, but this didn't always put
the text
G'day Kevin,
On Mon, 23 Mar 2009 18:48:16 -0400
wrote:
> Sorry to be so dense but the article that you suggest does not give
> any information on how the arguments are packed up. I look at the
> call:
>
> val <- .Internal(fmin(function(arg) -f(arg, ...), lower, upper, tol))
>
> and then with t
Try this:
> library(chron)
> x <- c("06/25/04", "06/25/2004", "03/03/59", "03/03/1959")
> chron(x)
[1] 06/25/04 06/25/04 03/03/59 03/03/59
On Mon, Mar 23, 2009 at 8:06 PM, Farrel Buchinsky wrote:
> How does one convert to a date format when survey respondents have
> used two different date form
Try this (haven't checked the speed):
f <- function(x) table(factor(x, c(-1, 0, 1)))
100 * prop.table(t(apply(m, 1, f)), 1)
On Mon, Mar 23, 2009 at 8:46 PM, Daren Tan wrote:
> I have a matrix containing -1, 0, 1, however certain rows will not
> have all 3 numbers. I have written some codes to c
G'day Carl,
On Mon, 23 Mar 2009 20:11:19 -0400
Carl Witthoft wrote:
> >From: Wacek Kusnierczyk
> >Date: Sun, 22 Mar 2009 22:58:49 +0100
>
>
> >just for fun, you could do this with multiassignment, e.g., using
> >the (highly experimental and premature!) rvalues:
>
> >source('http://m
Dear Ivan,
On 3/23/09, pfc_ivan wrote:
>
> Hello guys, I am stuck here:
>
> How do I make 1000 samples of n = 10 observations from an Exponential
> distribution and then compute the mean for all those 1000 samples?
>
The R Commander will do this. See the menus Distributions ->
Continuous -> Exp
Hello,
Since this an R package, I'm sending the email. I'm loading all the
jar files as:
library(rJava)
hadoop <- Sys.getenv("HADOOP")
allfiles <-
c(list.files(hadoop,pattern="jar$",full.names=T),list.files(paste(hadoop,"lib",sep=.Platform$file.sep,collapse=""),pattern="jar$",full.names=T))
allfi
Tena koe Daren
One alternative:
apply(m, 1, function(x) 100*summary(factor(x,
levels=-1:1))/length(x[!is.na(x)]))
Doubtless there are others.
HTH
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Daren T
Greg Snow imail.org> writes:
>
> There is also the clt.examp function in the TeachingDemos package.
>
An interactive demo can also be found at http://www.math.csi.cuny.edu/gWidgetsWWW/ex-clt.html>http://www.math.csi.cuny.edu/gWidgetsWWW/ex-clt.html.
__
I have a matrix containing -1, 0, 1, however certain rows will not
have all 3 numbers. I have written some codes to compute the frequency
table of how many -1s, 0s, 1s per row, but it is very ugly and not
efficient if there are more than 3 numbers. Please suggest.
m <- rbind(sample(0:1, replace=T,
I would assume the expression in the 'if' is giving an NA as a result:
> if (NA) 1 else 2
Error in if (NA) 1 else 2 : missing value where TRUE/FALSE needed
>
On Mon, Mar 23, 2009 at 10:38 AM, Michael Curran wrote:
> Hi list,
>
> I want to try Gibbs sampling as a method of estimating a markov-sw
Try using 'strsplit' to split your string on the '/' and then create a
series of 'if's to determine how you want to output the new string.
You will probably need this approach since you may have to check the
validity and ranges of the numbers.
On Mon, Mar 23, 2009 at 8:06 PM, Farrel Buchinsky wro
>From: Wacek Kusnierczyk
>Date: Sun, 22 Mar 2009 22:58:49 +0100
>just for fun, you could do this with multiassignment, e.g., using the
>(highly experimental and premature!) rvalues:
>source('http://miscell.googlecode.com/svn/rvalues/rvalues.r')
>if (TRUE)
> c(df1, df2) := list(4
How does one convert to a date format when survey respondents have
used two different date formats whilst entering their data. There were
clearly told to use mm/dd/ but humans being humans some entered
mm/dd/yy. There was even validity checks on the forms but I allowed
them to be overridden sin
(this post suggests a patch to the sources, so i allow myself to divert
it to r-devel)
Bert Gunter wrote:
> x a numeric vector, matrix or data frame.
> y NULL (default) or a vector, matrix or data frame with compatible
> dimensions to x. The default is equivalent to y = x (but more efficient).
Satindra Chakravorty wrote:
I am trying to use the plot() function in R but can't seem to generate any
plots. I am running R version 2.7.1 on a server machine to which I am
remotely connected on a Linux platform.
I typed
plot(trees$volume)
as a test and got back the ">" prompt with no plot in
Tena koe Rodrigo
You could always make up some data and then show us what you have tried to do.
I would guess you need to check out:
plot # to do the basic plot
lines # to add lines to the plot
points # to add points to the plot
arrows # can be used to give the whiskers
apply # to get the means
I am trying to use the plot() function in R but can't seem to generate any
plots. I am running R version 2.7.1 on a server machine to which I am
remotely connected on a Linux platform.
I typed
> plot(trees$volume)
as a test and got back the ">" prompt with no plot in my current window or
in any o
Using sqldf you only need two statements, infile <- file(...) and
DF <- sqldf("select min(a), max(b), mean(c), ... from infile group by id").
The file statement identifies the filename and the second reads it
into sqlite (without
going through R), summarizes it and then reads the summarized version
Sorry to be so dense but the article that you suggest does not give any
information on how the arguments are packed up. I look at the call:
val <- .Internal(fmin(function(arg) -f(arg, ...), lower, upper, tol))
and then with the help of this article I find do_fmin in optimize.c:
SEXP attribute_h
Hi list members.
Ill try to plot the abundance means of nine transects as lines, with five
points on each transect (A to I). I will also need to plot for each point,
its standard deviation (once each point will have tree replicates) as
whiskers. Another problem will be that all the points should
On 23/03/2009 6:06 PM, Dennis Fisher wrote:
Colleagues,
R version 2.8.1 in OS X
Within a function (which is already within a function), I am sourcing
a file. The syntax of the command is something like (this is just an
example; the actual code is much more complicated):
This code works j
These comparisons are very simplistic. In most contexts, it would
make much better sense to measure "accuracy" in standard error units,
rather than in number of digits.
There doubtless are specialist applications where the 15th digit (or
even the 10th!) are important. But the check of acc
Dear CE.KA,
If 'x' is your data, you could do something like the following:
matplot(as.factor(c('1990','1992')),t(x[,-2]),type='l',
xaxt='n',xlab='Year',ylab="Some label here",lty=1)
axis(1,c(1990,"","","",1992),c(1990,"","","",1992))
legend('topleft',paste('Observation',1:5),col=1:5,lt
On Monday 23 March 2009, David Reiss wrote:
> I have a very large tab-delimited file, too big to store in memory via
> readLines() or read.delim(). Turns out I only need a few hundred of those
> lines to be read in. If it were not so large, I could read the entire file
> in and "grep" the lines I n
Colleagues,
R version 2.8.1 in OS X
Within a function (which is already within a function), I am sourcing
a file. The syntax of the command is something like (this is just an
example; the actual code is much more complicated):
BIGFUNCTION <- function()
{
DATAFRAME
- Mensaje reenviado
> De: Kenneth Roy Cabrera Torres
> Para: CE.KA
> Asunto: Re: [R] Graphic with several curves
> Fecha: Mon, 23 Mar 2009 17:02:48 -0500
>
> try matplot(),
>
> but you have to transpose the matrix
>
> with option type="l".
>
> Example:
>
> data1<-data.frame(
Hello,
I have a dataframe with 40 columns and around 450,000 rows. The first
column in each row is a factor id and the remaining are numeric. Some
rows have the same ids. What I want to do is to merge each set of rows
sharing the same ids (id set) into one single row (summarizing row)
wit
I have a very large tab-delimited file, too big to store in memory via
readLines() or read.delim(). Turns out I only need a few hundred of those
lines to be read in. If it were not so large, I could read the entire file
in and "grep" the lines I need. For such a large file; many calls to
read.delim
Hi R users,
Imagine the folowing data frame
19901991 1992
1 5 20 6
2 15 1 11
3 3 14 22
4 20 8 55
5 10 3 14
Is there a way to build a graphic in which:
- 1 c
Hi R users and developers:
Several hours ago I post a problem with the aggregate() function
using the last patch version of R.
(R version 2.8.1 Patched (2009-03-18 r48193))
I do compile again the 2.8.1 plain version (not patched) and
now it works again, both the examples and my old scripts that
On 21/03/2009, at 3:19 AM, Ravi Varadhan wrote:
I also tried a number of other things including changing the
"family", and parameters in
"loess.control", but to no avail. I looked at the Fortran codes
from both loess and gam.
They are daunting, to say the least. They are dense, a
On Mon, Mar 23, 2009 at 2:35 PM, Lawrence Hanser wrote:
> Dear Colleagues,
> I have what Roger Kirk (Experimental Design: Procedures for the Behavioral
> Sciences, 1968) refers to as a randomized block factorial design. The anova
> table would look like this:
>
> df
> A
On Mon, Mar 23, 2009 at 2:18 PM, Kingsford Jones
wrote:
> On Mon, Mar 23, 2009 at 11:26 AM, Ben Domingue wrote:
>> Hello,
>> How do I get the standard deviations for the random effects out of the
>> lme object? I feel like there's probably a simple way of doing this,
>> but I can't see it. Usin
There is also the clt.examp function in the TeachingDemos package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Beha
Dear Colleagues,
I have what Roger Kirk (Experimental Design: Procedures for the Behavioral
Sciences, 1968) refers to as a randomized block factorial design. The anova
table would look like this:
df
A 3
Subj/A 103 (error term for A)
B
On Mon, Mar 23, 2009 at 1:18 PM, Kingsford Jones
wrote:
> On Mon, Mar 23, 2009 at 11:26 AM, Ben Domingue wrote:
>> Hello,
>> How do I get the standard deviations for the random effects out of the
>> lme object? I feel like there's probably a simple way of doing this,
>> but I can't see it. Usin
On Mon, Mar 23, 2009 at 11:26 AM, Ben Domingue wrote:
> Hello,
> How do I get the standard deviations for the random effects out of the
> lme object? I feel like there's probably a simple way of doing this,
> but I can't see it. Using the first example from the documentation:
>
>> fm1 <- lme(dis
Ivan:
While you're figuring out how to execute the CLT in R you may find my
automated examples informative.
http://StatisticalEngineering.com/central_limit_theorem.htm
Charles Annis, P.E.
charles.an...@statisticalengineering.com
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngi
On Mon, Mar 23, 2009 at 11:04 AM, GRANT Lewis wrote:
[snip]
> factors<-matrix(runif(400),nrow=40)
> returns1<-matrix(runif(40),nrow=1)
> returns2<-matrix(runif(2000),nrow=50)
> coef(summary(lm(t(returns1)~factors)))[1,4]
[snip]
> (coef(summary(lm(t(returns2)~factors)))[50])[1,4]
>
> Error in `
Thanks. That worked.
On Mon, Mar 23, 2009 at 2:17 PM, Phil Spector wrote:
> Suppose your matrix is called "x". Then
>
>x[,seq(1,ncol(x),n)]
>
> will give you what you want.
>
> - Phil Spector
> Statistical Computin
Hii baptiste,
I tried you example, but I get still one plot and not as you said two plots
...
I use R for Windows, could this have something to do with this problem ?
baptiste auguie-2 wrote:
>
>
> On 23 Mar 2009, at 11:52, johnhj wrote:
>
>>
>> I have still the same problem... As you said
Phil and all,
Thank you for the responses. That method worked great!
However, I did try to replace the a few of the "Variables" using the same
method and received an error...
test_data2_df[test_data2_df$Location == 'HSV','Variables'] = "YardSize"
Is there something special I need to do in
pfc_ivan wrote:
I tried using the for (i..) to make 1000 differents sets of numbers, but then
I dont know how to get the mean value of all of them... because I dont even
think 1000 different sets of numbers were made, because when i print it it
always shows me the same values, basically I dont kn
(a <- replicate(5,rnorm(10)))
colMeans(a)
should get you started.
HTH,
baptiste
On 23 Mar 2009, at 18:29, pfc_ivan wrote:
I tried using the for (i..) to make 1000 differents sets of numbers,
but then
I dont know how to get the mean value of all of them... because I
dont even
think 1000
Dear Colleagues,
I have what Roger Kirk (Experimental Design: Procedures for the Behavioral
Sciences, 1968) refers to as a randomized block factorial design. The anova
table would look like this:
df
A 3
Subj/A 103 (error term for A)
B
I tried using the for (i..) to make 1000 differents sets of numbers, but then
I dont know how to get the mean value of all of them... because I dont even
think 1000 different sets of numbers were made, because when i print it it
always shows me the same values, basically I dont know how to replica
homework?
On Mon, Mar 23, 2009 at 1:30 PM, pfc_ivan wrote:
>
> Hello guys, I am stuck here:
>
> How do I make 1000 samples of n = 10 observations from an Exponential
> distribution and then compute the mean for all those 1000 samples?
>
> Basically I need to prove the Central Limit theorem, which
Hi,
I have a matrix and I want to create a matrix that includes every nth column
of the original matrix. Does anyone know how I can go about doing that ? Or
if you have an idea how I can do the same thing with a data frame ? If you
can just point me towards the appropriate function I would apprecia
On 23 Mar 2009, at 17:39, Jason Rupert wrote:
I would like to replace a few varaibles within a data frame.
For example, in the dataframe below (contrived) I would like to
replace the current housesize value only if the Location is HSV.
However, I would like to leave the other values int
Dear Jason,
Try this:
index<- with(test_data2_df,Location %in% "HSV")
test_data2_df$HouseSize<-ifelse(index,1000,test_data2_df$HouseSize) # I
used 1000 as reference
test_data2_df
HTH,
Jorge
On Mon, Mar 23, 2009 at 1:39 PM, Jason Rupert wrote:
>
> I would like to replace a few varaibles withi
I would like to replace a few varaibles within a data frame.
For example, in the dataframe below (contrived) I would like to replace the
current housesize value only if the Location is HSV. However, I would like to
leave the other values intact.
I tried "ifelse", but I don't really need
Hello guys, I am stuck here:
How do I make 1000 samples of n = 10 observations from an Exponential
distribution and then compute the mean for all those 1000 samples?
Basically I need to prove the Central Limit theorem, which states:
http://www.nabble.com/file/p22664113/d175f06cbf200bd52a2c27a2
Hello,
How do I get the standard deviations for the random effects out of the
lme object? I feel like there's probably a simple way of doing this,
but I can't see it. Using the first example from the documentation:
> fm1 <- lme(distance ~ age, data = Orthodont) # random is ~ age
> fm1
Linear mix
Hello,
I am trying to fit several Gaussian distributions to a data set. I
need to know the mean, standard deviation and maximum of each
Gaussian. My data set is simply a list of signal strengths and no
other information. I do not know the number of Guassians to fit. I've
looked at some packages li
You need only one loop,
year <- 1951:2000
filelist <- paste("C:\\Documents and Settings\\Data\
\table_",year,".txt", sep="")
filelist
for (ii in seq_along(year)) {
assign(paste("table_", year[ii], sep=""),
read.table(file=ifile[ii], header=TRUE,
sep
Hi
Probably a very basic question:
I am regressing a matrix of 50 response variables against a matrix of 10
factors using the lm function. This gives me an object with the output
for 50 regressions, as required. How do I now "use" the data?
For example, below is some code that generates sample
See:
https://stat.ethz.ch/pipermail/r-help/2009-March/192422.html
On Mon, Mar 23, 2009 at 12:52 PM, Steve Murray wrote:
>
> Dear all,
>
> I am trying to read in and assign data from 50 tables in an automated
> fashion. I have the following code, which I created with the help of
> textbooks and
Look at the symbols function or the subplot function in the TeachingDemos
package (or my.symbols in TeachingDemos).
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-bou
Dear all,
I am trying to read in and assign data from 50 tables in an automated fashion.
I have the following code, which I created with the help of textbooks and the
internet, but it only seems to read in the final data file over and over again.
For example, when I type:> table_1951 I get th
Inline Below.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
650-467-7374
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Wacek Kusnierczyk
Sent: Sunday, March 22, 2009 2:16 AM
To: rkevinbur...@charter.net
Cc: r-help@r-p
On Mon, Mar 23, 2009 at 7:38 AM, Lo_Lo wrote:
> I'm ploting graphics and I'd like to save them as a .jpeg file for example,
> but with a given size (in inches or cm).
>
> I tryed the function windows() but I think it just changes the size of the
> window and not the size of the graph that you're s
Look at the textplot function in the gplots package and the addtable2plot
function in the plotrix package. Either of those make it easy to 'plot' a
matrix.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> --
Hi R users and developers.
I compile the
R version 2.8.1 Patched (2009-03-18 r48193)
On my UBUNTU linux distribution.
But hen I ask for the aggregate example it fails.
What am I missing?
example(aggregate)
aggrgt> ## Compute the averages for the variables in 'state.x77',
grouped
aggrgt> ## ac
Does it work on a sliding window ? Does it estimate the cospectrum (the real
part) and the quadrature spectrum (complex), the coherence squared, and the
phase difference between two vector time series ?
SOme time ago I started to read its author's thesis. It seemed to me strictly
tailored on the
Hi,
This is really more a stats question than a R one, but
Does anyone have any familiarity with using the mean prediction
squared error scaled by the variance of the response, as a 'scale
free' criterion for evaluating different regression algorithms.
E.g.
Generate X_train, Y_train, X_test
Thanks very much to everyone. I think I will use a combination of both
techniques.
On 2009-March-22 , at 20:08 , Duncan Murdoch wrote:
That's pretty hard to make bulletproof. Why not just put those
functions in a package, and use that package?
I know it will be impossible to make bullet
have you tried sowas? I know you had talked about it, but it may do
what you want. I have used it for the wavelet cross spectrum.
On Mon, Mar 23, 2009 at 9:47 AM, wrote:
> Please, does anyone know of an R packge to estimate multidimensional spectral
> measure of coherence within a moving time
Hi:
I'm trying to estimate a model which involves the estimation of double
integrals, so I'm using adapt procedure. I need to calculate the
integrals trough my 2000 size database, so I do it using a loop. My
code correctly estimates the integral for the first row, but for the
second R crashes. I t
isn't there a width height argument in the jpeg function?
?jpeg
I am probably wrong,
Stephen Sefick
On Mon, Mar 23, 2009 at 10:38 AM, Lo_Lo wrote:
>
> Hi there !
>
> I'm ploting graphics and I'd like to save them as a .jpeg file for example,
> but with a given size (in inches or cm).
>
> I tryed
global wavelet spectrum? There is something somewhere - I just can't
remember where off the top of my head.
Stephen Sefick
On Sun, Mar 22, 2009 at 2:21 PM, stvienna wiener wrote:
> Dear list,
>
> in short: I would like to calculate the mean frequency
> of a signal (e.g. time series) using the w
Hi there !
I'm ploting graphics and I'd like to save them as a .jpeg file for example,
but with a given size (in inches or cm).
I tryed the function windows() but I think it just changes the size of the
window and not the size of the graph that you're saving.
Then I tryed with the function par
Hi,
Is this what you want?
#
mat <- matrix(c(65,4,22,24,6,81,5,8,0,11,85,19,4,7,3,90),4,4)
rowmarg <- rep(100, nrow(mat)) # the row margin totals that you want
colmarg <- c(90, 120, 80, 110) # the column margin totals that you want
newmat <- loglin( outer(rowmarg, colmarg) / sum(rowmarg), ma
On 23 Mar 2009, at 11:52, johnhj wrote:
I have still the same problem... As you said I tried with
par(mfrow=c(2,1))
and par(mfrow=c(1,2)) but without success. Could R compiler be the
problem ?
Why not? But may I suggest you try first the following example I sent
you yesterday,
pn
Please, does anyone know of an R packge to estimate multidimensional spectral
measure of coherence within a moving time window ?
Some time ago I expeimented with a similar package that performs Cross Spectrum
Analysis on the whole signal though.
Unluckily I deal with non-stationary signals whose
The data I used was just an example to work upon.
My real dataset is a confusion matrix of 24x24 (and 17x17), so coding it
into a model with all different kinds of combinations seems tedious.
That's why I hoped to use the ipf() function as it accepts a matrix as
input.
Thanks for the suggestion b
Keon,
why not fit a loglinear independence model which as far as I know is the
same.
Gerard
Here's an example from Agresti - Intro to Cat Data analysis
Example: Alcohol, cigarette, marijuana use
|--+--+|
| Alcohol |
I have still the same problem... As you said I tried with par(mfrow=c(2,1))
and par(mfrow=c(1,2)) but without success. Could R compiler be the problem ?
Wills, Kellie wrote:
>
> par(mfrow=c(1,1)) will give you just one panel. Try par(mfrow=c(2,1)) or
> par(mfrow=c(1,2)).
>
>
> -Original
Dear R-users,
Can someone help understand why in the following example R would return a
negative value for the last parameter when I have constrained it to be positive
and greater than or equal to 0.01?
> nlminb(start=initial value,objective=some function,
> lower=c(-Inf,-Inf,-Inf,0.01),upper=
Bernardo Rangel Tura wrote:
On Fri, 2009-03-20 at 18:29 +, Helena Mouriño wrote:
Dear all,
Im having an awkward problem in R. When I write the command
Fisher.test(school.data,workspace=2e+07), where school.data is the matrix
corresponding to the data set,
I get the error message:
FEXAC
Hi list,
I would like to normalize a matrix (two actually for comparison) using
iterative proportional fitting.
Using ipf() would be the easiest way to do this, however I can't get my
head around the use of the function. More specifically, the margins
settings...
for a matrix:
mat <- matrix(c(6
See examples on ?toupper page.
On Mon, Mar 23, 2009 at 8:03 AM, Daren Tan wrote:
> I managed to find toupper() which translates all letters to uppercase.
> Is there a function to capitalize only the first letter of word or
> phrase ?
>
> Thanks
>
> __
>
I managed to find toupper() which translates all letters to uppercase.
Is there a function to capitalize only the first letter of word or
phrase ?
Thanks
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read th
subset(table, year %in% c(1995,1998,2000))-->table2
> Hello,
> I'm trying to subset a dataframe where I have many observation taken in
> different years.
> I would like to subset the dataframe (in this example called "table") to get
> a new dataframe containing only the
> observation of year 19
Hi,
you might be also interested in a general overview as given here:
http://cran.r-project.org/web/views/Optimization.html
Hope this helps,
Roland
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Michael Kubovy
> Sent: Mond
On Fri, 2009-03-20 at 17:23 -0400, Zheng, Xin (NIH) [C] wrote:
> Hi there,
>
> Is there any such package? I searched but found none. Thanks in advance.
>
> Xin Zheng
>
Hi Xin
Do you try package DBI?
--
Bernardo Rangel Tura, M.D,MPH,Ph.D
National Institute of Cardiology
Brazil
__
Hello,
I'm trying to subset a dataframe where I have many observation taken in
different years.
I would like to subset the dataframe (in this example called "table") to get a
new dataframe containing only the
observation of year 1995, 1998 and 2000.
I've tried to use subset and the or operator "
On Fri, 2009-03-20 at 18:29 +, Helena Mouriño wrote:
> Dear all,
>
> Im having an awkward problem in R. When I write the command
> Fisher.test(school.data,workspace=2e+07), where school.data is the matrix
> corresponding to the data set,
> I get the error message:
> FEXACT error 7.
> LDSTP
Romain Francois wrote:
> Yihui Xie wrote:
>> Hi Everyone,
>>
>> I want to get the commands history as a character vector instead of
>> just displaying them, but the function history() just returns NULL. I
>> checked the source code of 'history' and could not find a solution.
>> Anybody has an idea?
On 3/23/2009 5:44 AM, Rob Denniker wrote:
> What's the neat way to create a dummy from a list?
> The code below is not replicable, but hopefully self-explanatory...
>
> d$treatment<-rep(1,length(d))
>
> notreat<-c("AR", "DE", "MS", "NY", "TN", "AK", "LA", "MD", "NC", "OK", "UT",
> "VA")
>
> #i
1 - 100 of 118 matches
Mail list logo
