Erin,
Linux supports many scripting languages.
Which language are you interested in: Perl, PHP, Bash, Python, etc???
--
Noah
On 9/2/09 10:35 PM, Erin Hodgess wrote:
> Dear R People:
>
> I know that this is off topic, but could anyone recommend a good book
> on Linux scripting please?
>
> Any he
I have a data that looks like this:
http://dpaste.com/88988/plain/
How can I extract/subset the data frame
based on selected uniq ID.
Let's say I want the first K uniq ID.
I want to be able to specify the parameter "K" here,
(i.e. given K=3, we hope to extract dat$V2 = 0,1,2).
I'm stuck with thi
Hi folks,
I was wondering if anyone could confirm/deny whether there exists any
kind of package to facilitate zoomable graphs with multiple plots (eg,
plot(..) and then points(..)).I've tried zoom from IDPmisc, and
iplot from the iplot and iplot extreme packages, but as far I can
tell, neithe
Dear R-friends,
How do you test the goodness of prediction of a model, when you predict on a
set of data DIFFERENT from the training set?
I explain myself: you train your model M (e.g. glm,gam,regression tree, brt)
on a set of data A with a response variable Y. You then predict the value of
th
Dear R People:
I know that this is off topic, but could anyone recommend a good book
on Linux scripting please?
Any help would be much appreciated!
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.
Hi there,
require(maptools)
?elide
good luck
milton
On Thu, Sep 3, 2009 at 1:04 AM, Hemavathi Ramulu wrote:
> Hi everyone,
> I have coding for repeating pentagon as below:
>
> plot(0:11,type="n")
> for (i in 1:10 )polygon(rep(c(4,5,7,8,6)), i*c(.5,.3,.3,.5,.7), bor=2)
>
> which are increasing ve
On Sep 2, 2009, at 8:47 PM, Richard M. Heiberger wrote:
Almost certainly, abind is what you need for the task.
Please dput() your matlab objects and send that to the list.
That will make your example reproducible.
Rich
From the documentation and behavior, abind seems to really not want
to
Hi everyone,
The code below will only draw grid lines in the *first* plot because
'panel.first' has already been evaluated (and hence it's NULL) after
the first plot is drawn:
devAskNewPage(TRUE)
f = function(...) {
for (i in 1:5) plot(rnorm(10), ...)
}
f(panel.first = grid())
For i in 2:5,
Hi everyone,
I have coding for repeating pentagon as below:
plot(0:11,type="n")
for (i in 1:10 )polygon(rep(c(4,5,7,8,6)), i*c(.5,.3,.3,.5,.7), bor=2)
which are increasing vertically.
Now, I want to know how to rotate the pentagon, so that I will get pattern
like flower.
Basicly, repeating penta
annie Zhang wrote:
Hi, Frank,
You mean the backward and forward stepwise selection is bad? You also
suggest the penalized logistic regression is the best choice? Is there
any function to do it as well as selecting the best penalty?
Annie
All variable selection is bad unless its in the con
Almost certainly, abind is what you need for the task.
Please dput() your matlab objects and send that to the list.
That will make your example reproducible.
Rich
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
On Sep 2, 2009, at 10:50 PM, Peter Meilstrup wrote:
I'm trying to massage some data from Matlab into R. The matlab file
has a "struct array" which when imported into R using the R.matlab
package, becomes an R list with 3+ dimensions, the first of which
corresponds to the structure fields,
Hi, Frank,
You mean the backward and forward stepwise selection is bad? You also
suggest the penalized logistic regression is the best choice? Is there any
function to do it as well as selecting the best penalty?
Annie
On Wed, Sep 2, 2009 at 7:41 PM, Frank E Harrell Jr wrote:
> David Winsemiu
I'm trying to massage some data from Matlab into R. The matlab file
has a "struct array" which when imported into R using the R.matlab
package, becomes an R list with 3+ dimensions, the first of which
corresponds to the structure fields, with corresponding row names, and
the second and thi
David Winsemius wrote:
>
>
> On Sep 2, 2009, at 9:36 PM, annie Zhang wrote:
>
>> Hi, R users,
>>
>> What may be the best function in R to do variable selection in
>> logistic
>> regression?
>
> PhD theses, and books by famous statisticians have been pursuing the
> answer to that question
David Winsemius wrote:
On Sep 2, 2009, at 9:36 PM, annie Zhang wrote:
Hi, R users,
What may be the best function in R to do variable selection in logistic
regression?
PhD theses, and books by famous statisticians have been pursuing the
answer to that question for decades.
I have the sam
if i were you, i will use sqlite, as mentioned in your email already.
spss should be able to read sqlite data easily through odbc.
On Tue, Sep 1, 2009 at 11:11 AM, Fredrik Karlsson wrote:
> Dear list,
>
> I am leaving my old position and now need to convert my R data frames
> into a format that c
On Sep 2, 2009, at 9:36 PM, annie Zhang wrote:
Hi, R users,
What may be the best function in R to do variable selection in
logistic
regression?
PhD theses, and books by famous statisticians have been pursuing the
answer to that question for decades.
I have the same number of variable
Hi, R users,
What may be the best function in R to do variable selection in logistic
regression? I have the same number of variables as the number of samples,
and I want to select the best variablesfor prediction. Is there any function
doing forward selection followed by backward elimination in st
Hi there,
I think the option of 30 seconds is ok because it is less than each one
expent reading the messages :-) Just kiding...
bests
milton
On Wed, Sep 2, 2009 at 8:01 PM, Leo Alekseyev wrote:
> Thanks everyone for the useful suggestions. The bottleneck might be
> memory limitations of my
Hi Karen.
If you are running windows, try:
http://cran.r-project.org/bin/windows/base/
Case it is Vista, see some tips on the same link.
cheers
milton
On Wed, Sep 2, 2009 at 8:13 PM, Karen Federico wrote:
> I'm trying to download this program and I'm not sure how to do it. Can you
> help m
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of sugo
> Sent: Wednesday, September 02, 2009 4:51 PM
> To: r-help@r-project.org
> Subject: [R] diff of two timestamps
>
>
> Hi all,
>
> I have the following problem: I have a csv
Try this:
dat <- read.table(textConnection("Timestamp1 Timestamp2
05:24:43 05:25:05
15:47:02 15:47:22
18:36:05 18:36:24
15:21:24 15:22:04"), T)
closeAllConnections()
Reduce("-", lapply(dat, strptime, format = "%H:%M:%S"))
On Wed, Sep 2, 2009 at 8:50 PM, sugo wrote:
>
> Hi all,
>
> I have the f
I'm trying to download this program and I'm not sure how to do it. Can you
help me with that? Thanks.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the post
Try
require(chron)
dat <- read.table(textConnection("Timestamp1 Timestamp2
05:24:43 05:25:05
15:47:02 15:47:22
18:36:05 18:36:24
15:21:24 15:22:04"), T)
closeAllConnections()
str(dat)
dat[,1] <- times(dat[,1])
dat[,2] <- times(dat[,2])
# numeric
as.numeric(dat[,2] <- times(dat[,2]) )
# test t
Try this:
> Lines <- "Timestamp1;Timestamp2;
+ 05:24:43;05:25:05;
+ 15:47:02;15:47:22;
+ 18:36:05;18:36:24;
+ 15:21:24;15:22:04;"
>
> library(chron)
> DF <- read.csv(textConnection(Lines), sep = ";", as.is = TRUE)
> times(DF$Timestamp2) - times(DF$Timestamp1)
[1] 00:00:22 00:00:20 00:00:19 00:00:4
Thanks everyone for the useful suggestions. The bottleneck might be
memory limitations of my machine (3.2GHz, 2 GB) and the fact that I am
aggregating on a field that is a string. Using the suggested
as.data.frame(table(my.df$my.field)) I do get a speedup, but the
computation still takes 30 secon
Hi all,
I have the following problem: I have a csv-file consisting of timestamp
values (no dates), e.g.:
Timestamp1;Timestamp2;
05:24:43;05:25:05;
15:47:02;15:47:22;
18:36:05;18:36:24;
15:21:24;15:22:04;
I need a vector with the difference of the two timestamps, so I read the
data with the read.
You may want to try using isplit (from the iterators package). Combined with
foreach, it's an efficient way of iterating through a data frame by groups
of rows defined by common values of a columns (which I think is what you're
after). You can speed things up further if you have a multiprocessor sy
On Wed, Sep 2, 2009 at 2:39 PM, wrote:
> Kingsford,
>
> Thanks for the information. As you suggest, if I don't hear from anyone
> else about the degrees of freedom issue in a couple days, I'll try r-devel.
Determining denominator degrees of freedom for F tests in mixed models
is an area of curren
Take 0.6 seconds on my slow laptop:
> n <- 1e6
> x <- data.frame(a=sample(LETTERS, n, TRUE))
> system.time(print(tapply(x$a, x$a, length)))
A B C D E F G H I J K
L M N O P Q
38555 38349 38647 38271 38456 38352 38644 38679 38575 38730
table() and xtabs() are fast only because they are just doing counts. If you
want the general case, you need ?tapply. aggregate() is basically a wrapper
for lapply and so you may have the same performance issues with tapply . Try
it to see. They are essentially doing the sort of hash table you desc
table is reasonably fast. I have more than 4 X 10^6 records and a 2D
table takes very little time:
nUA <- with (TRdta, table(URwbc, URrbc)) # both URwbc and URrbc are
factors
nUA
This does the same thing and took about 5 seconds just now:
xtabs( ~ URwbc + URrbc, data=TRdta)
On Sep 2, 20
I have a data frame with about 10^6 rows; I want to group the data
according to entries in one of the columns and do something with it.
For instance, suppose I want to count up the number of elements in
each group. I tried something like aggregate(my.df$my.field,
list(my.df$my.field), length) but
I'm not too familiar with geese but it looks like the
summary(fit)$correlation would be a dataframe. Maybe try getting the
first (or whatever row you're interested in):
...
corr_gee<-summary(fit)$correlation[1,1]
se_corrgee<-summary(fit)$correlation[1,2]
est[i,]<-c(corr_gee, se_corrgee)
...
Scott
Do str(f1) . What does it return?
> paste("set @g1=", "244901_at")
[1] "set @g1= 244901_at"
works fine, so it must be your data is not what you think it is.
On Wed, Sep 2, 2009 at 1:03 PM, Fahim Md wrote:
> Hi there!!!
> I am having trouble with *paste* function. I dont know how to proceed. I
On Wed, Sep 2, 2009 at 1:12 PM, Patrick
Connolly wrote:
> On Wed, 02-Sep-2009 at 07:02AM -0700, Mark Knecht wrote:
>
> |> On Tue, Sep 1, 2009 at 9:06 AM, Mark Knecht wrote:
> |> > Hi,
> |> > I'm not understanding how the width & height parameters are
> |> > supposed to work. When I execute the fo
Hi there,
and how about this:
nsim<-500
est<-NULL
for(i in 1:nsim){
fit <- geese(x ~ trt, id=subject, data=data_gee, family=binomial,
corstr="exch", scale.fix=TRUE)
.
corr_gee<-summary(fit)$correlation[1]
se_corrgee<-summary(fit)$correlation[2]
est<-data.frame(rbind(est,cbind(corr_gee
Kingsford,
Thanks for the information. As you suggest, if I don't hear from anyone
else about the degrees of freedom issue in a couple days, I'll try r-devel.
Also, while I appreciate your explanation of the correlation matrix
produced by summary.gls, I'm afriad I don't have the statistical backg
William, thank you so much! /tmp has noexec in the fstab and doesn't allow any
file
to be executable. I temporary changed those settings and I was able to
install rJava and RJDBC.
Perhaps the installation routine shouldn't use /tmp and use /var/tmp instead.
Many thanks,
Matt
-Original Mes
Hi R-users,
I have a problem for updating the estimates of correlation coefficient in
simulation loop.
I want to get the matrix of correlation coefficients (matrix, name: est) from
geese by using loop(500 times) .
I used following code to update,
nsim<-500
est<-matrix(ncol=2, nrow=nsim)
for(i in
Hello, I'm new to R and I'm building a system that uses a cluster of
computers to parallelize computations. To handle cluster setup I'm using the
R package Snowfall, but knowledge of Snowfall may not be necessary to answer
my question. I'm using a socket-based approach to cluster setup, so all I
n
I am having trouble with paste function. I dont know how to proceed. I tried
many options but i failed miserably.
I am using a variable f1 to assign a string as below:
f1=dataLine[locAffyProbeID];
( the value of f1 is 244901_at )
Then I am using the paste function
paste("set @g1=", f1);
in
Thanks Erik and Henrique,
That's what I was after.
Jonas
On Tue, Sep 1, 2009 at 8:08 PM, Henrique Dallazuanna wrote:
> Try this:
>
> > sapply(vec.names, get)
>
> But for this example, you don't need for, try:
>
> > dat - 1
>
> On Tue, Sep 1, 2009 at 2:52 PM, jonas garcia <
> garcia.jona...@g
On Wed, 02-Sep-2009 at 07:02AM -0700, Mark Knecht wrote:
|> On Tue, Sep 1, 2009 at 9:06 AM, Mark Knecht wrote:
|> > Hi,
|> > I'm not understanding how the width & height parameters are
|> > supposed to work. When I execute the following 4 commands:
|> >
|> > X11()
|> > X11(width=20, height=20)
|
On 9/2/2009 1:28 PM, Phil Spector wrote:
Fahim -
Apparently dataline is a factor, so you'd need to use
paste('set g=',as.character(f1))
That shouldn't be necessary:
> f1 <- factor("abc")
> paste('set g=', f1)
[1] "set g= abc"
I think we need reproducible code to diagnose this one.
On 9/2/2009 1:31 PM, Peter Ehlers wrote:
Heinrich,
You could create your own function mywithin()
by inserting a couple of rev()'s in within.data.frame().
In within.data.frame(), replace the two commented lines
with those immediately following:
mywithin <-
function (data, expr, ...)
{
pare
I did something very similar in ggplot2 about a year ago. Use the
unique sampling location from the species count data (rownames) as a
factor column, and then use that in a geom_color or geom_shape add on
to qplot
untested:
library ggplot2
a <- metaMDS(foo, k=3)
b <- rownames(foo)
d <- data.fram
Good point, I failed to spot this kink. You might be interested in a recent
discussion on r-help and r-devel,
http://markmail.org/message/4hvdmwqjyqwprbwf
Best,
baptiste
2009/9/2 Sebastien Bihorel
> Hi Baptiste,
>
> Thank for the help. One thing though that I found while transposing your
> s
On Wed, Sep 2, 2009 at 9:27 AM, RINNER
Heinrich wrote:
> Dear R community,
>
> I am using function 'within' in R.2.9.1 to add variables to an existing
> data.frame. This works wonderful, except for one minor point: The new
> variables are added to the data in reverse order.
>
> For example:
> x <
I think
install.packages("akima")
library(akima)
example(interp)
would be most useful.
--
View this message in context:
http://www.nabble.com/%22simple%22-3-dimensional-plots--tp25247706p25262822.html
Sent from the R help mailing list archive at Nabble.com.
Hi all,
Is there a way in R to plot points using symbols as defined in another
vector without adding a separate points line for each symbol? For
example, I have the results from an ordination with ~ 35 points and an
associated vector that corresponds to different symbols in pch for the
35 symbols
Hello,
I'm using RJDBC/DBI to query a MS SQL DB that looks like:
YMDHMs Num1stsAboveBaseline
2009-05-18 00:001
2009-05-18 00:012
2009-05-18 00:022
...
The first column will get converted to POSIXct.
However, dbGetQuery converts the first column to a facto
Heinrich,
You could create your own function mywithin()
by inserting a couple of rev()'s in within.data.frame().
In within.data.frame(), replace the two commented lines
with those immediately following:
mywithin <-
function (data, expr, ...)
{
parent <- parent.frame()
#e <- evalq(enviro
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Payam Minoofar
> Sent: Wednesday, September 02, 2009 10:09 AM
> To: r-help@r-project.org
> Subject: [R] pruning data
>
> Hello everyone,
>
> I am trying to prune a data frame fo
Payam -
Take a look at na.omit . This is exactly what it
was written for.
- Phil
On Wed, 2 Sep 2009, Payam Minoofar wrote:
Hello everyone,
I am trying to prune a data frame for partial least squares analysis.
I need to delete an entire row if one cell i
On 9/2/2009 9:27 AM, RINNER Heinrich wrote:
Dear R community,
I am using function 'within' in R.2.9.1 to add variables to an existing
data.frame. This works wonderful, except for one minor point: The new variables
are added to the data in reverse order.
For example:
x <- data.frame(a = 1:3, b
Fahim -
Apparently dataline is a factor, so you'd need to use
paste('set g=',as.character(f1))
- Phil Spector
Statistical Computing Facility
Department of Statistics
Try this:
DF[complete.cases(DF),]
On Wed, Sep 2, 2009 at 2:09 PM, Payam Minoofar
wrote:
> Hello everyone,
>
> I am trying to prune a data frame for partial least squares analysis.
> I need to delete an entire row if one cell in the row contains a NA.
>
> Presently, I am running a loop that is su
Hi Baptiste,
Thank for the help. One thing though that I found while transposing your
syntax to my problem: lab must be of class expression for your syntax to
work. For instance, if one replaces the second elements of the lab
variables by a simple integer, lab is not more of class expression,
Dear Payam,
Here is a suggestion:
index <- apply(yourdata, 1, function(x) any( is.na(x) ) )
yourdata[ !index, ]
Above creates an index (TRUE) when any row of the data contains a missing
value. Then it filters up (extract) the rows that have complete
observations.
See ?any, ?is.na, ?"!" and ?appl
?is.na
?sum
?if
?else
I think that is how I would approach it, but I could be far off.
On Wed, Sep 2, 2009 at 12:09 PM, Payam
Minoofar wrote:
> Hello everyone,
>
> I am trying to prune a data frame for partial least squares analysis.
> I need to delete an entire row if one cell in the row contain
Hi there!!
I am having trouble with *paste* function. I dont know how to proceed. I
tried many options but i failed miserably.
I am using a variable f1 to assign a string as below:
f1=dataLine[locAffyProbeID];
( the value of f1 is *244901_at* )
Then I am using the paste function
paste("set g=",
Hello everyone,
I am trying to prune a data frame for partial least squares analysis.
I need to delete an entire row if one cell in the row contains a NA.
Presently, I am running a loop that is supposed to extract the rows
that are full of numbers into a second data frame and skips the rows
I write about R every weekday at the Revolutions blog:
http://blog.revolution-computing.com
In case you missed them, here are some articles from last month of
particular interest to R users.
http://bit.ly/11YkB0 listed seven reasons of an anthropology professor
for using R.
http://bit.ly/9sbno l
Try this:
transform(transform(x, c = a ^ 2, d = b ^2), e = c + d)
On Wed, Sep 2, 2009 at 10:27 AM, RINNER Heinrich <
heinrich.rin...@tirol.gv.at> wrote:
> Dear R community,
>
> I am using function 'within' in R.2.9.1 to add variables to an existing
> data.frame. This works wonderful, except for
Hi there!!!
I am having trouble with *paste* function. I dont know how to proceed. I
tried many options but i failed miserably.
I am using a variable f1 to assign a string as below:
f1=dataLine[locAffyProbeID];
( the value of f1 is *244901_at* )
Then I am using the paste function
paste("set @g
Your data ranges from 0.67 to 1.21, just to simplify things let's assume the
that the histogram will go from the pretty numbers of 0.65 to 1.25 for a total
width of 0.6. Now consider the simplest histogram consisting of 1 single bar
going from 0.65 to 1.25 (very uninteresting histogram, but goo
Martin,
Thanks for showing the timing tests. It is important
to see how the time (and memory usage) grows with
the size of the problem, where size may be the number
of rows or length of the lag.
Here is another function to toss in the hat. It uses no
loops and does all the sum by diff'ing a
Try this:
mapply(function(x, y)mitest[x, y], c("a", "b", "c"), c("b", "c", "b"))
or
diag(`[`(mitest, i = c("a", "b", "c"), j = c("b", "c", "b")))
On Wed, Sep 2, 2009 at 6:57 AM, Agustin Lobo wrote:
> Given:
>
> > mitest <- matrix(1:16,ncol=4)
> > dimnames(mitest)[[1]] <- c("a","b","c","d")
Hi,
Try this,
myplot <- function(subject) { plot(subject,
main=deparse(substitute(subject))) }
s1 <- c(200,200,190,180)
myplot(s1)
see ?deparse
HTH,
baptiste
2009/9/2 Marianne Promberger
> Dear list,
>
> I've written a function that plots subjects. Something like:
>
> myplot <- functi
Not sure I completely understand what you want. You might try looking at
the irt.ability() function in the MiscPscyho package for ability
estimates.
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Tessari
> Sent: Wednesday, Se
Hi,
Try this,
library(grid)
value <- c(0.1)
lab <- c("test",
expression(bquote(paste(.(value[1]*100), " and percentiles1",
sep=""))),
bquote(expression(.(value[1]*100)*" and percentiles2")),
bquote(paste(.(value[1]*100), " and percentiles3", sep="")) )
grid.newpage
Hi miksanta.
They seem to be ordered variables (with specific notation for NA).
Some are qualitative and some quantitative
On Wed, Sep 2, 2009 at 4:19 PM, wrote:
> Hello,
>
> i have this dataset
> http://www.umass.edu/statdata/statdata/data/pharynx.txt.
>
> the variables GRADE, T_STAGE anda N
Dear list,
I've written a function that plots subjects. Something like:
myplot <- function(subject) { plot(subject) }
Subjects are vectors, e.g. ...
s1 <- c(200,200,190,180)
... and plotting them works fine, e.g. ...
myplot(s1)
Now I want to have "s1" etc appear in the plot title, but I don't
R Import and Export manual on the R site is a start.
Also what kind of data are you trying to read?
given a simple csv file called sss.csv like this
a, b, c
1,2,3
4,5,6
on your C drive
you can read it into R with
read.csv("C:/sss.csv")
See ?read.table for more information
--- On Tue, 9/
Dear R-users,
I am trying to use the grid.text and expression functions to display
several character strings and plotmath text on a viewport. Some strings
can include a variable portion (PI.limits in the following example),
which I thought could be implemented by combining the bquote and the
Alexander Shenkin wrote:
> Though, from my limited understanding, the 'apply' family of functions
> are actually just loops. Please correct me if I'm wrong. So, while
> more readable (which is important), they're not necessarily more
> efficient than explicit 'for' loops.
Hi Allie -- This uses a
Dear R Users,
I am using the LLTM and the LRSM functions in the eRm package to do repeated
measurements where there are 2 measurement points for a list of 10 items. I
am trying to get ability estimates but am having trouble. I don't think
that it is appropriate to use the pmat function since t
Hi Steve,
Thanks for asking - I forgot to mention I use windows XP.
And the only support foreach has for windows (as far as I read from the
manuals so far) - is for the snow package.
Cheers,
Tal
On Wed, Sep 2, 2009 at 6:57 PM, Steve Lianoglou <
mailinglist.honey...@gmail.com> wrote:
> Hi Tal
Hi Tal,
On Sep 2, 2009, at 11:52 AM, Tal Galili wrote:
Hello dear R community.
I just started playing with the snowfall package (a wrapper for the
snow
package), and found it very convenient.
(See also this great website:
http://www.imbi.uni-freiburg.de/parallel/ )
I was wondering if it is
Hello dear R community.
I just started playing with the snowfall package (a wrapper for the snow
package), and found it very convenient.
(See also this great website:
http://www.imbi.uni-freiburg.de/parallel/ )
I was wondering if it is possible to connect snowfall with the foreach
package (since
On Tue, 1 Sep 2009, dolar wrote:
Would like some tips on how to avoid loops as I know they are slow in R
If I understand your criterion (and calling your data.frame 'dat'):
criterion <- as.matrix(dist(dat$a)) <= 5 & outer(dat$a,dat$a,">=")
criterion %*% as.matrix(dat[, c("b","c")])
b
Hi,
See: ?system.file
-steve
On Sep 2, 2009, at 11:39 AM, jbryer wrote:
I too am looking to do the same thing. Anyone have any insight as to
this can
be done?
Thanks,
Jason
Chris Stubben wrote:
I wrote a package which includes a number of genome sequencing
project
statistics on th
Crudely but I think it works
x <- data.frame(aa <- mtcars$mpg)
b <- ggplot(x, aes(aa)) + geom_histogram(aes(y=..density..)) +
stat_function(fun=dnorm, args=list(mean=mean(x$aa), sd=sd(x$aa)))
b
--- On Wed, 9/2/09, Gundala Viswanath wrote:
> From: Gundala Viswanath
> Subject: [R] How
I too am looking to do the same thing. Anyone have any insight as to this can
be done?
Thanks,
Jason
Chris Stubben wrote:
>
> I wrote a package which includes a number of genome sequencing project
> statistics on the web like http://www.ncbi.nlm.nih.gov/genomes/lproks.cgi.
> I included some g
I forgot to mention that the ellipse has a rotation. It's horizontal
axis is not parallel to the x-axis.
Just repeating, when I try to draw a circuit keeping the center same
as the ellipse, and radius equal to the first value returned by
> sqrt(eigen(cov.trob(mydataforellipse)$cov)$values)
taking
Dear R community,
I am using function 'within' in R.2.9.1 to add variables to an existing
data.frame. This works wonderful, except for one minor point: The new variables
are added to the data in reverse order.
For example:
x <- data.frame(a = 1:3, b = 4:6)
y <- within(x, {
c = a^2
d =
Loris Bennett writes:
I get the same problem using R version 2.9.2.
I would be very grateful if anyone could shed some light on this
issue.
Regards
Loris
> loris.benn...@fu-berlin.de (Loris Bennett) writes:
>
>> Hi,
>>
>> I am getting the following error
>>
>># (R-2.9.1/src/library/me
Jari, thanks for the quick answer.
> sqrt(eigen(cov.trob(mydataforellipse)$cov)$values)
what will this return?
For my data, I get:
> sqrt(eigen(cov.trob(r)$cov)$values)
[1] 1.857733e-05 4.953181e-06
Is this Left hand value the major or the semi major length?
I also try to plot a circuit keepi
Hi,
Thank you for your responses.
Héctor, I had also discovered the solution you suggested. However, when I
install files from a list (such as pckg.list), failure in the installation of
one package in the list results in none of the packages later in the list being
installed. After moving th
Hello,
I have been having difficulty getting boxplot to give the output I want -
probably a result of the way I have been handling the data.
The data is arranged in columns: each date has two sets of data. The number
of data points varies with the date, so each column is of different length.
I
Given:
> mitest <- matrix(1:16,ncol=4)
> dimnames(mitest)[[1]] <- c("a","b","c","d")
> dimnames(mitest)[[2]] <- c("a","b","c","d")
> mitest
a b c d
a 1 5 9 13
b 2 6 10 14
c 3 7 11 15
d 4 8 12 16
I can do:
> mitest[cbind(c(1,2,3),c(2,3,2))]
[1] 5 10 7
but using the names does not work:
> m
Another advantage of the apply family of functions is that
they determine the size and type of their output in an
efficient way, which is sometimes tricky when you write
the loop yourself.
- Phil Spector
Statistica
That is not very complex with densities instead of counts.
library(ggplot2)
ggplot(mtcars, aes(x = mpg)) +
geom_histogram(aes(y = ..density..), fill = "red") +
stat_function(
fun = dnorm,
args = with(mtcars, c(mean = mean(mpg), sd = sd(mpg)))
I've been trying to compile 2.9.2 under solaris 10 the last couple days without
success. configure runs fine and I'm using GNU Make. We're trying to build
under 64 bit (and I'm wondering if this might be part of our problem). Here is
the output from the make...
ld: warning: file /usr/local/
Currently, I am doing it this way.
x <- mtcars$mpg
h<-hist(x, breaks=10, col="red", xlab="Miles Per Gallon",
main="Histogram with Normal Curve")
xfit<-seq(min(x),max(x),length=40)
yfit<-dnorm(xfit,mean=mean(x),sd=sd(x))
yfit <- yfit*diff(h$mids[1:2])*length(x)
lines(xfit, yfit, col="blue", lwd=
If you can do it- try a for loop and another solution to prove this to
yourself. A for loop can get a little unwieldy for a novice like me
to understand the code, but doable. The simpler the better, but they
are not terribly slow. I have run into a couple of situations where a
vectorized solutio
> Would like some tips on how to avoid loops as I know they are slow in R
They are not slow. They are slower than vectorised equivalents, but
not slower than apply and friends.
Hadley
--
http://had.co.nz/
__
R-help@r-project.org mailing list
https:/
Thanks a million Thierry. I solved the problem
with new installation. I apologize for troubling you
in this trivial matter.
- G.V.
On Wed, Sep 2, 2009 at 10:26 PM, ONKELINX,
Thierry wrote:
> I'm using the latest version: 0.8.3 on R 2.9.2
>
>
> -
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