> Dario Strbenac
> on Wed, 18 May 2011 12:00:11 +1000 writes:
> I am using PAM with k = 10 clusters, but I only get one cluster
> ID for all my observations. I couldn't find any discussion about
> this in the help file, or mailing lists. Is there a reasonable
> explan
On Tue, May 24, 2011 at 12:31 AM, Rekha wrote:
> S . Am getting error still now if i use both version of the codes .
>
> i have to call any library?
No. The code you've written here doesn't require loading any other R libraries.
Please copy and paste the exact code you've entered along with the
S . Am getting error still now if i use both version of the codes .
i have to call any library?
On Tue, May 24, 2011 at 9:22 AM, Steve Lianoglou <
mailinglist.honey...@gmail.com> wrote:
> Hi,
>
> On Mon, May 23, 2011 at 11:41 PM, Rekha wrote:
> > Hello All,*
> >
> > *I want to draw a histogram
Geoffrey:
There may be something for this in one of the packages dealing with dates.
If not, here's one (incomplete) idea, based on something I used for a similar
issue a little while ago. Essentially, make a data frame that ranks each
weekday over a period in ascending order. This data frame
Dear List,
I have just got a laptop with win 7 64 bit installed on it. I
installed R and quite a few packages I use. When I try to start
BiodiversityR the library loads without any problems but when starting
the GUI it gives me the following error message:
Sourced: BiodiversityGUI.R
Error : .onAt
Janko,
Thanks a lot for your reply. Option 2 is exactly what I was looking for.
Mil gracias!
Mateo
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I wrote a little function called bgfun that adds gridlines and a background,
but it's not working with I plot using the formula.
I have some theories on what's happening, but even if my theory is right, I
don't know how to fix it.
Someone have a straightforward silver bullet?
Thank you,
Gene
I would read the datasets into a list first, something like this which will
make a list of dataframes:
filenames <- dir() # where only filenames you want to read in are in this
directory
dataframelist <- lapply(filenames, read.csv, header = TRUE, sep = ",")
You should be able to put the whol
Hi Scott,
Thanks for this.
Got some questions below:
Thanks
Hugh
Date: Mon, 23 May 2011 17:32:52 -0500
From: scttchamberla...@gmail.com
To: h_a_patie...@hotmail.com
CC: r-help@r-project.org
Subject: Re: [R] Reading Data from mle into excel?
I would read the datasets into a list fir
I am trying to implement a maximum likelihood estimation using maxLik but
keep getting this message "Error in is.function(func) : 'func' is missing"
I am new to R and cannot figure out what's wrong. Any advice will be much
appreciated.
Thanks
[[alternative HTML version deleted]]
Hi,
On Mon, May 23, 2011 at 11:41 PM, Rekha wrote:
> Hello All,*
>
> *I want to draw a histogram with density curve. *
>
> *For that simply i created a data called*"x" *and i used the function called
> * hist(x, col = "blue", freq = FALSE),** *from this function i got a
> histogram*.
>
> *Af
Hello All,*
*I want to draw a histogram with density curve. *
*For that simply i created a data called*"x" *and i used the function called
*hist(x, col = "blue", freq = FALSE),** *from this function i got a
histogram*.
*After that , i tried this function* ** lines(density(x), col = "red",
Hi:
Here's one way to do the pairwise tables. I'm restricting attention to
the variables with only a few levels, but the idea should be clear
enough.
# Put the variable names into a vector
vars <- names(infert)[c(1, 3:6)]
# Use expand.grid() to generate all pairs of variables
# It's important to
Hi CAR,
On Mon, May 23, 2011 at 10:52 AM, CAR wrote:
> Hi everybody!!
>
> I'm trying to assign boolean values to a column in a matrix depending on the
> number of times present in another column. I have no clue, I have been 2
> days looking for a way, if somebody knows what kind of functions I ne
I think cognizance should be taken of fortune("very uneasy").
cheers,
Rolf Turner
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guid
Hi:
This isn't too hard to do. The strategy is basically this:
(1) Create a list of file names. (See ?list.files for some ideas)
(2) Read the data files from (1) into a list.
(3) Create a function to apply to each data frame in the list.
(4) Apply the function to each data frame.
(5) Extract the
untested
x <- lapply(names(infert),function(a)table(infert[[a]]))
extend this to the rest of your problem.
Sent from my iPad
On May 23, 2011, at 20:33, "Sparks, John James" wrote:
> Dear R Helpers,
>
> First, I apologize for asking for help on the first of my topics. I have
> been looking a
Dear R Helpers,
First, I apologize for asking for help on the first of my topics. I have
been looking at the posts and pages for apply, tapply etc, and I know that
the solution to this must be ridiculously easy, but I just can't seem to
get my brain around it. If I want to produce a set of table
Dear R Helpers,
First, I apologize for asking for help on the first of my topics. I have
been looking at the posts and pages for apply, tapply etc, and I know that
the solution to this must be ridiculously easy, but I just can't seem to
get my brain around it. If I want to produce a set of table
On 24/05/11 11:23, Richard Friedman wrote:
> Dear R-list,
>
> In the R-book, p.464, Michael Crawley recommends that error
> bars for bar plots of normally distributed continuous response
> variables with categorical explanatory variables be given by
> 1/2 of the least significant difference, wh
Dear R-list,
In the R-book, p.464, Michael Crawley recommends that error
bars for bar plots of normally distributed continuous response
variables with categorical explanatory variables be given by
1/2 of the least significant difference, where the least significant
difference is defines a
On 2011-05-23 07:40, Rashid Bakirov wrote:
Hello all R gurus,
I have a following problem which I hope someone will help me to solve.
I have a data.frame in form similar to below.
testframe<-data.frame("Name"=c("aa","aa","aa","aa","aa","bb","bb","bb","bb","bb"),"Value"=c(1,100,1,1,1,100,10
Hi:
Perhaps the cld() function in the multcomp package is what you're looking for.
Dennis
On Mon, May 23, 2011 at 9:40 AM, Adrienne Keller
wrote:
> Is there code in R to automatically assign letters to different groups that
> are found to be statistically significant using a post-hoc test follo
Hi Mateo,
not sure if I totally get what you're after, but maybe this helps:
SharpeRatio.annualized <- function(roc){
print("I'm computing the Sharpe Ratio")
return()
}
MyF <- function(Tic, price){
print("Option 1")
expr <- expression(Ratio.Tic <- SharpeRatio.annualized(ro
On Mon, May 23, 2011 at 4:05 PM, Heiman, Thomas J. wrote:
> Hi,
>
> I tried to attach these files before as .csv and they did not go through..
> This time they are .txt files.. I am trying to get the attached following
> two timeseries (these are small subsets of the whole thing) into R so I ca
cameron.bracken wrote:
>
> You need to look for a file (it may not be in your project directory) that
> contains "___LOCK" in its file name and delete it.
>
Specifically, the rogue lockfile should be in the same directory as
options('tikzMetricsDictionary'). Or, if you haven't set a permanent
l
On 24/05/11 01:24, user84 wrote:
Hi,
could anyone tell me how predict() predicts the new value(s), of a MA(1)
arima-modell.
its really easy to make it with an AR(1), knowing the last term, but how can
i or R know the last error?
I think what you're asking here is this:
Suppose X_t = a_t +
I get the following error:
Error in backSpline.npolySpline(sp) : spline must be monotone
Has anyone had this error before? any ideas on a workaround?
>
> vols=read.csv(file="C:/Documents and Settings/Hugh/My
> Documents/PhD/Swaption vols.csv"
+ , header=TRUE, sep=",")
> X<-ts(vols[,2])
> #X
>
Hello,
I am stuck in a relatively simple procedure and was wondering if anybody
knows the answer. I am a relatively new R user.
How do I use an argument of a custom function in the name of a dataset in R?
For example, I have the function:
MyF <- function(Tic, price){
x
Hi there,
I ran the following code:
vols=read.csv(file="C:/Documents and Settings/Hugh/My Documents/PhD/Swaption
vols.csv"
, header=TRUE, sep=",")
X<-ts(vols[,2])
#X
dcOU<-function(x,t,x0,theta,log=FALSE){
Ex<-theta[1]/theta[2]+(x0-theta[1]/theta[2])*exp(-theta[2]*t)
Vx<-theta[3]^2*(1-exp(-2*t
Not sure if you ever figured this out, but I was just directed to this post.
You need to look for a file (it may not be in your project directory) that
contains "___LOCK" in its file name and delete it. filehash creates this to
prevent multiple writes to the same file at the same time and somehow
Thanks for the tip!!! I read the R manual and some other basic helps. The
problem here is I dont know even the name of what I´m looking for
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Sent from the R hel
On May 23, 2011, at 4:31 PM, David Winsemius wrote:
>
> On May 23, 2011, at 4:32 PM, Marc Schwartz wrote:
>
>> On May 23, 2011, at 3:13 PM, Adrienne Keller wrote:
>>
>>> I am making a barplot using barplot2 from gplots and would like to
>>> format the names of my categorical variables (tree spe
On May 23, 2011, at 4:32 PM, Marc Schwartz wrote:
On May 23, 2011, at 3:13 PM, Adrienne Keller wrote:
I am making a barplot using barplot2 from gplots and would like to
format the names of my categorical variables (tree species) on the x-
axis so that the genus name is above the species name
On May 23, 2011, at 5:16 PM, David Winsemius wrote:
On May 23, 2011, at 4:31 PM, Ben Bolker wrote:
Heiman, Thomas J. mitre.org> writes:
tmp <- read.table("baltimore.csv", sep = " ") ##This is timeseries 2
z <- zoo(tmp[, 2:20], as.Date(as.character(tmp[, 1]), format = ))
In this file you
Hi,
I perform latent class analysis on a matrix of dichotomous variables to
create an indicator of class/category membership for each observation. I
would like to know whether there is a function that selects the best fit in
terms of number of classes/categories.
Currently, I am doing this with t
On May 23, 2011, at 4:31 PM, Ben Bolker wrote:
Heiman, Thomas J. mitre.org> writes:
tmp <- read.table("baltimore.csv", sep = " ") ##This is timeseries 2
z <- zoo(tmp[, 2:20], as.Date(as.character(tmp[, 1]), format = ))
In this file you have no separators in the dates so the format would
Adrienne Keller umontana.edu> writes:
> I am making a barplot using barplot2 from gplots and would like to
> format the names of my categorical variables (tree species) on the x-
> axis so that the genus name is above the species name (to save room).
> My code so far is:
[snip]
Try "Dia
Heiman, Thomas J. mitre.org> writes:
> tmp <- read.table("baltimore.csv", sep = " ") ##This is timeseries 2
> z <- zoo(tmp[, 2:20], as.Date(as.character(tmp[, 1]), format = "%y %m %d"))
>
> tmp1 <- read.table("baltimorefludata.csv", sep = " ") ##This is timeseries 1
> z2 <- zoo(tmp[,2], as.Date(
On May 23, 2011, at 3:13 PM, Adrienne Keller wrote:
> I am making a barplot using barplot2 from gplots and would like to
> format the names of my categorical variables (tree species) on the x-
> axis so that the genus name is above the species name (to save room).
> My code so far is:
>
> sp
Thank you very much, everyone! Extremely helpful!
I really liked these two solutions:
# reshape is really cool + it gives me the value I need!
library(reshape2)
solution1<-subset(melt(x, id = c('y', 'z')), value > 0)
# This one is also very good, just need one additional loop - to
include variab
Ramnath R gmail.com> writes:
>
> Hi,
>
> Anybody know what are the common Standard statistical methods used for the
> analysis of a dataset,and
> anybody know which of these methods give similar results
>
Sorry, but this question is the statistical equivalent of
"how long is a piece of stri
I am making a barplot using barplot2 from gplots and would like to
format the names of my categorical variables (tree species) on the x-
axis so that the genus name is above the species name (to save room).
My code so far is:
speciesnames<-c("Dialium guianensis", "Inga alba", "Tachigali
ver
No attachment. But if your date is in the MMDD then the format syntax is
...format="%Y%m%d").
Best,
--- On Mon, 5/23/11, Heiman, Thomas J. wrote:
From: Heiman, Thomas J.
Subject: [R] getting time series into r
To: "R-help@r-project.org"
Date: Monday, May 23, 2011, 2:32 PM
Hi,
I am tr
Hi,
I tried to attach these files before as .csv and they did not go through.. This
time they are .txt files.. I am trying to get the attached following two
timeseries (these are small subsets of the whole thing) into R so I can merge
them using zoo.
tmp <- read.table("baltimore.csv", sep = "
Hi David,
I have attached them to this note as .txt files.. Sorry about that..
Sincerely,
tom
Thomas Heiman, PhD
Info Systems Eng, Sr
The MITRE Corporation | Center for Enterprise Modernization
Office: 703-983-2951 | thei...@mitre.org
-Original Message-
From: David Winsemius [mailto:
Hello,
I have a few questions concerning the DCC-GARCH model and its programming in
R.
So here is what I want to do:
I take quotes of two indices - S&P500 and DJ. And the aim is to estimate
coefficients of the DCC-GARCH model for them. This is how I do it:
library(tseries)
p1 = get.hist.quote(ins
On May 23, 2011, at 2:32 PM, Heiman, Thomas J. wrote:
Hi,
I am trying to get the attached following two timeseries
Nothing attached. The mail-server refuses files with csv extension but
would have accepted them with .txt extension. It's just a dumb
machine, after all.
--
David.
(these
Date: Mon, 23 May 2011 14:43:59 +0100
From: arraystrugg...@gmail.com
To: r-help@r-project.org
Subject: [R] RGL package installation problem on Centos
Dear R users,
I have installed the latest version of R from source on Centos (using
configure a
On 11-05-23 5:55 AM, Pierre Bruyer wrote:
Hello everybody,
I search a function which returns the contour of map with levels like
contourLines, but I would like this function return the border of the map too,
because the function contourLines cannot consider the corner of the map and it
is not
Hi,
I have two groups of data of different size:
group A: x1, x2, , x_n;
group B: y1, y2, , y_m; (m is not equal to n)
The two groups are independent but observations within each group are
not independent,
i.e., x1, x2, ..., x_n are not independent; but x's are independent from y's
On 11-05-23 8:04 AM, Mike Marchywka wrote:
I think I just saw a thread go by on " how do I make interactive PDF from
rgl output" but I ignored it at the time and can't seem to find consensus result
on google. Essentially I wanted to create pdf or other output to
share a 3D plot and let viewers
On 11-05-23 9:43 AM, john herbert wrote:
Dear R users,
I have installed the latest version of R from source on Centos (using
configure and make install).
This seemed to work fine, with no Errors reported and R at the command line
starts R.
However, if I try and installed the package rgl using;
i
Hi,
I am trying to get the attached following two timeseries (these are small
subsets of the whole thing) into R so I can merge them using zoo.
tmp <- read.table("baltimore.csv", sep = ",") ##This is timeseries 2
z <- zoo(tmp[, 2:20], as.Date(as.character(tmp[, 1]), format = "%y %m %d"))
tmp1
On Mon, 2011-05-23 at 09:58 -0400, Ravi Varadhan wrote:
> Why does this not find a better solution?
>
> > x <- seq(0,2*pi, length=1000)
> > x <- cbind(x/(2*pi), sin(x))
>
> > fit1 <- principal.curve(x, plot = TRUE, trace = TRUE, maxit = 100,
> + start = cbind(sort(x[,1]), rep(1, nrow(x
> Star
Hi,
Anybody know what are the common Standard statistical methods used for the
analysis of a dataset,and
anybody know which of these methods give similar results
Ram
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
htt
Hi:
On Mon, May 23, 2011 at 7:52 AM, CAR wrote:
> Hi everybody!!
>
> I'm trying to assign boolean values to a column in a matrix depending on the
> number of times present in another column. I have no clue,
Sorry, I don't have time to help. But have you read "The Introduction
to R" tutorial wher
On Mon, May 23, 2011 at 1:30 PM, Heiman, Thomas J. wrote:
> Hi,
>
> I am trying to get the following two timeseries (these are small subsets of
> the whole thing) into R so I can merge them using zoo.
>
The email may have messed up your data. Try attaching it to your post
as a .txt file. It ma
Dear all, I would like to draw a 3D plot as shown
here http://en.wikipedia.org/wiki/File:NaturalLogarithmAll.png, for this
function "f = exp[ 1 - x^2 - y^2]" (this function is some arbitrary!). I am
aware of different 3D plotting system in R, however it would be great if I can
get that kind of
Thanks Ian, not too late at all. I need to do some testing but your
idea seems like it should work. Thanks, Bryan
On May 23, 2011, at 12:44 PM, Ian Gow wrote:
This might be a little late, but one option for this issue might be
to add
a line like
Sys.setenv("DROPBOX_PATH" = "/Users/Person
Hi,
I am trying to get the following two timeseries (these are small subsets of the
whole thing) into R so I can merge them using zoo.
Timeseries 1=[
Thank you for the email. The data is unbalanced, meaning that some days are
missing. So the sequence of days could be something like Tuesday,
Wednesday, Friday, Monday. The diff function would produce 1, 2, 3. But I
would like it to produce 1, 2, 1 since there is really only 1 day between
Frida
Hi Everyone,
I need to run several simple linear regressions at once, using the
following data. Response variables: Bird species (sp 1, sp2, sp3...spn).
Independent variable: Natprop - proportion of natural area. covarate:
Effort = hours). One single linear regression would be: lmSp1 <- lm(sp1~
na
Hi all,
I need to run several simple linear regressions at once, using the
following data. Response variables: Bird species (sp 1, sp2, sp3...spn).
Independent variable: Natprop - proportion of natural area. covarate:
Effort = hours). One single linear regression would be: lmSp1 <- lm(sp1~
natprop
Dear R-help list,
I have a problem storing results from a bootstrap loop. What I doing is
creating an empty matrix before the loop, then run the loop. Before the end
of the loop I have a command line which defines which results inside the
loop will have to go in the empty matrix. Here is the code I
zcatav wrote:
>
> How can i add data values or proportion values on bar or pie charts?
>
You can add any text or numbers to any place of a graph with this command:
> text(45,20,"some text")
where: 45 is position on the X-axis and 20 - on the Y. Note that numbers
denote the MIDDLE of the text,not
Is there code in R to automatically assign letters to different groups
that are found to be statistically significant using a post-hoc test
following an ANOVA? For example, let's say I found that relative rates
of tree growth were statistically significant between tree species
using an ANOV
Hi everybody!!
I'm trying to assign boolean values to a column in a matrix depending on the
number of times present in another column. I have no clue, I have been 2
days looking for a way, if somebody knows what kind of functions I need to
use I will really apreciate it.
example:
Color Type Mar
Hi,
could anyone tell me how predict() predicts the new value(s), of a MA(1)
arima-modell.
its really easy to make it with an AR(1), knowing the last term, but how can
i or R know the last error?
It would also help if somebody could tell me how to find the "open" source
of the function predict().
Hello all R gurus,
I have a following problem which I hope someone will help me to solve.
I have a data.frame in form similar to below. >
testframe<-data.frame("Name"=c("aa","aa","aa","aa","aa","bb","bb","bb","bb","bb"),"Value"=c(1,100,1,1,1,100,100,100,100,1))
Name Value
1aa 1
Dear R users,
I have installed the latest version of R from source on Centos (using
configure and make install).
This seemed to work fine, with no Errors reported and R at the command line
starts R.
However, if I try and installed the package rgl using;
install.packages("rgl")
I get the following
Hello,
How can i add data values or proportion values on bar or pie charts?
Thanks in advance.
--
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Sent from the R help mailing list archive at Nabble.com.
__
In fact "temp.data$" is not necessary, but the command still not run.
Thanks for your response, I'm also tring other point of view (suggested by
r-helpers).
For example ?ddply and ?lmList.
If you are interested I will keep you updated.
Francesco
> Date: Fri, 20 May 2011 13:04:41 -0400
> Subjec
Hello everybody,
I search a function which returns the contour of map with levels like
contourLines, but I would like this function return the border of the map too,
because the function contourLines cannot consider the corner of the map and it
is not adapted to fill polygon after that.
Thanks
Hi,
I am new to R and has been depending mostly on the online tutotials to learn
R. I have to deal with zero inflated negative binomial distribution. I am
however unable to understand the following example from this link
http://www.ats.ucla.edu/stat/r/dae/zinbreg.htm
The result gives two blocks.
Hi
I'm using the lattice function to plot catch data for two types of catch
between two communities over time using the xyplot. I've used
*"smooth"*when specifying
* type=c( )* . Here is a sample of my code below:
> xyplot(ns+ rc~wk|location, type=c("p", "smooth"), xlab="Week",
+ main="Native sh
Your request seems bizarre to me. If you really want to ignore the actual time
intervals, just do your analysis without the actual dates.
If you just want forward-looking time intervals, then put them in the correct
index locations.
?diff
c(diff(DF$DATE),1)
Jim,
Thank you for your help! That is precisely what I was looking for. From your
help I was able to edit the output and then print it to a txt file (because
I didn't want to print it all in the R console due to the thousands of
lines).
R is a very powerful language, but it is rather difficult fo
Geoff,
I think you could write an if loop to solve this.
You could write:
for(i in 1:num_obs){
if(DAYS[i]=='Monday'){
DF$DAYS.BETWEEN[i]<-1
}
}
Where 'num_obs' is the total number of temperature observations you have.
This would only be correct if you had no missing data on Fridays. Also, if
This might be a little late, but one option for this issue might be to add
a line like
Sys.setenv("DROPBOX_PATH" = "/Users/PersonA/Dropbox/")
to Person A's .Rprofile (on my Mac, this is in my home directory) and the
equivalent (mutatis mutandis) to Person B's .Rprofile and then use
Sys.getenv("D
You may have to tune the curve fitting parameters. Have a look at
arguments 'smoother' and '...' in help("principal.curve"), i.e. by
default principal.curve() uses "smoother" function smooth.spline() in
each iteration and principal.curve() passes down any additional
arguments to it that you give (
Sorry - please ignore. I've rerun-it from scatch, and it worked this time!
D.
On Mon, May 23, 2011 at 11:48 AM, Dimitri Liakhovitski
wrote:
> Sorry for no code - but it's a more of a general question.
> I have read in a data frame ("|"-delimited, .txt).
> daily<-read.table(file="filename.txt",sep
Sorry for no code - but it's a more of a general question.
I have read in a data frame ("|"-delimited, .txt).
daily<-read.table(file="filename.txt",sep="|",header=T)
One of the variables is a factor with 110 levels:
>str(daily$dma_id)
Factor w/ 110 levels "500","501","503",...
108 levels of this
Check the Help archives, google, search cran, as this has been
requested many times. IIRC, Greg Snow's gtools package has a function
that does this.
-- Bert
On Mon, May 23, 2011 at 8:39 AM, Tal Galili wrote:
> Thanks Bery and Richard.
> So how would you suggest to create a histogram function wit
try this -- it scales to the maximum value:
set.seed(242)
z = rnorm(30)
hist_z <- hist(z)
hist_z$counts / hist_z$density # the relation is 15
# why is this 15 ??
# So I can now do:
hist(z)
# change in this statement - scale to the max
y <- density(z)$y * max(hist_z$counts) / max(density(z)$y)
#is
Thanks Bery and Richard.
So how would you suggest to create a histogram function with an added
density plot on top of it?
Here is what I've came up so far (ANY suggestion for improvement will be
welcomed!):
#-
# function
hist.plus.dens
... Which makes the cumulative (discrete) distribution 1, as it should be.
-- Bert
On Mon, May 23, 2011 at 7:59 AM, Tal Galili wrote:
> Thanks Richard!
>
> Contact
> Details:---
> Contact me: tal.gal...@gmail.com | 972-52-72758
Hello, I have some unbalanced panel data that is measured on weekdays only
(i.e., excluding Saturday and Sunday). I would like to get the number of
days between dates such that the number of days between a Friday and a
Monday is 1 (and not 3). Here is some code to illustrate my problem:
library(
Thanks Richard!
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
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Hello,
I am using the function postForm to submit queries to a website. When I submit
one query I get the query contents back, but when I submit a different query to
the same site I get the query results from the previous search. I don't seem to
have this problem using GET format searches.
I
Hi Timothy,
and thanks for the answer. Loops where exactly what I was trying to
avoid as much as possible. My initial idea was that that once I had
recursive indexes at my disposal (which were retrieved over recursive
loops), I could simply use it in a similar manner as we do with indexes
(t
> In order to do this I can use the relation between count and density, but
I
> would like to know if there is a way for me to predict it upfront.
In the code for hist.default, you'll see the line
dens <- counts/(n * diff(breaks))
> Here is an example:
>
> set.seed(242)
Hi all,
I'm looking to add a "density" smoother on top of a hist when Freq=T.
In order to do this I can use the relation between count and density, but I
would like to know if there is a way for me to predict it upfront.
Here is an example:
set.seed(242)
z = rnorm(30)
hist_z <- hist(z)
hist_z$cou
> LIST OF CONVENTIONS/STYLES FOR R:
>
> [1] R coding standards in the R Internals manual
> http://www.cran.r-project.org/doc/manuals/R-ints.html#R-coding-standards
>
> [2] Bioconductor coding standards
> http://wiki.fhcrc.org/bioc/Coding_Standards
>
> [3] Google R style
> [http://google-styleg
Hi,
we want to calculate a linear regression that results from three test
series. (The "Average linear regression" over 3 replication.)
Each test serie consists of the values for 3 times (0min, 25min,
50min) and is repeated three times.
We need such a regression for each combination of subject
Why does this not find a better solution?
> x <- seq(0,2*pi, length=1000)
> x <- cbind(x/(2*pi), sin(x))
> fit1 <- principal.curve(x, plot = TRUE, trace = TRUE, maxit = 100,
+ start = cbind(sort(x[,1]), rep(1, nrow(x
Starting curve---distance^2: 1499.5
Iteration 1---distance^2: 3.114789
Itera
Dear list members,
I am running a rather long forach loop while using 16 cores on a cluster
simultaneously using the package doMC. After approximately 12000 calculations I
get the following error message:
Fehler in calcs.iter(i) :
task 5166 failed - "Indizierung auÃerhalb der Grenzen"
cal
Shuguang Sun gmail.com> writes:
> How could I use function in-visible to user in my code? For example
> 'survfitKM' or 'survfit.formula' in package survival. These functions
> are in-visible to user.
?":::"
e.g.
survival:::survfit.formula
__
R-help
Dear R user,
How could I use function in-visible to user in my code? For example
'survfitKM' or 'survfit.formula' in package survival. These functions
are in-visible to user.
Thanks,
Shuguang
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R-help@r-project.org mailing list
https://stat.ethz.ch/ma
On Thu, 2011-05-19 at 06:43 -0700, guy33 wrote:
> Hey all,
>
> I can't seem to get the princurve package to produce correct results, even
> in the simplest cases. For example, if you just generate a 1 period
> noiseless sine wave, and ask for the principal curve and plot, the returned
> curve is
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