Dear R helpers,
I have one basic doubt about the value of pi. In school, we have learned that
pi = 22/7 (which is = 3.142857). However, if I type pi in R, I get pi =
3.141593. So which value of pi should be considered?
Regards
Vincy
[[alternative HTML version deleted]]
God this listserve is awesome. Thanks to everyone for their ideas.
I'll speed memory test tomorrow and change the code. Thanks again!
Matt
On Sun, May 29, 2011 at 6:44 PM, Ian Gow iand...@gmail.com wrote:
Not a new approach, but some benchmark data (the perl=TRUE speeds up Jim's
suggestion):
Dear Vincy,
I hope that in school you also learned that 22/7 is an approximation.
Please consult your local mathematician for a proof that pi != 22/7.
A quick search will provide you with volumes of information on what pi
is, how it may be calculated, and calculations out to thousands of
digits.
There is an urban legend that says Indiana passed a law implying pi = 3.
(Because it says so in the bible...)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Joshua Wiley
Sent: Monday, 30 May 2011 4:10 PM
To: Vincy Pyne
Cc:
On May 28, 2011, at 20:08 , Natalie Stephenson wrote:
Hi ...
Please could you help with probably a very simple problem I have. I'm
completely new to R and am trying to follow a tutorial using R for Force
Distribution Analysis that I got from ...
On Sun, May 29, 2011 at 11:53 PM, bill.venab...@csiro.au wrote:
There is an urban legend that says Indiana passed a law implying pi = 3.
(Because it says so in the bible...)
Apparently the Fortran language has a DATA statement just for this
purpose. This is allegedly a quote from an early
If you want the weighted variance added to the data frame, don't melt
the data first.
library(Hmisc)
dat-data.frame(ids=c(A,A,A,B,B,B),rate=c(12,14,17,10,8,5),
sample=c(100,80,60,50,40,45))
library(plyr)
library(Hmisc)
ddply(dat, 'ids', transform, wtdvar = wtd.var(rate, sample))
Dear All,
I have 11 variables, and i would like generate combinations of those
variables by fours variables i,e 330 possibility (choose(11,4)).?
After that, make a regression analysis with this 330 possibility ?
is there a program ( or package) to do that ?
Any help would be appreciated
On 29.05.2011 23:19, Chris Chapman wrote:
Uwe -- thank you. No, this occurs on three different machines: two at
work (a Lenovo laptop running Win7-32, plus an HP workstation running
Win7-64) ... and I just tried another Compaq desktop machine at home
running WinXP-32, with the same result.
I
On 29.05.2011 22:52, Alexander Engelhardt wrote:
Hello,
I'm trying to plot a series of pages in a pdf with one main caption for
each page.
Each page should then have two plots in one row, preferably with an own
caption.
I can't plot the main caption in a window, and subsequently plot the two
Thank you very much for your help - everybody, and especially Joshua!
On Mon, May 30, 2011 at 1:21 AM, Joshua Wiley jwiley.ps...@gmail.com wrote:
Hi Philipp,
## Read in your data
## posting the output of dput() would have made our lives easier
d1 - read.table(textConnection(x y
1 a X
2 b
Hi:
Use the PolynomF package. Once you've installed it from CRAN, try the
following code:
library(PolynomF)
example(PolynomF)
The example contains a block of code to construct a list of Hermite
polynomials which are stored in a list object named H. After you've
run the example, try
deriv(H)
On 30-May-11 07:06:57, Peter Langfelder wrote:
On Sun, May 29, 2011 at 11:53 PM, bill.venab...@csiro.au wrote:
There is an urban legend that says Indiana passed a law implying
pi = 3.
(Because it says so in the bible...)
Apparently the Fortran language has a DATA statement just for this
Hello all,
I'm working with the svm (libsvm) implementation from library(e1071).
Currently I'm trying to extend recursive feature elimination (R-SMV) to
work with multiclass classification.
My problem is that if I run svm for a 3 class problem I get a 2-D vector
back from
model$coefs, can
Am 30.05.2011 10:39, schrieb Uwe Ligges:
On 29.05.2011 22:52, Alexander Engelhardt wrote:
Hello,
I'm trying to plot a series of pages in a pdf with one main caption for
each page.
Each page should then have two plots in one row, preferably with an own
caption.
I can't plot the main caption in
Thanks a lot for your hint
marco
2011/5/27 Erich Neuwirth erich.neuwi...@univie.ac.at
arr-1:72
dim(arr)-c(2,4,3,3)
apply(arr,1:2,sum)
[,1] [,2] [,3] [,4]
[1,] 297 315 333 351
[2,] 306 324 342 360
apply(arr,3:4,sum)
[,1] [,2] [,3]
[1,] 36 228 420
[2,] 100 292
Thanks a lot for your help
marco
2011/5/27 Thomas Lumley tlum...@uw.edu
On Fri, May 27, 2011 at 10:06 AM, marco milella vru...@gmail.com wrote:
Hi to everybody
I have an array with dimensions 2,4,3,3.
Wanting to sum the matrices in the first two dimensions, I'm trying to
use
the apply
For my problem I have
#Constraints
b11-0
b21-0
b12-0
b22-0
n1=0
n2=0
n1=1
n1=1
n1+n2=1
In order to use spg I set
Amat-matrix(rbind(c(rep(0,10)),c(rep(0,10)),c(rep(0,2),1,rep(0,7)),
c(rep(0,3),1,rep(0,6)),c(rep(0,10)),c(rep(0,10)),
c(rep(0,6),1,rep(0,3)),c(rep(0,7),1,rep(0,2)),
That's the beauty of this R forum. This forum is full of knowledgeable wizards
and replies received along-with the related discussions pertaining to a simple
harmless question like this enriches us tremendously. Thanks a lot for all your
comments. I am sticking to the value of 'pi' as provided
For an engineer, assuming that pi = 3.142 will
probably enable him to build a very satisfactory
bridge. Assuming that pi = 3.14159265358979323844
will give the circumference of the Earth's orbit
to one millionth of a millimetre.
And just a few more decimal places will get you the
On Fri, May 27, 2011 at 6:12 PM, Bert Gunter gunter.ber...@gene.com wrote:
Jonathan et. al:
Yes, of course, but I'd say Type III error: Right answer to wrong question.
The real question (imho only obviously) is: What data structure
should be used? -- and the standard answer for this sort of
Dear all,
I am stuck with a problem that might be trivial for most of you (and
therefore is a bit embarrassing for me...):
I want to calculate a generalized least squares regression using two
time series (Y depending on X) with an autoregressive correlation
structure of order two (the data
thank u very much Wiliam Dunlap,
your function is very helpful, i been trying to find the right algo,,,and i
saw yours, it works perfectly.
i even checked with system.time function, and it's clear the cum.reset is
the best.
system.time(ave(J, rev(cumsum(rev(is.na(J, FUN=cumsum))
utilisateur
Hi
I am totally new to R and I would like to know how to perform both row and
column normalization (so that both rows and columns sum to zero and have
standard deviation of 1) using R on gene expression data.
Thanks for the help.
--
View this message in context:
Hi,
I have a problem when indexing the corpus. I used the following syntax:
Setwd (c :/)
Library (tm)
Txt = Corpus (DirSource (.); readerControl = list (language = frensh))
an error message comes:
Messages d'avis :
1: In readLines(y, encoding = x$Encoding) :
ligne finale incomplète
Date: Sun, 29 May 2011 23:09:47 -0700
From: jwiley.ps...@gmail.com
To: vincy_p...@yahoo.ca
CC: r-help@r-project.org
Subject: Re: [R] Value of 'pi'
Dear Vincy,
I hope that in school you also learned that 22/7 is an approximation.
Please
Dear ALL
I want to simulate data from Multivariate normal distribution.
GE.N-mvrnorm(25,mu,S)
S -matrix(rep(0,1),nrow=100)
for( i in 1:100){sigma-runif(100,0.1,10);S
[i,i]=sigma[i];mu-runif(100,0,10)}
for (i in 1:20){for (j in 1:20){if (i != j){S [i,j]=0.3*sigma[i]*sigma[j]}}}
for (i in
On 30/05/2011 6:17 AM, rgui wrote:
Hi,
I have a problem when indexing the corpus. I used the following syntax:
Setwd (c :/)
Library (tm)
Txt = Corpus (DirSource (.); readerControl = list (language = frensh))
Capitalization is important in R, so when asking a question, please cut
Hello Steve,
Thanks again for your time.
Since this is part of learning for me, I have set up another system for it
to avoid getting into much problem with my laptop and hence my work.
Whatever I learn in this new system about the installation I can safely
apply to my laptop.
I have tried to get
This is not accepted as good statistical practice, and the resulting
inferential quantities will violate all known statistical principles.
Frank
Mohamed Lajnef wrote:
Dear All,
I have 11 variables, and i would like generate combinations of those
variables by fours variables i,e 330
In other words, Abandon hope, all ye who enter...
-- Bert
(Sorry, couldn't resist. Berlioz's March to the Gallows from
Symphonie Fantastique also played in my head when I read Frank's
comment, but I couldn't put that in the email.)
On Mon, May 30, 2011 at 5:52 AM, Frank Harrell
Dear list,
I would like to set one specific Reference Class field to be of an
arbitrary class. Is there a class that all R objects inherit from? I
thought that ANY was something like this, but obviously that's not true:
inherits(1:3, ANY)
[1] FALSE
Regards,
Janko
[[alternative HTML
Hi
I wonder if you can help me- im trying to do research in R and i cant find
anywhere that explains step by step how to do single, double or triple
exponential smoothing on a time series dataset.
Any help, links or other would be greatly appreciated.
[[alternative HTML version deleted]]
Hello,
I have a dataframe with e.g. three different curves (columns), which I want
to have averaged.
I use the row.Means function, which is working good. However as I have a
time series,
some parts of the three columns have sometimes NA. I want however that if
only one or two columns
are
Hi,
I need some help with this one: how do I check whether a vector is
already present in a list of vectors.
I have seen %in% recommended in a similar case but that obviously does
not work here.
c(1,2,3) %in% list(c(1,2,3), c(4,5,6))
returns
[1] FALSE FALSE FALSE
which makes sense since 1,
Does anyone know of an R implementation of Rao's (1965) described 'Analysis of
Dispersion' which is the generalization of the General Linear Model, Y = XB, to
multiple Y variables and the further use of the method to 'Test of Additional
Information' in one set of Y variables corrected for a
Hi,
this works: apply(mtcars[c(Mazda RX4, Merc 450SLC, Maserati Bora), ],
1, plot)
How do you add labels and additional edit in such a plot? For instance I
want
mpg cyl disp hp dratwt qsec vs am gear carb to be plottet below the
X-axis
--
View this message in context:
Try this:
list(c(1,2,3), c(4,5,6)) %in% list(c(1,2,3))
On Mon, May 30, 2011 at 10:36 AM, Marcin Wlodarczak
mwlodarc...@uni-bielefeld.de wrote:
Hi,
I need some help with this one: how do I check whether a vector is
already present in a list of vectors.
I have seen %in% recommended in a
Hi,
I am trying to install the package for DECORANA onto R, I have tried the CCA
and it does not seem to work. Does anyone know?
Tawanda
On Fri, Nov 13, 2009 at 9:53 AM, kvarpun jallouli.med.am...@gmail.comwrote:
Duncan Murdoch-2 wrote:
On 11/12/2009 11:37 AM, kvarpun wrote:
Duncan
On 30.05.2011 15:36, Marcin Wlodarczak wrote:
Hi,
I need some help with this one: how do I check whether a vector is
already present in a list of vectors.
I have seen %in% recommended in a similar case but that obviously does
not work here.
c(1,2,3) %in% list(c(1,2,3), c(4,5,6))
You
You almost solved your own problem with that last statement. Instead
of comparing apples and oranges, you need to compare oranges and
oranges:
list(c(1,2,3)) %in% list(c(1,2,3), c(4,5,6))
[1] TRUE
list(c(1,2,3)) %in% list(c(1,2,9), c(4,5,6))
[1] FALSE
Sarah
On Mon, May 30, 2011 at 9:36 AM,
Hi Luisa,
it was really difficult to manage to understand what you need. Assuming
that i've done it correctly this should be similar to what you want:
A.matrix - matrix(rpois(n=28,lambda=2),nrow=7)
M.matrix - matrix(0,nrow=dim(A.matrix)[2],ncol=dim(A.matrix)[2])
for(i in 1:dim(df)[2])
for(j
Here is the wiki entry on the Indiana Pi Bill
http://en.wikipedia.org/wiki/Indiana_Pi_Bill
Ravi.
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
bill.venab...@csiro.au [bill.venab...@csiro.au]
Sent: Monday, May 30, 2011
I have not been able to figure out the format for specifying limits, at,
labels in scales - list(relation='free', limits=..., at=..., labels=...)
when there is more than one paneling variable, e.g., xyplot(y ~ x | a*b).
Thanks for any guidance.
Frank
-
Frank Harrell
Department of
Hi Everybody
I have started to learn how to use the XLConnect package and I think it is
going to be very helpful to help me to operate between Excel users.
Is there a function that can export a dataframe to a Excel sheet and
automatically create a named region for every column (excluding the
I hope this modified example will work
set.seed(100)
disease-paste(rep(c('y','n'),50))
gender-c(paste(rep(c('m','f'),25)), paste(rep(c('f','m'),25)))
mcp-rnorm(100, mean=1000,sd=600)
age-rnorm(100,mean=32,sd=20)
dat-data.frame(disease=disease,sex=gender,Dr_age=age,MCP=mcp)
Hello,
I just installed Rv2.13,lme4 v0.999375-40, and matrix 0.999375-50. When I
try to launch lme4 I get the following error message:
library(lme4)
Error in inDL(x, as.logical(local), as.logical(now), ...) :
function 'cholmod_start' not provided by package 'Matrix'
Error: package/namespace
Frank:
1. Note that x and y scales can be specified separately by scales =
list(x = list(...), y = list(...))
2. For at, labels, etc. ?xyplot says:
The location of tick marks along the axis (in native
coordinates), ** or a list as long as the number of panels describing
tick locations for
On May 30, 2011, at 11:18 AM, Frank Harrell wrote:
I have not been able to figure out the format for specifying limits,
at,
labels in scales - list(relation='free', limits=..., at=...,
labels=...)
when there is more than one paneling variable, e.g., xyplot(y ~ x |
a*b).
Thanks for any
Thanks very much David and Bert. That did it.
Frank
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context:
http://r.789695.n4.nabble.com/Syntax-for-lattice-scales-argument-with-a-b-tp3560973p3561131.html
Sent from the R help mailing list archive
On May 30, 2011, at 11:30 AM, 1Rnwb wrote:
I hope this modified example will work
set.seed(100)
disease-paste(rep(c('y','n'),50))
gender-c(paste(rep(c('m','f'),25)), paste(rep(c('f','m'),25)))
mcp-rnorm(100, mean=1000,sd=600)
age-rnorm(100,mean=32,sd=20)
Hi, I'm looking for help extracting some information of the zillow website.
I'd like to do this for the general case where I manually change the address
by modifying the url (see code below). With the url containing the address,
I'd like to be able to extract the same information each time. The
On 2011-05-29 23:08, Matthew Keller wrote:
God this listserve is awesome. Thanks to everyone for their ideas.
I'll speed memory test tomorrow and change the code. Thanks again!
Since you're dealing with a vector of ~ 1e8 elements, you might
find that (at a probably small cost of time) you can
Hi all,
I'm working on a dataframe (webs) with two variables ($size, $vspe), and two
grouping factors ($type, $mutu). I would like to do a lattice plot wherein
the levels of $mutu would be different shapes, and the levels of $type
different colors. Additionally, I'd like to add a regression line
Hi,
Am not sure if my code itself is correct. Here's what am trying to do:
Minimize integration of a function of gaussian distributed variable 'x' over
the interval qnorm(0.999) to Inf by changing value of parameter 'mu'. mu is
the shift in mean of 'x'.
Code:
# x follows gaussian distribution
#
On 05/30/2011 04:14 PM, Uwe Ligges wrote:
On 30.05.2011 15:36, Marcin Wlodarczak wrote:
Hi,
I need some help with this one: how do I check whether a vector is
already present in a list of vectors.
I have seen %in% recommended in a similar case but that obviously does
not work here.
c(1,2,3)
You can put print statements into your integrand to see
what is happening:
integrand-function(x){
+ on.exit(print(cbind(x, fx))) ;
+ fx -
(e*pnorm((qnorm(p)+sqrt(R)*x)/sqrt(1-R))*dnorm(x))^2/dnorm(x+mu)
+ fx
+ }
mu-c(-1)
output-optim(par=mu, fx2, method=BFGS)
x
Hi,
I am trying to transform a data matrix into a vector and have not be able to
accomplish want I am looking for. The setup is as follows. I start with a 3 x
3 matrix:
5 1 3
3 3 2
1 2 4
I would like to transorm it into a 27 x 1 vector of the follwing form
5
5
5
1
1
1
3
3
3
.
.
.
1
1
1
2
2
tmp - matrix(c(5,3,1, 1,3,2, 1,2,4), 3, 3)
tmp
[,1] [,2] [,3]
[1,]511
[2,]332
[3,]124
rep(t(tmp), each=3)
[1] 5 5 5 1 1 1 1 1 1 3 3 3 3 3 3 2 2 2 1 1 1 2 2 2 4 4 4
On Mon, May 30, 2011 at 4:22 PM, Charles Ellis charl...@mairesearch.comwrote:
Hi,
I
Hi,
I used GAM function with this code:
library(mgcv)
modgam=gam(Z~X+Y,data=table)
pred=predict(modgam,type=response)
vis.gam(modgam,view=c(X,Y),plot.type=contour,zlim=c(1,16))
My problem is on my figure, the predicted values begin by 0 to 16 with an
interval equal 2:
1- I want that the
Hi Charles,
Try
rep(c(tmp), each = 3)
HTH,
Jorge
On Mon, May 30, 2011 at 4:22 PM, Charles Ellis wrote:
Hi,
I am trying to transform a data matrix into a vector and have not be able
to accomplish want I am looking for. The setup is as follows. I start with
a 3 x 3 matrix:
5 1 3
3 3
Hello, I have a C++ code for 2D random walks with traps and I want to convert
it in a R code with its syntaxs, can anyone help???
It's easy for me to adapt the body but I want help with the beginig (variable
declaration) and th end exporting the output to a file ( like write.table() or
Hello,
I would like to create a group variable that is based on the values of three
variables:
For example,
dat - data.frame(A=c(1,1,1,1,1,2,2,2,2,2),
B=c(1,1,1,5,5,5,9,9,9,9),
C=c(1,1,1,1,1,2,2,7,7,7))
dat
A B C
1 1 1 1
2 1 1 1
3 1 1 1
4 1
Hi,
There are probably much better ways, but try this
transform(dat, group = as.numeric(factor(paste(A,B,C, sep=
HTH,
baptiste
On 31 May 2011 09:47, Mendolia, Franco fmendo...@mcw.edu wrote:
Hello,
I would like to create a group variable that is based on the values of three
On Mon, 30 May 2011 16:47:45 -0500,
Mendolia, Franco fmendo...@mcw.edu wrote:
Hello, I would like to create a group variable that is based on the
values of three variables:
For example,
dat - data.frame(A=c(1,1,1,1,1,2,2,2,2,2),
B=c(1,1,1,5,5,5,9,9,9,9),
Brilliant! Works like a charm. Thanks both.
Cheers,
Charles
From: Richard M. Heiberger [mailto:r...@temple.edu]
Sent: Monday, May 30, 2011 4:36 PM
To: Charles Ellis
Cc: r-help@R-project.org
Subject: Re: [R] Transforming a data matrix into a vector
tmp - matrix(c(5,3,1, 1,3,2, 1,2,4), 3, 3)
Hello,
interesting would be to estimate what would have happened,
if the law would have been installed.
What consequences would that have caused?
Could it make the earth flat again?
And which shape would the earth have, if pi = 3?
And how would a football-ball look like?
(Like american
Read the data using scan():
#
# a1 a2 a3 a4
# ----
# b1 b2 b3 b1 b2 b3 b1 b2 b3 b1 b2 b3
# --- --- ------ --- ------ --- ------
Hmm, I think the difference would be much less dramatic. The rest of
the world would just have to update its Midwest/Indiana list to:
Weird Things about Indiana:
1) Daylight Savings Time (to save candles?)
2) Still has horses, carts, and buggies
3) Does not believe in math (really just an
This is really a question about the help file for gl.
The arguments are
gl(n, k, length = n*k, labels = 1:n, ordered = FALSE)
'n' is the number of factor levels. That seems to be easy enough
'k' is called the number of replications. This is perhaps not the best way
to express what it is. k
Pi is an irRATIOnal number, meaning that it is not equal to the ratio of any
integers (whole numbers). Hence, 22/7 is ONLY an approximation. The
built-in value for pi in R is also just an approximation (pi has no terminal
digit on the right of the decimal point so any finite number of digits will
I was referring to the height of the pdf. The 1st distribution is the simple
uniform between 0 and 1, the second is also uniform, but on a discontinuous
region. The first will generate numbers between 0 and 1 with equal
probability, the second will be similar but only up to 0.99 with the very
Greetings -
I'm battling POSIXct, as per the code below. My input is actually an XL
file, but the weird results below correctly model what I am seeing in my
program.
Before I punt and use lubridate or timeDate, could anyone please help me
understand why POSIXct forces my variable back to
On May 30, 2011, at 10:20 PM, Galen Moore wrote:
Greetings -
I'm battling POSIXct, as per the code below. My input is actually
an XL
file, but the weird results below correctly model what I am seeing
in my
program.
Before I punt and use lubridate or timeDate, could anyone please
Perhaps because the timezone is specified as a character string and not a
date-time object complete with timezone.
From the help filr for as.POSIXct.numeric:
origin:a date-time object, or something which can be coerced by
as.POSIXct(tz=GMT) to such an object.
Note the coercion.
Bill
I think that has something to do with it. I'll keep chewing on things and
see what I can make work.
I've reached a point where I get the right answer for the wrong reasons
;-)
Many thanks,
Galen
-Original Message-
From: bill.venab...@csiro.au [mailto:bill.venab...@csiro.au]
Sent:
Thank you, David.
I am not subtracting the seconds to convert POSIXct's GMT to MDT, and don't
understand why I should need to.
Any hints, however, as to why
dateP - as.POSIXct(tstamp, origin=1970-01-01, tzone=MDT)
returns a date in the correct tz in my first instance below, yet
dateP2 -
I want to use the rtmvt from the {tmvtnorm} package using the gibbs
algorithm but how to i specify the nested function rtmvnorm to use gibbs as
well?
Right now I am using the code:
for (i in 1:g){
for (j in 1:n){
sgamma[,,i,j] = rtmvt(n=50,
I once knew someone who thought that a 1-sided upper 99% confidence limit
for the mean with n=7 was calculated by multiplying the standard error of
the mean by pi.
-Don
On 5/30/11 6:00 PM, Bentley Coffey bentleygcof...@gmail.com wrote:
Pi is an irRATIOnal number, meaning that it is not equal to
Indeed!
pt(pi, df=6)
[1] 0.9899863
Simon.
On 31/05/11 14:38, MacQueen, Don wrote:
I once knew someone who thought that a 1-sided upper 99% confidence limit
for the mean with n=7 was calculated by multiplying the standard error of
the mean by pi.
-Don
On 5/30/11 6:00 PM, Bentley
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