i tried to run your code, this is how/why you got NaN:
mu-0.2
sig-0.2
S0-100
j-0.2
dt-1/252
int-0.1
i-0
is.nan
function (x) .Primitive(is.nan)
k-rnorm(1,0,1)
k
[1] 0.3214954
theta-ifelse((k(int*dt)),1,0)
theta
[1] 0
m-rnorm(1)
m
[1] -0.4525731
gam-qnorm(m,0,1)
Warning
On Fri, 2008-11-21 at 22:37 -0800, subbudas wrote:
hello everyone ,
i have written some code in R for jump diffusion model.
the code generates answer as
NaN
There were 50 or more warnings (use warnings() to see the first 50)
my code is
mu-0.2
sig-0.2
S0-100
j-0.2
dt-1/252
int-0.1
Hi,
I am residing in Taipei, Taiwan. My windows has English R (2.7.2)
installed. I noticed that the CRAN mirror for Taipei is Taiwan
(Taipeh), Is Taipeh a Deutsch or English name?
Thanks,
cruz
__
R-help@r-project.org mailing list
On Fri, Nov 21, 2008 at 5:50 AM, Gerit Offermann [EMAIL PROTECTED] wrote:
Dear list,
thanks to your help I managed to find means of analysing my data.
However, the whole data set contains 264 variables. Of which some are
factors, others are not. The factors tend to be grouped, e.g.
I need one more clarification here :
Here I did :
fn - function(i) return(list(i, i^2))
ss = sapply(1:4, fn)
Here the object ss should be a matrix object :
is.matrix(ss)
However I feel it lacks some the matrix object properties. For example the
syntax min(ss[1,]) generates an error :
Error in
megh wrote:
I need one more clarification here :
Here I did :
fn - function(i) return(list(i, i^2))
ss = sapply(1:4, fn)
Here the object ss should be a matrix object :
is.matrix(ss)
However I feel it lacks some the matrix object properties. For example the
syntax min(ss[1,]) generates an
Try str(ss) to see what it really looks like. You probably want:
fn - function(i) c(i, i^2)
On Sat, Nov 22, 2008 at 4:54 AM, megh [EMAIL PROTECTED] wrote:
I need one more clarification here :
Here I did :
fn - function(i) return(list(i, i^2))
ss = sapply(1:4, fn)
Here the object ss
I want to draw following plot, given here
http://www.2shared.com/file/4327128/830b82c4/pic.html
for the following data :
dat - cbind(rnorm(100), sample(c(1:4), 1000, T))
colnames(dat) - c(data,level)
Here x-axis should be on data and y-axis is for level and z-axis should
be to display the
Hi R users,
I am doing some image analysis and I'd like to know if anyone knows of:
1. a way to specify a rectangle (using xy coordinates) within an image to
have R evaluate using xy pixel coordinates and
2. a way to get brightness readings for that part of the image.
If there's some way to
I have question: how can I put the value on the bar chart.
This my code:
barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber)
This's my data:
Year Hr Family Tuberize
1 2007 20 A 0.26
2 2007 20 B 6.08
3 2007 20 C 0.00
4 2007 20 D 0.27
5 2008 20
List,
I would like to set a variable to hold, say, the size of my plots in a
Sweave document. i.e. something like the following in my '.Rnw' file:
==
smallPlotSize = 4
fig1, echo=false, results=hide,
Dear R guru,
I am Saikat Sarkar working as a researcher of Economics in Tampere
University, Finland. I am trying to estimate some Garch related tests with
Bayesian analysis by R programme.
I am not good in R but trying to survive.
Anyway I have the coding but not working properly. I have tried
A short addendum, resulting from an off-list discussion:
The reason why Colleen's code failed was raising a negative base to a
fractional exponent in the third state equation for certain sets of
parameters, esp. fractional values of beta.
Old versions of odesolve broke down, and recent
I thought I'd share a few workaround routes I've considered (my
attempt at using Amos' Fortran routines failed miserably -- if anyone
is interested i can explain what I tried),
- Ryacas seems to provide a very simple way to evaluate bessel
functions with complex argument,
2008/11/22 RON70 [EMAIL PROTECTED]:
I want to draw following plot, given here
http://www.2shared.com/file/4327128/830b82c4/pic.html
for the following data :
dat - cbind(rnorm(100), sample(c(1:4), 1000, T))
colnames(dat) - c(data,level)
Here x-axis should be on data and y-axis is for level
SM == Stavros Macrakis [EMAIL PROTECTED]
on Fri, 21 Nov 2008 14:44:37 -0500 writes:
library(Kendall) Kendall(1:3,1:3)
SM WARNING: Error exit, tauk2. IFAULT = 12 tau = 1,
SM 2-sided pvalue =1
SM I believe Kendall tau is well-defined for this case and
SM the reported
You need to also provide the data that your code is using since the
error message indicates that the problem is probably in the way that
the object 'a' is defined and there is no indication of what it looks
like. You should either provide the output of str(a), or the output
of 'dput(a)' so we
Hi
Wondering if anyone knows of a package that does Nested Clade Analysis?
Thanks
--
View this message in context:
http://www.nabble.com/Nested-Clade-Analysis-tp20637180p20637180.html
Sent from the R help mailing list archive at Nabble.com.
__
Here is a function I have used. I got it from the list, and
unfortunately don't remember who to credit:
c.Factor -
function (x, y)
{
newlevels = union(levels(x), levels(y))
m = match(levels(y), newlevels)
ans = c(unclass(x), m[unclass(y)])
levels(ans) = newlevels
class(ans)
On 21/11/2008 2:30 PM, Rajarshi Guha wrote:
Hi, I'm using rgl to generate a 3D surface plot and I'm struggling to
get the lighting correct. Currently the surface gets plotted, but is
very 'shiny'. On rotating the view, I get to see parts of the surface
- but overall I don't see much detail
On Thu, 2008-11-20 at 19:43 -0800, Andrew J. Rominger wrote:
Dear list,
First off, let me offer my apologies, I know this is a very basic
question. After amassing a large number of objects (from multiple
projects) in one working directory, I'd like to be able to start using
different
You can try this also:
unlist(list(f1, f2))
On Fri, Nov 21, 2008 at 3:15 PM, udi cohen [EMAIL PROTECTED] wrote:
Hi all,
I hope it's not too trivial for the list - I'm trying to concatenate
two factor arrays, and obtain the following:
f1-factor(c(a,a,b))
f1
[1] a a b
Levels: a b
On 22/11/2008 2:01 AM, Kyle Matoba wrote:
List,
I would like to set a variable to hold, say, the size of my plots in a
Sweave document. i.e. something like the following in my '.Rnw' file:
==
smallPlotSize = 4
fig1,
Try something like the code below. Unfortunately I cannot remember who wrote
the code.
===
my.values=10:15
x - barplot(my.values, ylim=c(0,11))
text(x, my.values, wibble, pos=3) # always does what you want
#
Thanks a lot for the pointer to rgl.pop() - that works (as does
looking at the examples!)
On Nov 22, 2008, at 10:28 AM, Duncan Murdoch wrote:
On 21/11/2008 2:30 PM, Rajarshi Guha wrote:
Hi, I'm using rgl to generate a 3D surface plot and I'm struggling
to get the lighting correct.
Following the example in
https://stat.ethz.ch/pipermail/r-help/2006-January/086985.html
You might want to try something like:
barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber,
+ panel = function(y,x,...){
+ panel.barchart(x,y,...)
+
On second look I see that although values appear in every panel, they
are the same in every panel as well. Some sort of use of the
subscripting facility probably needs to be employed.
--
David Winsemius
On Nov 22, 2008, at 1:05 PM, David Winsemius wrote:
Following the example in
To follow Dieter's comment,
You can in fact fit a data to a sine in Excel using LINEST. I've done
it. I don't recommend it :-) .
What I did was create columns containing sin(x) and cos(x) , roughly
speaking, and fit using
LINEST([y=values],{sines, cosines},...)
Ya need the cosines or
No, it only needed closer attention to the example:
This gets you to the point where you need to fix the y scale settings
but the values are properly cycled through.
barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber,
panel = function(y,x,...){
On Sat, Nov 22, 2008 at 9:04 AM, Martin Maechler
[EMAIL PROTECTED] wrote:
SM I believe Kendall tau is well-defined for this case...
The real question is *WHY* there needs to be a separate package 'Kendall'
when R itself does everything you want and does not show any problems?
Thanks for
Dear R community,
I am trying to apply a simple operation to several dataframes (e.g.
nrow) and cannot get the looping to work. My objective is to get an output
that indicates me the number of rows for every dataframe.
c
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1 1 11 21 31 41 51 61 71 81 91
2 2
?get
names - c('c', 'cc')
for (i in names){
print(nrow(get(i)))
}
On Sat, Nov 22, 2008 at 4:06 PM, Georg Ehret [EMAIL PROTECTED] wrote:
Dear R community,
I am trying to apply a simple operation to several dataframes (e.g.
nrow) and cannot get the looping to work. My objective is to
On Nov 21, 2008, at 11:28 AM, Douglas Bates wrote:
As tempting as it may be to want to have several dials and knobs on
statistical models to tune their behavior, we still need to be careful
to specify a mathematical model that is consistent.
-- Douglas Bates
Yet another
I need help determining the unique columns of a matrix and the numbers of each
unique column. For example, let's say I have a matrix with 6 columns, 2 of
these are filled with the value of 1 and the other 4 are filled with the value
of 0. I would then like to have a command that tells me what
It seems that there is no Hong Kong in maps' world cities.
world.cities[substr(world.cities$name,1,3)==Hon,]
namecountry.etcpop latlong capital
14623 Honami Japan 26040 33.61 130.68 0
14624 Honaz Turkey 8073 37.75
Goal:
Suppose you have a vector that is a discrete variable with values ranging
from 1 to 3, and length of 10. We'll use this as the example:
y - c(1,2,3,1,2,3,1,2,3,1)
...and suppose you want your new vector (y.new) to be equal in length to the
possible discrete values (3) times the length
On Sat, Nov 22, 2008 at 12:00 PM, zerfetzen [EMAIL PROTECTED] wrote:
Goal:
Suppose you have a vector that is a discrete variable with values ranging
from 1 to 3, and length of 10. We'll use this as the example:
y - c(1,2,3,1,2,3,1,2,3,1)
...and suppose you want your new vector (y.new) to
Try this:
outer(y, sort(unique(y)), ==)+0
On Sat, Nov 22, 2008 at 3:37 PM, zerfetzen [EMAIL PROTECTED] wrote:
Goal:
Suppose you have a vector that is a discrete variable with values ranging
from 1 to 3, and length of 10. We'll use this as the example:
y - c(1,2,3,1,2,3,1,2,3,1)
...and
On Sat, 22 Nov 2008 10:00:18 -0800 (PST)
zerfetzen [EMAIL PROTECTED] wrote:
Goal:
Suppose you have a vector that is a discrete variable with values
ranging from 1 to 3, and length of 10. We'll use this as the example:
y - c(1,2,3,1,2,3,1,2,3,1)
...and suppose you want your new vector
On Sat, Nov 22, 2008 at 10:20 AM, jim holtman [EMAIL PROTECTED] wrote:
c.Factor -
function (x, y)
{
newlevels = union(levels(x), levels(y))
m = match(levels(y), newlevels)
ans = c(unclass(x), m[unclass(y)])
levels(ans) = newlevels
class(ans) = factor
ans
}
This
You are right. union used 'unique(c(x,y))' and I am not sure if
'unique' preserves the order, but the help page seems to indicate that
an element is omitted if it is identical to any previous element ;
this might mean that the order is preserved.
On Sat, Nov 22, 2008 at 11:43 PM, Stavros
One way is to 'paste' together the values in a column and then use
'table' to count them.
'duplicated' can probably do the same thing with the MARGIN option to
find the duplicated one. You still them have to find the original
ones.
On Sat, Nov 22, 2008 at 3:42 PM, Salas, Andria Kay [EMAIL
On Sun, 23 Nov 2008, jim holtman wrote:
You are right. union used 'unique(c(x,y))' and I am not sure if
'unique' preserves the order, but the help page seems to indicate that
an element is omitted if it is identical to any previous element ;
this might mean that the order is preserved.
It
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