Hi, I have following optimization problem:
Min: x1 + x2 +...+ x7
subject to:
x1 + x2 >= 80
x2 + x3 >= 65
x3 + x4 >= 40
all xi are ***positive integer***.
Can somebody help me in this optimization problem?
Thanks for your help
__
R-help@r-project.org
Dear all, I am looking for some procedure to apply 'ifelse' condition on
function. I have created an alternative to lapply() function with exactly same
set of arguments named lapply1(), however with different internal codes.
Therefore I want something like, if (some condition) then call lapply1(
Dear all, while executing some function, there are some custom messages popping
up onto the R console and I do not want to see them. While looking into the
corresponding codes of those function, I see that those are coming from
message() function.
Is there any way to stop those messages coming
Let say i have a square matrix and applied the 'vech' operator to stack the
lower triangular elements into a vector:
> Mat <- matrix(1:25, 5)
> Mat
[,1] [,2] [,3] [,4] [,5]
[1,] 1 6 11 16 21
[2,] 2 7 12 17 22
[3,] 3 8 13 18 23
[4,] 4 9 14 19 24
Let say, I have a character vector of arbitrary length:
Vector <- c("a", "b", "c")
Using that vector I would like to create a matrix (with number of columns as 2)
with all pairwise combinations of those elements, like:
Vector <- c("a", "b", "c")
Mat <- rbind(c("a", "b"), c("a", "c"), c("b",
Dear all, I have following kind of character vector:
Vec <- c("344426", "dwjjsgcj", "123sgdc", "aagha123", "sdh343asgh", "123jhd51")
Now I want to split each element of this vector according to numeric and string
element. For example in the 1st element of that vector, there is no string
elemen
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
Hi all, I am to find some way on how I can tell R to use this small number
10^-20 as zero by default. This means if any number is below this then that
should be treated as negative, or if I divide something by any number less than
that (in absolute term) then, Inf will be displayed etc.
I have
\D+)(\\d+)(\\D+)(\\d +)",
c)[[1]]
[1] "ABCFR" "34564" "." "354" "IJVEOJC" "3434""."
"36453"
Can you please tell me how can I modify that?
Thanks,
On Sun, Feb 13, 2011 at 11:10 PM, Gabor Grot
Please consider following string:
MyString <- "ABCFR34564IJVEOJC3434"
Here you see that, there are 4 groups in above string. 1st and 3rd groups
are for english letters and 2nd and 4th for numeric. Given a string, how can
I separate out those 4 groups?
Thanks for your time
[[alternative
I need some help in defining a "print" method for my new S4 class
definition. So fer I have worked like this:
setClass("MyClass", sealed=F, representation(slot1 = "list",#create a
new class
slot2= "vector",
slot3 = "vector",
slot4 = "vector"))
I am looking for an elegant way how I can test the equality of lengths of
multiple vectors.
For example, this is working fine:
> length(rnorm(4)) == length(rnorm(5))
[1] FALSE
However this is not:
> length(rnorm(4)) == length(rnorm(5)) == length(rnorm(6))
Error: unexpected '==' in "length(rnorm(
Dear all, I need to download an excel file from net, on which I have address
like http://www.2shared.com/file/MMSMWv4B/MyData.html. Can I somehow
directly download this file into my R workbook?
Thanks,
[[alternative HTML version deleted]]
__
R-
Dear all, can somebody point me from where to download "rcompgen" package?
CRAN does not seem to hold that.
Installing this package through install.packages() tells this package is not
available.
Thanks
[[alternative HTML version deleted]]
__
R
Hi there, can anyone tell me how to extract to values of a particular slot for
some S4 object? Let take following example:
> library(fOptions)
> val <-GBSOption(TypeFlag = "c", S = 60, X = 65, Time = 1/4, r = 0.08, b =
> 0.08, sigma = 0.30)
> val
Title:
Black Scholes Option Valuation
Call:
Dear friend, I have to construct some recursive algorithm for which I used some
for loop like:
res <- vector(length=1)
res[1] = 0
for (i in 2:(1+1)) res[i] <- res[i-1]*some function
I have noticed that this is taking too much time. Is there any way to speed up
things?
Thanks,
I dont know whether I am missing something or not:
> head(read.zoo(file="f:/dat1.txt", header=T, sep=",", format = "%m/%d/%Y
> %H:%M:%S"), tz="GMT")
data.open data.high data.low data.close
2010-10-15 73.7 73.7 73.7 73.7
2010-10-15 73.8 73.8 73.8
ects do not work if the index entries in ‘order.by’
are not unique
Is it a bug or a rule that for any function, placing of it's arguments matter?
Thanks,
--- On Sat, 10/16/10, Megh Dal wrote:
> From: Megh Dal
> Subject: Re: [R] Problem with merging two zoo objects
> To: &
th merging two zoo objects
> To: "Megh Dal"
> Cc: r-h...@stat.math.ethz.ch
> Date: Saturday, October 16, 2010, 12:11 AM
> On Fri, Oct 15, 2010 at 2:20 PM, Megh
> Dal
> wrote:
> > Dear all, I have following 2 zoo objects. However when
> I try to merge those
Dear all, I have following 2 zoo objects. However when I try to merge those 2
objects into one, nothing is coming as intended. Please see below the objects
as well as the merged object:
> dat11
V2 V3 V4 V5
2010-10-15 13:43:54 73.8 73.8 73.8 73.8
2010-10-15 13:44:15 7
again the same against
"envir". By putting so, what I am going to tell R?
Thanks,
--- On Thu, 10/14/10, Joshua Wiley wrote:
> From: Joshua Wiley
> Subject: Re: [R] Query on save.image()
> To: "Megh Dal"
> Cc: r-h...@stat.math.ethz.ch
> Date: Thursday, O
Can anyone please tell me how can use save.image() function if it is placed
within a function (i.e. some level up from the base level environment)? Here I
experimented with following codes:
#rm(list=ls())
fn <- function() {
x <- rnorm(5)
save.image("f:/dat.RData")
}
fn()
However I se
Suppose I have following arbitrary matrix:
> set.seed(1)
> mat <- matrix(rnorm(6), 3, 2)
> mat
[,1] [,2]
[1,] -0.6264538 1.5952808
[2,] 0.1836433 0.3295078
[3,] -0.8356286 -0.8204684
Now I want to make a simple object like (character type):
"-0.6264538,1.5952808;0.1836433,0.3
Hi, I was trying to split the following matrix "dat":
> set.seed(1)
> dat <- matrix(rnorm(4*16), 4, 16)
> dat
[,1] [,2] [,3][,4][,5][,6]
[,7] [,8][,9] [,10] [,11]
[1,] -0.6264538 0.3295078 0.5757814 -0.62124058 -0.016
Hi, I want to split a text to seperate numerical and non-numerical portions of
that. For example suppose I have a text "abc 3456" and I want to split in 2
parts like "abc" & "3456".
Is there any function to do that?
Thanks,
__
R-help@r-project.org ma
Hi, is there any way to say: "this class 'x' is a S3 class?" For example what
is the type of class "data.frame"? Is it a S3 class or S4?
How can I get a complete list of all S3 classes currently available?
Thanks,
__
R-help@r-project.org mailing list
bject of class "numeric" is not valid for slot "y1" in an
object of class "b"; is(value, "character") is not TRUE
Here my questions are:
1. Why I am getting the prototype object in next to previous example for slots
x & y?
2. Why just previous exampl
Here I am having problem to define a subclass, specially if I define that
subclass after defining initialize() method for its superclass. Here is my code:
> setClass("a", representation=list(x="numeric", y="numeric"),
> prototype=list(x=rnorm(10), y=rnorm(10)))
[1] "a"
> setMethod("initialize",
Thanks Duncan, I understood. Your explanation is really great. Thank you so
much for your time.
--- On Wed, 7/21/10, Duncan Murdoch wrote:
> From: Duncan Murdoch
> Subject: Re: [R] Objects within environment
> To: "Megh Dal"
> Cc: r-h...@stat.math.ethz.ch
> Date: Wed
7/21/10, Duncan Murdoch wrote:
> From: Duncan Murdoch
> Subject: Re: [R] Objects within environment
> To: "Megh Dal"
> Cc: r-h...@stat.math.ethz.ch
> Date: Wednesday, July 21, 2010, 4:48 PM
> On 21/07/2010 5:57 AM, Megh Dal
> wrote:
> > Hi all, I
Hi all, I have following environments loaded with my current R session:
> search()
[1] ".GlobalEnv""package:stats" "package:graphics"
"package:grDevices"
[5] "package:utils" "package:datasets" "package:methods" "Autoloads"
[9] "package:base"
How can I find the ob
Dear all, I need to download some data from this webpage:
http://markets.ft.com/ft/markets/researchArchive.asp
Notable thing here is that there are some "fields" to be selected to get the
desired data. Is there any R facility to do this directly?
Obviously I can do it manually and then just co
Hi all, can somebody help me to split a time series (zoo) object on monthwise.
For example, suppose I have following time series object:
library(zoo)
dat1 <- zooreg(rnorm(300), start=as.Date("2009-01-01"), frequency=1)
>From dat1, I want to create a list-object dat2 like:
dat2[[1]] <- all obser
Dear all,
Please forgive me if there is a duplicate post; my previous mail perhaps didnt
reach the list...
Let say I have following time series
library(zoo)
> dat1 <- zooreg(rnorm(10), start=as.Date("2010-01-01"), frequency=1)
> dat1[c(3, 7,8)] = NA
> dat1
2010-01-01 2010-01-02 2010-01-03
Hi all, good morning,
My question is not really R related rather a practical problem and wondering if
stat-gurus here can show some light how that can be solved with some
statistical/mathematical tool.
I have some 10 items on which 10,000 viewers put their views based on some 12
attributes say
n wrote:
From: jim holtman
Subject: Re: [R] Handling character string
To: "Megh Dal"
Cc: "Erik Iverson" , r-h...@stat.math.ethz.ch
Date: Saturday, June 12, 2010, 10:18 PM
This is probably what you want:
> sub("^[[:space:]]*", "", " Now is the ti
'', str)" etc, none is working if space is at the 1st
position.
What would be the correct approach?
Thanks,
--- On Sat, 6/12/10, Erik Iverson wrote:
From: Erik Iverson
Subject: Re: [R] Handling character string
To: "Megh Dal"
Cc: r-h...@stat.math.ethz.ch
Date: Saturday, June
Dear all,
Is there any R function to say these 2 character strings "temp" and " temp"
are actually same? If I type following code R says there are indeed different :
> "temp" == " temp"[1] FALSE
Is there any way out?
[[alternative HTML version deleted]]
___
Hi all, Here I am trying to implement the switch() function to choose value of
a variable depending on the value of an input variable :
temp1 <- "1"
temp1.name <- switch(temp1,
1 == "aa",
2 == "bb",
Dear falks, here I have written following function :
fn <- Vectorize(function(x = 1:3, y = 3:6) {
x <- matrix(x, nrow=1)
y <- matrix(y, ncol=1)
dat <- apply(x, 2, function(xx) {
apply(y, 1, function(yy) {
retu
] [RGL] Need help to modify current plot
To: "Megh Dal"
Cc: r-h...@stat.math.ethz.ch
Date: Tuesday, May 18, 2010, 3:51 PM
Megh Dal wrote:
> Dear folks, I have created a plot on RGL device :
> x = 1:6
> y = seq(-12, 5, by=1)
> z = matrix(0, length(y), length(x))
> z[13,3
Dear folks, I have created a plot on RGL device :
x = 1:6
y = seq(-12, 5, by=1)
z = matrix(0, length(y), length(x))
z[13,3] = 1; z[13,4] = 1.011765
surface3d(x, y, t(z), col=rainbow(1000))
grid3d(c("x-", "y-", "z"))
Now I want to draw 2 lines along x=3 & x=4, over the surface (with different
c
Hi all, previously I submitted this thread through Nabble which seems fail
therefore sending it again
suppose I have written following function :
> fn = function(x) return(x+x^2)
> fn
function(x) return(x+x^2)
Here you see, if I type only the function name all inside information of this
Can anyone please tell me how to define a "list". Suppose I want to define a
list object "result" with length n then want to fill each place of "result"
with different objects. For e.g.
i=1
result[1] = rnorm(1)
i=2
result[2] = rnorm(2)
...
i=n
result[n] = rnorm(n)
What wo
ector
> To: [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED]
> Date: Sunday, October 12, 2008, 1:06 AM
> Hi Megh,
> Try this:
>
> x<-c(100,96,88,100,100,96,80,68,92,96,88,92,68,84,84,88,72,88,72,88)
> x1 = sample(x, 5, replace=FALSE)
>
> x[ ! x %in% x1]
>
> HTH,
I have 2 vecros :
x<-c(100,96,88,100,100,96,80,68,92,96,88,92,68,84,84,88,72,88,72,88)
x1 = sample(x, 5, replace=FALSE)
Now i want to get remaining values of vector "x" those are not member of vector
"x1". Can anyone please tell me how to do that?
__
R
Hi,
I am trying to draw a Q-Q plot, however got following error.
> library(sn)
> library(car)
> dat1 = rst(1000, 0, 1, 0, 2)
> qq.plot(dat1, "st", 0, 1, 0, 9)
Error in plot.window(...) :
invalid value specified for graphical parameter "las"
Can anyone please tell me why this error is coming?
$year
[1] "2008"
$month
[1] "06"
$day
[1] "23"
$`svn rev`
[1] "45970"
$language
[1] "R"
$version.string
[1] "R version 2.7.1 (2008-06-23)"
--- On Thu, 10/2/08, Uwe Ligges <[EMAIL PROTECTED]> wrote:
> From: Uwe Ligges <[EM
I have drawn a 3D scatter plot :
library(mnormt)
library(scatterplot3d)
dat = cbind(rmnorm(3, rep(0,2), diag(2)), 1:3)
scatterplot3d(dat)
Now I want to do 2 things :
1 : In the Z-axis (i.e. height), I want to see only numbers 1,2,3, etc NOT,
1,1.5,2,2.5.
2. I want to add two Horizo
When I try to load "fBasics" package, I get following error/warning :
> library(fBasics)
Loading required package: fImport
Loading required package: fSeries
Loading required package: fBasics
Loading required package: fImport
Loading required package: fSeries
Loading required package: fBasics
Loadi
I want to generate a valid variance-covariance matrix. One way could be to
generate some random sample from multivariate normal distribution and then
calculate cov. matrix. Another way could be to sample from wishart distribution
itself. However both cases need a valid i.e. PD covariance matrix.
I have following matrix :
a = matrix(rnorm(36), 6)
Now I want to replace the lower-triangular elements with it's upper-triangular
elements. That is I want to make a symmetric matrix from a. I have tried with
lower.tri() and upper.tri() function, but got desired result. Can anyone please
tell m
t;- rnorm(6)
> dat3 <- rnorm(7)
>
> lis <- lapply(paste("dat", 1:3, sep =
> ""), get)
> lis
> output <- sapply(lis, sd)
> print(output)
> print(str(output))
>
>
>
> On Fri, Sep 19, 2008 at 1:44 AM, Megh Dal wrote:
>
> &g
norm(7)
>
> lis <- lapply(paste("dat", 1:3, sep =
> ""), get)
> lis
> sapply(lis, sd)
>
>
> I hope it helps.
>
> Best,
> Dimitris
>
>
> Megh Dal wrote:
> > [My previous message rejected, therefore I am sending
> sa
[My previous message rejected, therefore I am sending same one with some
modification]
I have 3 vectors with object name : dat1, dat2, dat3
Now I want to create a loop, like :
for (i in 1:3)
{
cat(sd(dati))
}
How I can do this in R?
Regards,
I have following dataset:
> res
[,1] [,2] [,3]
[1,] 19464 1.27
[2,] 19465 1.27
[3,] 19466 1.27
[4,] 19467 1.27
[5,] 19468 1.52
[6,] 19469 1.52
[7,] 1946 10 1.52
[8,] 1946 11 1.52
[9,] 1946 12 1.62
[10,] 19471 1.62
[11,] 19472 1.62
[12,] 1947
I have following
" 1975 01 7711.16"
Here I need to identify where the is there and then concatenate rest of
the digits without , i.e. I want to have "1975017711.16". Is there any R
function?
Regards,
__
R-help@r-project.org mailing list
https://sta
t; Subject: Re: [R] Compiling date
> To: "David Scott" <[EMAIL PROTECTED]>
> Cc: "Dr Eberhard W Lisse" <[EMAIL PROTECTED]>, "Megh Dal" <[EMAIL
> PROTECTED]>, [EMAIL PROTECTED]
> Date: Tuesday, September 9, 2008, 11:37 PM
> Is this Month-D
Hi,
I have following kind of dataset (all are dates) in my Excel sheet.
09/08/08
09/05/08
09/04/08
09/02/08
09/01/08
29/08/2008
28/08/2008
27/08/2008
26/08/2008
25/08/2008
22/08/2008
21/08/2008
20/08/2008
18/08/2008
14/08/2008
13/08/2008
08/12/08
08/11/08
08/08/08
08/07/08
However I want to use
gt; Cc: [EMAIL PROTECTED]
> Date: Thursday, July 24, 2008, 5:42 PM
> On 7/24/2008 7:34 AM, Megh Dal wrote:
> > I have created following interactive plot :
> >
> > library(TeachingDemos)
> > plott = function(x)
> >{
> > return(hist(rnorm(as.integer(10
I have created following interactive plot :
library(TeachingDemos)
plott = function(x)
{
return(hist(rnorm(as.integer(1000, 10, x)), xlab=NA))
}
tkexamp(plott, list(x=list('slider',from=1,to=40, resolution=0.1, init=2)),
plotloc='top')
Here everything works fine, but the problaem is wh
--- On Tue, 7/22/08, hadley wickham <[EMAIL PROTECTED]> wrote:
> From: hadley wickham <[EMAIL PROTECTED]>
> Subject: Re: [R] scatter plot using ggplot
> To: [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED]
> Date: Tuesday, July 22, 2008, 10:00 PM
> On Tue, Jul 22, 2008
I used ggplot to create a scatter plot :
library(ggplot)
library(mnormt)
Sigma = matrix(c(1, 0.6, 0.6, 1), 2, 2)
x = rmnorm(20, c(0,0), Sigma)
xx = x[order(x[,1]),]
y = xx[,1]
z = xx[,2]
qplot(z, y, type="point", main="x-y plot", xlab="x", col="blue")
However I want following:
1. Plot color must
Hi,
following data is taken from
http://www.economagic.com/em-cgi/data.exe/var/west-texas-crude-long. Problem
with this data is when you copy it from this site you would get something like
that :
1946 063331.27
1946 079991.27
1946 087771.52
1946 096661.52
They should be interpret in follo
; to a function separately.
>
> Try this:
>
> funs <- c("Mean", "SD")
> f <- function(fun) aggregate(data1, as.yearqtr,
> get(fun))
> do.call(cbind, sapply(tolower(funs), f, simplify = FALSE))
>
>
> On Sun, Jul 6, 2008 at 10:43 PM, Megh D
ndieck <[EMAIL PROTECTED]>
> Subject: Re: [R] Error in defining function
> To: [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED]
> Date: Monday, July 7, 2008, 1:23 AM
> On Sun, Jul 6, 2008 at 3:19 PM, Megh Dal
> <[EMAIL PROTECTED]> wrote:
> > Can anyone please tell me
Can anyone please tell me why I am getting this error?
library(zoo)
Z.index <- as.Date(sample(12450:15500, 3000))
Z.data <- matrix(rnorm(300), ncol = 1)
data1 <- zoo(Z.data, Z.index)
fnc = function(data1)
{
selection2 = select.list(c("Mean"), multiple = F)
Mean = function(dataa) me
I have time series observation on daily frequencies :
library(zoo)
SD=1
date1 = seq(as.Date("01/01/01", format = "%m/%d/%y"), as.Date("12/31/02",
format = "%m/%d/%y"), by = 1)
len1 = length(date1); data1 = zoo(matrix(rnorm(len1, mean=0, sd=SD*0.5), nrow =
len1), date1)
plot(data1)
Now I want
Hi,
I have one question on expand.grid() function.
When I write following syntax :expand.grid(c("u", "l"), c("u", "l"), c("u",
"l")) I get following as desired :
Var1 Var2 Var3
1uuu
2luu
3ulu
4llu
5uul
6lul
7
Hi,
I got following error in write.table() :
> write.table(dataa, file="c:/data1.csv", row.names=F, col.names=T, sep=",")
Error in file(file, ifelse(append, "a", "w")) :
cannot open the connection
In addition: Warning message:
In file(file, ifelse(append, "a", "w")) :
cannot open file 'c:/d
please provide the dput output as
requested.
On Sun, Jun 1, 2008 at 9:59 AM, Megh Dal wrote:
>> packageDescription("zoo")$Version
> [1] "1.1-1"
>
>
> Gabor Grothendieck wrote:
>
> You need to unambiguously specify what your data
> looks like. Ple
scription("zoo")$Version
On Sun, Jun 1, 2008 at 9:37 AM, Megh Dal wrote:
> I got following:
>
>> z <- zoo(data1[,3], as.yearmon(data1[,1], "%b-%y"))
>> head(z)
> Jan 0001 Jan 0002 Jan 0003 Jan 0004 Jan 0005 Jan 0006
> 1206.68 782.45 1187.00 13
ad(z)
0001-01-01 0002-01-01 0003-01-01 0004-01-01 0005-01-01 0006-01-01
1206.68 782.451187.001398.771883.231431.80
Nowhere 1980 is coming. Any better suggestion?
Gabor Grothendieck <[EMAIL PROTECTED]> wrote: On Sun, Jun 1, 2008 at 8:47 AM,
Gabor Grothendieck
wrote
Hi all,
I have following monthly time series :
> head(data1)
V1 V2 V3
1 Nov-80 NA 1007.44
2 Dec-80 NA 982.05
3 Jan-81 NA 994.25
4 Feb-81 NA 996.31
5 Mar-81 NA 939.91
6 Apr-81 NA 923.32
Now I want to convert it to a 'zoo' object. I wrote following syntax :
ss = zoo(data1[,3], as.
Still I did not find any suggestion. Is my problem not elaborate enough?
Megh Dal <[EMAIL PROTECTED]> wrote: I think I should be clear exactly what I
want :
take following example :
a = b = seq(1, 5, by=500)
v = matrix(0, nrow=length(a), ncol=length(a))
for (i in 1:le
I got following error while I was using solve.QP() in my problem:
> Dmat = matrix(c(0.0001741, 0.0001280, 0.0001280, 0.0002570), nrow=2)
> dvec = t(c(0,0))
> Amat = matrix(c(-1,1,0,-1,0, 1,0,1,0,-1), nrow=5)
> bvec = c(-2, 1, 1, -5, -5)
> solve.QP(Dmat,dvec,Amat,bvec=bvec)
Error i
uot;green"), alpha=0.7,aspect=c(1,1,0.5))
Here you see that the surface is the plot of a x'Cx for different values of
components of x. And the red region is the portion of that plot that satisfy 0
wrote: Megh Dal wrote:
> Hi,
>
> I want to find solution of function :
Forgot to send one copy to R help. Sorry
Megh Dal <[EMAIL PROTECTED]> wrote: Date: Wed, 7 May 2008 02:45:09 -0700 (PDT)
From: Megh Dal <[EMAIL PROTECTED]>
Subject: Re: [R] Solution of function
To: Berwin A Turlach <[EMAIL PROTECTED]>
Hi Berwin,
Thanks for having
Hi,
I want to find solution of function : f(x,y) = x'Cx - a under constraints :
0 < x,y < p
0 < x-y< q
where a, p,q are given constants and x = (x, y) and C is a 2X2 matrix (given)
Can anyone suggest me any R function to do that?
-
Hi all,
I have following problem :
a = b = seq(1, 5, by=500)
v = matrix(0, nrow=length(a), ncol=length(a))
for (i in 1:length(a))
{
for (j in 1:length(a))
{
d = c(17989*a[i], -18109*b[j])
v[i,j] = t(d) %*% matrix(c(0.0001741, 0.0001280, 0.0001280, 0.000
I have a TS of monthly observations.
head(data4)
1991(1) 1991(2) 1991(3) 1991(4) 1991(5) 1991(6)
12.00864 11.94203 11.98386 12.01900 12.19226 12.15488
Now I want to make 11 dummy variables indicating months. Therefore I did
followings :
For Jan :
rep(c(rep(0,0), 1, rep(0,
Hi all,
I feel there is a incompleteness in mAr.est function in mAr package for VAR
estimation. I does not cheak whether there is multicolinearity in data set.
Here I used mAr.est function for following dataset :
> head(log(data1)
+ )
V1 V2 V3 V4 V5
g
Previously I tried with image() function [however there is also no option to
put a color pallet :( ], there is a option 'zlim' to trim the not-required
values. Is there any option in levelplot() function as well? I already gone
through it's help file, however got nothing on th
plot(Z) etc
However I can not reproduce this example in my problem. How I can change my
data matrix 'dat' to 'Z' ?
If anyone show me some light it would be great for me.
Regards,
Megh Dal <[EMAIL PROTECTED]> wrote:
Roy Mendelssohn <[EMAIL
Hi all,
Suppose I have following dataset :
library(zoo)
SD = 1
date1 = seq(as.Date("01/01/90", format = "%m/%d/%y"), as.Date("12/31/08",
format = "%m/%d/%y"), by = 1)
len1 = length(date1); data1 = zoo(matrix(rnorm(len1, mean=0, sd=SD*0.5), nrow =
len1), date1)
Now I want to extr
Suppose I have following dataset :
> head(data1)
Date Return
1 03/31/00 0.14230650
2 04/28/00 -0.03276228
3 05/31/00 -0.06527890
4 06/30/00 -0.04999873
5 07/31/00 -0.01447902
6 08/31/00 0.22265729
Now I convert it to zoo object :
> data11 = zoo(data1[,2], as.Date(data1[,1],
Hi all,
Suppose I have to letters 'u' and 'd'. Now I want to find all combinations
like that :
uu
ud
du
.
dd
This type of combination generally required for Option Pricing in
financial/derivative market. Here generally 'u' means: up-move and 'd' means
down mo
(-4*pi, 4*pi, len=27)
r <- sqrt(outer(x^2, y^2, "+"))
image(x, y, r, col=gray((0:32)/32))
colors <- colorRampPalette(c('red', 'yellow', 'blue')) # create you
color spectrum
image(x,y,r, col=colors(100))
On Thu, Feb 28, 2008 at 9:28 PM, Megh Dal wrote:
>
ng link]. If anyone here tell me how to do that, it will be good for
me.
Regards,
Megh Dal <[EMAIL PROTECTED]> wrote: Hi all,
Can anyone here please tell me whether is it possible to produce a chart
displayed in http://www.datawolf.blogspot.com/ in R for visualizing
multivariate tim
else(dim(data1)[1] > 1,
cov, var))
Still I am getting error. What to do?
Henrique Dallazuanna <[EMAIL PROTECTED]> wrote:
Perhaps something like this:
lapply(split(data1, format(index(data1), "%m")), cov)
On 27/02/2008, Megh Dal wrote:
> let create a 'zoo' obje
let create a 'zoo' object :
library(zoo)
date.data = seq(as.Date("01/01/01", format = "%m/%d/%y"), as.Date("06/25/02",
format = "%m/%d/%y"), by = 1)
len = length(date.data)
data1 = zoo(matrix(rnorm(2*len), nrow = len), date.data )
head(data1)
Now I want to create an 3 dimensio
Hi, I would like to ask here one stat related question. Suppose Z ~
Skew-Normal(0,1,1). Now I want to find a variable f: Z -> Y which has Y~
Skew-Normal(0,1,lambda) distribution. Can anyone give me some light how to do
that?
Regards,
-
[[a
Hi, I would like to ask here one stat related question. Suppose Z ~
Skew-Normal(0,1,1). Now I want to find a variable f: Z -> Y which has Z ~
Skew-Normal(0,1,lambda) distribution. Can anyone give me some light how to do
that?
Regards,
-
[[
Hi all,
Can anyone here please tell me whether is it possible to produce a chart
displayed in http://www.datawolf.blogspot.com/ in R for visualizing
multivariate time series? If possible how?
Regards,
-
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-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Attiglah, Mama
Sent: 30 January 2008 11:32
To: Megh Dal; [EMAIL PROTECTED]
Subject: Re: [R] Multiplying each row of a big matrix wit
I have a big matrix 'ret'. I want to multiply each row of it with a 2nd vector
'pos', resulting result, I want to save in a vector named 'port'. I wrote
following code:
> pos
[1] 2593419 2130220 6198197 1673888 198 1784732 2052120 -7490228
-5275000
> dim(ret)
[1] 500 9
Hi all T gurus,
I would like to test if my dataset is indeed from N(0, 0.011908969).
K.S. test gives following result:
> ks.test(data, "pnorm", 0, 0.011908969)
One-sample Kolmogorov-Smirnov test
data: data
D = 0.1092, p-value = 1.318e-05
alternative hypothesis: two-s
Hi all,
Suppose I have a population of 3 alphabets : A, B, C. From this population,
number of ways that any 2 can be chosen is 3 i.e. AB, AC, and BC.
Is there any R function to generalize this process, for any number of
alphabets/numbers and for any sub-sample size?
Thanks and r
Hi all,
I have a time series like that :
DateValue
01/03/05 -0.008471364
01/04/05 -0.008153802
01/05/05 -0.000780031
01/06/05 -0.000130064
01/07/05 -0.000650576
01/08/05 -0.000130166
01/10/05 -0.004174282
01/11/05 0.01027384
01/12/05 -0.006099558
01/13/05 -0.0
Hi all,
I have a time series like that :
DateValue
01/03/05 -0.008471364
01/04/05 -0.008153802
01/05/05 -0.000780031
01/06/05 -0.000130064
01/07/05 -0.000650576
01/08/05 -0.000130166
01/10/05 -0.004174282
01/11/05 0.01027384
01/12/05 -0.006099558
01/13/05 -0.0
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