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
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, c)); Mat # number of
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
element. Therefore
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
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
IJVEOJC 3434.
36453
Can you please tell me how can I modify that?
Thanks,
On Sun, Feb 13, 2011 at 11:10 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Sun, Feb 13, 2011 at 10:27 AM, Megh Dal megh700...@gmail.com wrote:
Please consider following string:
MyString
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
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]]
__
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]]
__
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,
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
: [R] Problem with merging two zoo objects
To: Megh Dal megh700...@yahoo.com
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 megh700...@yahoo.com
wrote:
Dear all, I have following 2 zoo objects. However when
I try to merge
I have compared dat11 and x using str() function, however did not find
drastic difference:
str(dat11)
‘zoo’ series from 2010-10-15 13:43:54 to 2010-10-15 13:49:51
Data: num [1:7, 1:4] 73.8 73.8 73.8 73.8 73.8 73.8 73.7 73.8 73.8 73.8 ...
- attr(*, dimnames)=List of 2
..$ : chr [1:7] 7 6 5
Thanks Gabor for pointing to my old version. However I got one more question
why the argument tz= is sitting there? As you are not passing any explicit
value for that, I am assuming it is redundant. Without any tz argument, I
got following:
head(read.zoo(file=f:/dat1.txt, header=T, sep=,,
matter?
Thanks,
--- On Sat, 10/16/10, Megh Dal megh700...@yahoo.com wrote:
From: Megh Dal megh700...@yahoo.com
Subject: Re: [R] Problem with merging two zoo objects
To: Gabor Grothendieck ggrothendi...@gmail.com
Cc: r-help@r-project.org
Date: Saturday, October 16, 2010, 7:20 AM
I dont know
the time component. Where I am going wrong?
Thanks,
--- On Sat, 10/16/10, Gabor Grothendieck ggrothendi...@gmail.com wrote:
From: Gabor Grothendieck ggrothendi...@gmail.com
Subject: Re: [R] Problem with merging two zoo objects
To: Megh megh700...@yahoo.com
Cc: r-help@r-project.org
Date: Saturday
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 see
the same against
envir. By putting so, what I am going to tell R?
Thanks,
--- On Thu, 10/14/10, Joshua Wiley jwiley.ps...@gmail.com wrote:
From: Joshua Wiley jwiley.ps...@gmail.com
Subject: Re: [R] Query on save.image()
To: Megh Dal megh700...@yahoo.com
Cc: r-h...@stat.math.ethz.ch
Date: Thursday
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):
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 -0.01619026
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 mailing
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
Still no reply on my query. Is it not understandable? Can I reformulate
better my question?
Thanks,
--
View this message in context:
http://r.789695.n4.nabble.com/Having-problem-to-define-a-subclass-please-help-me-tp2308170p230.html
Sent from the R help mailing list archive at Nabble.com.
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, a,
understanding,
all defined law for super-class should be inherited by it's sub-class,
therefore no need to define again.
I would be really grateful if someone clarify those.
Thanks
--- On Fri, 7/30/10, Megh Dal megh700...@yahoo.com wrote:
From: Megh Dal megh700...@yahoo.com
Subject: Having problem
Hi all, I have following environments loaded with my current R session:
search()
[1] .GlobalEnvpackage:stats package:graphics
package:grDevices
[5] package:utils package:datasets package:methods Autoloads
[9] package:base
How can I find the objects under a
: [R] Objects within environment
To: Megh Dal megh700...@yahoo.com
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 have following environments loaded with my
current R session:
search()
[1] .GlobalEnv
Thanks Duncan, I understood. Your explanation is really great. Thank you so
much for your time.
--- On Wed, 7/21/10, Duncan Murdoch murdoch.dun...@gmail.com wrote:
From: Duncan Murdoch murdoch.dun...@gmail.com
Subject: Re: [R] Objects within environment
To: Megh Dal megh700...@yahoo.com
Cc
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
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
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
attribute through a binary variable like 1 - positive, -1 -
negative.
Is their any a mathematical/statistical tool available to answer thise type of
question. If anyone help me out I would be very grateful.
Thanks for your attention.
Megh,
__
R
/12/10, Erik Iverson er...@ccbr.umn.edu wrote:
From: Erik Iverson er...@ccbr.umn.edu
Subject: Re: [R] Handling character string
To: Megh Dal megh700...@yahoo.com
Cc: r-h...@stat.math.ethz.ch
Date: Saturday, June 12, 2010, 2:36 AM
Megh Dal wrote:
Dear all, Is there any R function to say these 2
...@gmail.com
Subject: Re: [R] Handling character string
To: Megh Dal megh700...@yahoo.com
Cc: Erik Iverson er...@ccbr.umn.edu, 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 time)
[1] Now is the time
You need
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) {
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
Murdoch murdoch.dun...@gmail.com
Subject: Re: [R] [RGL] Need help to modify current plot
To: Megh Dal megh700...@yahoo.com
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
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
I need to split a given matrix in a sequential order. Let my matrix is :
dat - cbind(sample(c(100,200), 10, T), sample(c(50,100, 150, 180), 10,
T), sample(seq(20, 200, by=20), 10, T)); dat
[,1] [,2] [,3]
[1,] 200 100 80
[2,] 100 180 80
[3,] 200 150 180
[4,] 200 50 140
Is there any R function to delete a CSV file saved in disk? For example here
I have created following R function :
asd - function(a = 5) {
dat - read.delim(file=F:/Test.csv, header=T, sep=,)
res - dat[1,1]*a
# Here I want to delete Test.csv
return(res)
}
Here I am looking
Hi, I was trying to run your code in my VBA editor, however could not
succeed. The execution stumbled in the line Call Rinterface.StartRServer
itself. I have RCOM package installed into my R environment. Do I need to
install anything else to run that? Would guys here guide me?
Thanks
--
View
Dear all,
I am looking for some kind of interactive plot to draw a histogram for a
normal distribution with different sample size. In that plot there would be
some sort of scroll-bar which will take min value 10 and maximum value of
10,000 as sample size (n) from a standard normal distribution.
Hi, I have a zoo object with monthly frequency :
library(zoo)
dat - zooreg(rnorm(50), as.yearmon(2000-01-01), frequency=12)
Now I want to make a zoo object with daily frequency from dat wherein
value for a each day for a particular month will be value of dat at that
particular month.
Is there
idea please?
Megh wrote:
Hi, I have a zoo object with monthly frequency :
library(zoo)
dat - zooreg(rnorm(50), as.yearmon(2000-01-01), frequency=12)
Now I want to make a zoo object with daily frequency from dat wherein
value for a each day for a particular month will be value of dat
+
scale_colour_discrete(legend = F)
Hadley
On Mon, Dec 7, 2009 at 11:17 AM, Megh megh700...@yahoo.com wrote:
Here I have following code :
dat = rnorm(100)
ggplot() + geom_histogram(aes(dat, fill=..count..)) +
scale_fill_gradient(Count, low=green, high=red) +
opts(legend.position=none
Dear all,
Please consider following date as.Date(2009-02-01). If I subtract 1
then it will give last day, similarly if I subtract 2 it will give 2nd
last day. But what about if I want to get last month, 2md last month
i.e. 2009-01-01 or 2008-12-01 etc or even year?
Is there any automated way
library(ggplot2)
m - ggplot(movies, aes(x=rating))
m + geom_histogram()
Now in x-axis instead of numbers like 2,4,6,8,10..., I want to write like
2%,4%,6%,8%,10%...
Is there any way to do that through GGPLOT ? Your help will be highly
appreciated.
Thanks,
--
View this message in context:
Here I have following code :
dat = rnorm(100)
ggplot() + geom_histogram(aes(dat, fill=..count..)) +
scale_fill_gradient(Count, low=green, high=red) +
opts(legend.position=none)
# Above is without any legend
## Now I want to place two points with legends
dat1 - data.frame(c(0,0), c(1,0)); Label
of them. My question is how I can remove the
color-pallet. Your help will be highly appreciated.
Thanks,
Megh wrote:
Thanks Ista for your mail. Here I wanted to have control on color-pallet.
It is because, here my entire plot window is subdivided in 3 sub-plots
horizontally, on basis
In my Excel file, I have data in following format :
Feb-07 38.49
Mar-07 39.95
Apr-07 37.47
May-07 35.77
Jun-07 32.96
Jul-07 33.27
I tried to copy this data as a time series using following code :
library(zoo)
dat - read.zoo(file=clipboard, format=%m-%y)
However getting following error :
Let consider following plot :
p - ggplot(mtcars, aes(mpg, wt))
p + geom_point(colour=grey50, size = 4) + geom_point(aes(colour =
cyl))
Now I want R to hide the color-pallet on cyl, placed in the right edge
completely. Can anyone please guide me how to do that?
Thanks,
--
View this
?
Thanks,
Ista Zahn wrote:
There was a recent discussion of the ggplot2 mailing list about a
similar issue. The first question is how will people know what the
colors mean if you remove the legend?
-Ish
On Tue, Dec 1, 2009 at 11:41 PM, Megh megh700...@yahoo.com wrote:
Let consider
Is there any R function to calculate automatically the last day of a
particular month? For example sep2009 should be converted to last day of
September of 2009?
Thanks
--
View this message in context:
http://www.nabble.com/How-to-get-last-day-of-a-month--tp25425645p25425645.html
Sent from the
There is an object LETTERS which displays all letters from a to z. Is
there any similar object whicg displays the months as well in
chronological order? like jan, feb,...,dec
Thanks,
--
View this message in context:
I have following object :
date2
[,1] [,2]
[1,] apr 1992
[2,] aug 1992
[3,] dec 1992
[4,] feb 1992
[5,] jan 1992
[6,] jul 1992
[7,] jun 1992
[8,] mar 1992
[9,] may 1992
[10,] nov 1992
[11,] oct 1992
[12,] sep 1992
[13,] apr 1993
[14,] aug 1993
[15,] dec 1993
); a
Can I go ahead with this procedure? Somebody please validate this?
Thanks
megh wrote:
Hi, I have following kind of model : Y = X1 * a1 + X2 * a2 + error
Here sampled data for Y, X1, X2 are like that :
Y - replicate(10, matrix(rnorm(2),2), simplify = F)
X1 - replicate(10, matrix
Hi, I have following kind of model : Y = X1 * a1 + X2 * a2 + error
Here sampled data for Y, X1, X2 are like that :
Y - replicate(10, matrix(rnorm(2),2), simplify = F)
X1 - replicate(10, matrix(rnorm(4),2), simplify = F)
X2 - replicate(10, matrix(rnorm(4),2), simplify = F)
My goal is to
Y is 2 dimensional vector, X1 and X2 are 2x2 matrices.
Ted.Harding-2 wrote:
On 21-Aug-09 16:28:26, megh wrote:
Hi, I have following kind of model : Y = X1 * a1 + X2 * a2 + error
Here sampled data for Y, X1, X2 are like that :
Y - replicate(10, matrix(rnorm(2),2), simplify = F)
X1
Hi,
I have created a list object like that :
x = vector(list)
for (i in 1:5) x[[i]] = rnorm(2)
x
Now I want to do two things :
1. for each i, I want to do following matrix calculation : t(x[[i]]) %*%
x[[i]] i.e. for each i, I want to get a 2x2 matrix
2. Next I want to get x[[1]] + x[[2]] +
Thanks for your suggestions. I need one more thing :
x = y = vector(list)
for (i in 1:5) x[[i]] = rnorm(2); y[[i]] = rnorm(2)
Here I want to get t(x[[i]]) %*% y[[i]] for each i. Can anyone please help
me?
Regards,
Jorge Ivan Velez wrote:
Hi megh,
Perhaps?
# Data
x = vector(list
Let say, I have an arbitrary vector :
i=1
assign(paste(dat,i,sep=), rnorm(5))
Now I want to update that dat1 vector by ommiting last 2 elements i.e.
dat1 = dat1[c(1:3)]
However here my problem is, as dat1 depends on another variable i, I
cannot use above syntax directly. I want to automate
Hi all, whenever I try to plot a histogram using qplot() function of
ggplot2 library, I get error like this :
qplot(rnorm(1000), geom=histogram, binwidth=0.2, main = , xlab=,
ylab=)
Error in scale[[1]] : subscript out of bounds
However if I remove ylab= argument, then it is working fine. Am I
Suppose, I have following
vec - vector(list, length=3)
for (i in 1:3) vec[[i]] - matrix(rnorm((i+3)*2), (i+3))
vec
Now I want to change the type of vec from list to a matrix with (4+5+6)
rows and 2 columns. How can I do that?
Thanks and regards,
--
View this message in context:
I want to create a number of vectors like :
vec1 - rnorm(1)
vec2 - rnorm(2)
vec3 - rnorm(3)
and so on...
Here I tried following :
for (i in 1:10) paste(vec, i, sep=) - rnorm(i)
However obviously that is not working. Here vectors I need to be seperated
i.e I do not want to create a
Hi all, can you please clarify me what is the wrong with following codes :
set.seed(30)
z = matrix(rnorm(10), 5, 2)
apply(z, 1, function(x) sum(z[x,1]*1, z[x,2]*3))
However I can not get the desired result. For example, sum(z[1,1]*1,
z[1,2]*3) gives -5.822442 which is actually correct. Am I
A=matrix(c(1,1,1,1,5,5,1,5,14),nrow=3)
t(eigen(A)$vector) %*% A %*% eigen(A)$vector
mat1 = t(eigen(A)$vector)
mat2 = diag(eigen(A)$values)# this is your diagonal matrix
Manli Yan wrote:
Hi everyone:
I try to use r to do the Cholesky Decomposition,which is A=LDL',so far I
only
To calculate Percentile for a set of observations Excel has percentile()
function. R function quantile() does the same thing. Is there any
significant difference btw percentile and quantile?
Regrads,
--
View this message in context:
Yes, I aware of those definitions. However I wanted to know the difference
btw the words Percentile and quantile, if any. Secondly your link
navigates to some non-english site, which I could not understand.
Dieter Menne wrote:
megh megh74 at yahoo.com writes:
To calculate
Hi, I am trying to create a vector of length 10 (say), wherein each element
will be average of random sample of size 100, from a distribution, say
Normal. Can anyone please tell me without creating a for loop, how I can
do that?
Regards,
--
View this message in context:
No, it is not homework. I obviously could do that using a for-loop, and that
I already did. However I thought whether there could be a better approach as
it was looking very messy and unprofessional.
Uwe Ligges-3 wrote:
megh wrote:
Hi, I am trying to create a vector of length 10 (say
Hi,
suppose I have three vectors like :
l1 = 1:4
l2 = 4:9
l3 = 16:67
now I want to construct a loop like :
for (i in 1:3)
{
count1[i] = length(li) # i.e. it will take l1, l2, l3 according to
value of i
}
Can anyone please tell me how to do that?
Regards,
--
View this message in
.
uniroot(f,c(0,100),tol=1/10^12)
$root
[1] 50
$f.root
[1] 1.337393e+31
$iter
[1] 4
$estim.prec
[1] 5.186962e-13
albyn
Quoting megh megh700...@yahoo.com:
I have a strange problem with uniroot() function. Here is the result :
uniroot(th, c(-20, 20))
$root
[1] 4.216521e-05
Sorry, in my previous post I forgot to include strike price, which is K =
1160
megh wrote:
Thanks for this reply. Here I was trying to calculate implied volatility
using BS formula. This is my code :
oo = 384.40 # traded option price
uu = 1563.25 # underlying price
tt = 0.656
Is there any R function which calculate the Orthogonal Complement of a mxn
matrix (with full column rank)?
Thanks in advance
--
View this message in context:
http://www.nabble.com/Orthogonal-Complement-tp21406355p21406355.html
Sent from the R help mailing list archive at Nabble.com.
I have a strange problem with uniroot() function. Here is the result :
uniroot(th, c(-20, 20))
$root
[1] 4.216521e-05
$f.root
[1] 16.66423
$iter
[1] 27
$estim.prec
[1] 6.103516e-05
Pls forgive for not reproducing whole code, here my question is how f.root
can be 16.66423? As it is finding
Is there any R function find the order of an operation? I am looking for a
general R function to find for example how many basic operations (i.e.
addition, multiplication and exponentiation) are performed when two square
matrices are multiplied.
Regards,
--
View this message in context:
Dear all,
I would like to draw a 3-D horizontal cylinder preferably in RGL device
(because this gives the look from different angles). Basic idea is from
http://www.tau.ac.il/cc/pages/docs/sas8/insight/chap18/sect3.htm.
Below is the description exactly what I want to do.
Please see at figure
Hi all,
I my c: drive I have possibly 1,000 notepad files, with .txt extension. They
are named as the dates on which they were saved i.e. 1st file name is
Volume_4-18-2008, 2nd one is Volume_4-21-2008, 3rd one
Volume_4-22-2008 and so on
Also, content of each file are in same format
total
1000 text files in that directory and therefore I create a vector like
sel.no - c(1:1000). Next I use the i-th element of the vector sel.no to
access the i-th file?
With regards,
Charles C. Berry wrote:
On Mon, 15 Dec 2008, megh wrote:
Hi all,
I my c: drive I have possibly 1,000
https://svn.r-project.org/R/trunk/src/main is not opening for last few days.
Is there any problem with this or the materials were moved to different
address?
Duncan Murdoch-2 wrote:
On 11/19/2008 10:13 AM, megh wrote:
In the optim() function there is a syntax :
res - .Internal(optim(par
,
fn - function(i) return(i^2)
sapply(1:4, fn)
[1] 1 4 9 16
Hope this helps,
baptiste
On 20 Nov 2008, at 16:31, megh wrote:
I have written following codes, with intention to get a list with
values
1,2,9,16 :
fn - function(i) return(i^2)
lapply(1:4, fn, i)
However
I have written following codes, with intention to get a list with values
1,2,9,16 :
fn - function(i) return(i^2)
lapply(1:4, fn, i)
However I got following error :
Error in FUN(1:4[[1L]], ...) : unused argument(s) (1)
Can anyone please tell me what will be the correct code here?
Regards,
--
In the optim() function there is a syntax :
res - .Internal(optim(par, fn1, gr1, method, con, lower,
Here how can I see the inside-codes of .Internal function ?
Regards,
--
View this message in context:
http://www.nabble.com/How-to-look-within-.Internal---tp20581897p20581897.html
Sent from
I get following result, while I write this code :
lapply(1:5, function(i) c(1,2,3)^i)
[[1]]
[1] 1 2 3
[[2]]
[1] 1 4 9
[[3]]
[1] 1 8 27
[[4]]
[1] 1 16 81
[[5]]
[1] 1 32 243
This is fine. However my goal is : each element of this list should depend
on previous element like :
lis #
I have got one post here
http://tolstoy.newcastle.edu.au/R/help/04/10/5221.html
here it is suggested to start R with a flag --no-restore-data .
Can anyone please tell me how to start R like that?
megh wrote:
I am getting error unable to restore saved data in .RData, whenever I
start R
I am getting error unable to restore saved data in .RData, whenever I start
R console. I have reinstalled R with latest version but still getting same
error. Once I click OK the R window crashes. Can anyone suggest me about
the courses to do?
Regards,
--
View this message in context:
Is there anything like goto loop, which exists in most computer programs?
e.g. I am looking for this kind of stuff :
if(i 6) goto step-02
Any idea?
Regards,
--
View this message in context:
http://www.nabble.com/%22Goto%22-loop-tp20263445p20263445.html
Sent from the R help mailing list
I am looking fpr a algo to find matrix inverse. Till time I am aware of
Gauss-Jordan Elimination procedure to find the same. Are there any other
algo. as well? What does R use to find the inverse?
--
View this message in context:
Can anyone please tell whether there is any R function to act as VEC and
VECH operator on Matrix? Yes of course, I can write a
user-defined-function for that or else, I can put dim(mat) - NULL. However
I am looking for some R function.
Your help will be highly appreciated.
Regards,
--
View
?
a - matrix(1:9, 3) * 10
b - matrix(1:9, 3)
kronecker(b,a)
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
project.org] On Behalf Of megh
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 would
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?
__
1 - 100 of 159 matches
Mail list logo