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,
do not work if the index entries in ‘order.by’
are not unique
In either case, I am missing the "time" component. Where I am going wrong?
Thanks,
--- On Sat, 10/16/10, Gabor Grothendieck wrote:
> From: Gabor Grothendieck
> Subject: Re: [R] Problem with merging two zoo obj
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: &
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=",",
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]
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
Still no reply on my query. Is it not understandable? Can I reformulate
better my question?
Thanks,
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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
r put their views on
those 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 ve
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?
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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
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 14
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 l
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
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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.
. Any 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
>
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
geom_point(aes(x, y, colour = Label), data = dat1, size = 4) +
> scale_fill_gradient("Count", low = "green", high = "red")
>
>
> Unfortunately not. In the future, you will be able to do +
> scale_colour_discrete(legend = F)
>
> Hadley
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 a
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))
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,
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ht
take care 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
> horiz
?
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 wrote:
>
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,
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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 err
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
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Sent from the
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"
[1
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,
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<- Reduce("rbind", Y)
a <- solve(temp1%*%temp2) %*% (temp1%*%temp3); 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 ar
"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 :
>>
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 calcu
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?
>
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]] +...
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 au
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 worki
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,
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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 cre
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 mi
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 f
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 yahoo.com> writ
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,
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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 v
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,
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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,
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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 # underly
n is related to the machine epsilon, used
> in the default value for "tol". try fiddling with the tol argument.
>
>> 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
Is there any R function which calculate the Orthogonal Complement of a mxn
matrix (with full column rank)?
Thanks in advance
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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,
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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 18
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 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 fo
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 :
>>
&
> fn() and causes an error. Just use lapply(1:4, fn), or better yet,
> sapply,
>
> > 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 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,
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Sent fr
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 # 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", wh
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,
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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,
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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?
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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,
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riginal Message-
>> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
>> project.org] On Behalf Of megh
>> Sent: Wednesday, October 22, 2008 3:22 AM
>> To: r-help@r-project.org
>> Subject: [R] A matrix automation problem
>>
>>
>> [I am really sorry if it is
[I am really sorry if it is double posted, I doubt me previous post could not
reach forum due to some problem with net]
Suppose I have a matrix :
a = matrix(1:9, 3)
>From this matrix, I construct 9 additional matrices :
i = 1:9
bi = a * i
Now combining all those 9 new matrices, I construct a
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
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