Please make your examples minimal. That includes not using
huge variable names that make it harder to work with.
This is not a zoo object with one column. Its is a zoo object
based on a vector.
zoo(matrix(1:3, 3, 1))
is not the same as
zoo(1:3)
The former has one column but the latter does n
Unfortunately that's wrong. Look at ?zoo . The first
argument must be a numeric vector, matrix or factor. A data
frame is not allowed.
On 11/14/06, Horace Tso <[EMAIL PROTECTED]> wrote:
> Mark, it's not the most elegant solution but here it is,
>
> I have a zoo with a vector of numeric values,
First, read the last line of every message to r-help. The code below
is not self contained nor reproducible.
You can only put a column name on a zoo object that has columns!
That means its data portion must be an nx1 matrix, not a vector.
On 11/14/06, Leeds, Mark (IED) <[EMAIL PROTECTED]> wrote
A workaround if you need to multiply an its object by a number would
be to convert the its object to zoo, do the multiply on the zoo object
and then convert the resulting zoo object back to its. Using im from
Martin's example:
library(its)
library(zoo)
im
as.its(3 * as.zoo(im))
On 11/14/06, Mart
Try:
V1 <- c("Welfare_Group_1024", "Welfare_Group_1536", "Welfare_Group_160")
V2 <- c("xxxWelfare_Group_1024", "yWelfare_Group_1536",
"zWelfare_Group_160")
lcs <- function(ff) { L <- ff[1]; for(f in ff) L <- lcs2(f,L); L }
lcs(V1)
lcs(V2)
where lcs2 was posted here:
http://finzi.psych.upenn
Here are a couple of possibilities:
as.data.frame(t(t(df) * vct))
df * rep(vct, each = nrow(df))
On 11/13/06, RDM <[EMAIL PROTECTED]> wrote:
> # Newbie alert
> # I am wanting to multiply the rows in a dataframe by a vector.
> # However, the default behavior appears to be for the vector to
On 11/13/06, Roger Bivand <[EMAIL PROTECTED]> wrote:
> On Mon, 13 Nov 2006, Luiz Rodrigo Tozzi wrote:
>
> > Hi
> >
> > I want to know if there is any possibility of "executing" the content of a
> > vector, for example:
> >
> > example=c("Test",1,0,0,0,"seq(14,42,by=2)",0,0,1)
> >
> > i want to know
This is slightly faster but not by much:
ad <- function(m, a) { diag(m) <- diag(m) + a; m }
R <- ad(P %*% P - t(P) - P, 1)
On 11/12/06, YONGWAN CHUN <[EMAIL PROTECTED]> wrote:
> Hello,
>
>
> I wonder by chance if there is a way to reduce computing time for matrix
> addition or subtraction. With
Try this:
plot(0, main = quote(P(X ~ "|" ~ A ~ intersect(B
On 11/10/06, stortignauz <[EMAIL PROTECTED]> wrote:
>
> Hi all,
>
> I'm trying to write an axis label
> that sounds P( X | K and Xb ) (probability of X given K and Xb )
> but I need the intersection symbol
> (the "\cap" in latex)
> A
Try:
legend("topleft", expression(R[adj]^2 == 0.66))
On 11/9/06, CG Pettersson <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> W2k, R2.4.0
>
> I want to place a legend in a regression plot, stating the adjusted
> R-square value. After some struggle with the coding I am nearly there, but
> only nearly
Aside from the answers to use text you could use letters instead of
numbers which would allow you to continue to use a single character
and might have advantages in terms of saving space on the chart:
pch = letters[trial_no]
pch = c(letters, LETTERS)[trial_no]
pch = c(1:9, letters, LETTERS)[trial_
You are using two different x's and one has nothing to do with the other.
All of your examples are simply internally inconsistent so there is no
reason to think they would work.
On 11/9/06, Carmen Meier <[EMAIL PROTECTED]> wrote:
> Gabor Grothendieck schrieb:
> > Please pro
Please provide a complete self contained example. I can't follow the
partial code below; however, its likely you are plotting one thing
and creating axes using another so there is no reason it should
come out right.
On 11/8/06, Carmen Meier <[EMAIL PROTECTED]> wrote:
> Gabor Grothe
Your code plots x which has nothing to do with xt.
On 11/8/06, Carmen Meier <[EMAIL PROTECTED]> wrote:
> Yes this one works, but (sorry) in the OP there was a plot without data
> to define the range
> So I tried to use your working suggestion in that manner:
>
> library(zoo)
> library(chron)
> tim
You could approximate it with splines and then integrate that:
x <- 0:10/10
f <- function(x) 1 + x + x^2 + x^3 + x^4
y <- f(x)
fs <- splinefun(x, y)
integrate(fs, 0, 1)
integrate(f, 0, 1)
On 11/8/06, Xiaofan Cao <[EMAIL PROTECTED]> wrote:
> Hi everyone,
>
> I'm trying to integrate f(x) over x whe
You indicated that one of the solutions did not work for you but
I am not clear which one you were referring to. To avoid confusion
I have deleted all my and your comments except for the following
which works for me if I paste it into a fresh session using
Edit | Paste commands only
menu in Win
<- 12
> xt <- times(seq(mn, mx, length = n))
> xt <- times(unique(sub("..$", "00", xt)))
> plot(z, xaxt = "n")
> axis(1, xt, sub(":00$", "", xt))
> R.version.string
[1] "R version 2.4.0 Patched (2006-10-24 r39722)"
&
works fine in any configuration.
> but I would prefer an "automatic" input.
> That means that I would like to use the datafield[1] for the minimum
> time and the datafield[max] (means the last one) for the maximum time.
> divided into x steps.The datafiled size could reach more tha
There is a discussion and code in the R News 4/1 help desk article.
On 11/8/06, Ivan Kalafatic <[EMAIL PROTECTED]> wrote:
> Does anyone know where I can find any tool for Windows that converts
> dates from ordinary formats into POSIX?
> I need it to import sime time series from Excel into R and us
Try this:
plot(z, xaxt = "n")
xt <- paste("23", seq(5, 50, 5), sep = ":")
axis(1, times(paste(xt, 0, sep = ":")), xt)
On 11/8/06, Carmen Meier <[EMAIL PROTECTED]> wrote:
> Hi to all,
> I have some problems to get the times-scale to the x-axis the times are
> coming from an excel sheet f. e
> [1]
In addition to the reshape solution posted one could use melt from
the reshape package:
melt(df1, id = 1:2)
On 11/7/06, Thorsten Muehge <[EMAIL PROTECTED]> wrote:
>
> Hello Experts,
> how do I reformat a data frame in the way described below:
>
>
> df1:
> IDdesc resist thick temp
> 1
Try this:
library(gsubfn)
s <-c("ABC", "this WOUld be gOOD")
strapply(s, "[A-Z]{3}", perl = TRUE)
See:
http://code.google.com/p/gsubfn/
On 11/7/06, Lapointe, Pierre <[EMAIL PROTECTED]> wrote:
> Gregexpr - extract results with lapply
>
> Hello,
>
> I need to extract sequences of three upper case
I personally follow those recommendations.
On 11/5/06, Joe Byers <[EMAIL PROTECTED]> wrote:
> Gabor,
>
> Thank you very much. That is a wonderful article and will help very
> much. Might I ask which date schema do you prefer?
>
> Thank you
> Joe
>
>
> Gabor
There are a few versions for Windows here:
http://tolstoy.newcastle.edu.au/R/help/06/02/21634.html
On 11/5/06, Mauricio Cardeal <[EMAIL PROTECTED]> wrote:
> Hello !
>
> Please, how can I clear the RConsole screen ? Is there any "clear" command ?
>
> Thanks,
> Mauricio
___
See the help desk article in R News 4/1 for a discussion on how to choose.
On 11/5/06, Joe W. Byers <[EMAIL PROTECTED]> wrote:
> I have been working with R extensively for several months. I switched
> from SAS and Matlab to R. My question is
>
> Can anyone explain the benefits and detractions of
To plot this with classic graphics convert data frame from
long to wide format using one of the three alternatives below
and then plot:
# 1. subset
DFm <- subset(DF, gender == "m")
DFf <- subset(DF, gender == "f")
DFw <- cbind(DFm[1], xBar.m = DFm[,3], xBar.f = DFf[,3])
# 2. reshape
DFw <- reshap
It looks into the data argument first and if not found there it looks into
environment(formula). Thus to pick it out of the 'with' try this:
with(KK, glm(fo, data = environment())
Also you could try this:
environment(fo) <- KK
glm(fo)
On 11/3/06, vito muggeo <[EMAIL PROTECTED]> wrote:
> Dear
First create your running sum function and then tapply it to the
groups:
runsum <- function(x, k = 3) diff(c(rep(0, k), cumsum(x)), k)
DF$back <- unlist(tapply(DF$plays, DF$Userid, runsum))
Note that there is a survey of running sum/mean/etc. here:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/
Try this:
> L <- list(matrix(1:10, 10))
> L
[[1]]
[,1]
[1,]1
[2,]2
[3,]3
[4,]4
[5,]5
[6,]6
[7,]7
[8,]8
[9,]9
[10,] 10
> plot(L[[1]])
If your object is somehow different then please display it like this:
dput(x, control = "all")
where x i
You could use the file= argument on cat to avoid the two calls to sink:
cat(tmp, file = tmp.file)
On 11/2/06, [EMAIL PROTECTED]
<[EMAIL PROTECTED]> wrote:
> Hello,
>
> thanks a lot for your help on splitting the string to get a numeric vector.
> I'm now writign the string to a tempfile and read
aggregate can also be used:
aggregate(d, list(d = d), length) # output is data frame
or using zoo:
library(zoo)
aggregate(zoo(as.vector(d)), d, length) # output is a zoo object
If what you actually have is a zoo object such as this one:
z <- zoo(1:10, d)
then the last statement co
Try this:
f <- function(DF, vnam) {
if (is.character(DF)) DF <- get(DF)
head(DF[[vnam]])
}
f(iris, "Species")
f("iris", "Species")
On 11/1/06, Jonathan Greenberg <[EMAIL PROTECTED]> wrote:
> If I want to pass a data frame variable name and a specific component name
> of the data frame to a
Try this where m is the matrix:
100 * colMeans(m > 5 & m < 9, na.rm = TRUE)
On 11/1/06, antonio rodriguez <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Having a matrix F.zoo (6575,189) with NA's in some columns I'm trying to
> extract from each column the percent of days within an specific range,
> so I'
If the x values are unique, as in this example, then
xyplot.zoo can do it without a panel function (or
perhaps more accurately the default panel function
in xyplot.zoo can do it). Note that a zoo object is
a vector or matrix with a time index attached to it.
You don't really have to be familiar wi
Try:
rep(1:3, length = 17)
or
as.numeric(gl(3, 1, 17))
or
y <- 1:17
replace(y, TRUE, 1:3) # ignore warning
On 11/1/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
>
>
> i need a strange algorithm that i can easily do in a loop but need
> o do without looping.
>
> suppose i have a vector of
If you are willing to use grid you can set the color of the tick marks,
labels and place a box around the chart like this:
x <- 1:10
y <- x*x
# create box around plot
library(ggplot)
qplot(x, y)
# black border around plot
grid.edit("grill-border", height = unit(.99, "npc"),
width = unit(.99,
I also have to deal with fixed format files from time to time.
Generally I have no control over the format in those cases.
On 10/31/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> "Archaic" it may be, but I still have to deal with fixed format data
> files on a daily basis.
>
> David L. Reiner
I think you will have to ensure that lt has the same set of times as
the prior plot to do it that way. Since time(lt) is a subset of
time(aggfxdata) it would be good enough to do this:
par(new = TRUE)
lt2 <- merge(zoo(, time(aggfxdata)), lt)
plot(lt2, type = "l", lty = 2)
On 10/31/06, Leeds, Mar
On 10/31/06, Robert Lischke <[EMAIL PROTECTED]> wrote:
> Hi there,
>
> is there a way to make the following code generic, so that i can paste()
> the columns in tmp1 depending on the dimensions given in embed()?
>
> > tmp1 <- embed(as.vector(shrinkTable$AOIname), dimension = 2);
> > tmp1
> [,
But what if you wanted to get the subset of those rows containing
heights over 250 -- you would want zero rows to be returned:
On 10/31/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> I've just found some behaviour which strikes me as odd, but I'm not sure
> whether it's a bug or a feature. If
Try this:
A <- list(matrix(c(2,3,3,1,2,2), 3, dimnames = list(NULL, letters[1:2])),
matrix(c(3,3,2,5,1,3), 3, dimnames = list(NULL, letters[3:4])))
B <- list(matrix(c(1:5, 20*(1:5), 10+20*(1:5)), 5,
dimnames = list(NULL, letters[5:7])),
matrix(c(1:5, 10*(1:5), 10+
It should be zoo, not as.zoo. We are constructing a zoo object
from scratch here, not converting a time series object of a different
class to zoo.
On 10/30/06, Leeds, Mark (IED) <[EMAIL PROTECTED]> wrote:
> oops, for the pasting to work,
> You would have to get rid of the blank line that occurs d
I don't know the answer to your question but if you are using dates
with no times and don't need time zones (both of which appear to
be the case here) then you could use "Date"
class and avoid the issue altogether. See the help desk article in
R News 4/1 where there is a discussion of how to choos
Sorry, here it is fixed up. The c should have been list:
Lines <- "conid person construct
1 1 offence
2 1 insight"
read.table(textConnection(Lines), header = TRUE,
colClasses = list(person = NULL))
On 10/30/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Here is a
Try this. The first line breaks it up into lists and the second
line drops any list that is not greater than 1 in length:
out <- tapply(seq(x), x, function(x)x)
out[sapply(out, length) > 1]
On 10/30/06, Søren Merser <[EMAIL PROTECTED]> wrote:
> Hi
> Say I've this vector with several duplicates
>
lt change slightly but not into desired one.
>
>
>
> Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> See the colClasses argument of read.table.
>
> e.g.
>
> read.table("myfile", header = TRUE, colClasses = c(person = NULL))
>
> assuming you don't want
See the colClasses argument of read.table.
e.g.
read.table("myfile", header = TRUE, colClasses = c(person = NULL))
assuming you don't want the column labelled person in the header.
On 10/30/06, Amir Safari <[EMAIL PROTECTED]> wrote:
>
>
>Dear R users,
> Sometimes it is needed to read on
Move to the appropriate directory first (or else move your
file to where you are now). For example,
1. if the file is in \ then
setwd("/")
2. or
getwd()
which shows where you are now and then you can move your
file to that spot.
3. Another possibility is:
read.delim(file.choose())
On 10/29/06, Matthew Walker <[EMAIL PROTECTED]> wrote:
> On Sun, 2006-10-29 at 18:47 -0500, Duncan Murdoch wrote:
> [snip]
> > > Hi Duncan,
> > >
> > > Thanks for your reply.
> > >
> > > How do you know that (i) boxplot.stats lives in the grDevices namespace?
> >
> > getAnywhere(boxplot.stats) star
On 10/29/06, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 10/29/2006 6:08 PM, Matthew Walker wrote:
> > On Sat, 2006-10-28 at 08:03 -0400, Duncan Murdoch wrote:
> >> On 10/27/2006 11:06 PM, Matthew Walker wrote:
> >>> Hi everyone,
> >>>
> >>> I think I have found a minor issue with the R function
t;
> 5. Finally, the function traceback() lists the call-stack. How do I
> know which one of the two "boxplot.stats" functions was called?
I think you need to refer to the search list or your knowledge
of how it works in other cases.
>
> Thanks,
>
> Matthew
>
>
On 10/29/06, antonio rodriguez <[EMAIL PROTECTED]> wrote:
> Gabor Grothendieck escribió:
> Thanks Gabor! Now it's OK (nice the output of dput. I was reluctant to
> paste all data because is 6575 days long!)
>
You can write the following, say, to display just a bit:
dput(
sum(table(idx))[idx]]
aggregate(z, starts, mean)
By the way, dput(v, control = "all") will output variable v
in a form easily pastable by someone else into their session.
On 10/29/06, antonio rodriguez <[EMAIL PROTECTED]> wrote:
> Gabor Grothendieck escribió:
> > Try
On 10/29/06, Marc Schwartz <[EMAIL PROTECTED]> wrote:
> On Sun, 2006-10-29 at 10:31 -0600, tom soyer wrote:
> > Hi,
> >
> > I noticed that as.Date() could not convert date string to date type if the
> > dates are very old. For example, if the date string is "1-Mar-50", then
> > as.Date() would conv
On 10/29/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> On 10/29/06, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> > On 10/29/2006 8:23 AM, Gabor Grothendieck wrote:
> > > On 10/29/06, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> > >> On 10/29/2006
On 10/29/06, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 10/29/2006 8:23 AM, Gabor Grothendieck wrote:
> > On 10/29/06, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> >> On 10/29/2006 5:52 AM, Duncan Murdoch wrote:
> >>> On 10/29/2006 1:12 AM, Jesse Al
On 10/29/06, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 10/29/2006 5:52 AM, Duncan Murdoch wrote:
> > On 10/29/2006 1:12 AM, Jesse Albert Canchola wrote:
> >> Here's how the problem started. In Rgui.exe, I attempted to get more
> >> visible output in the buffer (I wanted to be able to scroll u
Its because of the frequency. The lower the frequency number
the higher will be the numbers on the scale axis.
> as.ts(aggfxdata[,"bid"])
Time Series:
Start = 1144713660
End = 1144716960
Frequency = 0.0167
[1] 118.4800 NA 118.4760 NA 118.4600 118.4367 118.4171 118.3810
[
ds <- time(z)[cumsum(table(idx))[idx]]
# average over each run using the time of the end of run for the result
# replace ends with starts if that is preferred
aggregate(z, ends, mean)
On 10/28/06, antonio rodriguez <[EMAIL PROTECTED]> wrote:
> Gabor Grothendieck escribió:
> >
On 10/28/06, antonio rodriguez <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Having an zoo object I can subset it to obtain the days where I have the
> values within some range:
>
> is.zoo(z)
> TRUE
>
> subset(z[,1], z[,1]>=5 & z[,1]<= 10) #Yields: Year(day)
>
> 1988(13) 1988(14) 1988(16) 1988(20) 1988(
On 10/28/06, Michael Prager <[EMAIL PROTECTED]> wrote:
> Thanks to Gabor G., Duncan M., and Hong O. for helpful
> replies. I've made some progress, but have two questions.
>
> Can anyone explain *how* R CMD searches for latex? I have
> provided a batch file (shell script) and and alias (I use a
>
There are pointers to information in the links section of:
http://code.google.com/p/batchfiles/
and some XP batch files that facilitate using R and
building packages on XP. Regarding LaTeX, use MiKTeX.
On 10/28/06, Michael Prager <[EMAIL PROTECTED]> wrote:
> I would like to start building R p
The edit operations does not change the boxplot.stats that you
are debugging. It creates a new boxplot.stats (i.e. now there
are two) and the new one does not have debugging turned on.
Try
getAnywhere("boxplot.stats")
and it finds two. If you remove the one you just created using rm debugging
r
gt; so my guess is that if I just unique it first, everything will work
> fine. Oh geez. Thanks so much.
>
>
> -Original Message-
> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
> Sent: Friday, October 27, 2006 10:03 PM
> To: Leeds, Mark (IED)
> Cc: r-help@stat.math.
zoo objects must have unique index values but the last two in the head
output below are the same:
> head(fxdata)
bid ask
2006-04-03 03:30:00 27.21 27.26
2006-04-03 03:46:42 27.21 27.26
2006-04-03 03:46:54 27.25 27.26
2006-04-03 03:57:08 27.55 27.26
2006-04-03 04:00:00 27.50
Try this:
gmtDiff <- function(time) time - as.POSIXct(format(time), tz = "GMT")
gmtDiff(Sys.time())
gmtDiff(as.POSIXct("2006-10-27", tz = "GMT"))
which both give me the correct answer currently.
The expression after the minus sign comes from the table at the end
of the help desk article in R Ne
I don't have specific experience with this but strapply
of package gsubfn can extract information from a string by content
as opposed to delimiters. e.g.
> library(gsubfn)
> strapply("abc34def56xyz", "[0-9]+", c)[[1]]
[1] "34" "56"
On 10/27/06, roger koenker <[EMAIL PROTECTED]> wrote:
> I have a
There is an example here using plot
http://www.mayin.org/ajayshah/KB/R/html/g5.html
and an example here using xyplot.zoo:
library(zoo)
?xyplot.zoo
that would also work with xyplot.ts
On 10/26/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> Hello All,
>
> Thank you in advance.
>
This is easy to do with the proto package. Here is a basic version:
library(proto)
ml <- proto(start = 1, end = 5, val = NULL, it = function(.) {
if (is.null(.$val)) return(.$val <- .$start) else .$val <- .$val + 1
if (.$val > .$end) return(FALSE) else .$val
})
# test 1
while(ml$it()) ca
Note the R command, prompt, which will generate an outline of the help page for
you. Also you might want to look through the existing help files for
many examples.
On 10/25/06, Weiwei Shi <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am wondering how I can write a help page for my own function so when
>
With the style of whole objects encouraged for R that code looks
like this:
rowString <- LETTERS[1:3]; cols <- 3 # test data
mat <- outer(rowString, formatC(1:cols, flag = "0", width = 2), paste, sep = "")
mat <- paste(mat, "csv", sep = ".")
sapply(mat, function(el) { if (exists(el)) { DF <- read
In terms of non-R software there is a review of software packages
in this JSS article:
http://www.jstatsoft.org/v07/i09/JSS_055.pdf
On 10/25/06, pucicu <[EMAIL PROTECTED]> wrote:
>
> Hello!
>
> As far as I know, there is not yet any package for R, which can compute the
> RQA measures you mentione
Read the ?Startup help file.
On 10/24/06, Grateful Frog <[EMAIL PROTECTED]> wrote:
> Hi!
>
> Thanks for your reply.
>
> Indeed, I'm on Linux.
>
> I tried to do what you said, but nothing happened.
>
> I made a file called First.R in my work directory. It contains this code
>
> .First<-function(){
;- cumsum(r$lengths)
idx.first <- idx.last - r$lengths + 1
# and the first and last times of each run being
tt <- time(nhtemp)
cbind(tt[idx.first], tt[idx.last])
On 10/24/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> On 10/24/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
On 10/24/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> # Consider the builtin in nhtemp "ts" series
> # Then the runs below above and below the mean are
> r <- rle(as.vector(nmtemp > mean(nhtemp)))
> r
>
> # with the first and last indexes of ea
# Consider the builtin in nhtemp "ts" series
# Then the runs below above and below the mean are
r <- rle(as.vector(nmtemp > mean(nhtemp)))
r
# with the first and last indexes of each
# runs being given by:
idx.last <- cumsum(r$lengths)
idx.first <- idx.last - r$lengths + 1
# and the first and la
Here is a third way to do the loop:
rows <- match(substring(filenames, 1, 1), LETTERS)
cols <- as.numeric(substring(filenames, 2, 3))
m[cbind(rows, cols)] <- sapply(DFs, F)
On 10/24/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> On 10/24/06, Gabor Grothendieck <[EM
On 10/24/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Try something like this (untested):
>
> # replace with your function that inputs a data frame and outputs a number
> F <- function(x) length(x)
> # replace with setwd to appropriate path
> setwd("
Try something like this (untested):
# replace with your function that inputs a data frame and outputs a number
F <- function(x) length(x)
# replace with setwd to appropriate path
setwd("/")
filenames <- dir(pattern = "[.]csv$")
# DFs is a list of data frames; names(DFs) will be the filenames
DFs
Or as factors:
blockrowfac <- gl(12, 4 * 16)
blockcolfac <- gl(4, 16, 4 * 16 * 12)
On 10/24/06, Petr Pikal <[EMAIL PROTECTED]> wrote:
> Hi
>
> If you just want a specific sequences added to your data and you have
> your data ordered as shown why not to use simply
>
> blockrow <- rep(1:12, each=64
On 10/23/06, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 10/23/2006 10:11 AM, Alberto Monteiro wrote:
> > This must be dumbest question ever asked, but...
> >
> > When I ask help.search("tcltk"), I get a reference to tcltk-package.
> >
> > When I ask help("tcltk-package"), I get "rtfm(s) in 'R_H
erfere with text
> size cex.
>text(xy[,1]+.5, xy[,2]+.2, i, col = colors(), cex = 0.7)
>if (n > 0)
> colors()[identify(xy, n = n, labels = colors(), plot = FALSE)]
> }
> # test
> printColorSampler(0)
> printColorSampler(1)
> printColorSampler(pch
quot;, ylab = "")
text(xy[,1]+.5, xy[,2]+.2, i, col = colors(), cex = 0.7)
if (n > 0)
colors()[identify(xy, n = n, labels = colors(), plot = FALSE)]
}
# test
printColorSampler(0)
printColorSampler(1)
On 10/22/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
&g
I have removed the dots, vectorized it and changed the
argument to the number points to be identified (default 0):
getColorName <- function(colorNumber) colors()[colorNumber]
printColorSampler <- function(n = 0) {
i <- seq(colors())
k <- ceiling(sqrt(length(i)))
xy <- cbind(floor(i/k)*2,
Try this using builtin data set BOD:
medsq <- function(p, DF) median((p[1] + p[2] * DF[[1]] - DF[[2]])^2)
init <- coef(lm(demand ~ Time, BOD))
optim(init, medsq, DF = BOD)
library(MASS)
lqs(demand ~ Time, BOD, method = "lms")
On 10/22/06, Pedro Mardones <[EMAIL PROTECTED]> wrote:
> Does anyon
?which.max
On 10/21/06, tom soyer <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I noticed that max(x) returns the maximum value of a vector, but the
> function doesn't give the user the option of retrieving the row index number
> instead. If I used max(x) to find the maximum value of vector x, then is
> th
Actually I think its last _observation_ carried forward.
On 10/21/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> The zoo package has na.locf (last occurrernce carried foward)
> for this purpose:
>
> > x
> [1] 1 NA NA NA NA 2 NA NA NA NA 3 NA NA
> > library(
The zoo package has na.locf (last occurrernce carried foward)
for this purpose:
> x
[1] 1 NA NA NA NA 2 NA NA NA NA 3 NA NA
> library(zoo)
> na.locf(x)
[1] 1 1 1 1 1 2 2 2 2 2 3 3 3
On 10/21/06, Dennis Fisher <[EMAIL PROTECTED]> wrote:
> Colleagues
>
> After reading in some clinical data, I
ich we see that the only difference in their internal representation
is the addition of the names attribute.
On 10/21/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> On 10/21/06, tom soyer <[EMAIL PROTECTED]> wrote:
> > Gabor,
> >
> > Thanks for the code exampl
or(a)
> > is.vector(a)
> [1] TRUE
> > a[2]
> [1] -7.2875
>
> As you can see, although coef() returns a vector already, only after
> as.vector(a) is used, did a[2] include the slope without the name of the
> slope. Why is that, and what happened to the name of the slope
Using the builtin BOD data frame:
as.vector(coef(lm(demand ~ Time, BOD)))[2]
On 10/21/06, tom soyer <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am trying to get R to return just the slope of a linear regression line,
> but it seems that R has to return both the slope and the name of the slope.
> For
27;
> } # BTW, could this syntax be more succint?
> axis(1, x, labels=xtickLabels)
>
>
> I see only the first tick label (-100) and nothing else.
>
> Thanks, Darren
>
>
>
>
>
> On 8/11/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> > Try this
I run XP, not Vista, but have always installed R in
c:\Program Files\R\R...
and have never had a problem with the space in Program Files
so I doubt that that would be the problem.
On 20 Oct 2006 22:31:09 +0200, Peter Dalgaard <[EMAIL PROTECTED]> wrote:
> "Charles Annis, P.E." <[EMAIL PROTECTED]>
Try summaryBy in package doBy. e.g. using the built in dataset CO2:
summaryBy(uptake ~ Plant, CO2, FUN = c(mean, min, max))
On 10/20/06, Jonathan Greenberg <[EMAIL PROTECTED]> wrote:
> Is there a way to calculate, say, the mean, min and max using aggregate
> using one line of code? Or do I need
ttt is a list of functions so each function in ttt is passed in turn to
the anonymous function as argument f.
On 10/20/06, James Foadi <[EMAIL PROTECTED]> wrote:
> Many thanks to those who have answered my question.
> Could I ask Gabor and Peter the meaning of:
>
> > sum(sapply(ttt,function(f) f(x
Create a function f which, given Species s, calculates the residual sum
of squares and then sapply it to all Species. In your case Species is
the month and the levels of the Species correspond to the months 1:12
# calculate rss for Species s
f <- function(s) {
iris.lm <- lm(Sepal.Width ~ Sepal
Try this:
L <- list(function(x) 1, function(x) x, sin, cos)
sumL <- function(x) sum(sapply(L, function(f) f(x)))
sumL(pi) # pi
On 10/20/06, James Foadi <[EMAIL PROTECTED]> wrote:
> Dear all,
> I have looked for an answer for a couple of days, but can't come with any
> solution.
>
> I have a set
This has been discussed a number of times in the past, e.g.
see this thread:
http://tolstoy.newcastle.edu.au/R/e2/help/06/09/0980.html
On 10/20/06, Paul Brewin <[EMAIL PROTECTED]> wrote:
> Dear R-listers,
>
> I would like to know if there is a way to programmatically generate
> parameter start va
Try this:
sub("(-[^-]*)-", "\\10", c("1-1-1", "1-1-2", "1-2-1"))
On 10/19/06, ronggui <[EMAIL PROTECTED]> wrote:
> I have a string vector like these:
> [1] "1-1-1" "1-1-2" "1-2-1"
>
> And I wanna replace the second "-" with "0", that is, I wanna get the result
> like: [1] "1-101" "1-102" "1-201"
Or since that messes up the values:
u <- unique(t(apply(DF, 1, function(x) as.numeric(factor(x, levels =
unique(x))
DF[rownames(u), ]
On 10/19/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> If DF is a data frame containing the rows then:
>
> unique(t(apply(DF, 1, func
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