Check xpd= in ?par
On 2/5/06, ivo welch [EMAIL PROTECTED] wrote:
Dear R wizards: First, thank you for all the responses to my earlier
queries. Will keep me busy tomorrow morning. Can I add one graphics
question to my ever changing set of bothering questions, please?
plot( c(0,1), c(0,1)
Try
?seq.Date
On 2/6/06, Sumanta Basak [EMAIL PROTECTED] wrote:
Dear R-users,
First of all, I'm sorry if this is a simple question. I want a daily
date series which will be a sequence. E.g. starting from 07/03/1962 to
12/03/1997. Thanks in advance. I want to paste this date series with
Check out:
https://www.stat.math.ethz.ch/pipermail/r-devel/2006-January/036034.html
On 2/6/06, Robert Mcfadden [EMAIL PROTECTED] wrote:
Dear R Users
I would like to use optim function to optimize a function. I read help but I
couldn't find what I need: is it possible to get information after
Try this:
write.table(sprintf(%10.5f%10.5f, A, B), file = myfile.dat,
quote = FALSE, row = FALSE, col = FALSE)
plot(1:100, 1:100, log = xy)
grid() # only if you want a grid
?plot.default has info on the log= argument. Use ? with the other
commands to
It does break already. Try this:
plot(1:10, c(1:5,NA,7:10), type = l)
plot(c(1:5,NA,7:10), 1:10, type = l)
On 2/7/06, Denis Chabot [EMAIL PROTECTED] wrote:
Hi,
Sometimes data series (not necessarily time series) suffer breaks
where data were expected, but not collected. Often the regular
Also see ?bquote
On 2/8/06, Thomas Steiner [EMAIL PROTECTED] wrote:
I found it immediately after posting :(
substitute is my friend:
text(0.5,5,substitute(f[Sv] ==k*x^2, list(k=zf[1])))
__
R-help@stat.math.ethz.ch mailing list
In Word use a fixed font such as Courier rather than a proportional
font and it will look ok.
On 2/9/06, Tom Backer Johnsen [EMAIL PROTECTED] wrote:
I have just started looking at R, and are getting more and more irritated
at myself for not having done that before.
However, one of the things
I cannot explain it but I must have come across it since
I noticed in various places in some of my code I have
used, in terms of your example, the following:
plot(y ~ x, main = as.expression(bquote(m[1] == .(a
plot(y ~ x, main = list(bquote(m[1] == .(a
both of which work as expected.
I have made a few more improvements:
expand.grid.id - function(id, ...) {
vars - list(...)
nv - length(vars)
lims - sapply(vars,length)
stopifnot(length(lims) 0, id = prod(lims), length(names(vars)) == nv)
res - structure(vector(list,nv), .Names = names(vars))
if (nv 1) for(i in nv:2) {
Yes, the R2HTML route is probably the quickest. Its just one line
of code (plus the call to load in R2HTML). Try this where iris
is a data set built into R:
library(R2HTML)
HTML( iris, file(clipboard,w), append=FALSE )
Now paste the clipboard into Excel and from there into Word.
(If you
On 2/9/06, Berton Gunter [EMAIL PROTECTED] wrote:
Folks:
R 2.2.0 on Windows.
I find the following somewhat puzzling:
a-1; x-0:1; y-x
## following works fine:
plot(x,y ,main= bquote(n[1] == .(a) ))
## following produces an error:
plot(y~x ,main= bquote(n[1] == .(a) ))
Error in
On 10 Feb 2006 12:39:43 +0100, Peter Dalgaard [EMAIL PROTECTED] wrote:
Gabor Grothendieck [EMAIL PROTECTED] writes:
Yes, the R2HTML route is probably the quickest. Its just one line
of code (plus the call to load in R2HTML). Try this where iris
is a data set built into R:
library
Given that this may very well be the most common use of the
R2HTML package I wonder if the R2HTML package developer would
be interested in providing an HTML2clip convenience wrapper
as part of the R2HTML package like this:
HTML2clip - function(x, file. = file(clipboard, w), append = FALSE,
On 2/10/06, Tom Backer Johnsen [EMAIL PROTECTED] wrote:
At 17:10 09.02.2006 -0600, you wrote:
Tom Backer Johnsen wrote:
There has been an incredible number of responses in a short time, with a
number of different suggestions. With hindsight, I must admit I have not
been quite clear, so
Check out:
http://addictedtor.free.fr/graphiques/graphcode.php?graph=116
On 2/11/06, mark shanks [EMAIL PROTECTED] wrote:
Hi,
I have a set of data in x,y coordinates across the range of -5 to 5 in each
dimension. I would like to obtain the frequency distribution of the
different points, and
On 2/11/06, Patricia J. Hawkins [EMAIL PROTECTED] wrote:
iw == ivo welch [EMAIL PROTECTED] writes:
iw * SUGGESTION: can we please offer the ?: operator ala C in
iw addition to ifelse()? This would make R code prettier.
R:
if (condition) xxx else yyy
if (condition) xxx else if (yyy) zzz
Could you walk us through, in detail, what that graph is showing?
On 2/12/06, Michael Prager [EMAIL PROTECTED] wrote:
Besides the answers you already have, you might look at my 4D graph
example (with code) on the R Graphics Gallery:
Assuming this test data:
iris2 - round(iris[1:10, 1:4])
Try this:
t(t(iris2) == c(iris2[1,])) + 0
On 2/13/06, Rick DeShon [EMAIL PROTECTED] wrote:
Hi All.
I'm new to R and trying to learn the data manipulation routines. At the
moment, this is the one that has me stumped.
Imagine n test
You can do it manually by reading in the headers separately:
headers - read.table(test, header = FALSE, nrow = 1, sep = ;, as.is = TRUE)
read.table(test, header = FALSE, skip = 1, sep = ;, col.names = headers)
On 2/13/06, Diethelm Wuertz [EMAIL PROTECTED] wrote:
I have a file named test.csv
Try this:
coef(A)
On 2/13/06, Quin Wills [EMAIL PROTECTED] wrote:
The question is general for all functions, but here is a specific example -
# For the logistic regression of the following correlated variables:
C - c(457, 1371, 4113, 12339, 37017, 111051, 333153, 999459)
E -
On 2/14/06, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 2/14/2006 8:56 AM, Wolfram Fischer wrote:
I defined three functions:
fun0 - function( x=1:5, y=1:5, ... ) xyplot( y ~ x, ... )
fun1 - function( x=1:5, y=1:5, ... ) fun2( y ~ x, ... )
fun2 - function( ... ) xyplot( ... )
The
Check out ?stars
On 2/14/06, Rainer Hahnekamp [EMAIL PROTECTED] wrote:
Hello,
I'm trying to create a simple network diagram like that on
http://www.pauck.de/marco/photo/infrared/comparison_of_films/diagram3.gif.
Unfortunately I've not yet found any function that could do this.
I think
On 2/15/06, Laetitia Marisa [EMAIL PROTECTED] wrote:
Hello everyone,
How can I specify in tcltk file browser the initial directory to
MyComputer in Windows where Drives and Partition are accessible?
And just a little question if anyone knows, is there a way to use the
function choose.files
Read up about vignettes and Sweave which allows
one to intersperse R code and latex.
On 2/14/06, Edgar Urdas [EMAIL PROTECTED] wrote:
how do i perform the stuff i just recorded?!
__
R-help@stat.math.ethz.ch mailing list
plot.zoo takes the nc= argument which specifies the number of columns
it uses, e.g.
library(zoo)
library(tseries)
data(USeconomic)
z - as.zoo(USeconomic)
plot(z, nc = 2)
On 2/15/06, Søren Højsgaard [EMAIL PROTECTED] wrote:
I am trying to plot two 3-dimensional time series in one window (such
data.frame(OneVector, TwoVector,
sapply(apropos(^N[0-9]*$), get, simplify = FALSE))
On 2/20/06, Daniele Medri [EMAIL PROTECTED] wrote:
Hi all,
I need to create a data.frame from a variable number of vectors.
The number of these vectors could change so I need a dinamic way to
group all in
Try this:
# test data
tt - c(23:05:02, 23:10:02, 23:15:03, 23:20:03, 23:25:03,
23:30:03, 23:35:03, 23:40:03, 23:45:04, 23:50:04, 23:55:03,
23:55:03)
x - c(0.575764, 0.738379, 0.567414, 0.663436, 0.599834, 0.679571,
0.88141, 0.868848, 0.969271, 0.878968, 0.990972, 0.990972)
library(zoo)
Your expression is trying to sum a character vector
containing the names of the variables that begin with ABC.
Try this and try executing each portion of it to understand
it better:
ABC1 - ABC2 - 1
do.call(sum, lapply(apropos(glob2rx(ABC*)), get)) # 2
On 2/20/06, [EMAIL PROTECTED] [EMAIL
If f is long then you can get some savings like this:
do.call(f, mtcars) # note: used f rather than f
This does not solve the whole problem but its a step.
On 2/20/06, hadley wickham [EMAIL PROTECTED] wrote:
A problem that I've encountered when using do.call a lot is very large
stack
See:
?rowSums
On 2/20/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
I have a dataframe called data with 5 records (in rows) each of
which has been scored on each of many variables (in columns).
Five of the variables are named var1, var2, var3, var4, var5 using
headers. The other
The results are actually correct if you consider daylight savings time.
For example, try this and note that the difference is 23 hours, not 24 hours:
as.POSIXct(2004-04-05) - as.POSIXct(2004-04-04)
You can address this by either using Date or chron classes or adding
the tz = GMT argument on
I am not sure whether this is desirable but here is another way just
in case:
paste(setdiff(a, a[duplicated(a)]))
You could replace paste with as.character if you prefer or
could remove it entirely if you want the result as a factor.
On 2/22/06, Robin Hankin [EMAIL PROTECTED] wrote:
Hi.
I
The barplot solution already presented is probably what you want
but just in case here is a zoo solution:
library(zoo)
z - merge(zoo(x2), zoo(x1, seq(x1)+.5))
plot(z, type = h, plot.type = single, col = 1:2, lwd = 5)
Or a similar solution without zoo:
plot(c(x1, x2) ~ c(seq(x1)+.5, seq(x2)),
You could do it directly like this:
structure(split(ch.mat, col(ch.mat)),
row.names = 1:nrow(ch.mat), .Names = 1:ncol(ch.mat),
class = data.frame)
On 2/23/06, John M. Miyamoto [EMAIL PROTECTED] wrote:
Dear R-Help,
Suppose I have a character matrix, e.g.,
(ch.mat -
Just source the file:
source(mywinbugsfile.R)
head(y)
On 2/23/06, Jeffrey Moore [EMAIL PROTECTED] wrote:
For those who use WinBUGS (or for those who are just familar with this
format), I have a text file that looks like this (which is how R would
export data if you used the structure
Or even
aggregate(DF[3:4], DF[1:2], sum)
On 2/24/06, Marc Schwartz (via MN) [EMAIL PROTECTED] wrote:
On Fri, 2006-02-24 at 08:18 -0800, Matt Crawford wrote:
I am having trouble doing the following. I have a data.frame like
this, where x and y are a variable that I want to do calculations
This was just discussed recently. Try:
library(gtools)
?mixedorder
On 2/24/06, mtb954 mtb954 [EMAIL PROTECTED] wrote:
I'm trying to sort a DATAFRAME by a column ID that contains
alphanumeric data. Specifically,ID contains integers all preceeded
by the character g as in:
g1, g6, g3, g19,
the graph, without having to
manually quote all the data items in the array?
Thank you.
Sangeetha
= Original Message From Gabor Grothendieck [EMAIL PROTECTED]
=
Try this:
# test data
tt - c(23:05:02, 23:10:02, 23:15:03, 23:20:03, 23:25:03,
23:30:03, 23:35:03, 23:40:03, 23:45:04, 23:50
And also, read 1. R News 4/1 help desk article on dates and times,
2. the zoo vignette:
library(zoo)
vignette(zoo)
and 3. re-read the Introduction to R manual:
http://cran.r-project.org/doc/manuals/R-intro.pdf
On 2/24/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
See:
library(zoo
Create another variable that gives the run number and aggregate on
both the habitat and run number removing the run number after
aggregating:
runno - cumsum(c(TRUE, diff(as.numeric(transect[,2])) !=0))
aggregate(transect[,1], list(obs = transect[,2], runno = runno), sum)[,-2]
This does not give
was the key point... and then
cumsum for polishing. Really great and also elegant (concise). I like it!
Thanks a lot!!!
Cheers,
Patrick
Gabor Grothendieck a écrit :
Create another variable that gives the run number and aggregate on
both the
habitat and run number removing the run number
' observed come simply from the seed in
rpois(length(habitats),2)
It is unlikely it is the same on your and my computer...
Cheers,
Patrick
Gabor Grothendieck a écrit :
We are just comparing the difference to 0 so it does not matter if its
positive
or negative. All that matters is whether
Try this:
Xm - as.matrix(X)
X.lm - lm(c(Xm) ~ factor(col(Xm)) + factor(row(Xm)))
sum(resid(X.lm)^2)
or if the idea was to do it without using lm try replacing
your calculation of X.predicted with this:
X.predicted -
outer(operator.adjustment, machine.adjustment, +) + mean(mean(X))
On
Here is one further idea:
Xm - as.matrix(X)
Xm. - scale(Xm, scale = FALSE)
Xm.. - t(scale(t(tmp), scale = FALSE))
sum(Xm..^2)
On 2/27/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
Xm - as.matrix(X)
X.lm - lm(c(Xm) ~ factor(col(Xm)) + factor(row(Xm)))
sum(resid(X.lm)^2
Try:
crossprod(x)
or
t(x) %*% x
On 2/28/06, ronggui [EMAIL PROTECTED] wrote:
This is the code:
x-matrix(rnorm(20),5)
y-list()
for (i in seq(nrow(x))) y[[i]]-t(x[i,,drop=F])%*%x[i,,drop=F]
y[[1]]+y[[2]]+y[[3]]+y[[4]]+y[[5]]
How can I do it without using for loops?
Thank you in advance!
You could use a bootstrapped confidence interval. You can find
R code examples of using bootstrapped confidence intervals
for correlation coefficients (and also an example of the
alternative Fisher Transform) in the proto vignette:
library(proto)
vignette(proto) # see 3.2
On 2/28/06, McGehee,
Use as.matrix to convert your data frame to a matrix
and suppose we have this test data as a matrix:
mat - matrix(seq(30*7), 30, 7)
Then try this:
library(zoo)
mat2 - coredata(rapply(zoo(mat), 12, prod))
See:
library(zoo)
vignette(zoo)
and the various zoo help files for more
Try this:
lapply(z, [, 2:3, TRUE)
On 2/28/06, Federico Calboli [EMAIL PROTECTED] wrote:
Hi All,
I have a list of matrices:
x
[,1] [,2]
[1,]14
[2,]25
[3,]36
y
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 18 21 24 27 30 33
[2,] 19 22 25 28
I think the idea of the prior respondents was that R_DATADIR
would be set outside R and the application and just fetched from
an environment variable inside the application so that the application
is independent of it.
If, in fact, your R code knows the data directory anyways, you could
just do
Or use gl which directly forms a factor:
group - gl(2, 5, 30)
time - gl(3, 10)
subject - gl(10, 1, 30)
On 2/28/06, John Vokey [EMAIL PROTECTED] wrote:
Christian,
You need, first to factor() your factors in the data frame P.PA,
and then denote the error-terms in aov correctly, as follows:
Are there any R packages that relate to the
following data reduction problem fo finding
maximally independent variables?
Currently what I am doing is solving the following
minimax problem: Suppose we want to find the
three maximally independent variables. From the
full n by n correlation
(DF), 3)
maxcor - function(x) max(as.vector(as.dist(cor(DF[,x]
names(DF)[z[which.min(apply(z, 1, maxcor)),]]
Gabor Grothendieck a écrit :
Are there any R packages that relate to the
following data reduction problem fo finding
maximally independent variables?
Currently what I am doing
Also check out ?append and the after= argument, in particular.
On 3/1/06, Ben Bolker [EMAIL PROTECTED] wrote:
Christian Hoffmann christian.hoffmann at wsl.ch writes:
Hi,
In a second try I will ask this list to give me some useful pointers.
[There is a general problem with the list,
In case others are interested I did get a reply offlist
regarding the escouf function in the pastecs package.
See:
library(pastecs)
?escouf
Also see pages 47-52 of
system.file(doc/pastecs.pdf, package = pastecs)
(in French).
On 3/1/06, Gabor Grothendieck [EMAIL PROTECTED] wrote
I just asked a similar question yesterday except I was using
correlation-based distances rather than Euclidean. Check
out that thread:
https://www.stat.math.ethz.ch/pipermail/r-help/2006-March/087782.html
On 3/1/06, Matt Sakals [EMAIL PROTECTED] wrote:
Dear all,
I have point data from an
Package lpSolve might help.
On 3/1/06, Mark [EMAIL PROTECTED] wrote:
Dear R community,
I have a dataframe with 500,000 rows and 102 columns. The rows
represent spatial polygons, some of which overlap others (i.e., not
all rows are independent of each other).
Given a particular row, the
Do you mean you wish to create a vector without the trailing NAs
but with the others? If so, try:
library(zoo)
head(x, length(na.locf(x, rev = TRUE)))
or
library(zoo)
head(x, length(na.locf(rev(x
On 3/2/06, Robert Lundqvist [EMAIL PROTECTED] wrote:
I wonder if anyone could help me find
Note that this still has the restriction that getdots must be used
directly in the function which is to be deparsed so one could not,
for example, put it in another utility function which deparses
the function and extracts the first component. To do that you
need to pass the frame:
f -
This might help:
http://zoonek2.free.fr/UNIX/48_R/03.html
On 3/3/06, Dan Bolser [EMAIL PROTECTED] wrote:
Hi,
Since I started to make some 'final' plots of my data I found that I
have tons of questions related to 'the little things'. Rather than
bother the list with all the questions (ahem),
It will change the coefficients from lm and therefore their t-values but it
will not change the residuals, fitted values, R-squared, F
statistic, etc. For example, try this:
set.seed(1)
x - 1:10
y - x + rnorm(10)
lm0 - lm(y ~ x) # without scale
lms - lm(y ~ scale(x)) # with scale
Try bquote (I may not have followed precisely what you
want but hopefully this gives the idea). Note that you
may need to use a few judiciously placed ~ and
phantom(), as shown, in order to maintain syntax:
plot(NA, xlim = c(0, 100), ylim = c(0, 100))
avar1 - 10
amath1 - bquote(slope ==
Try this (note that your x and y do not have the same length
and in this case the expression will recycle the shorter one
and give a warning):
z - c(rbind(x, y))
On 3/5/06, Ajay Narottam Shah [EMAIL PROTECTED] wrote:
Suppose one has
x - c(1, 2, 7, 9, 14)
y - c(71, 72, 77)
If you make the levels the same does that give what you want:
levs - c(LETTERS[1:6], 0)
tmp1 - data.frame(col1 = factor(c(A, A, C, C, 0, 0), levs))
tmp2 - data.frame(col1 = factor(c(C, D, E, F), levs), col2 = 1:4)
merge(tmp2, tmp1, all = TRUE, sort = FALSE)
merge(tmp1, tmp2, all = TRUE, sort =
I think you will need to reorder it:
out - merge( cbind(tmp1, seq = 1:nrow(tmp1)), tmp2, all.x = TRUE, sort = FALSE)
out[out$seq, -2]
On 3/6/06, Gregor Gorjanc [EMAIL PROTECTED] wrote:
Gabor Grothendieck wrote:
If you make the levels the same does that give what you want:
levs - c
Actually we don't need sort = FALSE if we are reordering it anyways:
out - merge( cbind(tmp1, seq = 1:nrow(tmp1)), tmp2, all.x = TRUE)
out[out$seq, -2]
On 3/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
I think you will need to reorder it:
out - merge( cbind(tmp1, seq = 1:nrow(tmp1
On 3/6/06, Gregor Gorjanc [EMAIL PROTECTED] wrote:
But I want to get out
A NA
A NA
C 1
C 1
0 NA
0 NA
That's what I get except for the rownames. Be sure to
make the factor levels consistent. I have renamed the data frames
tmp1a and tmp2a to distinguish them from the ones in your
post
POSIXlt stores its data in a 9 component structure so its
length is always 9. See R News 4/1 help desk article for more info.
To get what you are looking for try using POSIXct, not POSIXlt:
length(as.POSIXct(times))
On 3/6/06, Jason Horn [EMAIL PROTECTED] wrote:
I have a vector of POSIX
= TRUE)
out - out[out$seq, -2]
rownames(out) - rownames(tmp1a)
out
col1 col2
1A NA
2A NA
3C1
4C1
50 NA
60 NA
On 3/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
On 3/6/06, Gregor Gorjanc [EMAIL PROTECTED] wrote:
But I want to get out
A NA
A NA
and reseting of rownames.
Like the prior solution, it assumes that the elements of tmp2a$col1
are unique.
On 3/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Sorry, I mixed up out and outa in the last post. Here it is correctly.
levs - c(LETTERS[1:6], 0)
tmp1a - data.frame(col1 = factor(c(A, A, C
The following are nearly identical to what others have already
written but just in case:
A - c(B)
or
A - B[,1]
or if B were already a vector, b, in the first place, rather than a matrix:
b - 1:2
A - b
On 3/6/06, Michael [EMAIL PROTECTED] wrote:
Hi all,
I want to substract vector B from
POSIXct, not POSIXlt, are used in data frame columns.
See ?POSIXct where this is mentioned and try:
A$date - as.POSIXct(A$date)
Also be sure to read R News 4/1 for more about dates and times.
On 3/6/06, Andrew Athan [EMAIL PROTECTED] wrote:
I'm sure this is just the result of a basic
Try this:
m - diag(4-abs(-4:4)) # test matrix
wm - which.max(m)
c(row(m)[wm], col(m)[wm]) # c(5,5)
On 3/6/06, Michael [EMAIL PROTECTED] wrote:
Hi all,
I want to use which.max to identify the maximum in a 2D array/matrix and I
want argmin and return the row and column indices.
But
Try this:
iris.by - do.call(rbind, by(iris[,-5], iris[,5,drop=FALSE],colSums))
do.call(rbind, iris.by)
On 3/7/06, Vivek Satsangi [EMAIL PROTECTED] wrote:
Folks,
I know that I can do the following using a loop. That's been a lot
easier for me to write and understand. But I am trying to force
Sorry, that should be:
iris.by - by(iris[,-5], iris[,5,drop=FALSE],colSums)
do.call(rbind, iris.by)
On 3/7/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
iris.by - do.call(rbind, by(iris[,-5], iris[,5,drop=FALSE],colSums))
do.call(rbind, iris.by)
On 3/7/06, Vivek Satsangi [EMAIL
Only and GMT are really guaranteed to work on all systems
since the time zones are system dependent but try: CDT6CST
and see if that works on your system.
On 3/7/06, Jason Horn [EMAIL PROTECTED] wrote:
Whoops,
[EDIT]
as.POSIX(x, tz=UTC) ... works, gives UTC times
as.POSIX(x, tz=EST) ...
systems are.
Thanks!
On Mar 7, 2006, at 9:00 AM, Gabor Grothendieck wrote:
Only and GMT are really guaranteed to work on all systems
since the time zones are system dependent but try: CDT6CST
and see if that works on your system.
On 3/7/06, Jason Horn [EMAIL PROTECTED] wrote
Another solution (ex #3) is to return a data frame with at least two columns
and then remove the dummy one using drop = FALSE:
# ex #1. error
iris.by - by(iris, iris[,5, drop = FALSE], length)
do.call(rbind, iris.by)
# ex #2. ok but no column heading
iris.by - by(iris, iris[,5, drop=FALSE],
POSIX objects are UTC internally but can be displayed
with respect to other time zones.
By the way, do you really need time zones in the first
place? Read R News 4/1 help desk article which will
help you determine which date/time class is
suitable for your application and also has a table
of
I agree that that is the best solution although not to the
question as stated which said: I want to use which.max
On 3/7/06, Michael [EMAIL PROTECTED] wrote:
I think this is the best solution! Thank you!
On 3/7/06, Petr Pikal [EMAIL PROTECTED] wrote:
Hi
If you do not insist on
If your class is a subclass of data.frame then I think
it ought to be a special sort of data frame so all you
need to do is the following in which case you get subscripting
for free by inheritance:
# constructor
myobj - function(...)
structure(data.frame(...), class = c(myobj,
Just to get something reproducible lets assume the
objects of class myobj each consists of a one-element
list that contains a data frame. Then try this:
# constructor
myobj - function(...)
structure(list(value = data.frame(...)), class = myobj)
$.myobj - function(obj, x) .subset2(obj,
On 3/7/06, hadley wickham [EMAIL PROTECTED] wrote:
Just to get something reproducible lets assume the
objects of class myobj each consists of a one-element
list that contains a data frame. Then try this:
Thanks for that - it makes sense. Every time I try to use inheritance
in R, I
The problem is that x[[1]] in the definition of as.data.frame.myobj
invokes [[.myobj whereas we want to extract the first element of the
list in the internal representation of x -- which is not the same.
Try this instead:
as.data.frame.myobj - function(x) .subset2(x, 1)
lm(y ~ x, z)
Call:
Another possibility is eapply where I have used naCount
from Henrik's solution:
prop - function(x)
list(class = data.class(x), dim = dim(x), size = object.size(x), NAs
= naCount(x))
do.call(rbind, eapply(.GlobalEnv, prop))
On 3/8/06, Henrik Bengtsson [EMAIL PROTECTED] wrote:
library(R.oo)
Check out:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/52957.html
for a similar problem.
On 3/8/06, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
Dear All,
I'm trying to read a text data file that contains several records separated
by a blank line. Each record starts with a row that
- matrix(unlist(matlis), 4, length(matlis))
S1 - matrix(rowSums(S), 2, 2)
The code works ,but I want to ask if there is any other more better
ways to do it? It seems that this kind of computation is quite common.
2006/2/28, Gabor Grothendieck [EMAIL PROTECTED]:
Try:
crossprod(x
arguments one could have done this instead of the
two lines involving result above:
do.call(psum, mm) # if psum analog to pmax were available
do.call(+, mm) # if + allowed 2 arguments
On 3/8/06, Thomas Lumley [EMAIL PROTECTED] wrote:
On Wed, 8 Mar 2006, Gabor Grothendieck wrote
On 3/8/06, Thomas L Jones [EMAIL PROTECTED] wrote:
From Tom:
The subject is debugging a program written in the R language,under
Windows. (Sorry, but I do not know either the Apple OS or *nix.) A
computer program will usually not work on the first try, if only
because of the risk of typos.
Can you create a small self contained reproducible example
that does not work? The reproducible example I provided earlier on
this thread worked fine.
One idea is to check what the class is of the output of
your .GGobiCall. If it were of class ggobiDataset then
it would in turn be calling
I think the problem is that your ggobiDataset objects are also
data.frame objects. They must NOT be. For example,
note how the following example fails once we add data.frame
to the class vector of x: The reason is that ordinary inheritance
is not used by model.frame; rather, it uses
Try this:
x - 1:20
c(rowsum(x, gl(length(x), 5, length(x)))
If the length of x is not a mulitple of 5 there will be a stub at the
end containing the sum of less than 5 elements.
On 3/10/06, Amir Safari [EMAIL PROTECTED] wrote:
Hi R Users,
I don't know how much is difficult my problem
The argument to legend should be of class date, not character.
Also please post reproducible code including definitions of all
variables used and library calls.
On 3/10/06, Bryan Sykes [EMAIL PROTECTED] wrote:
Hi:
I'm trying to plot dates on the x-axis of a code, but the legend is not being
Check out:
http://www.sciviews.org/_rgui/projects/Editors.html
http://ess.r-project.org
On 3/10/06, Ana Patricia Martins [EMAIL PROTECTED] wrote:
Hi to all,
I initiate in R - Linux and I've some problems to find an editor with R
interface as like RWinEdt for WinEdt.
Anyone know one?
Is anyone using vi for R?
I use vim (actually gvim) on Windows XP. It has syntax highlighting
for R, tex and many other formats.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
zoo has the yearmon class which will convert a date to a year +
0/12 to 11/12 to represent the month, disregarding the day, so:
library(zoo)
12 * as.numeric((as.yearmon(2006-03-07) - as.yearmon(2006-02-06))) # 1
On 3/10/06, Smith, Phil [EMAIL PROTECTED] wrote:
Hi R-people:
I need a function
to
change the date string to a Date class before you call as.yearmon? i.e.
12*
as.numeric(as.yearmon(as.Date(2006-03-07))-as.yearmon(as.Date(2006-02-07)))
That returns 1 in Windows XP
Regards
Francisco
From: Gabor Grothendieck [EMAIL PROTECTED]
To: Smith, Phil [EMAIL PROTECTED]
CC: r-help
Try
RSiteSearch(shade area under curve)
On 3/11/06, Albert Sorribas [EMAIL PROTECTED] wrote:
I'm trying to obtain a plot of given function, say the density of a
normal, in which I could indicate the area under the curve between two
points (a,b).
I.e. the P(aXb). Could anyone indicate me
y is of class matrix and $ is used for data frames, not matrices.
Use y[,x1] or else create a data frame, data.frame(x1, x2, x3).
On 3/11/06, Sander Oom [EMAIL PROTECTED] wrote:
Dear R-users,
I need to do an SQL like, conditional, operation on a data frame using
an ifelse construction.
I am not sure I understand the question but is the situation that
you have two vectors: x and y such that for each level of y
x is constant so that for each level of y you want to find that value
of x? In that case:
x - c(A, A, A, B, B)
y - c(1,1,2,3,3)
unique(data.frame(x,y))
or
tapply(x, y,
Where does one find A2Rplot that is called in that example?
A2R does not appear to be on CRAN. I did find this:
http://addictedtor.free.fr/Download/A2R.zip
which is a zip file containing some R code but it is not a package,
which the line library(A2R) seems to need, and it does not include
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