On 8/17/06, Martin Maechler <[EMAIL PROTECTED]> wrote:
> >>>>> "Gregor" == Gregor Gorjanc <[EMAIL PROTECTED]>
> >>>>> on Fri, 11 Aug 2006 00:27:27 + (UTC) writes:
>
>Gregor> Gabor Grothendieck gmail.com> writes:
&g
Try:
mapply(rbind, a, b, SIMPLIFY = FALSE)
On 8/17/06, Domenico Vistocco <[EMAIL PROTECTED]> wrote:
> Dear helpeRs,
>
> suppose I have two lists as follows:
>
> a = list(1:5,5:9)
> b = lapply(a,"*",2)
>
>
> I would like to rbind-ing the two lists, that is I would like to use
> something as rbind
Also check out the displaylist:
http://tolstoy.newcastle.edu.au/R/help/04/05/0817.html
On 8/17/06, Lothar Botelho-Machado <[EMAIL PROTECTED]> wrote:
> -BEGIN PGP SIGNED MESSAGE-
> Hash: SHA1
>
> Thank you,
>
> It seems that a list of plots is just possible using lattice plots. But
> that'
The following reshapes mat so we can take the means of the columns
of the resulting 3d array and then transposes it back to the original
orientation:
t(colMeans(array(t(mat), c(100, 448, 24
You might want to try it on this test set first where anscombe
is an 11x8 data set built into R. He
Its a FAQ
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-do-lattice_002ftrellis-graphics-not-work_003f
On 8/16/06, Nick Desilsky <[EMAIL PROTECTED]> wrote:
> Hi,
>
> running the following code by itself runs as expected.
> --
Try this after displaying the xyplot:
# this fit.curve returns the whole nls object, not the coefs
fit.curve<-function(tab) {
nls(y1 ~ 100/(1+exp(((-log(81))/a)*(x1-b))),
start=list(a=min(tab$x1[tab$y1>76],na.rm=T)-max(tab$x1[tab$y1<15],na.rm=T),b=tab$x1[abs(tab$y1-50)==min(abs(tab$y1-50),n
Try this:
x <- seq(-100,1000,25)
y <- x * x
plot(x, y, xaxt = "n")
axis(1, x[x %% 100 == 0])
On 8/11/06, Darren Weber <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm stuck on creating a plot with x tick labels for every Nth tick
> mark - how is that done? I don't see a simple solution to this in
> he
The approach here is to perform the repetition on the indices (or rownames)
rather than on the data frame directly. Using the builtin data frame BOD
any of the following would work:
BOD[gl(nrow(BOD), 2),]
BOD[rep(1:nrow(BOD), each = 2),]
BOD[rep(rownames(BOD), each = 2),]
On 8/11/06, Horace Tso
quot;), class = "data.frame", row.names = c("1", "2", "3",
"4", "5", "6", "7", "8", "9", "10", "11", "12", "13"))
histogram(~ unlist(DF[,-1]), type = "density&
ax(hist(DF$A1, breaks = breaks, plot = FALSE)$intensities)
ymax2 <- max(hist(DF$A2, breaks = breaks, plot = FALSE)$intensities)
ylim <- c(0, max(ymax1, ymax2))
# draw the two histograms and two densities
hist(DF$A1, ang = 45, col = "red", ylim = ylim,
breaks = breaks, freq
>From your description I assume you want both histograms
and the densities all on the same chart. With existing R
graphics I am not sure that there really is a simple way to
do that.
That aside, note that the hist function returns a list of
components that includes
- breaks, defining the breakpo
Here are three ways:
xx <- as.Date("2006-01-05")
# 1. use as.POSIXlt
as.POSIXlt(xx)$mday
as.POSIXlt(xx)$mon + 1
as.POSIXlt(xx)$year + 1900
# 2. use format
as.numeric(format(xx, "%d"))
as.numeric(format(xx, "%m"))
as.numeric(format(xx, "%Y"))
# 3. use month.day.year in chron package
library(chro
Try:
lm(Sepal.Length ~., iris)
On 8/10/06, r user <[EMAIL PROTECTED]> wrote:
> I am using R in a Windows environment.
>
> I have a basic question regarding lm().
>
> I have a dataframe "data1" with ncol=w.
>
> I know that my dependent variable is in column1.
>
> Is there a way to write the regres
A matrix M can be thought of as a linear transformation which maps
input vector x to output vector y:
y = Mx
The eigenvectors are those "directions" that this mapping preserves.
That is if x is an eigenvector then y = ax for some scalar a. i.e.
y lies in the same one dimensional space as x.
On 8/9/06, John McHenry <[EMAIL PROTECTED]> wrote:
> Hi WizaRds,
>
> In MATLAB you can do
>
> x=1:10
>
> and then specify
>
> x(2:end)
>
> to get
>
> 2 3 4 5 6 7 8 9 10
>
In R you could do the above via:
x[-1]
> or whatever (note that in MATLAB the parenthetic index notation is used, not
> bra
That's not a valid specification. See the description of the index.cond
argument in ?xyplot and in particular this part:
If 'index.cond' is a list, it has to be as long as the number of
conditioning variables, and the 'i'-th component has to
be a valid
Dmitris has already provided the solution but just throught I would'
mention that your third alternative can be written:
apply(mymatrix, 1, fun2, bb = bb)
(assuming fun2 has arguments idx and bb) which is not
nearly so ugly so you might reconsider whether its ok
for you to just pass bb.
On 8/9/
Try this:
DF[unlist(tapply(rownames(DF), DF$id, function(x) c(x, x[1]))),]
On 8/9/06, Leonardo Lami <[EMAIL PROTECTED]> wrote:
> Hi all,
> I have a simple question:
> I have a data.frame like this:
>
> id x y
> 1 50 1647685 4815259
> 2 50 1647546 4815196
> 3 50 1647454 4815294
>
Check out the machining learning task view at:
http://cran.r-project.org/src/contrib/Views/
On 8/9/06, Christian Miehle <[EMAIL PROTECTED]> wrote:
> Hallo,
>
> Ich bin auf der Suche nach umgesetzten evolutionären Algorithmen in R. Leider
> habe ich kein entsprechendes Package oder Funktionen die
1. Use the x, y and corner components to the key= list to specify
the legend position, and
2. pass the panel.number in the panel function and test that as shown
in the panel function below.
Alternately you can place the horizontal line on afterwards using
trellis.focus/trellis.unfoc
I agree. Also, sending a copy to the poster means that they are
likely to get it first which seems like a desirable courtesy.
On 8/8/06, Marc Schwartz (via MN) <[EMAIL PROTECTED]> wrote:
> [Re-sending to the list only for archiving, as my original reply had too
> many recipients and I cancelled i
Try this:
# mat is test matrix
mat <- matrix(1:25, 5)
mat[2,2] <- mat[3,4] <- NA
crossprod(!is.na(mat))
On 8/7/06, Adam D. I. Kramer <[EMAIL PROTECTED]> wrote:
> Hello,
>
> I'm using a very large data set (n > 100,000 for 7 columns), for which I'm
> pretty happy dealing with pairwise-deleted cor
Although this is probably not directly applicable to this problem
I might mention here that merge.zoo does support left and right
joins and that handles problems similar to this. z3t, z3ft,
z3tf and z3f below have times of both unioned, the times of
z2, the times of z1 and the times of both z1 and
Also RSiteSearch("ts.plot.2Axis")
On 8/7/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Try:
>
> RSiteSearch("Horses and Hounds")
>
>
> On 8/7/06, Sonal Darbari <[EMAIL PROTECTED]> wrote:
> > Hi,
> >
> > What commands
Try:
RSiteSearch("Horses and Hounds")
On 8/7/06, Sonal Darbari <[EMAIL PROTECTED]> wrote:
> Hi,
>
> What commands are needed to get an output like this:
>
>
> 1. On X-Axis : 2 Indices ex. S&P500 and DOW JONES
> 2. Their repective dates
>
>
> If I use the "plot" command, I get one output & if
t that and it says:
>
>
> Partitioning Clustering:
>
>Functionkmeans()
>from package stats provides several algorithms for computing
> partitions with respect to Euclidean distance.
> Hence why
There are many clustering functions in R and R packages and some
take distance objects whereas others do not. You likely read about
hclust or some different clustering function. See ?kmeans for the
kmeans function and also look at the CRAN Task View on clustering for
other clustering functions:
Also check out CrossTable in the gmodels package.
Regarding your other question, assuming we have
tab<-table(x,y) as in Philippe's post, the fraction of
pairs in x and y that match can be calculated via
any of these:
sum(x==y) / length(x)
sum(diag(tab)) / sum(tab)
library(e1071)
classAg
Try this:
Lines <- "story,datepub
story10,1 April 1999
story 90,1 March 2002
story 37,10 July 1985
"
DF <- read.csv(textConnection(Lines))
DF[order(as.Date(DF$datepub, "%d %B %Y")),]
On 8/6/06, Bob Green <[EMAIL PROTECTED]> wrote:
>
> I am hoping for some advice regarding ordering a dataframe,
Here are three ways:
# read in data
Lines <- "object1 object1 78
object1 object2 45
object1 object3 34
object1 object4 45
object2 object2 89
object2 object3 32
object2 object4 13
"
DF <- read.table(textConnection(Lines))
# 1 - xtabs
xt <- as.matrix(xtabs(V3 ~., DF))
# 2 - reshape
wide <- reshape
for (ch in x$children) recurse(ch)
}
recurse(grid.get("pretty"))
On 8/3/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> If you are willing to use grid then you could create only the sex
> factor in the left strips since its already in the desired position
>
Sorry, you wanted a ylab=, not a main=. Try using xyplot in lattice:
library(lattice)
xyplot(1~1, ylab = expression(atop(phantom(0)^14*C*"-glyphosate line",
"line2")))
On 8/4/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Use atop:
>
> plot
Use atop:
plot(1, main = expression(atop(" "^14*C*"-glyphosate line", "line2")))
On 8/4/06, Andrew Kniss <[EMAIL PROTECTED]> wrote:
> I've tried several different ways to accomplish this, but as yet to no
> avail. My y-axis for a plot has a rather long label, and thus I have
> been using "/n"
When specifying a column name with [ the name must be quoted (unlike
when using it with $):
a[a$y > 0.5, "y"] <- 1
On 8/4/06, Sander Oom <[EMAIL PROTECTED]> wrote:
> Dear R users,
>
> When you do:
> > x <- rnorm(10)
> > y <- rnorm(10)
> > z <- rnorm(10)
> > a <- data.frame(x,y,z)
> > a$x
> [1
Just sending this to you. One thing that might be easy to do
yet give a lot of flexibility is to:
1. put meaningful names on the grobs. Even with just this it would be
possible to do a getNames() in grid and then from inspection grid.edit
the appropriate one(s).
2. create a routine that retrieve
ut.widths settings with no luck. It
> seems the spaces are calculated based on the number of conditioning
> variables, in this case 2 (sex+smoker).
>
>
> Thanks in advance...
> -Sam
>
>
> -Original Message-
> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
> S
On 8/3/06, John Kane <[EMAIL PROTECTED]> wrote:
>
> --- Don MacQueen <[EMAIL PROTECTED]> wrote:
>
> > You don't need to find out the column index. This
> > works:
> >
> > Df[5,'bat'] <- 100
> >
> > -Don
> >
>
> Thanks, I'd tried
> Df[5, bat] <- 100 :(
>
> I never thought of the ' ' being neede
On 8/2/06, Walker, Sam <[EMAIL PROTECTED]> wrote:
> How do I change the font size in the facet labels along the edges of the
> plot?
>
> For example (from the ggplot help file):
> p<-ggplot(tips, sex ~ smoker, aesthetics=list(x=tip/total_bill))
> gghistogram(p)
>
> In this plot, the facet l
If you set it through par.settings then it will affect both the
drawing and the legend:
xyplot(Sepal.Length ~ Petal.Length, iris, groups = Species, auto.key = TRUE,
par.settings = list(superpose.symbol = list(pch = "*", cex = 1)))
On 8/2/06, Kaushik Katari <[EMAIL PROTECTED]> wrote:
> I am doi
Assuming this data:
s <- structure(list(L.qol.0 = 83, L.qol.0.08 = 86, L.qol.0.17 = 89,
L.qol.0.25 = 92, L.qol.0.5 = 91, L.qol.0.42 = 87, L.qol.0.34 = 90),
.Names = c("L.qol.0", "L.qol.0.08", "L.qol.0.17", "L.qol.0.25",
"L.qol.0.5", "L.qol.0.42", "L.qol.0.34"),
class = "data.frame", r
On 8/2/06, Sergio Martino <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I would like to realize in R a structure like the fortran common ie a way to
> declare some variable that can only be accessed by all the functions which
> need to.
>
> Browsing the archive it seems that the simplest way is to declare t
Try something along these lines assuming that the current
directory is \Program Files\R\R-2.3.1pat. Note use
of paste to create the command line to pass to pipe:
# search for indicated string in each of the files and
# for each match output the file name
Files <- c("CHANGES", "COPYING", "NEWS", "
This refers to the windows command pedump.exe found in the Rtools
collection at:
http://www.murdoch-sutherland.com/Rtools/
On 8/2/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
>
> Hello Sir,
>
> I am just wondering that pedump is a command of 'R' because in could not
> find in the 'R' help
3 as required.
for (j in 1:3) print(environment(formula(mod[[j]]))$i)
# following two lines give same answer
# showing prediction works
predict(mod[[2]], list(x = 1:10))
fitted(lm(y ~ poly(x,2)))
On 8/1/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Actually in thinking about this s
Actually in thinking about this some more that still gets you
into a mess if you want to do prediction at anything other
than the original points.
On 8/1/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> A simple way around this is to pass it as a data frame.
> In the code below the
A simple way around this is to pass it as a data frame.
In the code below the only change we made was to change
the formula from y ~ poly(x, i) to y ~ . and pass poly(x,i)
in a data frame as argument 2 of lm:
# test data
set.seed(1)
x <- 1:10
y <- x^3 + rnorm(10)
# run same code except change the
There is a global option setting for na.action. See ?na.action .
That does not completely address your question but might
help with lm, glm, etc.
You could define your own wrapper functions if you know ahead of time
which functions with na.rm= args you need. e.g.
my.max = function(..., na.rm = g
Its easiest to just check the source. biplot is a generic which calls
biplot.princomp which calls biplot.default which in turn calls plot so
try this and examine the source:
stats:::biplot.default
On 7/31/06, Patrick Connolly <[EMAIL PROTECTED]> wrote:
> I'm attempting to modify how biplot draw
Try this:
m <- 5; beta <- c(-1, 1)
plot(1, main = bquote(atop(beta == .(deparse(beta)),
"Expected number of true positives" == .(m
On 7/31/06, Juan Lewinger <[EMAIL PROTECTED]> wrote:
> Dear R users,
>
> Two questions:
>
> 1) Is there a way to simplify the mtext() line below ?
>
> beta=c(
= "light grey"))
# re-plot panel over rectangle
do.call("panel.xyplot", trellis.panelArgs())
trellis.unfocus()
nevertheless, as a point of general interest I would still be
interested to know
what a general grid-based solution might be.
On 7/30/06, Gabor Grothendieck &l
Try:
?sensory
str(sensory)
dput(sensory)
lapply(sensory, class)
lapply(sensory, dim)
to see what it looks like inside. Seems that sensory is a data frame
consisting of two columns each of which is a matrix except that each
has a class of "AsIs". Thus try this (where I(...) creates objects of
cl
ere. This would also be helpful for adding grid
lines afterwards or other lines, rectangles, etc.
On 7/30/06, Paul Murrell <[EMAIL PROTECTED]> wrote:
> Hi
>
>
> Gabor Grothendieck wrote:
> > I am trying to create a lattice plot and would like to later, i.e. after
> > the pl
I assume the 3 is supposed to be a subscript. Try this:
b1 <- x <- y <- 1
plot(x,y, main = bquote("Results for " ~ beta[3] ==.(b1)))
On 7/30/06, Marco Boks <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I need to plot the beta as the symbol, followed by the index 3 as the title
> of a graph.
>
> This
Please provide reproducible examples (as discussed at end of each
posting):
Lines <- "Apple S 21.0
Apple A 21.6
Apple O 43.0
Orange A 45.0
Orange O 64.0
Orange S 32.5
Mango M 40.3
Mango A 32.6
Mango S 24.6
"
tb <- read.table(textConnection(Lines))
# alternative 1 - create a matrix
t
Jul 2006 17:20:29 -0400,
> "Gabor Grothendieck" <[EMAIL PROTECTED]> wrote:
>
> > I am trying to create a lattice plot and would like to later, i.e. after
> > the plot is drawn, add a grey rectangle behind a portion of it. The
> > following works except that the rect
I am trying to create a lattice plot and would like to later, i.e. after
the plot is drawn, add a grey rectangle behind a portion of it.
The following works except that the rectrangle is on top of and
obscures a portion of the chart. I also tried adding col = "transparent"
to the gpar list but tha
Or even easier,
x <- scan("clipboard", what = "")
at least on Windows.
On 7/28/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Enter this at the console
>
> x <- scan(what = "")
>
> and after pressing the enter after the right pare
Enter this at the console
x <- scan(what = "")
and after pressing the enter after the right paren,
do a paste and then press enter twice.
On 7/27/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> I am new to R, so please forgive me if there is an obvious answer to
> this question. I have d
Just read them in and throw them away:
read.table("myfile.dat", ...whatever...)[-c(10, 12), ]
On 7/27/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> I am reading the data using "read.table". However, there are a few rows I
> want to skip. How can I do that in an easy way? Su
How large is your data?
If its large you may or may not have problems. If its small you probably
won't. Try prototyping the most data intensive portion in R before you
commit significant resources.
S Plus can time stamp objects and R cannot although you could come
up with some workarounds for
Try this where gr and theta are as in your post:
xyplot(1~1|gr,
main = as.expression(bquote(theta == .(theta))),
strip = strip.custom(factor.levels = expression(theta, beta))
)
On 7/27/06, Valentin Todorov <[EMAIL PROTECTED]> wrote:
> Unfortunately this does not work for lattice
Use the notation x[ , 1, drop = FALSE]
See ?"["
On 7/27/06, Neuro LeSuperHéros <[EMAIL PROTECTED]> wrote:
> Transpose vector extracted from a matrix
>
> Hello,
>
> I am doing a recursive analysis that uses every line (vector) of a matrix in
> a loop. In the model, I need to transpose those vectors
Try this where f and A2 are as in your post:
out <-f(A2[A2>0])
replace(matrix(0, length(A2), ncol(out)), A2 > 0, out)
On 7/27/06, Robin Hankin <[EMAIL PROTECTED]> wrote:
> Hi
>
>
> I have a little vector function that takes a vector A of strictly
> positive integers
> and outputs a matrix
The complexity of the function should not matter.
Here is another example of this technique:
http://tolstoy.newcastle.edu.au/R/help/04/06/1430.html
On 7/27/06, Heinz Tuechler <[EMAIL PROTECTED]> wrote:
> At 06:10 27.07.2006 -0400, Gabor Grothendieck wrote:
> >If you are willing t
See ?match.call
On 7/27/06, Armstrong, Whit <[EMAIL PROTECTED]> wrote:
> I see that plot.default uses deparse(substitute(x)) to extract the
> character name of an argument and put it on the vertical axis.
>
> Hence:
> foo <- 1:10
> plot( foo )
>
> will put the label "foo" on the vertical axis.
>
>
Assuming the problem is to partition the 10x10 matrix x into 25 two by two
squares and then average each of those squares, try this:
apply(array(x, c(2,5,2,5)), c(2,4), mean)
On 7/27/06, Vladimir Eremeev <[EMAIL PROTECTED]> wrote:
> Dear r-help,
>
> I have a matrix, suppose, 10x10, and I need
If you are willing to write fu2[Var] <- 3 instead of fu2(Var) <- 3
then this workaround may suffice:
fu2 <- structure(NA, class = "fu2")
"[<-.fu2" <- function(x, ..., value) { print(match.call()[[3]]); fu2 }
# test
fu2[Var] <- 3 # prints "Var"
On 7/27/06, Heinz Tuechler <[EMAIL PROTECTED]> wro
With the lattice package it would be done like this (where
the panel.points function places large red pluses on
the plot):
xyplot(Consumption ~ Quarter, group = Year, data, type = "o")
trellis.focus("panel", 1, 1)
panel.points(1:4, mean.per.quarter, pch = "+", cex = 2, col = "red")
trellis.unfocus
Here is a minor simplication:
my.hmtest <- structure(list(
estimate = t(t(out.data.mat[,"estimate",drop=FALSE])),
conf.int = out.data.mat[,2:3],
ctype = "Dunnett"),
class = "hmtest")
plot(my.hmtest)
On 7/26/06, Gabor Grothendieck <[EMA
Look through
multcomp:::plot.hmtest
to find out which components of an hmtest object are actually used.
Now look at what an hmtest object looks like by doing this
dput(Dcirec)
or looking through the source of the function that produces hmtest
objects. With this information in hand we can cons
If you are using grep then I think you have it right. Note that
"this" %in% trg
is also available.
On 26 Jul 2006 11:16:25 -0400, Allen S. Rout <[EMAIL PROTECTED]> wrote:
>
>
>
> Greetings, all.
>
> I'm fiddling with some text manipulation in R, and I've found
> something which feels counter
See
?filter - simple and exponential are special cases
?runmean - in package caTools (the fastest)
?rollmean - in zoo package
?embed - can write your own using embed as basis
?sma - in package fSeries, also see ewma in same package
Probably other functions in other packages too.
On 7/26/06, [EMA
Try these:
# 1
library(Hmisc)
summary(y ~ ind, dat, fun = range, overall = FALSE)
# 2
# or with specified column names
f <- function(x) c(head = head(x,1), tail = tail(x,1))
summary(y ~ ind, dat, fun = f, overall = FALSE)
# 3
# another approach using by - same f as above
do.call(rbind, by(dat$y,
This was just discussed yesterday. See the thread:
https://www.stat.math.ethz.ch/pipermail/r-help/2006-July/109931.html
On 7/26/06, Marco Boks <[EMAIL PROTECTED]> wrote:
> I am a newbie, and I am afraid this may be a rather trivial question. However
> I could not find the answer anywhere.
>
>
>
Here are a few possibilies:
x <- c(4, 12, 20)
rep(x, each = 3) + 0:2
rep(x, each = 3) + sequence(rep(3, length(x))) - 1
c(sapply(x, seq, length = 3))
On 7/25/06, etienne <[EMAIL PROTECTED]> wrote:
> I need sequences that have gaps in them, such as the
> following:
>
> 4 5 6 | 12 13 14 | 20 21
<- strrev(gsub(pat, "\\1\\", text, perl = TRUE))
strrev(gsub("\\b(\\w+)(?=.*>r<\\1)", ">r<\\1", out, perl = TRUE))
On 7/23/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> The following requires more than just a single gsub but it does solve
Try:
plot(1:10, main = bquote("Results for" ~ theta == .(theta)))
On 7/25/06, Adrian Dragulescu <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> I want to have a title that will look something like:
> "Results for \theta=2.1", given that I have a variable theta=2.1, and
> \theta should show on the s
Regarding having to do a lot of backtracking one can just
look at the relative comparison of speeds and we see
that they are comparable in speed.
In fact the bottleneck is not the backtacking but strapply.
I had coded the regexp version for compactness of code but if we replace
the strapply with c
And if lattice is ok then try this:
library(lattice)
xyplot(Consumption ~ Quarter, group = Year, data, type = "o")
On 7/24/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Try:
>
> matplot(levels(data$Quarter), matrix(data$Consumption, 4), type = "o")
>
Try:
matplot(levels(data$Quarter), matrix(data$Consumption, 4), type = "o")
On 7/24/06, John McHenry <[EMAIL PROTECTED]> wrote:
> Hi WizaRds,
>
> I'd like to overplot UK fuel consumption per quarter over the course of five
> years.
> Sounds simple enough?
>
> Unless I'm missing something, the f
l.adj<-as.vector(mult.comp$p.value.adj)
> out.data.mat<-cbind(isoforms,estimate,lower,upper,p.val.raw,p.val.bon,p.val.adj)
> write.table(out.data.mat, file=filename.csv, sep=",", qmethod="double",
> col.name=NA)
>
> Thanks ../Murli
>
> _
Try:
RSiteSearch("finding peaks")
On 7/24/06, Tauber, Dr E. <[EMAIL PROTECTED]> wrote:
> Dear R-users,
>
> We are monitoring the activity of animals during a few days period. The
> data from each animal (crossing of infra-red beam) are collected as a
> time series (in 30 min bins). An example is
A . (dot) matches any character and $ matches the end of string so
this replaces the last two characters with the empty string:
sub("..$", "", x)
On 7/24/06, Wade Wall <[EMAIL PROTECTED]> wrote:
> Hi all,
> I am looking for a function in R to trim the last two characters of an
> 8 character s
On 7/24/06, Peter J. Lee <[EMAIL PROTECTED]> wrote:
> I'm aware that S N Krishna asked the same
> question. However, I have failed to implement the
> posted solution for running rank order
> correlations on multiple subsets of data using the by() function.
>
> Here is my problem:
>
> Take a set of
Just one more comment. It is possible to define length.POSIXlt yourself
in which case diff works with POSIXlt objects.
> length.POSIXlt <- function(x) length(x[[1]])
> diff(dts)
Time differences of 91, 92, 183, 91, 92, 182, 91, 92, 182 days
On 7/23/06, Gabor Grothendieck <[EMA
It depends on what information you want to save and how the
program on the other end needs it.
For the save version I would at least use ascii = TRUE to get it
in a more readable fashion.
Look at
file.show("mult_test.dat")
file.show("mult.out") # but use ascii=TRUE on your save statement.
to s
Here is another possibility:
rep(a, length = length(x))
On 7/23/06, Gregor Gorjanc <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Gabor Grothendieck wrote:
> > Try:
> >
> > foo2 <- function(x, a) cbind(x,a)[,2]
> >
>
> thank you for this. It does work to
Try:
foo2 <- function(x, a) cbind(x,a)[,2]
On 7/23/06, Gregor Gorjanc <[EMAIL PROTECTED]> wrote:
> Hello!
>
> I am writting a function, which should recycle one of its arguments if
> length of the argument is approprate i.e. something like
>
> foo <- function(x, a)
> {
> n <- length(x)
> if(le
Moving this to r-devel.
Looking at the diff.POSIXt code we see the problem is that it takes the
length of the input using length which is wrong since in the case
of POSIXlt the length is always 9 (or maybe length should be
defined differently for POSIXlt?). Try this which gives the same
problem:
The following requires more than just a single gsub but it does solve
the problem. Modify to suit.
The first gsub places <...> around the first occurrence of any
duplicated suffixes. We use the (?=...) zero width regexp
to circumvent the nesting problem.
Then we use strapply from the gsubfn pac
On 7/22/06, Nair, Murlidharan T <[EMAIL PROTECTED]> wrote:
> -Original Message-
> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
> Sent: Saturday, July 22, 2006 1:37 PM
> To: Nair, Murlidharan T
> Cc: R-help@stat.math.ethz.ch
> Subject: Re: [R] Multc
On 7/22/06, Nair, Murlidharan T <[EMAIL PROTECTED]> wrote:
> Here it is again, hope this is more clear
>
> I am using the following data (only a small subset is given):
>
> Habitat Fungus.yield
> Birch 20.83829053
> Birch 22.9718181
> Birch 22.28216829
> Birch 24.23136797
> Birch 22.32147961
> Birc
On 7/22/06, Nair, Murlidharan T <[EMAIL PROTECTED]> wrote:
> I REALLY NEED HELP WITH THIS PLEASE
>
[]
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project
Check out this recent thread:
https://www.stat.math.ethz.ch/pipermail/r-help/2006-July/109731.html
On 7/22/06, ahmed el kenawy <[EMAIL PROTECTED]> wrote:
> hi
>
> i created two files in excel with a dbf format in a similar way. the first
> is opened in R, however, when i try to open the second.
The following assumes that within each component of vectorlist
the vector elements are unique. In that case the first two lines
define vectorlist and perform the grep, as in your post. Elements
of the intersection must occur n times where n is the number
of components of vectorlist that match the
See if this works:
read.csv("datafile.csv", row.names = 1, fill = TRUE)
On 7/21/06, Ahamarshan jn <[EMAIL PROTECTED]> wrote:
> Hi,
> I have a dataset saved in *.csv format, that contains
> 13 columns (the first column being the title name and
> the rest experiments) and about 2500 rows.
> Not a
Check this thread:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/40898.html
On 7/21/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> Hello all,
> I am currently working with rpart to classify vegetation types by spectral
> characteristics, and am comming up with poor classifications based on
As others have mentioned its not really a good idea
to modify the namespace of a package and writing
a wrapper as Duncan suggested is much preferable.
An intermediate approach that
is not as good as the wrapper but better than modifying
the namespace is to copy the objects of interest
to your work
In some cases it may be sufficient to abbreviate the colnames:
> library(MASS); data(survey)
> head(survey)
Sex Wr.Hnd NW.Hnd W.HndFold PulseClap Exer Smoke Height M.I
1 Female 18.5 18.0 Right R on L92Left Some Never 173.00 Metric
2 Male 19.5 20.5 Left R on
Note that if you use mapply in the way I suggested, which is not
the same as in your post, then its just as fast. (Also the version
of mapply in your post gives different numerical results than
the for loop whereas mine gives the same.) like.mat is the for
loop version, like.mat2 is your mapply
1101 - 1200 of 3469 matches
Mail list logo
