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
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 all
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
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. i
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
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
Birch
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] Multcomp
On 7/22/06, Nair, Murlidharan T [EMAIL
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
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:
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(length(a) n) {
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 some extent, but not much better
than mine foo.
foo2(c(1, 2, 3
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 see
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 [EMAIL PROTECTED
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 data from
A . (dot) matches any character and $ matches the end of string so
this replaces the last two characters with the empty string:
sub(..$, , x)
On 7/24/06, Wade Wall [EMAIL PROTECTED] wrote:
Hi all,
I am looking for a function in R to trim the last two characters of an
8 character string in
Try:
RSiteSearch(finding peaks)
On 7/24/06, Tauber, Dr E. [EMAIL PROTECTED] wrote:
Dear R-users,
We are monitoring the activity of animals during a few days period. The
data from each animal (crossing of infra-red beam) are collected as a
time series (in 30 min bins). An example is
,estimate,lower,upper,p.val.raw,p.val.bon,p.val.adj)
write.table(out.data.mat, file=filename.csv, sep=,, qmethod=double,
col.name=NA)
Thanks ../Murli
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Sun 7/23/2006 10:11 PM
To: Nair, Murlidharan T
Cc: r
Try:
matplot(levels(data$Quarter), matrix(data$Consumption, 4), type = o)
On 7/24/06, John McHenry [EMAIL PROTECTED] wrote:
Hi WizaRds,
I'd like to overplot UK fuel consumption per quarter over the course of five
years.
Sounds simple enough?
Unless I'm missing something, the following
And if lattice is ok then try this:
library(lattice)
xyplot(Consumption ~ Quarter, group = Year, data, type = o)
On 7/24/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try:
matplot(levels(data$Quarter), matrix(data$Consumption, 4), type = o)
On 7/24/06, John McHenry [EMAIL PROTECTED
Regarding having to do a lot of backtracking one can just
look at the relative comparison of speeds and we see
that they are comparable in speed.
In fact the bottleneck is not the backtacking but strapply.
I had coded the regexp version for compactness of code but if we replace
the strapply with
Try:
plot(1:10, main = bquote(Results for ~ theta == .(theta)))
On 7/25/06, Adrian Dragulescu [EMAIL PROTECTED] wrote:
Hello,
I want to have a title that will look something like:
Results for \theta=2.1, given that I have a variable theta=2.1, and
\theta should show on the screen like
\\1, out, perl = TRUE))
On 7/23/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
The following requires more than just a single gsub but it does solve
the problem. Modify to suit.
The first gsub places ... around the first occurrence of any
duplicated suffixes. We use the (?=...) zero width
Here are a few possibilies:
x - c(4, 12, 20)
rep(x, each = 3) + 0:2
rep(x, each = 3) + sequence(rep(3, length(x))) - 1
c(sapply(x, seq, length = 3))
On 7/25/06, etienne [EMAIL PROTECTED] wrote:
I need sequences that have gaps in them, such as the
following:
4 5 6 | 12 13 14 | 20 21 22
This was just discussed yesterday. See the thread:
https://www.stat.math.ethz.ch/pipermail/r-help/2006-July/109931.html
On 7/26/06, Marco Boks [EMAIL PROTECTED] wrote:
I am a newbie, and I am afraid this may be a rather trivial question. However
I could not find the answer anywhere.
I am
Try these:
# 1
library(Hmisc)
summary(y ~ ind, dat, fun = range, overall = FALSE)
# 2
# or with specified column names
f - function(x) c(head = head(x,1), tail = tail(x,1))
summary(y ~ ind, dat, fun = f, overall = FALSE)
# 3
# another approach using by - same f as above
do.call(rbind, by(dat$y,
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,
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 counterintuitive
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
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 [EMAIL PROTECTED] wrote:
Look through
multcomp
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()
On
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] wrote:
Dear
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 the
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.
However, for
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 to write fu2[Var] - 3
Try this where f and A2 are as in your post:
out -f(A2[A20])
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 M each of
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 that are
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
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
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? Suppose I
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 done
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 paren,
do a paste and then press enter twice.
On 7/27/06, [EMAIL
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 that
The reason I explicitly specified in the problem that the rectangle should
not be drawn first is that the xyplot is issued as part of a
larger routine that I don't want to modify.
On 7/29/06, Sebastian P. Luque [EMAIL PROTECTED] wrote:
Hi Gabor,
On Sat, 29 Jul 2006 17:20:29 -0400,
Gabor
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
tmat
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 code works
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 plot is drawn, add a grey rectangle behind a portion of it.
The following works
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
-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 [EMAIL PROTECTED] wrote:
Thanks. That's helpful
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 draws
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 =
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
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 only change we
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 some more that still gets you
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 using
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, NEWS)
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 the
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, row.names = 1)
#
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 doing a
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 labels are
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 needed.
Right -- the quotes are
, in this case 2 (sex+smoker).
Thanks in advance...
-Sam
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Wednesday, August 02, 2006 6:04 PM
To: Walker, Sam
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] ggplot facet label font size
On 8/2/06, Walker, Sam
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
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$x
[1] 1.37821893
Use atop:
plot(1, main = expression(atop( ^14*C*-glyphosate line, line2)))
On 8/4/06, Andrew Kniss [EMAIL PROTECTED] wrote:
I've tried several different ways to accomplish this, but as yet to no
avail. My y-axis for a plot has a rather long label, and thus I have
been using /n to break it
Sorry, you wanted a ylab=, not a main=. Try using xyplot in lattice:
library(lattice)
xyplot(1~1, ylab = expression(atop(phantom(0)^14*C*-glyphosate line,
line2)))
On 8/4/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Use atop:
plot(1, main = expression(atop( ^14*C*-glyphosate line
(grid.get(pretty))
On 8/3/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
If you are willing to use grid then you could create only the sex
factor in the left strips since its already in the desired position
but when displaying it output a factor.level, i.e. label of A.
(my.strip.left
Here are three ways:
# read in data
Lines - object1 object1 78
object1 object2 45
object1 object3 34
object1 object4 45
object2 object2 89
object2 object3 32
object2 object4 13
DF - read.table(textConnection(Lines))
# 1 - xtabs
xt - as.matrix(xtabs(V3 ~., DF))
# 2 - reshape
wide - reshape(DF,
Try this:
Lines - story,datepub
story10,1 April 1999
story 90,1 March 2002
story 37,10 July 1985
DF - read.csv(textConnection(Lines))
DF[order(as.Date(DF$datepub, %d %B %Y)),]
On 8/6/06, Bob Green [EMAIL PROTECTED] wrote:
I am hoping for some advice regarding ordering a dataframe, by date.
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)
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:
:
Functionkmeans()
from package stats provides several algorithms for computing
partitions with respect to Euclidean distance.
Hence why I am using a euclidean distance matrix. Why is this incorrect?
Gabor Grothendieck [EMAIL
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. SP500 and DOW JONES
2. Their repective dates
If I use the plot command, I get one output if I use it again, I
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 are needed to get an output like this:
1. On X-Axis : 2 Indices ex. SP500 and DOW
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
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 correlations
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 it.]
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
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 dieser
Try this:
DF[unlist(tapply(rownames(DF), DF$id, function(x) c(x, x[1]))),]
On 8/9/06, Leonardo Lami [EMAIL PROTECTED] wrote:
Hi all,
I have a simple question:
I have a data.frame like this:
id x y
1 50 1647685 4815259
2 50 1647546 4815196
3 50 1647454 4815294
4 50
Dmitris has already provided the solution but just throught I would'
mention that your third alternative can be written:
apply(mymatrix, 1, fun2, bb = bb)
(assuming fun2 has arguments idx and bb) which is not
nearly so ugly so you might reconsider whether its ok
for you to just pass bb.
On
That's not a valid specification. See the description of the index.cond
argument in ?xyplot and in particular this part:
If 'index.cond' is a list, it has to be as long as the number of
conditioning variables, and the 'i'-th component has to
be a valid
On 8/9/06, John McHenry [EMAIL PROTECTED] wrote:
Hi WizaRds,
In MATLAB you can do
x=1:10
and then specify
x(2:end)
to get
2 3 4 5 6 7 8 9 10
In R you could do the above via:
x[-1]
or whatever (note that in MATLAB the parenthetic index notation is used, not
brackets as in R).
A matrix M can be thought of as a linear transformation which maps
input vector x to output vector y:
y = Mx
The eigenvectors are those directions that this mapping preserves.
That is if x is an eigenvector then y = ax for some scalar a. i.e.
y lies in the same one dimensional space as x.
Try:
lm(Sepal.Length ~., iris)
On 8/10/06, r user [EMAIL PROTECTED] wrote:
I am using R in a Windows environment.
I have a basic question regarding lm().
I have a dataframe data1 with ncol=w.
I know that my dependent variable is in column1.
Is there a way to write the regression formula
Here are three ways:
xx - as.Date(2006-01-05)
# 1. use as.POSIXlt
as.POSIXlt(xx)$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(chron)
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
,
breaks = breaks, freq = FALSE, density = 10)
lines(density(DF$A2), col = blue)
On 8/11/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
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
[,j], border = j, lwd = j, lty = j,
breaks = breaks, col = transparent, ...)
panel.densityplot(DF[,j], col = j, ...)
})
On 8/11/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
The code below was missing the breaks= argument to hist.
I had not noticed because
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
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
help(plot)
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))),
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.
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.
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's a good
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 applied
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 ggrothendieck at gmail.com writes:
Here are three ways:
xx - as.Date(2006-01-05)
# 1
Use the first few rows of iris as test data and try this
where isnum is 1 for each numeric column and NA for
others.
irish - head(iris)
isnum - ifelse(sapply(iris, class) == numeric, 1, NA)
iris.data - data.matrix(iris)
rbind(iris, colMeans(iris.data) * isnum, sd(iris.data) * isnum)
On 8/17/06,
The parameter names are axis.text$font and axis.text$cex .
Try issuing the command:
trellis.par.get()
to get a complete list.
Here is an example:
histogram(1:10, par.settings = list(axis.text = list(font = 2, cex = 0.5)))
On 8/17/06, Debarchana Ghosh [EMAIL PROTECTED] wrote:
Hi All,
I'm
Here are two solutions. In both we break up DF into rows
which start with 1.
In solution #1 we create a new data frame with the required sequence
for A and zeros for B and then we fill it in.
In solution #2 we convert each set of rows to a zoo object z
where column A is the times and B is the
(rbind, by(DF, cumsum(DF$A == 1), f))
Error in zoo(, time(as.ts(z)), z, fill = 0) :
unused argument(s) (fill ...)
Unable to figure out the cause.
Thanks,
Sachin
Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are two solutions. In both we break up DF into rows
which start with 1
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