In R version 2.4.0 Under development (unstable) (2006-08-08 r38825)
one can do this:
as.data.frame(cbind(x.1,x.2),stringsAsFactors = FALSE)
On 8/18/06, Tom Boonen [EMAIL PROTECTED] wrote:
Dear List,
why does as.data.frame(cbind()) transform numeric variables to
factors, once one of the
Try this:
result - boxplot(Petal.Length ~ Species, iris)
if (length(result$out))
text(result$group, result$out, match(result$out, iris$Petal.Length),
pos = 4, col = red)
If the outliers can be non-unique then match is not enough.
In that case assume that the nth occurrence of
any value in
, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
result - boxplot(Petal.Length ~ Species, iris)
if (length(result$out))
text(result$group, result$out, match(result$out, iris$Petal.Length),
pos = 4, col = red)
If the outliers can be non-unique then match is not enough.
In that case
On 8/19/06, Charles Annis, P.E.
[EMAIL PROTECTED] wrote:
Much gratitude to Professor Ripley, Peter Dalgaard, Marc Schwartz, and Roger
Bivand.
__
Roger Bivand wrote that ... strsplit() returns a list - one list component
for each repeated.measures.columns element. Just pick
Try this. It uses Matthias' trick for getting rid of lag 1.
The it defines a new local plot.acf
and then
defines a new local plot.acf which
# test data
set.seed(1)
x - rnorm(1000)
# run acf
x.acf - acf(x)
# remove lag 0 -- see Matthias' post
x.acf$acf[1] - NA
# plot with custom x axis
First 4 colums of builtin data frame anscombe as a matrix:
as.matrix(anscombe[1:4])
On 8/19/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Dear all,
I have a question about how to get a matrix by combining a large number of
columns from a data file. Suppose I read a file which have 1000
Try these
get(names(AF)[2])
AF[Second] # this one different than the rest
AF[[Second]]
AF[, Second]
AF$Second
On 8/20/06, Murray Jorgensen [EMAIL PROTECTED] wrote:
I would like a function to strip quotes off character strings. I should
work like this:
A - matrix(1:6, nrow =
Try this. gl(2,50) is such that the first 50 points are series 1
and the second 50 points are series 2. The scales= argument
defines the positions of the tick marks and the xlim= argument
defines the x axis limits. The layout puts the panels on top
of each other rather than side by side. strip
Look at oma= and mar= parameters to par for controlling the
space when using mfrow=. e.g.
opar - par(oma = c(6, 0, 5, 0), mar = c(0, 5.1, 0, 2.1), mfrow = c(2,2))
for(i in 1:4) plot(1:10)
par(opar)
On 8/20/06, Anupam Tyagi [EMAIL PROTECTED] wrote:
I think information can be enhanced by using
the same scale of grid.points
from one panel to next even if the scale of the data
changes? For example: c(seq(601:700),seq(6510,7000,
by=10)) ~ seq(601:700) | gl(2,50).
--- Gabor Grothendieck [EMAIL PROTECTED]
wrote:
Try this. gl(2,50) is such that the first 50 points
are series 1
On 8/21/06, Sergio Martino [EMAIL PROTECTED] wrote:
Thanks for your fast replay and sorry for my late one (I was on holidays)
The structure I would like to emulate (the fortran common statement) is a
different from what you are describing.
The examples scoping and the links to OO programming
On 8/21/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
On 8/21/06, Sergio Martino [EMAIL PROTECTED] wrote:
Thanks for your fast replay and sorry for my late one (I was on holidays)
The structure I would like to emulate (the fortran common statement) is a
different from what you
Maybe an R-packages posting guide with an example and
an automatic append of a one or two line summary at
the end of each article posted - as already done on r-help.
On 8/21/06, Martin Maechler [EMAIL PROTECTED] wrote:
Hi Göran,
GB == Göran Broström [EMAIL PROTECTED]
on Mon, 21 Aug 2006
Here are two solutions:
A - 1:8
B - c(1,2,4,7,8)
C - c(5,3,10,12,17)
# solution 1 - assignment with subscripting
DF - data.frame(A, B = A, C = 0)
DF[A %in% B, C] - C
# solution 2 - merge
DF - with(merge(data.frame(A), data.frame(B, C), by = 1, all = TRUE),
data.frame(A, B = A, C =
Its not part of state.x77. Its a completely separate variable.
Try ls(package:datasets) and notice its in the list
or try ?state.region and note that its a variable in datasets.
On 8/21/06, John Kane [EMAIL PROTECTED] wrote:
I was looking ?aggregate and ran the first example
this. Column A is something like
this
A - c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6,
7,8,9,10,11,12)
i.e it repeats. Rest all is same. How can I modify your solution to take
care of this issue.
Thanx in advance.
Sachin
Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here
Try
library(MASS)
y.lmr - lm.ridge(y ~., longley)
coefs - print(y.lmr)
On 8/21/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Dear all,
I am trying to calculate the fitted values using a ridge model
(lm.ridge(), MASS library). Since the predict() does not work for lm.ridge
object, I want
, simplicity and comprehensiveness (there is no
example of aggregate.data.frame).
On 8/21/06, John Kane [EMAIL PROTECTED] wrote:
--- Gabor Grothendieck [EMAIL PROTECTED]
wrote:
Its not part of state.x77. Its a completely
separate variable.
Try ls(package:datasets) and notice its in the
list
On 8/22/06, Sergio Martino [EMAIL PROTECTED] wrote:
Hi
Thanks again. I hope not to waste to much of your time.
I delete some lines of your answer
Each time myfun is run a new environment is created to hold
its local variables. The parent of that environment is e in
this example by
Here are two solutions. seq(length = ...) instead of
just seq(...) is so that v can possibly contain zeros.
# data
v - 3:5
# solution 1 - rbind/lapply
f - function(n) {
s = seq(length = n)
replace(rep(NA, max(v)), s, s)
}
do.call(rbind, lapply(v, f))
# solution 2 - loop
mat -
,]123 NA NA
[3,]1234 NA
[4,]1234 NA
[5,]1234 NA
[6,]12345
[7,]12345
[8,] NA NA NA NA NA
rksh
On 22 Aug 2006, at 13:55, Gabor Grothendieck wrote:
Here are two
Here is a solution that uses gsub with a negative lookahead perl-style
regexp to do it:
VECTOR - c(1,4,2,6,5,0,11,10,4,3,6,8,6)
e -
([[:digit:]]+),(?=([[:digit:]]+),([[:digit:]]+),([[:digit:]]+),([[:digit:]]+))
out - gsub(e, \\1\\2\\3\\4\\5 , paste(VECTOR, collapse = ,), perl = TRUE)
Try this:
# data
DF - structure(list(id = c(1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4,
4, 4, 4, 4, 4, 4), date = structure(c(8, 9, 10, 11, 3, 7, 8,
10, 4, 1, 2, 3, 8, 9, 10, 11, 3, 5, 6), .Label = c(01/07/2006,
01/08/2006, 01/09/2006, 02/09/2006, 03/09/2006, 06/09/2006,
11/08/2006, 22/08/2006,
With lattice graphics:
library(lattice)
d1 - rnorm(100)
d2 - runif(100)
densityplot(~ d1 + d2, auto.key = TRUE)
On 8/23/06, Antje [EMAIL PROTECTED] wrote:
Hello,
I was wondering if I can plot two curves I get from density(data) into
one plot. I want to compare both.
With the following
You could also look at p.arrows in sfsmisc package which is
entirely in R.
On 8/23/06, B. Chasco [EMAIL PROTECTED] wrote:
There is a function called arrows() which is an .Internal function. How
difficult is it to modify that function to return the xy coordinates for
the line segments that
Try bquote:
y.hat - sigma.hat - 1.1
plot(1)
lab - bquote(hat(y) == .(y.hat) * , ~ hat(sigma)^2 == .(sigma.hat))
text(1, 1, lab, pos = 4)
On 8/23/06, Jeff D. Hamann [EMAIL PROTECTED] wrote:
I can't believe I'm having such a hard time with this and I haven't been
able to find out how to solve
There is some online material at:
http://zoonek2.free.fr/UNIX/48_R/all.html
http://pj.freefaculty.org/R/statsRus.html
On 8/24/06, Raphael Fraser [EMAIL PROTECTED] wrote:
I am new to R and am looking for a book that can help in learning to
program in R. I have looked at the R website suggested
On 8/24/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here is one more solution . It uses the reshape package.
Its faster than using reshape but not as fast as xtabs;
however, it is quite simple -- just one line and if that
matters it might be useful:
library(reshape)
system.time(w4
The covariance has the same sign as the
correlation so lets calculate the sample covariance
of the vector T1 = (X,Y) with T2 = (Y,Z) where we ignored
the third component in each case due to use=complete.
cov(T1, T2) = XY + YZ - (X+Y)/2 * (Y+Z)/2
X, Y and Z are random variables so we take
Look through the output of trellis.par.get() for the right parameters
or when all else fails use grid (which we use below for the
box around the panel since I could not locate the parameter):
library(lattice)
library(grid)
x - 1:12
g - gl(3,4)
lwd - 3
xyplot(x ~ x | g, type = l, lwd = lwd,
Try par.settings=
You can find examples via:
RSiteSearch(par.settings) .
On 8/24/06, Gattuso, Jean-Pierre [EMAIL PROTECTED] wrote:
Hi:
I am using the following command:
xyplot(dat6$CO3*1e6 ~ dat6$irradiance, data=dat6, group=ref,
xlab=list(label=expression(paste(Irradiance (, mu,
Here is a way to automate finding the lwd= parameters.
library(lattice)
# test data
x - 1:12
g - gl(3, 4)
lwd - 3
# set parameters
par - trellis.par.get()
par - lapply(par, function(x) replace(x, names(x) == lwd, lwd))
xyplot(x ~ x | g, type = l, par.settings = par)
On 8/24/06, Chuck Cleland
That should read finding and setting. Chuck already showed how
to find them.
On 8/24/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here is a way to automate finding the lwd= parameters.
library(lattice)
# test data
x - 1:12
g - gl(3, 4)
lwd - 3
# set parameters
par - trellis.par.get
There was some discussion here:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/73646.html
On 8/24/06, Jason Liao [EMAIL PROTECTED] wrote:
I recently coded a recursion algorithm in R and ir ran a few days
without returning any result. So I decided to try a simple case of
computing binomial
Use as.formula to convert the character string to an object of class formula
and note that we want to set the formula's environment appropriately:
fit.main - function(resp, x, env = parent.frame()) {
fo - as.formula(paste(y, ~, paste(x, collapse = +)))
environment(fo) - env
Use cbind to create a two column matrix, mat,
and multiply that by the appropriate inflation factors.
Then use rowsum to sum the rows according to the
id grouping factor.
inf.fac - list(year1 = 1, year2 = 5, year3 = 10)
mat - cbind(s1 = df1$cat1 + df1$cat2, s2 = df1$cat3 + df1$cat4)
rowsum(mat *
Try sd(obj.tr) which will give a vector of standard deviations, one per column.
A column's entry will be zero if and only if all values in the column
are the same.
On 8/24/06, Muhammad Subianto [EMAIL PROTECTED] wrote:
Dear all,
I apologize if my question is quite simple.
I have a dataset (20
Non-Clinical Statistics
South San Francisco, CA
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Gabor
Grothendieck
Sent: Thursday, August 24, 2006 4:28 PM
To: Muhammad Subianto
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Check values
Try this and note we must pay attention to which environment
the expression is evaluated in.
fit.sum - function(x, env = parent.frame())
eval(parse(text = paste(x, collapse = +)), env = env)
# test
x1 - x2 - 1
fit.sum(c(x1,x2))
On 8/25/06, Maria Montez [EMAIL PROTECTED] wrote:
Thank you
You likely want the answer that Sarah has already given but in
addition you might also want to look at the thread below,
the point being that detaching a package still leaves portions:
https://www.stat.math.ethz.ch/pipermail/r-help/2006-July/109056.html
On 8/25/06, Sarah Goslee [EMAIL
You can replace the for with lapply like this:
VECTOR - c(0, 3, 6, 3, 11, 2, 11, 4, 3, 4, 7, 7, 6, 4, 8)
f - function(i) unique(tail(VECTOR, length(VECTOR)-i+1))[1:5]
out - do.call(rbind, lapply(seq(along = VECTOR), f))
na.omit(rbind(rep(NA, 5), out))
Note that a matrix with zero rows is
On 8/26/06, Patrick Connolly [EMAIL PROTECTED] wrote:
On Sat, 26-Aug-2006 at 09:57AM +0100, Patrick Burns wrote:
| sub.m - lapply(m, function(x)x[x2])
| sub.m
| [[1]]
| [1] 3 4
|
| [[2]]
| [1] 4 5
|
| [[3]]
| [1] 4
|
| sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
|
On 8/27/06, Atte Tenkanen [EMAIL PROTECTED] wrote:
Thanks!
I see, that do.call-function is used often in R-algorithms... Looking over
some extra do.call-examples seems useful. This tail-function is also new for
me.
Is there some reason to use seq(along = VECTOR) instead of
On 8/27/06, Paul Smith [EMAIL PROTECTED] wrote:
On 8/26/06, Mike Nielsen [EMAIL PROTECTED] wrote:
Yes.
Can R compute the expected value of a random variable?
Mike: thank you very much indeed for your so insightful and complete
answer.
Answers are often in proportion to the questions
Here are two ways:
1. use indexes:
lapply(seq(along = x), function(i) cbind(x[[i]], y[[i]]))
2. use mapply:
mapply(cbind, x, y, SIMPLIFY = FALSE)
On 8/28/06, Muhammad Subianto [EMAIL PROTECTED] wrote:
Dear all,
I have dataset
x - list(matrix(1:20, 5, 4),matrix(1:20, 5, 4),matrix(1:20, 5,
Try:
x - 1:10 # test data
plot(x ~ x, type = n)
u - par(usr)
rect(u[1], u[3], u[2], u[4], col = grey, border = red)
points(x ~ x)
On 8/28/06, Arun Kumar Saha [EMAIL PROTECTED] wrote:
Dear all R users,
Is there any effective way to change the body color of any plot? I am aware
of the function
1. Put the code from
www.microsoft.com/technet/scriptcenter/resources/qanda/jul05/hey0721.mspx
into, say, \bin\displayconfiguration.vbs, and then from R do this:
as.numeric(gsub(.* , , grep(resolution,
shell('cscript \\bin\\displayconfiguration.vbs', intern = TRUE),
value = TRUE)))
or we
Put your time series into a ts or zoo object. Now using
EuStockMarkets which is builtin data set in R. (You might
want to use align = right in rollapply.)
library(tseries)
library(zoo)
eu91 - window(EuStockMarkets, end = 1992) # use portion for test data
eu91.p.value - rollapply(eu91, 61,
example:
x - list(matrix(1:20, 5, 4),matrix(1:20, 5, 4),matrix(1:20, 5,
4),matrix(1:20, 5, 4),matrix(1:20, 5, 4))
y - list(c(1, -1, -1, 1, 1),c(1, 1, -1, -1, -1),c(1, 1, 1, 1, 1),c(1,
1, -1, 1, -1),c(-1, -1, -1, -1, -1))
## Thanks to Gabor Grothendieck for this trick.
## SIMPLIFY? SIMPLIFY
patience.
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Gabor Grothendieck
Sent: Monday, August 28, 2006 11:39 AM
To: [EMAIL
Sorry, that still woudl not be good enough. Try this:
as.numeric(sub(^[^:]*:, , grep(resolution,
shell('cscript \\bin\\displayconfiguration.vbs', intern = TRUE),
value = TRUE)))
On 8/28/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
One other idea. Replace gsub with sub and see
One other idea. Replace gsub with sub and see if that helps.
Maybe the output from the video driver has spaces in it.
On 8/28/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
This is what happens when I run it in R 2.3.1:
as.numeric(gsub(.* , , grep(resolution,
+ shell('cscript \\bin
-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Gabor Grothendieck
Sent: Monday, August 28, 2006 2:21 PM
To: [EMAIL PROTECTED]
Cc: r-help
Subject: Re: [R] screen resolution effects on graphics
This is what happens when I run it in R 2.3.1:
as.numeric(gsub(.* , , grep
I think there are problems with browseURL on Windows XP with IE.
Put the file in screen.htm and then run:
shell.exec(screen.htm)
it will come up -- it will block it and you will have to click to
unblock it; however,
that will interfere with any automatic procedure.
On 8/28/06, Charles Annis,
Try:
RSiteSearch(cofactor)
On 8/28/06, Jessica M. Maia [EMAIL PROTECTED] wrote:
Sorry but I wasn't very clear in my previous message.
I want to compute the adjoint of a real matrix A,
which is the transpose of the cofactor matrix of A.
There is an example here:
The error message says that md1 and other have different
number of rows. Please read the last line of every message
to r-help.
On 8/28/06, John Kane [EMAIL PROTECTED] wrote:
I am stuck on a simple problem where an example works
fine but the real one does not.
I have a data.frame where I
This can be done using a similar style but without a loop by
using sapply over the indexes rather than over the columns:
sapply(1:ncol(X), f)
f - function(i) {...whatever...}
...whatever... can now refer to colnames(X)[i] and X[,i]
On 8/28/06, Sundar Dorai-Raj [EMAIL PROTECTED] wrote:
Nick
b is nearly singular. Note that one of its eigenvalues is -2.935e-8
which is close to zero. We can use the generalized inverse from
MASS to get one solution, x, but any multiple of the eigenvector
corresponding to the near-zero eigenvalue when added to that
will also give a solution as shown:
]][[screen]][[width]]
height - ie[[document]][[ParentWindow]][[screen]][[height]]
ie$Quit()
On 8/28/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
I think there are problems with browseURL on Windows XP with IE.
Put the file in screen.htm and then run:
shell.exec(screen.htm)
it will come up
Put the data into a list, not into individual variables. Something
like this (untested):
L - sapply(symbols[-length(symbols)],
get.hist.quote, start = 2005-01-09, quote = Close, simplify = FALSE)
mat - do.call(cbind, L)
cor.mat - cor(mat, use = complete)
symnum(cor.mat)
On 8/29/06,
Try this:
xyplot(val ~ x, data = df, type = p,
col = as.numeric(df$f1), pch = as.numeric(df$f2))
key1 - list(border = TRUE, colums = 2, text = list(levels(df$f1)),
points = list(pch = 1:nlevels(df$f1))
)
key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)),
I have not used this but wink is a free demo-maker for Windows
available at:
http://www.debugmode.com/wink/
Click the red on green arrow on that page to a (very short) demo.
On 8/29/06, Gregor Gorjanc [EMAIL PROTECTED] wrote:
Hello!
I have found terrific demo or R package functionality at
Don't know. I would check whether the functionality is included in wink.
On 8/29/06, Gregor Gorjanc [EMAIL PROTECTED] wrote:
Gabor Grothendieck wrote:
I have not used this but wink is a free demo-maker for Windows
available at:
http://www.debugmode.com/wink/
Click the red on green
Try xyplot.zoo in the zoo package (be sure you have
the latest zoo version). One caveat is that xyplot.zoo is
under development and could change. Here we use the
builtin data set anscombe as an example:
# display the anscombe data set
anscombe
library(lattice)
library(zoo)
# create zoo object
Note that before entering this you need:
library(lattice)
library(grid) # to access the viewport function
On 8/29/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
xyplot(val ~ x, data = df, type = p,
col = as.numeric(df$f1), pch = as.numeric(df$f2))
key1 - list(border
Try this. First we reduce the data to a frequency matrix and
then plot it using classic and then lattice graphics:
zm - as.matrix(rowsum(z1[-9], z1[,9]))
barplot(zm, beside = TRUE, col = grey.colors(2))
legend(topleft, legend = levels(z1[,9]), fill = grey.colors(2))
library(lattice)
Try:
sprintf(%x, 109)
On 9/30/06, Romain Lorrilliere [EMAIL PROTECTED] wrote:
Hi,
do you know, a method to convert an decimal value (integer) to the
corresponding hexadecimal value ?
thinks for help.
Romain
--
Lorrillière Romain
UMR 8079 Laboratoire Ecologie, Systématique et
In each case, look around (help.search,
RSiteSearch) to see if you can find a function
that handles weights. For the case you mention,
medians, it can be done via quantile regression:
x - w - 1:5
library(quantreg)
coef(rq(x ~ 1, weight = w))
On 8/30/06, Rick Bischoff
(key1, draw = TRUE, vp = viewport(.9, .9))
draw.key(key2, draw = TRUE, vp = viewport(.75, .9))
On 8/30/06, Laurent Rhelp [EMAIL PROTECTED] wrote:
Gabor Grothendieck a écrit :
Note that before entering this you need:
library(lattice)
library(grid) # to access the viewport function
On 8/29
))
draw.key(key2, draw = TRUE, vp = viewport(.75, .9))
On 8/30/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
To handle conditioning on survey we provide a panel function
that subsets col and pch:
# define test data - df
# note that your val2 and val3 lines had a syntax
# so we have commented them
(levels(df$f1)),
points = list(pch = my.pch)
)
key2 - list(border = TRUE, colums = 2, text = list(levels(df$f2)),
lines = list(col = my.col)
)
draw.key(key1, draw = TRUE, vp = viewport(.9, .9))
draw.key(key2, draw = TRUE, vp = viewport(.75, .9))
On 8/30/06, Gabor Grothendieck [EMAIL
If you are converting them to 'its' anyways then after the
conversion to 'its' use the 'its' union command.
On 8/31/06, James Howison [EMAIL PROTECTED] wrote:
Hi,
I've examined the archives and found quite a few questions on
concatenating dataframes, but none that really addressed my issue,
Or using the same x:
setdiff(x, grep(Farrah|Common, x, value = TRUE))
[1] more last
On 8/31/06, jim holtman [EMAIL PROTECTED] wrote:
This finds the matching indices of Farrah and Common and then create a
set that does not include them:
x - c('Farrah', 'more', 'Common', 'last')
got.F -
On 8/31/06, Monica Pisica [EMAIL PROTECTED] wrote:
Hi again,
OK i came up with this after i got few good sugegstions.
First my data.frame actually looks like that (Thanks for clarifications to
Prof. Brian Ripley)
V1 V2 V3 V4
1 jan 3 1 7
2 mar 2 4 2
3 may 1 3 2
4 jul 3
Look at ?nargs
On 8/31/06, Lord Tyranus [EMAIL PROTECTED] wrote:
I create a function with R, I want to know how many argument number.
Thanks in advance.
--
Web Page
http://geocities.com/lord_tyranus_96/
__
R-help@stat.math.ethz.ch mailing list
Try using outer:
outer(1:3, a, ==)+0
On 8/31/06, z s [EMAIL PROTECTED] wrote:
Hi,
I am trying to convert a variable a = sample(1:3,100,rep = T) represents
choices into a 3X100 dummy varible b with corresponding element set to 1
otherwise 0.
eg.
a: 1 3 2 1 2 3 1 1
b: 1 0 0 1 0 0 1
Here is an additional way:
model.matrix(~ factor(a) - 1)
On 9/1/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try using outer:
outer(1:3, a, ==)+0
On 8/31/06, z s [EMAIL PROTECTED] wrote:
Hi,
I am trying to convert a variable a = sample(1:3,100,rep = T) represents
choices
Your data seems to have come through messed up but lets
assume you have a data frame with one column per tumor.
The convert your data to stacked form and call histogram:
DF - data.frame(T1 = 1:10, T2 = 6:15)
library(lattice)
histogram(~ values | ind, stack(DF))
On 9/1/06, Nair, Murlidharan T
A simple for loop would do it:
x - 1:2; a - list(4:5, 8:10) # test data
cat(x a\n, file = ) # only if you want a header
for(i in seq(along = x)) cat(x[i], a[[i]], \n, file = )
On 9/1/06, Zhu Wang [EMAIL PROTECTED] wrote:
Dear R helpers,
To illustrate my problem, here is a simplified
If n and k are small try brute force.
g - expand.grid(0:4, 0:4, 0:4)
g[rowSums(g) == 4,]
or more generally:
n - 4; k - 3
g - do.call(expand.grid, rep(list(0:n), k))
g[rowSums(g) == n,]
On 9/2/06, Maja Schröter [EMAIL PROTECTED] wrote:
Hello everyone,
I've a problem and dont know how to
Try:
prop.table(x, 2)
On 9/2/06, Alexander Nervedi [EMAIL PROTECTED] wrote:
Hi
I have
x-matrix(c(1,2,3,4), ncol = 2)
x
[,1] [,2]
[1,]13
[2,]24
I'd like these two be divided by their column totals. so 1/3 and 2/3 and 3/7
and 4/7. The obvious
x/colSums(x)
] [5]
[1,]16 11 16 1
[2,]27 12 17 1
[3,]38 13 18 -1
[4,]49 14 19 1
[5,]5 10 15 20 -1
Thanks you for any help.
Best wishes, Muhammad Subianto
#Gabor Grothendieck ggrothendieck at gmail.com
#Mon Aug 28 13:53:52 CEST 2006
)))
cbind2(x,y)
cbind2(x[[1]], y[[1]])
cbind2(x[[1]][[1]], y[[1]][[1]])
On 9/3/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
If z is the result then z[[i]] is formed from x[[i]] and y[[i]] using
the previous solution, viz.
z - list()
z[[1]] - mapply(cbind, x[[1]], y[[1]], SIMPLIFY = FALSE)
z
)
)
draw.key(key1, draw = TRUE, vp = viewport(.9, .9))
draw.key(key2, draw = TRUE, vp = viewport(.75, .9))
On 9/3/06, Laurent Rhelp [EMAIL PROTECTED] wrote:
Gabor Grothendieck a écrit :
In thinking about this a bit more we can use
panel.superpose/panel.groups to shorten it:
# define data
There is an example here:
http://zoonek2.free.fr/UNIX/48_R/04.html#1.2
(Scroll down until you see the graphic with the two y axes. The
code is in the box just above it.)
On 9/3/06, Chris Grobler [EMAIL PROTECTED] wrote:
Dear All,
Having a bit of trouble with plotting two y variables on
Regarding your other questions
On 9/4/06, Xiao Zhao [EMAIL PROTECTED] wrote:
Dear R users,
I am doing my project which I want to plot a piecewise function, I knew
that I can use the command segments to plot. But the problem is I want
to use my real data which needs me to sort of my data by
On 9/4/06, Mike Nielsen [EMAIL PROTECTED] wrote:
R won't do variable interpolation inside quotation marks as perl does.
Just as an aside, gsubfn in package gsubfn will do perl-style
(well, sort of) string interpolation:
library(gsubfn)
i - 1
gsubfn(x = data_files/file$i)
[1] data_files/file1
On 9/4/06, Paul Murrell [EMAIL PROTECTED] wrote:
Hi
Petr Pikal wrote:
Dear all
as I did not get any response on my post about abline and
plot(augPred)) I try again. I hope I do not break some posting guide
rules. I would try to contact package maintainer directly but there
is
Try:
text(5,5,expression(x[i * , * j]))
On 9/4/06, Kenneth Cabrera [EMAIL PROTECTED] wrote:
Hi R users:
How can I have several subscript number with a comma in a plot.
I would like to have the LaTeX equivalent of
x_{i,j}.
I try:
plot(1:10,1:10,type=n)
text(5,5,expression(x[i,j]))
Check out:
http://tolstoy.newcastle.edu.au/R/help/01c/0340.html
On 9/5/06, Terry Therneau [EMAIL PROTECTED] wrote:
Question:
I am trying to impliment a function in R that we use quite regularly in
Splus, and it fails due to a lack of the terms.inner function in R.
The substitute is?
Part
?biplot indicates cex= controls the size of the point labels. Is
that what you want? I suspect not since you say you've tried
it so if its something that ?biplot does not answer you may need
to examine the source. Enter this into R to display the source:
stats:::biplot.default
On
Here are a few possibilities:
a - matrix(1:24, 4) # test data
a / rep(a[1,], each = 4)
a / outer(rep(1, nrow(a)), a[1,])
a %*% diag(1/a[1,])
sweep(a, 2, a[1,], /)
On 9/6/06, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
I am trying to divide the columns of a matrix by the first row in the
And here is one more:
t(apply(a, 1, function(x) x/a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are a few possibilities:
a - matrix(1:24, 4) # test data
a / rep(a[1,], each = 4)
a / outer(rep(1, nrow(a)), a[1,])
a %*% diag(1/a[1,])
sweep(a, 2, a[1,], /)
On 9/6
This last one could also be written slightly shorter as:
t(apply(a, 1, /, a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
And here is one more:
t(apply(a, 1, function(x) x/a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are a few possibilities
Yet another one using the idempotent apply in reshape package
that eliminates the transpose:
library(reshape)
iapply(a, 1, /, a[1,])
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
This last one could also be written slightly shorter as:
t(apply(a, 1, /, a[1,]))
On 9/6/06, Gabor
In terms of speed Toby's original idea was actually the fastest.
Here they are decreasing order of the largest timing in each
row of system.time. I also tried it with a 100x10 matrix and
got almost the same order:
library(reshape)
system.time(for(i in 1:1000) iapply(a, 1, /, a[1,]))
[1] 11.51
This does not actually remove it but you could overwrite it with
an arrow the same color as the background and then plot a
new arrow:
x - 1:10
plot(x ~ x)
arrows(1, 1, 2, 2)
# revise it
arrows(1, 1, 2, 2, col = white)
arrows(2, 2, 3, 3)
On 9/6/06, Graham Smith [EMAIL PROTECTED] wrote:
I know
Make each pair of points a separate group using group= and specify
that both points and lines be used via type = b. Also set the
symbols in par.settings= so that they are accessed by both
the main plot and the legend:
xyplot(var ~ visit, group = symbols[patient], type = b,
auto.key =
Just one correction (although in this case it does not change the
output) -- use group = patient rather than group = symbol[patient]:
xyplot(var ~ visit, group = patient, type = b, auto.key = list(space
= right),
par.settings = list(superpose.symbol = list(pch = symbols)))
On 9/6/06, Gabor
You can use the 'zoo' or 'its' packages. For 'zoo' see the
documents listed at the end of:
http://cran.r-project.org/src/contrib/Descriptions/zoo.html
On 9/6/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
hi all
i have date and the return series like below, but the dates are not in
uniform
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