What is the reason, that the levels of the factor
returned by cut() are not marked as ordered levels?
> is.ordered( cut( breaks=3, sample(10 ) ) )
FALSE
> help(factor)
...
If 'ordered' is 'TRUE', the factor levels are assumed to be ordered.
...
Wolfram
__
Similar answer as last time
aggregate, tapply, by or ppackage doBy
HTH
PS. If you want to add some other text then subject in your post do
not use HTML posting as recommended in posting guide.
Petr
On 28 Nov 2006 at 22:04, David Kaplan wrote:
Date sent: Tue, 28 Nov 2006 22:04:0
The number of knots is really one of the modelling assumptions. Provided you
don't make it restrictively small then the results should not be very
sensitive to this assumption, since the actual degrees of freedom of each
smooth are determined by how heavily the smooth is penalized, rather than
As requested on last line of every message to r-help please
provide self contained reproducible code. Also please
space it properly since its very difficult to read in your post.
I have used the models defined in example(Puromycin)
in place of your undefined ones and made some changes
to keep the
Hi
it is not stated that the cut shall return ordered factor. If you
want you can use
ordered(cut( breaks=3, sample(10 ) ))
or modify code for cut.default to accept ordered switch.
HTH
Petr
On 29 Nov 2006 at 9:59, Wolfram Fischer wrote:
Date sent: Wed, 29 Nov 2006 09:59:36 +01
Wolfram Fischer wrote:
> What is the reason, that the levels of the factor
> returned by cut() are not marked as ordered levels?
I don't know, but you can always make it ordered with
ordered(cut(breaks = 3, sample(10)))
>> help(factor)
> ...
> If 'ordered' is 'TRUE', the factor levels a
> 1, A, 24, The Red House
> 2, A, 25, King's Home, by the Sea
> ...
> Is there a way i can specify to R that "King's Home, by the Sea" is one
> word?
Yes: It needs to be quoted in the file:
2, A, 25, "King's Home, by the Sea"
cu
Philipp
--
Dr. Philipp Pagel
Alexander Nervedi wrote:
> Hi
>
> I have data like
>
> 1, A, 24, The Red House
> 2, A, 25, King's Home, by the Sea
> ...
>
> I'd like to read this in as three variables. I first tried
>
> temp <-read.csv(addresses, sep = "," ) it worked but line 2 was broken
> after King's Home, and by the Sea as
Wolfram Fischer wrote:
> What is the reason, that the levels of the factor
> returned by cut() are not marked as ordered levels?
>
>
>> is.ordered( cut( breaks=3, sample(10 ) ) )
>>
> FALSE
>
>
It would arguably be the Right Thing, but there would be complications
in modeling, where orde
In the days when tables were calculated laboriously, it was common
practice to introduce several deliberate rounding errors in every table.
These were used to catch infringements of copyright (and recover
reproduction fees).
Because tables came (and probably still do come) from a very few sourc
Hello,
I do this
a=c(10.2,0.4,0.2,5.2,8.6,1.6,0.0,1.0,6.3,1.4,0.4,0.2)
b=matrix(a,nrow=4)
rownames(b)=c("20/04/2002","21/04/2002","22/04/2002","23/04/2002")
colnames(b)=c("p1","p2","p3")
barplot(b,beside=T,col=(c("red","orange","yellow")))
then I have
http://www.mif.vu.lt/~roka5178/barplot1.JPG
You need the transpose of the matrix (or create it differently in the
first place!). Changing the last line to
> barplot(t(b),beside=T,col=(c("red","orange","yellow")))
should do the trick.
On 29/11/06, Robertas Kavaliūnas <[EMAIL PROTECTED]> wrote:
> Hello,
> I do this
> a=c(10.2,0.4,0.2,5.2,8
Hello R experts
I am looking for:
- advice
- sample commands
- pointers in MASS 4E
- packages
That help to produce 1D and 2D response plots from quadratic RSM linear
regression models that have been produced in R.
Let me explain a little what I mean. For 1D plots I mean the variable range
(
The R memory docs say that the --*-vsize option takes an integer
argument and then 'G', 'M', 'K', or 'k'. When I start R using
R --max-vsize=10G
I receive the warning:
WARNING: --max-vsize=10G=10'M': too large and ignored
The system that I'm working on is a 64-bit Sun server with 40G of
memo
I have searched through the mailing lists and the docs, and I seem
completely unable to find a way to make R list all available
environment variables that it uses.
The reason that I want to do this is the following:
The default program for reading files when using openVignettes on my
system is g
Hi
I have a large binary square matrix (about 2500 x 2500) obtained from a
distance matrix on x (smaller simulation below), in which I want to find the
largest square block of 1's for all permutations of the rows and columns. I
then want to remove the rows and columns corresponding to this block a
Hi to all
I would to determinate whether bits is a binary code and I would to find
out the which bit is set to 1
bits <-"00110110"
I found to detect whether there are only numbers
all.digits(bits)
but is there any function to detect whether there are only 0 and 1 in
the string
And how could I g
Preprocess the file surrounding the last field with quotes. The
xx <- line is for purposes of making it self contained here and in
reality would
be replaced with the commented
line above it. It simply reads the text in.
The yy <- line
then surrounds the fourth field with double quotes assuming
a
Carmen Meier wrote:
> Hi to all
> I would to determinate whether bits is a binary code and I would to find
> out the which bit is set to 1
>
> bits <-"00110110"
> I found to detect whether there are only numbers
> all.digits(bits)
> but is there any function to detect whether there are only 0 and
Dear Carmen,
length(grep("[^01]", bits)) == 0
should do the trick, returning TRUE if the string contains only 0's and 1's.
I hope this helps,
John
John Fox
Department of Sociology
McMaster University
Hamilton, Ontario
Canada L8S 4M4
905-525-9140x23604
http://so
On 29 Nov 2006 at 12:08, Peter Dalgaard wrote:
Date sent: Wed, 29 Nov 2006 12:08:21 +0100
From: Peter Dalgaard <[EMAIL PROTECTED]>
To: Wolfram Fischer <[EMAIL PROTECTED]>
Copies to: r-help@stat.math.ethz.ch
Subject:R
Hi
gregexpr can be used
bits <-"0011011aaa0"
gregexpr("[01]", bits)
[[1]]
[1] 1 2 3 4 5 6 7 11
attr(,"match.length")
[1] 1 1 1 1 1 1 1 1
or
grep("[^01]", bits)
if you want to know that there is any other character then 0 or 1
gregexpr("1", bits)
if you want to know location of 1's
HTH
Try:
# TRUE if all 0 and 1
regexpr("^[01]*$", bits) > 0
# positions of 1s
gregexpr("1", bits)[[1]]
On 11/29/06, Carmen Meier <[EMAIL PROTECTED]> wrote:
> Hi to all
> I would to determinate whether bits is a binary code and I would to find
> out the which bit is set to 1
>
> bits <-"00110110"
>
--- Robertas Kavaliûnas
<[EMAIL PROTECTED]> wrote:
> Hello,
> I do this
>
a=c(10.2,0.4,0.2,5.2,8.6,1.6,0.0,1.0,6.3,1.4,0.4,0.2)
> b=matrix(a,nrow=4)
>
rownames(b)=c("20/04/2002","21/04/2002","22/04/2002","23/04/2002")
> colnames(b)=c("p1","p2","p3")
> barplot(b,beside=T,col=(c("red","orange","yell
Hi
I know that p-values doesn't appear anymore in the summary of a linear
mixed-model with lmer. However, if I do a mixed logistic regression with
lmer using family=binomial, the summary includes a p-values for fixed
effects.
Is it normal, could I use those p-values to interpret the fixed effect
Hi,
I am new to this, can anyone help.
I am working with matrices, with stocknames as colNames and dates as
rowNames.
The data coontains percentile ranking for the stocks.
Date IBM MSFT 20061122 12 4 20061115 12 4
20061108
12 4 20061101 12 4 20061025 12 4 2006
These retrieve and set environment variables:
?Sys.getenv
?Sys.putenv
On 11/29/06, Karin Lagesen <[EMAIL PROTECTED]> wrote:
>
> I have searched through the mailing lists and the docs, and I seem
> completely unable to find a way to make R list all available
> environment variables that it uses.
>
On Wed, 29 Nov 2006, Daniel Gatti wrote:
> The R memory docs say that the --*-vsize option takes an integer
> argument and then 'G', 'M', 'K', or 'k'. When I start R using
>
> R --max-vsize=10G
>
> I receive the warning:
>
> WARNING: --max-vsize=10G=10'M': too large and ignored
>
> The system tha
Roy and others,
I'm also trying to build R 2.4.0 on aix 5 (5.3 in my case, on an
Intellistation Power 285) and I have not had luck yet. I notice that
the only mentions of AIX on this list in the past two months are
people not having luck building it. So far there don't seem to be any
responses,
Boa tarde,
Meu nome é Wesley Alves, sou Designer Interativo há 5 anos visitei o site desta
empresa. A principio, achei interessante as informações e a disposições da
mesma, mas acredito que posso melhorar ainda mais o web site de
vocês,estruturando melhor a parte de design e disposição geral d
On 11/29/06, Julien <[EMAIL PROTECTED]> wrote:
> Hi
> I know that p-values doesn't appear anymore in the summary of a linear
> mixed-model with lmer. However, if I do a mixed logistic regression with
> lmer using family=binomial, the summary includes a p-values for fixed
> effects.
> Is it normal,
Hi,
This is my first post.
Problem:
I have a matrix 5 by 5 and it has 3 NAs in them. Now, when i use length () for
a particular column
which has NA it returns 5. I know this is true, but how do i change my matrix
such that the length
becomes 5 minus the number of NAs i have. I know NA has a l
On Wed, 29 Nov 2006, Karin Lagesen wrote:
>
> I have searched through the mailing lists and the docs, and I seem
> completely unable to find a way to make R list all available
> environment variables that it uses.
>
> The reason that I want to do this is the following:
>
> The default program for
try this:
mat <- matrix(1:25, 5, 5)
mat[sample(25, 3)] <- NA
#
mat
colSums(!is.na(mat))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +
Prof Brian Ripley wrote:
> But is this a 64-bit build of R? (Solaris is 32-bit by default.)
> R --max-vsize=10G works on 64-bit Solaris build for me, but I get that
> (somewhat garbled) message on a 32-bit build.
Thanks. That was it. Apparently they had compiled R with 32-bit
settings. They
Try creating a zoo object from your data frame
# data from your post
DF <- structure(list(Date = as.integer(c(20061122, 20061115, 20061108,
20061101, 20061025, 20061018, 20061011, 20061004, 20060927)),
IBM = as.integer(c(12, 12, 12, 12, 12, 12, 11, 12, 10)),
MSFT = as.integer(c(4, 4, 4, 4,
The problem is that this doesn't seem to give what I want. I did look
at this. Perhaps doBy will work - haven't tried it yet. But, any other
suggestions are much appreciated.
===
David Kaplan, Ph.D.
Professor
Department
Thanks!! That did solve my problem. I am loving it.
--- Dimitris Rizopoulos <[EMAIL PROTECTED]> wrote:
> try this:
>
> mat <- matrix(1:25, 5, 5)
> mat[sample(25, 3)] <- NA
> #
> mat
> colSums(!is.na(mat))
>
>
> I hope it helps.
>
> Best,
> Dimitris
>
>
> Dimitris Rizopoulos
> Ph
I have received a few private emails asking for some simple
demonstration of calling Java code from R using the rJava package
(which can be installed directly inside R). Here is one example for
convolving two vectors (an example in the R manual about linking C with
R).
First write a Java program
On Wed, 2006-11-29 at 08:48 -0800, Jason Liao wrote:
> I have received a few private emails asking for some simple
> demonstration of calling Java code from R using the rJava package
> (which can be installed directly inside R).
One could also examine the sources of the rcdk package which provides
Hi,
I have a densityplot like this:
x = c(rnorm(100,1,2),rnorm(100,2,4),rnorm(100,3,6))
f = sample(c("A","B","C","D","E"),300,replace=TRUE)
df=data.frame(x,f)
library(lattice)
attach(df)
densityplot(~x, groups=f)
And I want to add a legend with the colours for the factors. How can I do that?
How
Albert Vilella wrote:
> Hi,
>
> I have a densityplot like this:
>
> x = c(rnorm(100,1,2),rnorm(100,2,4),rnorm(100,3,6))
> f = sample(c("A","B","C","D","E"),300,replace=TRUE)
> df=data.frame(x,f)
> library(lattice)
> attach(df)
> densityplot(~x, groups=f)
>
> And I want to add a legend with the c
Hello!
I wish to generate random binary trees in R.
Unfortunately recursive code like:
t <- function(p)
{
list(ifelse(runif(1)https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contain
Are this legend colors correlated to the plot?
If I do a:
densityplot(~x, groups=f, plot.points=FALSE, auto.key=TRUE,col=heat.colors(5))
I get different colors in the legend than the plot...
On 11/29/06, Chuck Cleland <[EMAIL PROTECTED]> wrote:
> Albert Vilella wrote:
> > Hi,
> >
> > I have a
Steven Finch wrote:
> Hello!
>
> I wish to generate random binary trees in R.
> Unfortunately recursive code like:
>
> t <- function(p)
> {
>list(ifelse(runif(1) }
ifelse() is vectorized and evaluates for both cases!
Instead, try:
t <- function(p)
list(if(runif(1) < p) t(p) else 0,
Try specifying it at the par.settings= level since that is where both
the plot and the legend get it from:
set.seed(1)
DF <- data.frame(x = c(rnorm(100,1,2),rnorm(100,2,4),rnorm(100,3,6)),
f = sample(c("A","B","C","D","E"),300,replace=TRUE))
library(lattice)
densityplot(~ x, DF, groups = f
I would really appreciate it if anyone could determine what is going
on with the following command. It is only half-working and is losing
lots of data. For the life of me I cannot even see the pattern of what
it is losing and what it is not.
I am attaching the R data set which you can use with th
Albert Vilella wrote:
> Are this legend colors correlated to the plot?
They are if you rely on the colors in
trellis.par.get("superpose.line")$col
If you want different colors you might use trellis.par.set() to
temporarily change the colors:
x <- c(rnorm(100,-2,1),rnorm(100,0,1),rnorm(100,2
Chuck and others. Here is the problem I need help on. Thanks.
I want to follow-up on a question I had a week or so ago. I want to
aggregate a very large data set. Generally in a program such as SPSS,
one chooses a so-called "break variable" that is the variable, and the
Using the builtin data set CO2:
aggregate(CO2[4:5], CO2[1], mean)
calculates the means of variables 4 and 5 based combinations of the
levels of variable 1. Is that what you want?
On 11/29/06, David Kaplan <[EMAIL PROTECTED]> wrote:
> Chuck and others. Here is the problem I need help on.
I meant: calculates the means of variables 4 and 5 based on the levels of
variable 1. Is that what you want?
On 11/29/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Using the builtin data set CO2:
>
> aggregate(CO2[4:5], CO2[1], mean)
>
> calculates the means of variables 4 and 5 based com
David:
Is this what you are looking for?
set.seed(10)
tmp <- data.frame(group = gl(3,10), something = rnorm(30))
with(tmp, tapply(something, group, mean))
1 2 3
-0.4906568 0.3695915 -0.9129640
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAI
Hi,
I'm trying to do a filled.contour plot where some points are labelled as
NA. How do I could plot this kind of graphics, so NA points are coloured
black, keeping the levels of remaining points. NA's values represent
land points (meaningless), and what I want to plot is the levels of a
vari
David Kaplan wrote:
> Chuck and others. Here is the problem I need help on. Thanks.
>
>
>
> I want to follow-up on a question I had a week or so ago. I want to
> aggregate a very large data set. Generally in a program such as SPSS,
> one chooses a so-called "break varia
Hello,
This should be a fairly simple question. I have 2 matrix's and I want to get
the difference between them and then remove the difference from one of them.
So A vs B, anything I see in B I want to remove from A, and then print out
the final matrix.
I have some code here but at the moment I'm
Karin Lagesen medisin.uio.no> writes:
...
> The default program for reading files when using openVignettes on my
> system is ggv. ggv is broken. I would therefore like to use a
> different program, and I found the variable R_PDFVIEWER which should
> do the trick. However, so far no luck.
>
> Ther
Actually, I think this is a much better variation on Gabor's excellent
example:
library(nlme)
aggregate(Orthodont[c("distance","age")], by = list(Subject =
Orthodont["Subject"]), FUN =mean)
Chuck Cleland wrote:
> David Kaplan wrote:
>> Chuck and others. Here is the problem I need help on. Than
Mark,
A and B will probably be different sizes. Won't 'C<-A-B' just do the
subtraction of matrix A by matrix B?
So What I was thinking was that it would find 2 numbers one at position 6
and the other at position 9. If it finds roughtly these numbers in matrix B
then it will identify this
I am trying to construct a cloud plot viewable in 3D with the classic 3D
glasses (one red lens, one bluish lens). I've figured out how to overlay
two printings of cloud() (from the lattice package) with appropriate colors,
however the box frame seems to only print in black. Is there any way to
co
Hi,
I would like to estimate something like y = a + b*d2*y + c*d3*y where
the dummies are created from some vector d with three (actually many
more) levels using factor(). But either there is included the variable
y or d1*y. How could I get rid of these?
Example:
x = c(1,2,3,4,5,6,7,8)
y = c(3,6
Is this what you want?
lm(y ~ fd + fd:y)
On 29/11/06, Andreas Goesele <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I would like to estimate something like y = a + b*d2*y + c*d3*y where
> the dummies are created from some vector d with three (actually many
> more) levels using factor(). But either there i
Hi,
I would like to solve a system of coupled ordinary differential equations,
where there is a delay (time lag) term. I would like to use the "lsoda"
function "odesolve" package. However, I am not sure how to specify the
delay term using the syntax allowed by odesolve.
Here is an example
My problem is the following : I create 2 zoo objects and then I try to
subset one of them using logic. indicesthatpass is a vector of trues and
falses but when I send it into bckret, it returns an empty bckret.
Obviously it has something to do with bckret being a zoo object and if I
do the same sub
Hi guys:
I'm testing some statistic discriminant analysis methods in R.
But I couldn't find the package for these methods:
1. Discriminant analysis implemented with nearest neighbor
method,
2. logistic discriminant analysis
3. POLYCLASS algorithm( there
Hello,
Sorry to all for the lack of communication, and thanks to those with the
suggestions.
So I have two matrixes which are different sizes, same column number but
different row number.
What I would like to do is to compare A to B. In the 6 and 9th column there
are ref numbers (molecular mass
On Wed, 29 Nov 2006, Leeds, Mark (IED) wrote:
> My problem is the following : I create 2 zoo objects and then I try to
> subset one of them using logic. indicesthatpass is a vector of trues and
Nope, it's a matrix (because you want the coredata to be a matrix) but
should be a vector. (Maybe we co
Hi,
I would like to add an additional piece of information to my previous
posting:
y[1](t) = 0, for all t in the interval (-delay, 0)
This completes the specification of the problem.
Thanks,
Ravi.
---
Ravi Varadh
On 11/29/06, Eric Peterson <[EMAIL PROTECTED]> wrote:
> I am trying to construct a cloud plot viewable in 3D with the classic 3D
> glasses (one red lens, one bluish lens). I've figured out how to overlay
> two printings of cloud() (from the lattice package) with appropriate colors,
> however the b
Hi Everyone,
After searching the subject and not being successful, I was wondering
if any you could explain me the idea behind the following fact:
all(NULL == 2) ## TRUE
any(NULL == 2) ## FALSE
Thanks a lot,
Benilton
--
Benilton Carvalho
PhD Candidate
Department of Biostatistics
Johns Hopki
lsoda does not solve delay differential equations, which is what you
need to be able to do. A quick search on netlib (netlib.org) turned up
one match for "delay differential equations": ode/ddverk.f, which might
help (though not in R).
R. Woodrow Setzer, Ph. D.
National Center for Computational T
R-help,
Given a simple linear model, say lm(x ~ y + z), I would like to remove
model terms that are factors with only one level. Thus, if 'z' were a
factor with only one level, lm(x ~ y + z) becomes lm(x ~ y + 1).
Likewise, if both 'y' and 'z' are one-level factors, then the resulting
calculation i
Ok so something like this.
A
1 2 3 4 5
A v w x z
G x z y w
C y v z x
D x v w z
B
1 2 3 4 5
F z x y z
D x z w v
E x z w v
So here I would just be comparing col 1. I would compare A[1,] to B[1,] and
then I would see that A[1,4] has the same element as B[1,2], so I would
remove A[,4]. Making A:
A
Hi there,
Am I missing something? The "%p" format, which should match against
AM/PM, seems to be broken in R 2.3.1 running on Windows XP SP2. For
example, the following fails
> strptime("10AM", "%I%p")
[1] NA
whereas the this works as you would expect, but is ambiguous
> strptime("10", "%I
"David Barron" <[EMAIL PROTECTED]> writes:
> Is this what you want?
>
> lm(y ~ fd + fd:y)
Thanks for you fast reply. But it's not what I wanted.
To make it clearer I have to correct the first part of what I wrote:
> On 29/11/06, Andreas Goesele <[EMAIL PROTECTED]> wrote:
>> I would like to est
Thanks, Woodrow. I also found a DDE solver called "dde23" in Matlab written
by L.F. Shampine. I will see if I can use it in Scilab, since I don't have
Matlab.
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor,
Hi all
I have recently moved to an Intel Mac and installed the latest R (2.4.0) but
I am having trouble with a number of packages. When I try to load them I
often get the following message:
Error: package 'pspline' was built for powerpc-apple-darwin6.8
I have tried to update all the packa
Merely convention.
NULL == 2 <==> logical(0), that is, a logical vector of length 0. It makes
sense (at least to me) that any(logical(0)) is FALSE, since no elements of
the vector are TRUE. all(logical(0)) is TRUE since no elements of the vector
are FALSE.
I think these are reasonable and fairly
The attached file never went through. How does one attach a R workskpace (or
two dataframes)
"Farrel Buchinsky" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
>I would really appreciate it if anyone could determine what is going
> on with the following command. It is only half-workin
Thomas,
The Gamma distr. can be fitted via ML using:
Library("MASS")
GF <- fitdistr(given_data,"gamma")
sh <- GF$estimate[1]
ra <- GF$estimate[2]
Fitting via Moments, m:
var <- m[2] - m[1]*m[1]
sh <- m[1]*m[1]/var
sc <- m[1]/var
ra <- 1/sc
G_pdf <- dgamma(breaks,shape=sh,rate=ra,scale=1/ra)
H
Bert Gunter <[EMAIL PROTECTED]> writes:
> Merely convention.
There's a similar thread in the R-devel newgroup about
'prod(numeric(0)) surprise' started by Ben Bolker, 09 Jan 2006,
(https://stat.ethz.ch/pipermail/r-devel/2006-January/036041.html) and
the post by Andy Liaw
(https://stat.ethz.ch/pip
Thank you Bert,
Sorry if it sounded like a complain, this is not what I meant.
The situation I had was: given a vector "v" containing integers or
NA's, I wanted to know if after excluding the NA's all the remaining
observations were 2's.
Naively, I assumed that:
all(na.omit(v) == 2)
would
On Nov 29, 2006, at 7:54 PM, Martin Morgan wrote:
>> http://en.wikipedia.org/wiki/Empty_set
>> which says, in part:
>> Operations on the empty set
>
<...>
> 'any' is like 'sum', 'all' is like 'prod'.
>
> Martin
Thank you very much Martin.
It's clear now.
Benilton
_
Ok, so columns are fixed and my apologizes, for the incorrect subscripting.
So I have the matching down because I'm matching a number so I've used an if
statement with some fuzz. At the moment my code from my earlier post gets me
a long vector with the data in it but not a nice matrix that is reada
Paul : I haven't tested this but try it and let me know. Let x be the #
of the col in A and y be the # of the column in B.
colinA<-A[,x]
colinB<-B[,y]
matchingrows<-B[(colinB %in% colinA),colinB]
C<-A[-1.0*c(matchingrows),]
I think C should be the resulting matrix that you want but it's untest
Is this what you want?
> A <- read.table(textConnection("A v w x z
+ G x z y w
+ C y v z x
+ D x v w z
+ "), as.is=T)
>
> B <- read.table(textConnection("F z x y z
+ D x z w v
+ E x z w v
+ "), as.is=T)
>
> A
V1 V2 V3 V4 V5
1 A v w x z
2 G x z y w
3 C y v z x
4 D x v w z
> B
Jim,
Is there a way to make match fuzzy? I had a look at ?match but
couldn't see anything about it so I thought I would as the guru. Also can I
do something like
index <- is.na(match(A[,1 & 2], B[,1 & 2]))
?
> index <- is.na(match(A[,1 & 2], B[,1 & 2]))
> index
[1] TRUE TRUE TRUE TRUE TRU
So I guess I was thinking like a fuzzy if. I'm matching a number in my
matrix and I would like to see if the numbers are the same to 0.1 accuracy.
So in an if statement I would do something like:
if(any((A[i,6] - B[j,6]) <= 0.1))
is this possible with match?
Sorry for my lack of R knowledge.
I haven't seen any replies to this, so I will offer some
suggestions even though I'm not familiar with 'plm'.
1. The standard formula syntax in S-Plus and R is that "y ~ x" is
a request to estimate b0 and b1 in y = b0+b1*x+error. Since this
already includes "1", it is essentially
On Nov 29, 2006, at 7:58 PM, Benilton Carvalho wrote:
> But that's okay, just a matter of adding an extra test (already done).
>
I'm guessing it would be something like: all(is.na(v) | v==2)
> When I asked for clarification about the reasons for this, I assumed
> that:
> if all(v) is TRUE ==> any(
that was trivial and logical ...
You just have to download the source from the repository and then use the
function source() with it in your script ...
2006/11/24, Ben <[EMAIL PROTECTED]>:
>
> When I was running the function latex() I got an error about a not found
> object : n.rgroup ...
>
> I
On Wed, 29 Nov 2006, Feng Qiu wrote:
> Hi guys:
> I'm testing some statistic discriminant analysis methods in R.
> But I couldn't find the package for these methods:
Where have you been looking? MASS (the book) has the first two of these,
and using search.r-project.org would also give
Do you really want the 1? Normally the intercept is not explicitly
given unless there are no other terms in the formula so assuming you
only want it if you have otherwise removed all the variables then:
zap1 <- function(fo) {
L <- sapply(all.vars(fo[[length(fo)]]), function(x)
nlevels(ge
One correction (to the get):
Do you really want the 1? Normally the intercept is not explicitly
given unless there are no other terms in the formula so assuming you
only want it if you have otherwise removed all the variables then:
zap1 <- function(fo) {
L <- sapply(all.vars(fo[[length(fo)]])
On Thu, 30 Nov 2006, Edward Abraham wrote:
> Hi there,
>
> Am I missing something? The "%p" format, which should match against
> AM/PM, seems to be broken in R 2.3.1 running on Windows XP SP2. For
> example, the following fails
>
> > strptime("10AM", "%I%p")
> [1] NA
>
>
> whereas the this works
if I want to put fig1plot to the left, figYPplot to the right
figYAaplot on the bottom.
How to modify the following cod to do these?
\begin{figure}[H]
\centering
\begin{minipage}[t]{0.5\textwidth}
\centering
<>=
<>
@
%\caption{Caption 1}
\end{minipage}
\begin{minipage}[t]{0.5\textwidth}
\centeri
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