I assume this is (or was) a specification issue. I think write.dbf uses the
shapefile library (C not R library) so it applies to the use of shapefiles and
just happens to have been included in the foreign package because it has a
generic usefullness. (Is that a word?)
Since I very rarely care
ange is
that the small example I made up did not use the original data source, but was
typed in the same way I did x1 above. However I can't reproduce the error so it
may still be a case of finger trouble on my part.
Tom
> -Original Message-
> From: Prof Brian Ripley
Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Mulholland, Tom
> Sent: Friday, 19 May 2006 11:48 AM
> To: R-Help (E-mail)
> Subject: [R] Converting character strings to numeric
>
>
> I assume that I have missed something fundamental and
I assume that I have missed something fundamental and that it is there in front
of me in "An Introduction to R", but I need someone to point me in the right
direction.
> x1 <- "1159 1129 1124 -5 -0.44 -1.52"
> x2 <- c("1159","1129","1124","-5","-0.44","-1.52")
> x3 <- unlist(strsplit(x1," "))
>
You might also consider looking at the R-sig-Geo list which has lots of
discussion about issues relating to file formats and the best ways to get data
in and out the various packages that are used.
Tom
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of
Well I downloaded the data using the link in your message which suggests that
the code is right. I don't have its loaded (I assume it's from the irregular
time series package) so I can't test the code as you have it.
Have you tried checking to see it it is an issue with the internet (your
brow
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of
> [EMAIL PROTECTED]
> Sent: Wednesday, 14 September 2005 8:13 AM
> To: vittorio
> Cc: r-help@stat.math.ethz.ch
> Subject: Re: [R] Reading data from a serial port
>
snip
> Now all this of course is writt
I searched the help for "cosine distance" and this was the first hit
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/3946.html
Tom
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Raymond K Pon
> Sent: Tuesday, 13 September 2005 3:48 AM
> To: r-help@st
I cannot state this with the certainty that others might, but the Rd format is
a text format. If you want to produce something else then you need to choose an
alternative method. For instance, 1.4 of "Writing R Extensions" notes that
"Documents in 'inst/doc' can be in arbitrary format, however w
I incorrectly relied upon my memory
...
> and that
> John Fox did something
> http://ils.unc.edu/~jfox/powerpoint/introduction.html that I
> enjoyed reading.
The work is that of Jackson Fox
Tom
__
R-help@stat.math.ethz.ch mailing list
https://stat
For some reason (probably that our organisation has blocked the site) I could
not see the original articles that prompted the post. I however immediately
assumed that this was precipitated by Tufte and his comments about PowerPoint
(I recall seeing a good example of PowerPoint on his site)
http
I note that the axis help seems to refer to padj. After playing around it is
obvious that I don't know what is meant by this argument, so maybe I'm doing
something wrong. My practical soultion is
plot(1:50,axes = FALSE,ylab = "")
axis(2,at = 1:50,labels = rep("",50),las = 2,padj = 0)
text(rep(-4
require( car )
set.seed(12345)
nn <- sample( c( 2, 4 ), size=50, replace=TRUE )
rr <- recode( nn, "2='TWO';4='FOUR'" )
table( rr, exclude=NULL )
ss <- recode( nn, "2='Num2';4='Num4'" ) # Doesn't work as expected
table( ss, exclude=NULL )
ss <- recode( nn, "2='Num2';4='Num4'", TRUE ) #?
table( ss,
If this is a question about plotting 2 series on a plot then in general terms
you can plot 2 series as per this example in the list
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/5463.html
If it is about the use of rgev you should tell us the name of the package.
If you read the posting guide
Search the archives for zoom and you will find plenty of answers on this
question.
RSiteSearch("zoom")
Tom
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Henrik Andersson
> Sent: Tuesday, 26 July 2005 4:16 PM
> To: r-help@stat.math.ethz.ch
> Subjec
doubleEm <- function(p1,p2,p3,p4){return(p1 *p1,p2 * p2, p3 * p3, p4 * p4)}
suppressWarnings(doubleEm(1,2,3,4))
[[1]]
[1] 1
[[2]]
[1] 4
[[3]]
[1] 9
[[4]]
[1] 16
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Paul Roebuck
> Sent: Wednesday, 20 July
You will find previous discussion about pairs in this list. There are limits on
what has been included within the functionality, but you can write your own
panel functions. Here's a starting point.
x <- runif(100)
dim(x) <- c(20,5)
panel.cor1 <- function(x,y,...){
mycor<- cor(x,y)
if (mycor
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Henrik Bengtsson
> Sent: Tuesday, 28 June 2005 2:54 AM
> To: Spencer Graves
> Cc: r-help@stat.math.ethz.ch; Dirk Eddelbuettel
> Subject: Re: [R] How to convert "c:\a\b" to "c:/a/b"?
>
... snipped
> T
It's not so much a problem, as not working the way you expected.
cluster:::plot.partition is used to do the plotting. If you look at the code
for this you can see the difficulty in putting every possible permutation into
the code. If for example you want the silhouette plot to be red using col =
Have you found the file "Using Rmetasim"? In windows you can access this file
by using the help and selecting "browse" directory. There appears to be a
reasonable amount of information here. It looks to me as if you need to work
your way through these files until you understand what is going on.
It would have been helpful if you had written the code that actually showed the
issue you have rather than leaving it to us to try and reproduce.
That leaves generic options. If the screen is too small for you then try
plotting it to postscript or PDF device and setting the paper size to A3 or
Well I skipped to the end so pardon me if I've missed something. My first
reaction was to go and look at the excellent article by Gabor in RNews 2004-1
on dates (p.32 in particular)
> as.POSIXct(strptime("7/12/2001 10:32",format = "%d/%m/%Y %H:%M"))
[1] "2001-12-07 10:32:00 W. Australia Standard
What makes you think that there is a ylab parameter?
> args(map)
function (database = "world", regions = ".", exact = FALSE, boundary = TRUE,
interior = TRUE, projection = "", parameters = NULL, orientation = NULL,
fill = FALSE, col = 1, plot = TRUE, add = FALSE, namesonly = FALSE,
I don't think Spencer replied to your message. It wasn't addressed to you. He
was replying to a specific post on nomograms. However you might try being less
specific as a search on "spatial cluster" gave back
[R] mclust - clustering by spatial patterns
http://finzi.psych.upenn.edu/R/Rhelp02a/a
Without a small example to see what you are doing it is hard to respond. There
are plenty of examples in the help for legend showing placement all over the
place. So I am guessing that this might help (The only thing added from the
help was the par(xpd = TRUE)
x <- 0:64/64
y <- sin(3
I think you may wish to look at the plotrix package, assumming that you have
taken care of the issues involved in breaking an axis and that your plots don't
result in misleading information.
I think to use the axis break you would have to calculate your own labels and
rescale the data, as it lo
I don't know, but I do know that if you search the mailing list the answer is
there.
One place to start might be
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/50776.html
I found this by searching for "fortran" on the mailing list.
You might also read the posting guide. The link above refers
Well since I know nothing about this topic I have lurked so far, but here's my
two bob's worth.
Firstly I tried to make sense of Brian's initial reply. I have got no idea who
Bellman is and you have not referenced (his/her) work in a way I can access the
issues you refer to. So I assumed that's
I have snippets of code that I have either taken from examples or off of the
list. I apologise to those I have stolen it from but I didn't keep the proper
references.
n <- 100
xx <- c(0:n, n:0)
yy <- c(c(0,cumsum(rnorm(n))), rev(c(0,cumsum(rnorm(n)
plot (xx, yy, type="n", xlab="Time", yla
Just a little bit of trivia.
The 2001 Census in Australia had a significant group of people who responded to
the question of religion with the answer Jedi or Jedi Knight. Unfortunately the
Australian Bureau of Statistics is a bit fuddy duddy about the issue as they
see it as a trivialisation of
Lattice is not my forte but here goes.
?panel.bwplot specifically notes that "pch, col, cex: graphical parameters
controlling the dot". If you look at the code for panel.bwplot you will see
where the colours come from in which case you can probably set up your own
colour scheme using trellis.pa
I recall having this problem. I think I had a version that didn't work. Did you
download the htmlhelp.exe from http://www.murdoch-sutherland.com/Rtools/ . The
path does not matter as I have it in a folder on it's own. I think that I also
had a path to the version that didn't work and I had to ge
Fristly when you are using a package (in this case date) put it in your email.
Dates are stored as numbers and if I recall correctly as.date will be the
number of days since sometime in 1960. As with other objects there are
generally methods that will ensure that the correct printed format will
There's built in FTP in windowsXP, which I assume is what they are likely to
have. If they open file explorer there's an entry called "My Network Places".
The problem I had seeing if it would work, was that
ftp://cran.r-project.org/pub/R/ doesn't want to know me. Not that I was at all
sure, tha
t; -Original Message-
> From: Liaw, Andy [mailto:[EMAIL PROTECTED]
> Sent: Thursday, 21 April 2005 10:06 AM
> To: Mulholland, Tom
> Subject: RE: [R] Histogram
>
>
> Have you tried it? hist.factor() as defined would be the
> hist method for
> the factor class, so
Well you have not given us anything to go on really. Are there more than 94
columns? Does each column have a valid fieldname? RODBC is not guaranteed to
work in every possible scenario. If you have a look through the list you will
find there are specific limitations which are not immediately app
Of course Andy meant hist.factor(f)
In particular you should note that Andy uses the table function to "transform
... the data from characters to numbers"
Tom
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Liaw, Andy
> Sent: Thursday, 21 April 2005
What does str(tsx) give?
So you are feeding in something the function has no idea about.
Try rsiTA([EMAIL PROTECTED],14)
Tom
> -Original Message-
> From: Neuro LeSuperHéros [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, 20 April 2005 9:42 AM
> To: Mulholland,
Should you be using rsiTA(tsx[,2],14). If you look at the function you will see
it is expecting just the values you want in the calculation.
If you give it a matrix it treats the whole matrix as being price data.
Tom
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTE
nt it helps when you
indicate where you have gleaned your information from and what attempts you
have made to solve your problem.
Tom
* The posting guide explains all of this much better than I do.
> -Original Message-
> From: Owen Buchner [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, 1
I guess one of the reasons that you have not had a reply is that you have not
followed the posting guide. If you give the list something to work with (a
small reproducible example)
You use the word package which in R is very precise, but which I think you are
using to describe the file you are
Well I don't know how I can live with myself. I guess I can't wait for the site
to mirror itself in case someone thinks I'm yesterday's man. ;-)
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Achim Zeileis
> Sent: Monday, 18 April 2005 10:56 PM
...
>
Since it is not clear exactly what you require, I have assumed that what you
are looking for is the minimum value for each row or column depending upon
which is airports and which is the towers.
Is this what you are looking for
x <- runif(100)
dim(x) <- c(5,20)
apply(x,1,function(y) which(y ==
I think this might have been my code
mapply(paste,strwrap(levels(ncdata$Chapter),18,simplify = FALSE),collapse =
"\n")
Tom
> -Original Message-
> From: Jan P. Smit [mailto:[EMAIL PROTECTED]
> Sent: Thursday, 14 April 2005 5:15 PM
> To: Mulholland, Tom
> Cc:
This may not be the best way but in the past I think I have done something like
levels(x) <- paste(strwrap(levels(x),20,prefix = ""),collapse = "\n")
Tom
> -Original Message-
> From: Jan P. Smit [mailto:[EMAIL PROTECTED]
> Sent: Thursday, 14 April 2005 11:48 AM
> To: r-help@stat.math.eth
I'm not sure that this answers your questions but maybe they partly help.
p. 7 in An introduction to R notes
"For most purposes the user will not be concerned if the "numbers" in a numeric
vector
are integers, reals or even complex. Internally calculations are done as double
precision real
numb
mailto:[EMAIL PROTECTED]
> Sent: Thursday, 31 March 2005 2:36 PM
> To: Mulholland, Tom; r-help@stat.math.ethz.ch
> Subject: RE: [R] 2d plotting and colours
>
>
> Thank you.
>
> mycols <- c("brown","orange","tomato")
> plot(x,col = mycols
Does ?na.omit help
x <- kmeans(na.omit(data),centres)
of course if you have too many NAs you need to be sure that their removal does
not unduly influence the results.
Although I am a bit confused as I thought that agnes did not allow NAs. I
assume that you are running an alternative clustering
See upper.tri and lower.tri.
I think that you might also look for specific packages that function using
matrices, from what I have seen these often have the capacity to ignore the
diagonal or use just the upper or lower triangle. This is not an area that I
use very much, but I have seen various
e groups by colors
>
> That could be done by
>
> plot(x, col = cl$cluster)
>
> This means that we need to set the default colours , say col
> = cl$cluster =
> a set of group numbers say 1...10 should produce 10 distinct
> colours points
> grouped by colour.
&
I think you need to read the posting guide (see the bottom of each post made)
and once you have done this take some time to compose your message.
The issue is that I have too little information about what you have done. It
looks to me as if you are using postscript, but I am not sure if you have
And getting back to your question about the palette
there are a lot of ways to do this
assuming you have just started a session
palette()
# will give
#[1] "black" "red" "green3" "blue""cyan"
#[6] "magenta" "yellow" "gray"
palette(rainbow(24)) # There's also 'heat.colors' &
The error message states that you are passing a parameter called cex which has
not been used. If you look at ?plclust more closely you will see it does not
have cex parameter. However the S3 method for class hclust, plot, does?
So does this help?
hc <- hclust(dist(USArrests), "ave")
plot(hc,cex
Do you have a really good reason to be using 1.9.1. If not then just keep using
1.9.0. Did you check what changes were made in the 1.9 upgrade. Often you will
find useful information about this type of issue in the change log.
You have not told us anything about the machine you are using and thi
I'm afraid you have lost me. What is it that you want that reordering the
formula does not achieve.
bwplot(yield ~ year | site, data = barley) has sites next to each other. If the
lattice structure is your issue (it appears you wish to remove the structure
and replace it with a wider space) the
Are you talking about something other than 'update.packages'?
as the help notes
Description:
These functions can be used to automatically compare the version
numbers of installed packages with the newest available version on
CRAN and update outdated packages on the fly.
> -O
There are two issues here identifying the outliers and highlighting them.
I have only a basic grasp of both of these concepts but will give what I have
in case it helps. There appears to have been a move in the last 2 decades to
improve the concepts of what actually constitutes an outlier, Brian
s)
It's not really reuseable. I guess I could pass a formula and work out a better
method of subsetting dimensions (where certain factor levels are not used. But
maybe someone has an elegant method they could share.
Tom
> -Original Message-
> From: Mulholland, Tom
> Sent: Tues
I'm not sure how to best explain what I am after but here goes. I have a data
frame with 2 geographical factors. One is the major region the other is the
component regions.
I am trying to process all the regions at the same time without using "for". So
I need (think, I do) a list of matrices e
It seems to me that you are trying to do too much at a time. Firstly I think it
would be a good idea to get you code working before you try and make a package.
Some possibilities are that you write somethin meaningful rather than the first
thing that pops into your head. What sort of output are
I've used rimage to read in graphics files (jpeg.) If I recall correctly, I
think I had to install some libraries. What I can't recall is if it was any
faster than pixmap, as I was mainly concerned with the file format and the
forensic image processing possibilities.
Tom
> -Original Messa
type ?NaN and the help will tell you.
> -Original Message-
> From: Brett Stansfield [mailto:[EMAIL PROTECTED]
> Sent: Monday, 21 March 2005 12:10 PM
> To: R help (E-mail)
> Subject: [R] NaN
>
>
> Dear R
> What does NaN mean?
> I recently did a correlation on a batch of data for some
>
type ?cor and read the help file
In particular read it till you find the entry that tells you how to deal with
missing observations. You need to do this because as you get more competent you
will undoubdtedly come across other issues, so learning the format of the help
and realising why you nee
I'm not sure I am answereing your question, but here goes
# Create a vector with 2400 items
x <- runif(2400)
# Create a 600 by 4 matrix
y <- matrix(x,ncol = 4)
#If you needed the matrix to be done row by row
#y <- matrix(x,ncol = 4,byrow = T)
# Extract values from a particular part of the matri
Sorry about the wrong link. The archives are at
http://sourceforge.net/mailarchive/forum.php?forum=rpy-list
> -Original Message-
> From: Mike R [mailto:[EMAIL PROTECTED]
> Sent: Friday, 11 March 2005 2:44 PM
> To: r-help@stat.math.ethz.ch
> Subject: [R] howto: plot location, web search
>
If it's specifically R that I am looking for I use the following link
http://finzi.psych.upenn.edu/search.html
If I were going to use google I would go to the advanced page and insert
finzi.psych.upenn.edu into the "Domain" field so as to restrict the search. As
you have found out the letter R
I don't use MySQL but I have seen messages like this before. They often have
replies which indicate that you need to follow the instructions more closely. I
suggest you search the list for these previous posts as I'm sure there is help
there, somewhere!
Tom
> -Original Message-
> From:
I think I recall seeing a limited capability in the PBSmapping package.
Tom
> -Original Message-
> From: yyan liu [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, 9 March 2005 1:20 PM
> To: r-help@stat.math.ethz.ch
> Subject: [R] from long/lat to UTM
>
>
> Hi:
> Is there any function in R
Does strsplit fit what you want?
Tom
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> Sent: Thursday, 24 February 2005 11:46 AM
> To: r-help@stat.math.ethz.ch
> Subject: [R] a simple question
>
>
> Dear All,
>
> I need to separate one column which is in a form
?axis
where you will find
See Also:
'axTicks' returns the axis tick locations corresponding to
'at=NULL'; 'pretty' is more flexible for computing pretty tick
coordinates and does _not_ depend on (nor adapt to) the coordinate
system in use.
Tom
> -Original Message-
>
x <- c(1,2,3,4,5,8,9,10,11,12,15,16,17,18,19,22,23,24,33,34,35)
require(cluster)
pam(x,5)
Medoids:
[,1]
[1,]3
[2,] 10
[3,] 17
[4,] 23
[5,] 34
Clustering vector:
[1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 5 5 5
Objective function:
build swap
1.285714 1.047619
Available comp
barplot(matrix(c(x,y),ncol = 2),beside=T)
Does this help
?barplot notes
height: either a vector or matrix of values describing the bars which
make up the plot. If 'height' is a vector, the plot consists
of a sequence of rectangular bars with heights given by the
I think you may need to check the list for posts relating to "memory", "large
datasets", etc
http://cran.r-project.org/search.html
It would help those on the list, if you gave more details about the machine you
are using. Given the vintage of R you areusing is considered quite old, there
is th
x <- rep(seq(1,10),10)
y <- rep(seq(1,10),each = 10)
z <- matrix(100 + x*0.5 + y*0.5,ncol=10,byrow=T)
contour(z)
or back to where you were originally
x <- 1:10
y <- 1:10
# z <- 100:110
z <- 1:100
contour(matrix(z,ncol=10))
read the help on contour it states that
z: a matrix containing the val
My first thought was that all it looked a bit complicated for something that
should be straightforward.
I created a file called t.txt. I worked out the way I would have done it and
then I tested to see which was fastest. One little hiccup is that the two
objects are not identical and I though t
I'm sure others with more experience will answer this, but for what it is worth
my experience suggests that memory issues are more often with the user and not
the machine. I don't use Linux so I can't make specific comments about the
capacity of your machine. However it appears that there is oft
I am immediately reminded of something I read which goes
"A sufficiently trained statistician can read the vagaries of a Q-Q plot like a
sharman can read a chicken's entrails, with a similar recourse to scientific
principles. Interpreting Q-Q plots is more a visceral than an intellectual
exerci
I think Gabor gave you the answer, even if you didn't see it
x <- rep(c("02/27/92", "02/27/92", "01/14/92", "02/28/92", "02/01/92"),each = 5)
z <- data.frame(Date = strptime(x, "%m/%d/%y"))
z$Value <- trunc(runif(25) * 5) + 1
z$State <- c("A","B","C","D","E")[z$Value]
z <- z[order(z$Date),]
z
z[ag
In general if you are using a package, it helps to identify the package. I
assume it is the tree package.
Is this what you are after (from the example on the ?tree page)
> attr(ir.tr,"ylevels")
[1] "setosa" "versicolor" "virginica"
Tom
>
> -Original Message-
> From: NICOLAS DEIG [m
I am probably just displaying my ignorance, but I have obviously managed to
miss exactly what you are referring to.
Firstly I have to thank you for making me look closer at the article. I had
done so but I had obviously skipped over the Comparison table, which I would
have found useful in the p
This seems to work
toPOSIX <- function(x){
y <- x - as.numeric(ISOdate(2005,1,1))
z <- ISOdate(2005,1,1) + y
return(z)
}
test <- as.numeric(ISOdate(2005,3,1) )
toPOSIX(test)
But whether one should be doing this I don't know. There are certainly
functions that play aorund with the POSIX
What makes you trust any software?
There are some obvious points. First of all the code is open so if you know
enough you can actually read the code and make sure it does what you want.
Secondly you can replicate a process using two pieces of software and compare
the results. You can check the
Well I am not sure that can call a single figure a cluster. Sure it's not near
the others but how can you conceptually measure it's cluster properties. It
seems reasonable that there has to be some form of doubt about it.
Back to that Google search hit number 3 www.stat.ncu.edu.tw/teacher/
hung
The answers to your questions are in the text below. However these questions
are generally answered in the base documentation. You might try going through
the Keywords by Topic which on the windows system at least) is accessed by
going through the html help entry on the help menu. Once your brow
I hesitate to add this comment since it either completely confuses people or
they take to it very quickly.
The data that you are using is mostly categorical. I expect that tables will
have been used in the past and that to acertain extent the graphics are
suppossed to help with getting a quick
?points has this example
plot(-4:4, -4:4, type = "n")# setting up coord. system
points(rnorm(200), rnorm(200), col = "red")
points(rnorm(100)/2, rnorm(100)/2, col = "blue", cex = 1.5)
In general you might want to check out the keyword section of the help, in
particular the Graphics section which
I use windows, so I can't state that the same is true of other operating
systems.
There is a file which is in the etc folder of the R directory called Rconsole.
part of mine looks like
## Font.
# Please use only fixed width font.
# If font=FixedFont the system fixed font is used; in this case
temp <- table(rpois(100,5),rpois(100,1))
temp
temp[temp[,1] > 1,]
temp[temp[,1] == 1,]
temp[temp[,1] == 2,]
temp[temp[,1] == 3,]
Of course you have not made it obvious if you are using the term table to mean
any output that looks like a table, but in general terms you will find
information in th
Try searching this list for delphi. This was the first post I found. You may
find it helpful.
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/19154.html
Tom
> -Original Message-
> From: YiYao_Jiang [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, 19 January 2005 10:12 AM
> To: r-help@stat.m
Sorry I didn't read the question properly. Please disregard, my mind was
elsewhere.
Tom
> -Original Message-
> From: Mulholland, Tom
> Sent: Thursday, 13 January 2005 10:52 AM
> To: Dr Carbon; r-help@stat.math.ethz.ch
> Subject: RE: [R] Finding seasonal pe
You stand more chance if you do it yourself
https://stat.ethz.ch/mailman/listinfo/r-help
> -Original Message-
> From: Kevin Ita [mailto:[EMAIL PROTECTED]
> Sent: Thursday, 13 January 2005 2:25 AM
> To: R-help@stat.math.ethz.ch
> Subject: [R] Please unsubscribe me from you list
>
>
> Plea
You might find breakpoints in strucchange helpful
Tom
> -Original Message-
> From: Dr Carbon [mailto:[EMAIL PROTECTED]
> Sent: Thursday, 13 January 2005 6:19 AM
> To: r-help@stat.math.ethz.ch
> Subject: [R] Finding seasonal peaks in a time series
>
>
> I have a seasonal time series
I have often noted that "statistics can't prove a damn thing, but they can be
really useful in disproving something." Having spent most of 80s and half of
the 90s with the Australian Bureau of Statistics to find out how you collect
these numbers, I am disconcerted at the apparent disregard for m
points(y1,z1)
image (x,axes = FALSE,xlim =c(-0.25,1.25),ylim = c(-0.25,1.25))
points(y1,z1)
rect(0,0.1,0.5,0.9)
axis(2,at = seq(-0.25,1.25,length = 5),labels = seq(0,20, length = 5))
Tom
-Original Message-
From: Costas Vorlow [mailto:[EMAIL PROTECTED]
Sent: Wednesday, 12 January 200
There are indeed speed advantages in using colSums etc. However the
disadvantage is that the newbie doesn't always find the power inherent in the
apply, sapply, tapply and mapply. For many things that I do, the speed is the
least of my worries; although I take the point that using apply for mean
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/29712.html
This was the first message that I came across when searching the archives. From
my limited experience I would think that you need to understand what you are
doing rather than what R is doing. As the message explains R is flagging issues
apply(y,2,mean)
Is this what you are after. If it is I would suggest that you look at the
examples not just for this but for what I call the apply family sapply, tapply,
mapply. Once you get the hang of these they are really helpful.
Tom.
> -Original Message-
> From: Thomas Hopper [mai
> From: Peter Dalgaard [mailto:[EMAIL PROTECTED]
> Sent: Friday, 7 January 2005 4:46 PM
> To: Mulholland, Tom
> Cc: R-Help (E-mail)
> Subject: Re: [R] Basic Linear Algebra
>
...
> I can't. 4th element should be just 12.
I did indeed transcribe my solution incorre
I don't normally have to go anywhere near this stuff , but it seems to me that
this should be a straight-forward process in R.
For the purposes of this enquiry I thought I would use something I can work out
on my own.
So I have my matrix and the right hand results from that matrix
tdata <- ma
I thought this would be simple enough and in some ways it is, but I don't have
an answer I would call easy
The manual process would be to use barplot instead of screeplot, so that you
could return the x values. The other choice would be to hack the screeplot
function to return the values. (see
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