Hello, is there an up-to-date reference for how many people use R? I'm
giving an R demo and want to cite wonderful R usage stats. How many
people use it (or download it)? How often is R used in peer-reviewed
pubs, etc. Is there any whiz-bang citation that says something like R
is great and
Hi all,
I realize this is asking a lot of lattice, but I want to add a second y
axis inside a xyplot and have y1 and y2 have different ranges. Given dat
below, I can add a second y axis by overlaying a new plot with
par(new=T) and label axis 4 with standard graphics. I've seen an example
for
I have a time series with a one year lag, ar=0.5. The series has some
interesting events that disappear when the series is whitened (i.e.,
fitting an AR process and looking at the residuals). I'd like to remove
the autocorrelation in stages to see the effect on the time series. Is
there a way to
Thanks, I wasn't thinking real clearly when I pressed 'send'. All
figured out now. -A
-Original Message-
From: Wensui Liu [mailto:[EMAIL PROTECTED]
Sent: Monday, February 26, 2007 10:15 AM
To: Andy Bunn
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Partial whitening of time series?
andy
The original post is ambiguous: do you want to find the intersection or do
you want to find whether a prespecified set is in the
intersection? Patrick
provided you an answer to the latter while you provided an answer to the
former. Actually, I thought using table as you did (mod the need for
Suppose I have a list where I want to extract only the elements that occur
in every component. For instance in the list foo I want to know that the
numbers 2 and 3 occur in every component. The solution I have seems
unnecessarily clunky. TIA, Andy
foo - list(x = 1:10, y=2:11, z=1:3)
bar
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Dan Chan
Sent: Tuesday, April 04, 2006 1:42 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Selecting from a Dataframe
Hi,
I have a data frame with many fields and there are many records.
One of the
Is there is a way to make square bars in xyplot with type=h?
dat - data.frame(foo = rep(1:10,2), bar = rep(1:10,2))
xyplot(foo~bar, data = dat, type=h,lwd=20)
Thanks! Andy
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-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of tom wright
Sent: Wednesday, March 15, 2006 6:04 AM
To: R-Stat Help
Subject: [R] matrix indexing
Can someone please give me a pointer here.
I have two matrices
matA
A B C
1
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Arnau Mir
Sent: Tuesday, March 14, 2006 12:45 PM
To: r-help@stat.math.ethz.ch
Subject: [R] different values of a vector
Hello.
I have a vector of length 2771 but it has only 87 different values.
-Original Message-
From: Deepayan Sarkar [mailto:[EMAIL PROTECTED]
Sent: Friday, March 10, 2006 4:10 PM
To: Andy Bunn
Cc: R-Help
Subject: Re: add trend line to each group of data in: xyplot(y1+y2 ~ x |
grp...
On 3/10/06, Andy Bunn [EMAIL PROTECTED] wrote:
Although this should
Although this should be trivial, I'm having a spot of trouble.
I want to make a lattice plot of the format y1+y2 ~ x | grp but then fit a
lm to each y variable and add an abline of those models in different colors.
If the xyplot followed y~x|grp I would write a panel function as below, but
I'm
Hello all:
I have a character variable (foo) that contains the names of some numeric
variables. For my application, I'd like to cbind the numeric variables and
calculate the row mean using the character variable. I think I do this using
do.call but the function to call using do.call is eluding
Does this thread help?
https://stat.ethz.ch/pipermail/r-help/2006-February/086874.html
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Thomas Hoffmann
Sent: Wednesday, February 22, 2006 2:17 PM
To: r-help@stat.math.ethz.ch
Subject: [R] shaded
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED]
Sent: Tuesday, February 21, 2006 12:54 PM
To: r-help@stat.math.ethz.ch
Subject: [R] How to select only certain rows when making a new
dataframe?
Dear R-users,
I have two data
In the lattice plot below I want to fill-in the areas under each lines that
are greater than zero in gray. Is there a straightforward way to go about
this? Thanks, Andy
library(lattice)
foo - data.frame(Yrs=rep(1:50,4), Y=rnorm(200),
Id=unlist(lapply(letters[1:4],rep,50)))
-Original Message-
From: Sundar Dorai-Raj [mailto:[EMAIL PROTECTED]
Sent: Wednesday, February 15, 2006 1:40 PM
To: Andy Bunn
Cc: R-Help
Subject: Re: [R] shading under the lines in a lattice xyplot?
Andy Bunn wrote:
In the lattice plot below I want to fill-in the areas under
Has anybody implemented code to extract coefficients for a Butterworth
low-pass filter? I know Matlab has it implemented in the signal toolbox. I
want to make use of a 10 point Butterworth low-pass filter for smoothing.
In Matlab the code would look like this:
% Determine the filter coefficients
Hi all (really probably just Deepayan):
In the plot below I want to add text on either side of each violin plot that
indicates the number of observations that are either positive or negative.
I'm trying to do this with ltext() and I've also monkeyed about with
panel.text(). The code below is
Just wondering if anyone knows of any text mining projects in
R...I googled
a bit but didn't get anything...
RSiteSearch(text mining) turns up 85 hits...
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PLEASE
How can I modify the example below to put a dot at the mean of each violin
plot? I assume I use panel.points but that's as far as I can go.
bwplot(voice.part ~ height, singer,
panel = function(..., box.ratio) {
panel.violin(..., col = transparent,
How can I modify the example below to put a dot at the mean of each violin
plot? I assume I use panel.points but that's as far as I can go.
bwplot(voice.part ~ height, singer,
panel = function(..., box.ratio) {
panel.violin(..., col = transparent,
Look at ?all.equal and ?identical as well as searching the archives for
those terms. You'll find many an illuminating thread on precision, floating
point arithmetic and other wonders.
minX - 4.2
min0 - 4.1
sigmaG - 0.1
Diff - minX-min0
all.equal(Diff, sigmaG)
identical(Diff, sigmaG)
HTH, Andy
The data structures in R are still very puzzling to me. Can anyone tell
me how I can easily convert these two dataframes to one single dataframe
with two columns (mean and sd) with 7 rows?
meanprofile
V1 V2 V3 V4 V5 V6 V7
2292.001 2178.620
Does anyone know how I can set up R so that when I make a graphic, the
graphics window remains behind the console window? It's annoying to
have to reach for the mouse every time I want to type another line of
code (e.g., to add another line to the plot). Thanks.
What OS? In Windows with R GUI
Anthony. Look at ?predict.rpart, I think this might be the kind of table you
are looking for.
data(iris)
sub - c(sample(1:50, 25), sample(51:100, 25), sample(101:150, 25))
fit - rpart(Species ~ ., data=iris, subset=sub)
fit
table(predict(fit, iris[-sub,], type=class),
Some thing like this?
mat - matrix(1:9,3,3)
mat
mat[apply(mat[,2:3] 4,1,all),]
# or less cryptically
foo - mat[,2:3] 4
bar - apply(foo,1,all)
mat[bar,]
HTH, Andy
NB: No need to send this kind of message to r-devel
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL
Check out the dim vs length for vectors thread:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/50720.html
This thread goes through the bug-or-feature discussion which is always
entertaining from a socio-R perspective.
Also, note Dim with a capital D doesn't exist.
HTH, Andy
-Original
by some R-guru, but that's part of the fun of it all!
I hope that helps, Andy
-Original Message-
From: Tony Evans [mailto:[EMAIL PROTECTED]
Sent: Monday, August 22, 2005 8:03 PM
To: Andy Bunn
Subject: Re: [R] cbind and rbind
Sorry, I did read the guide but am very new to this.
I
Hello: I'm reading in a series of text files (100 files that are each 2000
rows by 6 columns). I wish to combine the columns (6) of each file (100) and
get the row mean. I'd like to end up with a data.frame of 2000 rows by 6
columns.
foo - list()
for(i in 1:10){
# The real data are read in
I think about half of my question in R can be solved with a judicious
do.call.
Thanks, Andy
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PLEASE do read the posting guide!
R w - list(rnorm(10), rnorm(10))
R x - ts(w, start = 1980)
Even though you don't get an error message this statement is
erroneous. ?ts discusses the valid possibilities.
So it does, might I suggest add something like this to ts:
if (is.list(data))
stop(Data must be a
some_df[1, ] is actually a data frame: see ?[.data.frame. It's hard to
see what else it could be, as columns of a data frame are of arbitrary
classes.
I see, I was confusing class and mode. However, since a list can be a ts
object as in this example:
R w - list(rnorm(10), rnorm(10))
R x -
Adam:
Providing a reproducible example would be a first step...
That's the problem, I can't. But I str has come to the rescue:
R str(rw)
Time-Series [1:307] from 1690 to 1996: 0.986 1.347 1.502 1.594 1.475 ...
R str(pg)
List of 264
$ : num 0.227
$ : num 0.189
$ : num 0.237
$ : num 0.235
This seems like a FAQ, but I can't figure it out.
I have a mv ts object:
R tsp(pg)
[1] 1982 20031
R dim(pg)
[1] 22 12
and a univariate ts:
R tsp(rw)
[1] 1690 19961
Yet, when I try to intersect them:
R tsp(ts.intersect(rw, pg))
[1] 1982 21761
the process goes awry.
How to I
Hi all:
I have acquired a 100s of data files that I need to preprocess to get them
usable in R. The files are fixed width (to a point) and contain 1 to 3 lines
of header, followed by a variable number of fixed width data lines (that I
can read with read.fwf). I want to read through the files and
x - readLines(...)
tmp - file()
writeLines(x[substr(x, 83, 86) == STD], tmp)
read.fwf(tmp, ...)
I wrapped this approach into function and it worked swimmingly. Thanks
all. -Andy
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Just want to offer my congratulations to the JGR developers as
the recepient
of the 2005 Chambers Award. Great job, guys!!
http://stats.math.uni-augsburg.de/JGR/
This feels like the future of R to me. It's simple, powerful, and elegant
just like R. As soon as the binary that works with 2.1
I have not tried JGR but regarding your three adjective describing R,
R is very powerful but I am not sure I would characterize it as simple
and elegant -- complex and practical seem nearer to the mark to me.
I take umbrage (and not in the sense of affording shade).
Take 'simple' to mean
PS 6 days to the big Jedi holiday!
I wonder if anybody who gave the matter any thought would be surprised that
the R-Help list is populated by ubergeeks.
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PLEASE do
I have a very simple query with regard to summarizing the number of factors
present in a certain snippet of a data frame.
Given the following data frame:
foo - data.frame(yr = c(rep(1998,4), rep(1999,4), rep(2000,2)), div =
factor(c(rep(NA,4),A,B,C,D,A,C)),
org =
The three was a typo, which I regret very much. I don't know why I didn't
think of apply. I was obsessed with doing it as a table.
Thanks for your response,
-Andy
-Original Message-
From: Tony Plate [mailto:[EMAIL PROTECTED]
Sent: Tuesday, April 26, 2005 2:00 PM
To: Andy Bunn
Cc: R
Hi Ronaldo: First, this script was discussed a few days ago. Note that under
2.1 update.packages() has changed for the better. Look at the NEWS file.
So, if you want to keep updating on Tuesdays, despite sensible advice to the
contrary offered in this thread
Well for comparatative reason I would would like to subsume both plots
into a unifying plot and save them in one file.pdf.
I tried to find an answer in FAQ and mailinglists archive, no luck.
Maybe did miss an appropiate answer to my question, so a pointer to
solve my problem woud be
Do you mean a second y axis? If so then something like this would
do it
Not exactly, I would also take advantage of overlaying the fitting
curve of LAD on OLS, since both rely on the same dataset. Maybe two
regression lines with varying shapes (i.E. straight versus dotted
line)
Like this?
The Tuesday update script came back to get me! I knew it would.
update.packages has changed (for the better) with this release. Look at the
NEWS file:
The 'CRAN' argument to update.packages(), old.packages(),
new.packages(), download.packages() and install.packages() is
This is inelegant, but works:
# (following the example)
nPts - 254
foo - sin((2 * pi * 1/24) * 1:nPts)
foo - foo + rnorm(nPts, 0, 0.05)
bar - ts(foo, start = c(1980,3), frequency = 24)
mean.in.i - numeric(length(start(bar)[1]:end(bar)[1]))
peak.ts - ts(rep(NA, length(foo)), start = c(1980,3),
How about this?
length(res[res 0, 1])
HTH, Andy
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Frederic renaud
Sent: Wednesday, January 05, 2005 11:08 AM
To: r-help@stat.math.ethz.ch
Subject: [R] count element in column
Hi,
I 've a matrix n*1
Arggh! start and end. Of course. How stupid of me. And I'm not going to
interpolate with smooth.spline, but thanks for the warning. Sloppy language.
Thanks, Andy
-Original Message-
From: Liaw, Andy [mailto:[EMAIL PROTECTED]
Sent: Friday, December 10, 2004 2:49 PM
To: 'Andy Bunn'; R
Preamble: Sorry for being dense.
Now that that's done, here are my questions.
I want to put a polygon on a plot of a time series. I'm going to add lines
from a smooth.spline interpolation and other annotation to it. But here's
the general idea:
# Start code
n - 121
dat - rnorm(n)
Something like this will work:
foo.df - data.frame(x = 1:8, y1 = c(1,3,5, NA, NA, NA, NA, NA),
y2 = c(NA, NA, NA, 7, 9, 11, NA, NA),
y3 = c(NA, NA, NA, NA, NA, NA, 13, 15))
plot(foo.df$x, foo.df$y1, ylim = c(0,20), type = n)
In addition to Sean's reply look at ?dist and other ways of creating
distance / similarity matrices for applications like Mantels Test. Package
vegan might be particularly useful.
HTH, Andy
R x - rnorm(10)
R y - dist(x)
R str(x)
num [1:10] -0.431 0.564 0.901 -1.407 -0.991 ...
R str(y)
It looks like factanal is unable to optimize from these starting values
(kinda like the error message says). So, factanal.fit.mle isn't converging
and you have problems with your analysis. Try putting control = list(trace =
T) in your code to see what happenens. E.g.,
R
R v1 -
See the function ?citation under 2.0.0
R citation()
To cite R in publications use:
R Development Core Team (2004). R: A language and environment for
statistical computing. R Foundation for Statistical Computing,
Vienna, Austria. ISBN 3-900051-07-0, URL http://www.R-project.org.
HTH,
Does this do what you want?
foo.df - data.frame(x = rnorm(12), y = runif(12), z = factor(rep(1:3,4)))
bar.mat - matrix(NA, nrow = ncol(foo.df)-1, ncol = nlevels(foo.df$z))
for(i in 1:(ncol(foo.df)-1))
{
bar.mat[i,] - xtabs(foo.df[,i] ~ foo.df$z)
}
bar.mat
There's probably a slicker way with
Using this function with 2.0.0 XP and Firefox 1.0 (I've rediscovered the
internet) produces a curious result.
myString - RSiteSearch(string = 'Ripley')
myString
[1]
http://finzi.psych.upenn.edu/cgi-bin/htsearch?config=htdigrun1;restrict=Rhe
There has to be a better (more readable) way, but this works...
set.seed(323)
foo.df - data.frame(A = round(runif(5)), B = round(runif(5)), C =
round(runif(5)))
foo.df
A B C
1 0 1 1
2 1 1 1
3 1 1 1
4 0 1 1
5 1 1 0
names.list - lapply( apply( foo.df, 1, function( x ) colnames(
foo.df )[
see ?cumsum
x - 1:10
cumsum(x)
max(cumsum(x))
HTH, Andy
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Sean Davis
Sent: Friday, November 19, 2004 1:09 PM
To: r-help
Subject: [R] Running sum
I have vector X of length N that I want to have a
Is there a way to calculate the number of months between dates?
StartDate - strptime(01 March 1950, %d %B %Y)
EventDates - strptime(c(01 April 1955, 01 July 1980), %d %B %Y)
difftime(EventDates, StartDate)
So, there are 61 months between 01 March 1950 and 01 April 1955. There are
364 months
Hi Mick:
I'm a little unsure if this is what you are after but does this do it?
foo.mat - matrix(rnorm(100), nrow = 10, ncol = 10)
plot(foo.mat[1,], type=l, xlab = Crud, ylab = More Crud)
plot(foo.mat[1,order(foo.mat[1,])], type=l, xaxt = n, xlab = Crud,
ylab = More Crud)
axis(1,
This isn't pretty but it's a way to do it:
foo - data.frame(x = c(1,0,1,1,0,2,4), y = as.factor(c(0,2,1,1,0,3,1)))
Zero2NA - function(x){
if(is.numeric(x)) { x[x == 0] - NA; }
return(x)
}
foo2 - as.data.frame(lapply(foo, Zero2NA))
foo
foo2
HTH, Andy
-Original Message-
From:
How about installing an operating system that knows its way around that much
RAM?
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Graham Law
Sent: Monday, November 15, 2004 9:12 AM
To: [EMAIL PROTECTED]
Subject: [R] Using R in parallel on a 2 processor
You can use title, but the result is unsatisfying:
fit - arima(lh, c(1,0,0))
tsdiag(fit)
title(junk)
Perhaps mtext with an appropriate par configuration?
HTH, Andy
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Andrew Kniss
Sent: Monday,
How about this?
a-c(1,7,4,5,9,11)
b-c(7,4,9)
a[!a %in% b]
b-c(7,4,9, 100, 20, 34, 54)
a[!a %in% b]
see ?match, too
HTH, Andy
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Alexander Sokol
Sent: Thursday, November 11, 2004 8:34 AM
To: [EMAIL
I've ported somebody else's rather cumbersome Matlab model to R for
colleagues that want a free build of the model with the same type of I/O.
The Matlab model reads a text file with the initial parameters specified as:
C:\Data\Carluc\Rportmore Params.R
# Number of years to simulate
nYears = 50;
I might well be wrong, but I don't think there is. I went about rewriting
partition.tree for rpart once but stopped after I realized that it was much
easier to add lines and segments to plots by hand using the coordinates from
the rpart object (I then added a third predictor to my dataset making
How can I specify the repositories for upgrade()?
x - packageStatus(repositories =
http://cran.us.r-project.org//bin/windows/contrib/2.0;)
upgrade(x, ask = FALSE)
Error in update[, 3] : incorrect number of dimensions
x, the object of class packageStatus, prints and summarizes fine. I also
ran
Ooops. Make that subject line 'upgrade', not 'update!' Sorry. -AB
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Andy Bunn
Sent: Tuesday, November 09, 2004 9:23 AM
To: R-Help
Subject: [R] Error in update() when repositories is specified.
How can I
How about using ylim?
foo - rnorm(100, 0, 1)
boxplot(foo, ylim = c(-5, 5))
HTH, Andy
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Scott Rifkin
Sent: Tuesday, November 09, 2004 2:43 PM
To: [EMAIL PROTECTED]
Subject: [R] Boxplot plot range
How
How about something like this?
my.func - function(y, x1, x2, x3, x4 = NULL){
my.formula - as.formula(y ~ x1 + x2 + x3 + x4)
if(is.null(x4)) { my.formula - as.formula(y ~ x1 + x2 + x3) }
outlm - lm(my.formula)
meanvec-c(mean(x1),mean(x2),mean(x3))
if(is.null(x4) == F) {
Welcome to R.
You can start by looking at the predict function for the regression model you are
using (e.g., ?predict.lm if you are using a linear model).
Then do something like so:
data(cars)
cars.lm - lm(dist ~ speed, cars)
dist.pred - predict(cars.lm)
plot(cars$speed, cars$dist, ylim =
Load the library first:
library(Malmig)
?mtx.exp
HTH, Andy
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Hi,
Somebody asked me to make sure that all the machines running the in our lab
(XP and Linux, both running 2.0) have R installed and that A) All the
packages are installed and B) kept up-to-date.
Obediently, I began to modify a shared Rprofile so that once a week it
checks for new packages and
I think you want something like so:
# make some data
foo.df - data.frame(x = 1:100, y = runif(100), age = rnorm(100, 10, 1))
# stick some real NAs in all columns
foo.df[c(2,78,32,56),] - NA
# make some errant NAs in the column age
foo.df$age[c(99, 26, 75, 3)] - NA
# eg
foo.df[1:5,]
# remove the
Something like this should work:
foo - read.table(text2read.txt, colClasses=c(character, NULL,
NULL))$V1
foo - gsub(i[0-9]-, , foo)
HTH, Andy
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PLEASE do read the posting
In addition, look at Laying Out Pathways With Rgraphviz in R News which
describes the Rgraphviz packages on Bioconductor.
http://cran.r-project.org/doc/Rnews/Rnews_2004-2.pdf
HTH,
Andy
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I made a quick search and was unable to find a general implementation
of the interior function for an arbitrary polygon; I'm a bit
surprised about that. Hopefully someone else can point to one,
otherwise please write one, and document it and contribute it to R.
It's a relatively standard
Anna:
That is the most current version of RWinEdt. Uwe Ligges is working on a
version for 2.0.0 so check back soon for a new release.
http://cran.r-project.org/contrib/extra/winedt/
Also, when inquiring about a specific package it is often helpful to contact
the maintainer directly and not
ifelse accomplishes this pretty easily (at least I think it does what you
want)
Look at ?apply too.
HTH, Andy
## Try this
foo.dat - data.frame(Var1 = rnorm(4, 1, 1),
Var2 = (rnorm(4, 1, 1) * 0.25))
plot(density(foo.dat$Var1 / foo.dat$Var2))
RatioOne -
You had it.
plot(1:5, main = This is a title\nIn 2 lines)
HTH, Andy
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You can do that easily with 'try'
?try
'try' is a wrapper to run an expression that might fail and allow
the user's code to handle error-recovery.
HTH, Andy
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Look at the function vegdist in the library vegan.
It does Bray-Curtis and other common ecological distance measures.
HTH, Andy
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PLEASE do read the posting guide!
Look at the help for:
?as.numeric
HTH, Andy
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Janet Gannon
Sent: Monday, March 15, 2004 7:04 AM
To: [EMAIL PROTECTED]
Subject: [R] Simple numeric as.is question
I am reading a list of numbers from
Look at ?Extract
myDF - data.frame(c1 = c(1,1,1,1,2,2,2,2),
c2 = c(23,34,45,45,78,65,45,70),
c3 = c(12,15,67,87,23,19,90,32))
mean(myDF[myDF$c2 == 45,3])
HTH, Andy
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See the splancs library and the functions
gridpts
pcp.sim
csr
in particular. However, what you want to do is not completely trivial.
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PLEASE do read the posting
?write.table
setwd(c:\\temp)
myDF - data.frame(A = rnorm(100), B = rnorm(100))
write.table(myDF, file = myDF.dat, sep = \t, quote = F, row.names =
F)
HTH, Andy
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If you mean to put a check in to see if the file exists then something
like this would work:
for(i in 1:3){
aFile - paste(file, i, .dat, sep = )
if(file.exists(aFile) == T){
bb - read.table(aFile, header = F)
x11()
plot(bb)
dev.off()
}
}
If you
?source
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Kissell, Robert [EQRE]
Sent: Thursday, March 04, 2004 3:19 PM
To: [EMAIL PROTECTED]
Subject: [R] Command Line Programs
Hi,
I have recently started using R again (switched from MatLab)
?points
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)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]
On Behalf Of Raphael Schoenle
Sent:
Try the FAQ
http://cran.r-project.org/faqs.html
Or one of the manuals
http://cran.r-project.org/faqs.html
But first read the posting guide
http://www.R-project.org/posting-guide.html
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Susan Lin
Sent:
You know...I almost added this to your original post because I thought
you might want to read something back into ArcInfro when you were done!
Look at ?files
The easiest thing to do in R is to keep your header as a text file in
the working directory and then append it to your output file using
Is everybody writing ArcGIS ASCII rasters recently?
The GIS community is hopelessly tied to ESRI. So many people have
invested their careers in learning Arc that switching to GRASS is an
institutional nightmare. Most of what I do now is outside of Arc! And
certainly outside of their almost
min(v)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of li xian
Sent: Wednesday, February 25, 2004 8:46 PM
To: [EMAIL PROTECTED]
Subject: [R] minimum value
suppose I have a vector called v,
how can I get the index of the minimum element of vector v?
No problem. I'm glad it worked. It would be pretty easy to package as a
function for gstat.
-Andy
-Original Message-
From: femke [mailto:[EMAIL PROTECTED]
Sent: Thursday, February 19, 2004 2:54 PM
To: Andy Bunn
Subject: Re: [R] importing ascii grids (for gstat)
Thanks very
Sorry for the previous posting. I found the function in ?complex. My
apologies. -Andy
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
If you have exported the grid from Arc using the asciigrid command then
you can read it in with scan or read.table. You can tell R to skip the
six lines of header info and to convert - to NA e.g.,
$ snep.tmin - read.table(file = tmin.asc, sep = , na.strings =
-, skip = 6)
Check the
Look at ?as.factor
HTH, Andy
## Example
# Make some data
foo.df - data.frame(X1 = rnorm(10), X2 = runif(10), Y = round(runif(10)
+ 1))
# Summarize it
summary(foo.df)
# Tak an extra close look at column Y
class(foo.df$Y)
# Change Y to a factor
foo.df$Y - as.factor(foo.df$Y)
# Look at it again
Marisa,
If the examples in R Data Import/Export can't help you you'll have to be
way way way more specific.
http://cran.r-project.org/doc/manuals/R-data.pdf
I'm assuming you are using windows. Here's an example. Try saving the
attached file to c:\temp and copying the commands below into R. This
Try this. Look at ?flsuh.console if you are on windows
HTH, Andy
#
i-0
while(i100){
## do a lot of commands
##print i
cat(i, \n)
## if using windows
flush.console()
i-i+1
}
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I had a reviewer request a citation for a package that I had neglected to
cite. I followed a similar format to that suggested by Thomas Lumley and
also referenced a related book by the package's author. The editor thought
it was nifty.
My two cents.
-Andy
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