Vegard Andersen vegard.andersen at ism.uit.no writes:
:
: Hello!
:
: My problem is that the Julian date behind my dates seems to be wrong. I
: will examplify my problem.
:
: t1 - 1998-11-20
: t2 - as.Date(t1)
: # Here t2 is correctly 1998-11-20, but
: date.mdy(t2)
: $month
: [1] 11
: $day
:
Darren Weber darrenleeweber at gmail.com writes:
:
: Hi,
:
: if we have a file called Rscript.R that contains the following, for example:
:
: x - 1:100
: outfile = Rscript.Rout
: sink(outfile)
: print(x)
:
: and then we run
:
: source(Rscript.R)
:
: we get an output file called
faisal99 at inf.its-sby.edu writes:
:
: hi everyone,
: I'm still a newbie in statistics,
:
: I have a question about beta distribution, that is,
:
: On the ref/tutorials I've found on the net, why beta distribution always
: have value p(x) more than 1?
Consider the uniform distribution on
Witold Eryk Wolski W.E.Wolski at ncl.ac.uk writes:
:
: Dear Rgurus,
:
: To my knowledge the best way to visualize the distribution of a discrete
: variable X is
: plot(table(X))
:
: The problem which I have is the following. I have to discrete variables
: X and Y which distribution I would
Xiyan Lon xiyanlon at gmail.com writes:
:
: Dear useR again,
: How can I read a dataset if lines in dataset did not have same
: elements (have different lengths), For example:
:
: 12, 4, 16, 1, 1, 3, 1, 1, 15, 5, 1, 1, 14, 1, 1
: 22, 13, 5, 1, 1, 3, 1, 1, 15, 5, 1,
achilleas.psomas at wsl.ch writes:
:
: Hello R-Helpers..
:
: I am still new in R and I have the following question..
: I am applying the function chull on a 2D dataset and have the convex hull
: nicely
: calculated and plotted.
: Do you know if there is a way to extract the coordinates of the
Bill Simpson William.Simpson at drdc-rddc.gc.ca writes:
:
: I want to flatten a matrix and unflatten it again. Please tell me how to
: do it.
:
: 1. given a matrix:
: x1 y1 z1
: x2 y2 z2
: ...
: xk yk zk
: convert it to a vector:
: x1, y1, z1, x2, y2, z2, ..., xk, yk, zk
:
: 2. given a
Michael S michael_shen at hotmail.com writes:
:
: Dear all-helpers:
:
: I create one package ,code like this:
: output -
: function(x,y)
: {
: zz -textConnection(foo,w)
: sink(zz)
: a -5
: b -6
: z -a*b
: z
: e -spss
: h -c(1,2,3)
: ls()
:
Michael S michael_shen at hotmail.com writes:
:
: dear ALL-R-helper:
: I modify my question in textconnection output:
: I wrote one function in Rgui:
: output - function(y){
: x - textConnection(foo,w)
: sink(x)
: a -5
: b -6
: z -a*b
: z
: e -spss
:
Pascal BLEUYARD p.bleuyard at opgc.univ-bpclermont.fr writes:
:
: Hi all,
:
: I need to compute some occurence matrix: given a zero matrix and a set
: of paired indexes, I want to store the number of occurences of each paired
: index in a matrix. The paired indexes are stores as an index
Can you provide a link. I did a google search and found something
on a Japanese site but it turned out that the writer had made a
mistake and it linked to wmf2eps, not eps2wmf.
Christophe Pallier pallier at lscp.ehess.fr writes:
:
: Hello Christoph!
:
: In the past, I used an utility called
Luis Tercero luis.tercero at ebi-wasser.uni-karlsruhe.de writes:
:
: I have imported a data frame that looks like this:
:
:Measurement.Date.and.Time Z.Average..nm. PDI
: 572 Dienstag, 22. Mrz 2005 11:05:59 366,4 0,468
: 573 Dienstag, 22. Mrz 2005 11:09:30 353,4
Gabor Grothendieck ggrothendieck at myway.com writes:
Luis Tercero luis.tercero at ebi-wasser.uni-karlsruhe.de writes:
:
: I have imported a data frame that looks like this:
:
:Measurement.Date.and.Time Z.Average..nm. PDI
: 572 Dienstag, 22. Mrz 2005 11:05:59
Seung Jun jun at cc.gatech.edu writes:
:
: Fold in Mathematica (or reduce in Python) works as follows:
:
: Fold[f, x, {a, b, c}] := f[f[f[x,a],b],c]
:
: That is, f is a binary operator, x is the initial value, and the results
: are cascaded along the list. I've found it useful for reducing
If the odometer order in your post is essential then
you could try this:
expand.grid(1:5, 1:4, 1:3)[,3:1]
If R's reverse odometer order is ok then you could
simplify it to this:
expand.grid(1:3, 1:4, 1:5)
On Mon, 28 Mar 2005 15:20:46 -0800, Ronnen Levinson [EMAIL PROTECTED] wrote:
This happened to me in R 2.1.0 (I forget which specific version
since I now have March 27th) on Windows XP which I traced to
package dataRep. Once I removed that package help.search
worked again.
On Mon, 28 Mar 2005 22:08:58 -0500, Gerard Tromp
[EMAIL PROTECTED] wrote:
Greetings!
OS:
I believe that the function that Recall executes is the
function in which Recall, itself, is evaluated -- not the
function in which Recall appears. In normal cases these are
the same but if you pass Recall to another function then
they are not the same. Here Recall is being passed to
sapply
Jaroslaw's article was great. In fact, it was used as the basis for
rapply and some optimized special cases that will be included in
the R 2.1.0 version of zoo (which have been coded but not yet
released).
Regarding numerically stable summation, check out the idea
behind the following which I
Try this:
data.matrix(df.f12)
On Apr 2, 2005 6:01 AM, Heinz Tuechler [EMAIL PROTECTED] wrote:
Dear All,
Assume I have a data.frame that contains also factors and I would like to
get another data.frame containing the factors as numeric vectors, to apply
functions like sapply(..., median) on
I just started using gmail and one thing that I thought would
be annoying but sometimes is actually interesting are the
ads at the right hand side. They are keyed off the content
of the email and in the case of your post produced:
http://www.visibone.com/regular-expressions/?via=google120
On Apr 5, 2005 1:36 PM, Paul Johnson [EMAIL PROTECTED] wrote:
I'm writing R code to calculate Hierarchical Social Entropy, a diversity
index that Tucker Balch proposed. One article on this was published in
Autonomous Robots in 2000. You can find that and others through his web
page at Georgia
Some other advantages of making your own package are:
- you can use help.search to search for your own functions even if you
don't load the package
- if you can't even remember where your functions are (and I often
can't) then you may not remember what they do either and packaging
them
? Is the input script file
anargument to R and therefore available in something like argv?
On Mar 18, 2005 8:00 PM, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Darren Weber darrenleeweber at gmail.com writes:
:
: Hi,
:
: if we have a file called Rscript.R that contains the following, for example:
:
: x - 1
On Apr 7, 2005 1:18 AM, Itay Furman [EMAIL PROTECTED] wrote:
On Tue, 5 Apr 2005, Gabor Grothendieck wrote:
On Apr 5, 2005 6:59 PM, Itay Furman [EMAIL PROTECTED] wrote:
Hi,
I have a data set, the structure of which is something like this:
a - rep(c(a, b), c(6,6))
x - rep(c(x, y
On Apr 7, 2005 8:43 AM, Gregory BENMENZER
[EMAIL PROTECTED] wrote:
hello,
I created a package with my functions, and i wand to hide the code of some
functions.
Could you help me ?
Grégory
There was some discussion on the list that there is work being
done on an R compiler. I don't
On Apr 8, 2005 9:05 AM, Paul Rathouz [EMAIL PROTECTED] wrote:
OK. Thanks. So, if you use table() on a factor that contains NA's, but
for which NA is not a level, is there any way to get table to generate an
entry for the NAs? For example, in below, even exclude=NULL will not
give me an
ronggui 0034058 at fudan.edu.cn writes:
: hi,usRs,i am studing the R programming,but i can not get the point abut the
difference between UseMethod
: and NextMehod.i have read the manual and try to find the solutin from
internet,but i still not master it
: well.so anyone can give me a guide?it
On Apr 11, 2005 6:22 AM, Wolfram Fischer [EMAIL PROTECTED] wrote:
What is the easiest way to change within vector of strings
each letter after a space into a capital letter?
E.g.:
c( this is an element of the vector of strings, second element )
becomes:
c( This Is An Element Of The
On Apr 11, 2005 7:22 AM, Gavin Simpson [EMAIL PROTECTED] wrote:
Dear List,
I'm using Sweave to produce a series of class handouts for a course I am
running. The students in previous years have commented about wanting
output within the handouts so they can see what to expect the output to
Try this:
# test data
set.seed(1)
DF - data.frame(x1 = rnorm(5), x2 = rnorm(5), x3 = rnorm(5))
DF
model.list - c(x2, x3)
# transform
for(v in model.list) DF[v] - floor(DF[v])
On 4/20/06, Chad Reyhan Bhatti [EMAIL PROTECTED] wrote:
Hello,
I am not sure what to write in the subject line, but I
R supports a number of databases and if you only need to work with a small
amount of data at once it should be readily do-able; however, R keeps objects
in memory and if you need large amounts at once then you could run into
problems. Note that S-Plus keeps objects on disk and has other
features
Try:
V1 - matrix(c(10, 20, 30, 10, 10, 20), nc = 1)
V2 - 4 * (V1 == 10) + 6 * (V1 == 20) + 10 * (V1 == 30)
or
V2 - matrix(c(4, 6, 10)[V1/10], nc = 1)
On 4/21/06, Sachin J [EMAIL PROTECTED] wrote:
Hi,
How can I accomplish this task in R?
V1
10
20
30
10
10
20
Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try:
V1 - matrix(c(10, 20, 30, 10, 10, 20), nc = 1)
V2 - 4 * (V1 == 10) + 6 * (V1 == 20) + 10 * (V1 == 30)
or
V2 - matrix(c(4, 6, 10)[V1/10], nc = 1)
On 4/21/06, Sachin J wrote:
Hi,
How can I accomplish this task in R?
V1
10
20
Here is a compact solution using approx:
DF$V2 - approx(c(10, 20, 30), c(4,6,10), DF$V1)$y
On 4/21/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
DF - data.frame(V1 = c(10, 20, 30, 10, 10, 20))
DF$V2 - with(DF, 4 * (V1 == 10) + 6 * (V1 == 20) + 10 * (V1 == 30))
DF$V3 - c(4, 6, 10)[DF$V1
It is matched by the first argument which is called from even though,
as the documentation indicates under the explanation of the last
form, it refers to the ending value. Note, for example, that
seq(from = 3) gives 1:3 and not 3:1.
Also the help file does say:
The interpretation of the
Use merge:
# test data
both - list(structure(list(Doc = c(5, 9, 7, 5, 7, 9), Query = c(1, 1,
1, 2, 2, 2), Rank = c(1, 2, 3, 1, 2, 3)), .Names = c(Doc, Query,
Rank), class = data.frame, row.names = c(1, 2, 3, 4,
5, 6)), structure(list(Doc = c(4, 5, 9, 8, 5, 7), Query = c(1,
1, 1, 2, 2, 2), Rank =
Here are a couple of possibilities using the builtin iris data set. Note
that although the coefficients come out the same, the degrees of
freedom, etc., would differ:
n - rep(1:3, 50)
lm(Petal.Length ~ Petal.Width, iris, weight = n)
Call:
lm(formula = Petal.Length ~ Petal.Width, data = iris,
RSiteSearch(clear screen)
will locate Windows code to send a ctrl-L to the screen that you
can modify.
On 4/24/06, Tolga Uzuner [EMAIL PROTECTED] wrote:
Hi,
Is there a way to send an ESC command to the console from within a
script window, without using the mouse ?
Thanks,
Tolga
You just need the much smaller cross product matrix X'X and vector X'Y so you
can build those up as you read the data in in chunks.
On 4/24/06, Sachin J [EMAIL PROTECTED] wrote:
Hi,
I have a dataset consisting of 350,000 rows and 266 columns. Out of 266
columns 250 are dummy variable
This works for me on my Windows XP system:
Sys.putenv(LANGUAGE=FR); Sys.setlocale(LC_ALL,FR)
On 4/24/06, Lapointe, Pierre [EMAIL PROTECTED] wrote:
Hello,
How do you change the language of the labels in a graph. In this example, I
want to get French labels by changing Sys.putenv. I should
, and accumulate (i.e., add) them.
There are examples in S Programming, taken from independent replies by the
two authors to a post on S-news, if I remember correctly.
Andy
From: Sachin J
Gabor:
Can you elaborate more.
Thanx
Sachin
Gabor Grothendieck wrote:
You just need the much
Use dQuote. Assuming you have a data frame with the column
as factors:
DF - data.frame(x = letters) # test data
levels(DF$x) - dQuote(levels(DF$x))
On 4/25/06, Jerry Pressnell [EMAIL PROTECTED] wrote:
I wish to place quotation marks around each element of the following
list;
A similar question was just asked. See:
http://tolstoy.newcastle.edu.au/R/help/06/04/25898.html
On 4/25/06, Erez [EMAIL PROTECTED] wrote:
Hello,
I'm working with large matrix data and i would like to know if
there is any way to reduce the size of it because even that I'm
increasing the
At least for this case I think you could get the effect without modifyiing
CrossTable like this:
as.CrossTable - function(x) structure(x, class = c(CrossTable, class(x)))
print.CrossTable - function(x) for(L in x) cat(L, \n)
by(warpbreaks, warpbreaks$tension, function(x)
On 4/25/06, Dr. Michael Wolf [EMAIL PROTECTED] wrote:
3. RDCOMCLient and Excel Manual
===
Do you know a good overview of using Excel VBA code via RDCOMClient (e. g.
sh$Select())? Are there people interesting in working out such a paper? I
could contribute some
On 4/25/06, hadley wickham [EMAIL PROTECTED] wrote:
The R Web site is working fine. Even if it is not relifted from a long
time, it is functional. So, this is the point... and it should remain,
at least, as functional as it is.
As an experienced user of the R website, this probably is true
Its not hard if you know what to do but if you don't then its a nuisance
to figure it out every time.
On 4/25/06, Gavin Simpson [EMAIL PROTECTED] wrote:
On Tue, 2006-04-25 at 14:09 -0400, Gabor Grothendieck wrote:
On 4/25/06, hadley wickham [EMAIL PROTECTED] wrote:
The R Web site
On Windows, right click the web page, choose Properties and
copy the url there.
On 4/25/06, Spencer Graves [EMAIL PROTECTED] wrote:
see inline
hadley wickham wrote:
The R Web site is working fine. snip
As an experienced user of the R website, this probably is true for
you. However,
Also, the proto and R.oo packages provide object oriented
ways of working with environments.
On 4/26/06, Prof Brian Ripley [EMAIL PROTECTED] wrote:
?new.env
?local
should help you. R works with environments, basically a frame plus an
enclosure.
On Wed, 26 Apr 2006, Anna Whitfield wrote:
A third possibility is to using the proto package and define
a proto object (an environment with special meaning for $)
containing the two components .x and g like this:
library(proto)
p - proto(.x = 2, g = function(.) { print(.$.x); .$.x - 3 })
p$.x # 2
print(p$g()) # 2, 3
p$.x # 3
or you can
Also, if you don;t need to create child objects which override .x,
and I don't think you do here, p could be further simplified to
this (only the print statement has been changed):
p - proto(.x = 2, g = function(.) { print(.x); .$.x - 3 })
On 4/26/06, Gabor Grothendieck [EMAIL PROTECTED] wrote
On 4/26/06, Daniel Yang [EMAIL PROTECTED] wrote:
--- I changed to format to text instead of html ---
Dear R-help,
This is my first R day. I want to ask some more beginner's questions.
Read the posting guide at the bottom of each post to r-help.
Q1. How can I obtain the covariance matrix
There is a reference sheet here:
http://www.rpad.org/Rpad/R-refcard.pdf
a function finder here:
http://biostat.mc.vanderbilt.edu/s/finder/find.html
and task views here:
http://cran.r-project.org/src/contrib/Views/
Also use of RSiteSearch and help.search from within R
can be helpful.
On
Maybe a separate web site that shows R off or maybe just
a pointer to the R Graph Gallery.
On 4/26/06, Duncan Murdoch [EMAIL PROTECTED] wrote:
Romain Francois wrote:
Dear R users and developpers,
My question is adressed to both of you, so I choose R-help to post it.
Are there any plans
On 4/26/06, Dimitri Szerman [EMAIL PROTECTED] wrote:
Dear R-mates,
# Here's what I am trying to do. I have a dataset like this:
id = c(rep(1,8), rep(2,8))
dur1 - c( 17,18,19,18,24,19,24,24 )
est1 - c( rep(1,5), rep(2,3) )
dur2 - c(1,1,3,4,8,12,13,14)
est2 - rep(1,8)
mydata =
@stat.math.ethz.ch
Subject: [R] copying previously installed libraries to R 2.3.0
hi all,
is there a new mechanism in R 2.3.0 for copying libraries from, say, R 2.2.1
to R 2.3.0? i ask because gabor grothendieck comments in his copydir.bat
(from gabor's batchfiles at:
http://cran.r-project.org
ov$vn1 is not a variable. It is the result of applying the $ function to
the ov and vn1 arguments.
For example, using BOD which is a data frame that comes with R,
rather than get(BOD$Time) use get(BOD)[[Time]]
On 4/26/06, Thomas Davidoff [EMAIL PROTECTED] wrote:
I don't understand what my
stl does this internally:
x - na.action(as.ts(x))
so
stl(x, s.window, na.action = f)
is the same as
stl(f(as.ts(x)), s.window)
e.g.
nottem[25] - NA # nottem is a built in data set in R
stl(nottem, per) # error
stl(nottem, per, na.action = na.contiguous)
library(zoo)
stl(nottem, per,
They can be found here:
http://dir.gmane.org/gmane.comp.lang.r.announce
Announcements about the development of the R Project for Statistical
Computing and the availability of new code. (read-only)
http://dir.gmane.org/gmane.comp.lang.r.deal
Learning Bayesian networks in R - the 'deal' package
Assuming you are aware that the dir pages at gmane provide
the key links for each group, googling for:
dir gmane r-help
gets it as the first hit.
On 4/27/06, Jose Quesada [EMAIL PROTECTED] wrote:
Thanks all,
It was very surprising that I couldn't find it. I searched
news.gmane.orgfor
Check out the sspir package and
http://www.jstatsoft.org/index.php?vol=16
On 4/27/06, Pablo Almaraz [EMAIL PROTECTED] wrote:
Hi all,
Does anyone have an example of an autoregressive (AR) time-series model
specified as a state space model in R? That is, I want to go beyond the
locally linear
You probably need to contact the developer of pamr but
short of investigating it, a workaround might be to put a copy
of myd2 into the global environment since it likely will at
least look there, e.g. add this line to:
assign(myd2, myd2, .GlobalEnv)
domat.
On 4/27/06, Tim Smith [EMAIL
Try this:
tapply(x, cut(x, 12), sd)
On 4/28/06, sumanta basak [EMAIL PROTECTED] wrote:
Hi R-Experts,
I have a vector of length 72. I want to break it into 12 parts and want to
take standerd deviation of each group. Please help me in this regard.
Thanks,
Sumanta.
Good point.
Following Andy's comment sd(matrix(sort(x), nc=12))
could also be used if you want them broken up
by 6 smallest, next 6 smallest, etc. although
there might be differences in the case of ties.
Using tapply here are a number of ways of breaking
it up (the first three give the same
Try this:
lapply(names(tslist), function(nm) acf(tslist[[nm]], main = nm))
On 4/28/06, Ulf Mehlig [EMAIL PROTECTED] wrote:
Hello r-help,
I have a couple of time-series of different length and I would like to
produce a simple overview plot showing the autocorrelation functions of
the
, R.matlab, R.oo. It will truncate new directories as
clim, haplo, hier, pls, R.
On 4/26/06, Thomas Harte [EMAIL PROTECTED] wrote:
hi all,
is there a new mechanism in R 2.3.0 for copying libraries from, say, R
2.2.1 to R 2.3.0? i ask because gabor grothendieck comments in his
copydir.bat
Here are three possibilities:
1. aggregate on the columns that you want to sum and aggregate on
the columns that you want to average and then merge them:
By - A[, 2, drop = FALSE]
merge(aggregate(A[, 3, drop = FALSE], By, sum),
aggregate(A[, 4, drop = FALSE], By, mean))
2. use by:
f -
Try this where DF is your data frame:
DF[rep(seq(nrow(DF)), each = 3), ]
On 4/29/06, Jiantao Shi [EMAIL PROTECTED] wrote:
Hi,
i have a dataframe like this,
SourceTreatDrugReplicate
control0A1
control10A2
control30A3
10A1
You might want to check the double check the list archives
https://www.stat.math.ethz.ch/pipermail/r-help/
to see if your posts got through or not just in case its just
some problem in displaying your own posts.
On 4/29/06, Farrel Buchinsky [EMAIL PROTECTED] wrote:
Gabor Grothendieck
If your package is called mypkg you could create a mypkg-package.Rd
file. e.g.
library(dyn)
library(help = dyn) # note that mypkg-package is listed
package?dyn
?dyn-package # same
and you could add one or more vignettes, e.g.
library(zoo)
library(help = zoo) # note that the 2 vignettes are
The times class in chron will give hours and minutes. e.g.
library(times)
plot(times(0:23/23), 0:23)
and you could modify chron:::axis.times for the others.
On 4/30/06, Duncan Murdoch [EMAIL PROTECTED] wrote:
I have a variable containing a measurement of a duration in seconds
which I would
On 4/30/06, Guojun Zhu [EMAIL PROTECTED] wrote:
Hi. I am starting to learn R for a course project. I
am relative OK c++ programer. I found the R is very
different. I have read the an introduction to R. I
have to say it is not very newbie friendly. It does
not explain many things
That should have been:
library(chron)
plot(times(0:23/24), 0:23)
On 4/30/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
The times class in chron will give hours and minutes. e.g.
library(times)
plot(times(0:23/23), 0:23)
and you could modify chron:::axis.times for the others.
On 4/30/06
Look at
?filter
?embed
rollmean in the zoo package
running in the gtools package
runmean in the caTools package
The last one is probably the fastest.
On 4/30/06, Guojun Zhu [EMAIL PROTECTED] wrote:
I have a big data.frame with abou 20 column and 60,000
rows for analyze. Let us say I had a
My understanding is that you want to replace each rate with its average
over the associated bin and then plot age against that. In that
case try this:
DF # test data
age rate bin
1 0.002 10.0 A
2 0.045 0.1 B
3 0.130 15.0 A
4 0.150 34.0 D
with(DF, plot(ave(rate, bin), age))
Or perhaps a bit simpler:
plot(age ~ ave(clock, bin), DF)
On 4/30/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
My understanding is that you want to replace each rate with its average
over the associated bin and then plot age against that. In that
case try this:
DF # test data
age
Try this:
# test data
fo - y ~ female + I(age^2) + female:black + (age + education) * female
# create a list of form list(y = as.name(z.y), ...) for use with substitute
L - sapply(all.vars(fo), function(nm) as.name(paste(z, nm, sep = .)))
do.call(substitute, list(fo, L))
On 5/1/06, Andrew
Using the built in data frame iris, which has 5 columns, regress
Sepal.Length against all other variables except the last one:
lm(Sepal.Length ~., iris[1:4])
On 5/1/06, Guojun Zhu [EMAIL PROTECTED] wrote:
I need to run a regression with 14 normal variables
and 20 dummy variables. All the data
Some functions that may be of help:
?aggregate.ts
?cbind
?merge
and in the zoo package
?as.yearmon
?as.yearqtr
?aggregate.zoo
?merge.zoo
On 5/1/06, Guojun Zhu [EMAIL PROTECTED] wrote:
I have two data sets about lots of companies' stock
and fiscal data. One is monthly data with about
You could have a list of args for each one like this:
# test data
x - list(data = c(1,3,5), points = c(2,4))
myfunc - function(x, plot.args = NULL, points.args = NULL) {
do.call(plot, c(list(x$data), plot.args))
do.call(points, c(list(x$points), points.args))
}
myfunc(x,
Grothendieck [EMAIL PROTECTED]
wrote:
Or perhaps a bit simpler:
plot(age ~ ave(clock, bin), DF)
On 4/30/06, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
My understanding is that you want to replace each
rate with its average
over the associated bin and then plot age against
Here are a few alternatives:
replace(a, is.na(a), 0) + b
ifelse(is.na(a), 0, a) + b
mapply(sum, a, b, MoreArgs = list(na.rm = TRUE))
On 5/1/06, John Kane [EMAIL PROTECTED] wrote:
This is a simple question but I cannot seem to find
the answer.
I have two vectors but with missing data and I
On 5/1/06, Murray Jorgensen [EMAIL PROTECTED] wrote:
The file TENSILE.DAT from the Hand et al Handbook of Small Data Sets
looks like this:
0.023 0.032 0.054 0.069 0.081 0.094
0.105 0.127 0.148 0.169 0.188 0.216
0.255 0.277 0.311 0.361 0.376 0.395
0.432 0.463
Using runmean from caTools the first one below does
it in under 1 second but will not handle NAs. The
second one takes under 15 seconds and handles
them by replacing them with linear approximations.
Note that k must be odd.
# 1
library(caTools)
set.seed(1)
system.time({
y -
, Guojun Zhu [EMAIL PROTECTED] wrote:
Sorry to bother you guys again. This is great. But
this is for 61 number and the second case will change
60 to 61. run* only accept odd number window. How
to get around it with 60? Any suggestion? Thanks.
--- Gabor Grothendieck [EMAIL PROTECTED]
wrote
Try this:
e - expression(glm(y ~ age))
eval(e)
or this:
chr - glm(y ~ age)
eval(parse(text = chr))
On 5/2/06, Andrew Gelman [EMAIL PROTECTED] wrote:
Hi, all. I'm trying to automate some regression operations in R but am
confused about how to evaluate expressoins that are expressed as
On 5/2/06, Berton Gunter [EMAIL PROTECTED] wrote:
Here are a few alternatives:
replace(a, is.na(a), 0) + b
ifelse(is.na(a), 0, a) + b
mapply(sum, a, b, MoreArgs = list(na.rm = TRUE))
Well, Gabor, if you want to get fancy...
evalq({a[is.na(a)]-0;a})+b
Note that the evalq can
and mapply solutions do not change a.
On 5/2/06, Berton Gunter [EMAIL PROTECTED] wrote:
Below.
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Tuesday, May 02, 2006 10:42 AM
To: Berton Gunter
Cc: John Kane; R R-help
Subject: Re: [R] Adding elements
Try this:
# regression of Sepal.Length on cols 2 and 4 using first 100 rows
iris.lm - lm(Sepal.Length ~ ., iris[,c(1,2,4)], subset = 1:100)
# now do it with next 50 rows
predict(update(iris.lm, subset = 101:150))
# double check - this gives same result as last line
predict(lm(Sepal.Length ~ .,
Try this (where you can replace textConnection(L) with name
of file containing data):
L - 01/02/1990 0.531 0.479
01/03/1990 0.510 0.522
01/06/1990 0.602 0.604
library(zoo)
z - read.zoo(textConnection(L), format = %m/%d/%Y)
plot(z, plot.type = single)
This will give more info on zoo:
, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
# regression of Sepal.Length on cols 2 and 4 using first 100 rows
iris.lm - lm(Sepal.Length ~ ., iris[,c(1,2,4)], subset = 1:100)
# now do it with next 50 rows
predict(update(iris.lm, subset = 101:150))
# double check - this gives same
:
It does not work though. How is the lag work? How
does the lag work? I read the help and do not quite
understand. Here is a test
y
[1] 1 2 3 4 5 6 7 8 9 10
coredata(lag(y,-1))
[1] 1 2 3 4 5 6 7 8 9 10
attr(,tsp)
[1] 2 11 1
--- Gabor Grothendieck [EMAIL PROTECTED
Column names in iris that contain the string Sepal:
grep(Sepal, names(iris), value = TRUE)
On 5/3/06, Farrel Buchinsky [EMAIL PROTECTED] wrote:
How does one create a vector whose contents is the list of variables in a
dataframe pertaining to a particular pattern?
This is so simple but I
On 5/3/06, Uwe Ligges [EMAIL PROTECTED] wrote:
Johannes Graumann wrote:
On Tuesday 02 May 2006 23:33, Uwe Ligges wrote:
Then please read ?plotmath and use it:
labels = expression( = 0.66, == 0.33, = -0.33, = -0.66)
Error in lab != : comparison is not allowed for expressions
Try this:
do.call(sprintf, c(%9.2f\t%d\t%d\t%8.3f, as.list(v[iv])))
On 5/3/06, Paul Roebuck [EMAIL PROTECTED] wrote:
How would one go about getting sprintf to use the
values of a vector without having to specify each
argument individually?
v - c(1, 2, -1.197114, 0.1596687)
iv - c(3, 1,
You can use as.is = TRUE arg to read.xls to get character
data rather than factors.
On 5/3/06, Knut Krueger [EMAIL PROTECTED] wrote:
I have got factor from read.xls:
is(factor_value)
[1] factor oldClass
[288] -0.32 0.180.180.18-0.32 0.180.68
[295] 0.680.18
As a workaround you could use pie3D in the plotrix package
with height=0 and theta=pi, e.g.
library(plotrix)
pie3D(1:3, height = 0, theta = pi,
labels = expression( = 1, == 2, = 3))
On 5/3/06, Johannes Graumann [EMAIL PROTECTED] wrote:
On Wednesday 03 May 2006 09:05, Uwe Ligges wrote:
Suppose we want to sum C over levels of A and that B is constant
within levels of A. Then:
DF - data.frame(A = gl(2,2), B = gl(2,2), C = 1:4) # test data
do.call(rbind, by(DF, DF$A, function(x) replace(x[1,], C, sum(x$C
On 5/3/06, Guenther, Cameron [EMAIL PROTECTED] wrote:
Hello,
I
See:
https://www.stat.math.ethz.ch/pipermail/r-devel/2006-May/037542.html
On 5/4/06, Dieter Menne [EMAIL PROTECTED] wrote:
Dear R-Core,
after switching to 2.3.0, all my trusted do.call constructs that worked in
2.2 and earlier fail. I noted that changes were introduced to do.call, but I
Assuming DF is your data frame, try this:
aggregate(DF[,0-1:2], DF[,1:2], sum)
On 5/4/06, YIHSU CHEN [EMAIL PROTECTED] wrote:
Dear R users:
I have a data frame as follows, where e1-e3 are indicator variables with
value equal 0 or 1.
St County e1 e2 e3
1 2 1 0 0
1
901 - 1000 of 3333 matches
Mail list logo
