Hi list members,
?else tells us
In particular, you should not have a newline between '}' and
'else' to avoid a syntax error in entering a 'if ... else'
construct at the keyboard or via 'source'.
but there's no syntax error when you break the line between } and
else in a function,
Hi,
I wonder whether there is any convenient function (or package) to
extract tables from a HTML page? e.g. from
http://www.google.com/finance/historical?q=SHE:002251
I know we can readLines('URL'), gsub('td...', '...', source), ...
and at last get the numbers; I'm writing to ask whether someone
Hi,
That's because the parser knows how to deal with that stuff. However,
when you type the same if/else at the command line, it will be parsed
line by line, and the evaluator will not wait for the else to evaluate
the if. Try to copy and paste your if/else to the command line.
Romain
Hi,
I'm trying to get standard errors for some of the variables in my data
frame. One of the questions on my survey is whether faculty coordinate
across curriculum to include Arts Education as subject matter. All the
responses are coded in zeros and ones obviously. For some of the other
Hello,
I would like to create a function that computes intraday returns of a financial
asset
on a calendar time basis, without making any loop. For instance, I want to get
price returns every 60 seconds.
The main problem is that the times series of prices is irregularly spaced in
time.
I have
Thanks a lot!
The way with zoo worked perfect.
Here is the code I've used finally:
data.input01 -read.csv(./1_15min.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
data.input02 -read.csv(./2_15min.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
Yihui Xie wrote:
I wonder whether there is any convenient function (or package) to
extract tables from a HTML page? e.g. from
http://www.google.com/finance/historical?q=SHE:002251
Try a search on R (I prefer markmail search)
http://r-project.markmail.org/search/?q=extract%20html
Hem wrote:
user_id website time
20google0930
21yahoo0935
20facebook1000
25facebook1015
61google0940
...
My problem is how to sort the data? So that, I can get information about
one
Benedikt Niesterok wrote:
Is there a function in r to find the best fitting model for a set of data?
I would like to know if my data are related exponentially,linearly or if
there is a logarithmic correlation between my x and y values.
There is no one-stop method to do this. I would
Dear R-helpers,
conducting different community ecology analyses my main aim is to find
groupings in the data and geographical borders between communities and
to prove them statistically.
So after conducting a global test (Mantel) I am running a NMS and
cluster analyses. These are followed by
Hi,
There is definitely a more elegant way of doing this which I don't know
about (without a for loop), but try this:
mat - matrix(NA, nrow = max(user_id), ncol = 2)
mat[,1] - 1:max(user_id) # 1st column of matrix is the user ID
for (i in 1:max(user_id)){
temp1 - subset(data, user_id = i)
temp2
Dear R users,
I have a question about the Package DierckxSpline. I have tried to find the
answer by myself but it didn't worked out.
I wondered if Dierckxspline can use different sets of y values in one time to
fit a line with knot. I have different sets of Y values representing the same
Hi,
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Joseph Magagnoli
Sent: Thursday, April 02, 2009 5:36 PM
To: r-help@r-project.org
Subject: [R] [R} seasonal differencing
Hi all,
I was wondering how to construct a
Thanks, Romain! So I think, for consistency, the following result
deparse(parse(text = '
+ f = function(x) {
+if (x) {
+1
+} # a new line here!
+else {
+2
+}
+ }
+ ')
+ )
[1] structure(expression(f = function(x) { if (x) {
[3] 1
Hi
I adopted an idea from Chemical Engineering where one guy used set of
functions to check which can fit data well and used Excel for it. So I
tried if I can do it in R.
It is very rough and needs data to be a two column data frame with x in
the first column together with list of formulas.
I have a data frame containing monthly observations of the 'density' of each US
state, recorded in variables named density.AL, density.AK, density.AZ,
and so on for all 50 states. The data frame (called d) also contains a variable
called Date which is encoded as a string in the format Jan-09,
Dear all,
I'm puzzled by the following example inspired by a recent question on
R-help,
cc - textConnection(user_id website time
20google0930
21yahoo0935
20facebook1000
25facebook1015
61google
you can always wrap the whole if/else statement into innocent braces or
parentheses, as in
y = {
if (x) 1
else 2 }
y = (
if (x) 1
else 2 )
it doesn't have to be a function, and there is no need for the
assignment either -- you just need to tell the parser
Hi,
Is this what you want?
d - data.frame(density.AL = seq(1, 10),
density.AK = seq(1, 10), # many others...
Date=letters[1:10]) # dummy example
library(reshape)
melt(subset(d, Date == b), id=Date)
BTW, I spotted a few awkward things in your code,
st - c(AL, AK)
Hi,
There seems to be something wrong with your function getinfo.
Try first to substitute your dbGetQuery by a print statement, so that you
can see what is going on:
getinfo=function(t){print(the SQL query)}
allinfo=sapply(c(1985:2007),getinfo)
If these results look good, you can change your
Dear all,
For my PhD study I'm looking for relevant courses/workshops (short term)
in ecological data anlysis with R in Europe. After 2 days searching I'm
convinced that google is probably not the right medium to find this
information. If anyone can help me I will be most grateful.
Best regards
Dear list,
I often need to convert several variables from numeric or integer into
factors (before plotting, for instance), as in the following example,
d - data.frame(
x = seq(1, 10),
y = seq(1, 10),
z = rnorm(10),
a = letters[1:10])
d2 -
within(d,
If the question is specific to getting stock data from
google finance then check out getSymbols.google
in the quantmod package.
Also note that there exists an r-sig-finance list for questions
pertaining to R and finance.
On Fri, Apr 3, 2009 at 2:18 AM, Yihui Xie xieyi...@gmail.com wrote:
Hi,
Hi,
I'm trying to extract a histogram over the results from a bootstrap. However
I keep receiving the error message Error in hist.default(boot.lrtest$ll,
breaks = scott) : 'x' must be numeric.
The bootstrap I'm running looks like:
boot.test - function(data, indeces, maxit=20) {
+ y1 -
baptiste auguie wrote:
Dear list,
I often need to convert several variables from numeric or integer into
factors (before plotting, for instance), as in the following example,
d - data.frame(
x = seq(1, 10),
y = seq(1, 10),
z = rnorm(10),
a = letters[1:10])
d2 -
Dear All,
Sorry to bother you again.
I have a model:
coxfita=coxph(Surv(rem.Remtime/365,rem.Rcens)~all.sex,data=nearma)
and I'm trying to do a plot of Schoenfeld residuals using the code:
plot(cox.zph(coxfita))
abline(h=0,lty=3)
The error message I get is:
Error in plot.window(...) : need
Excellent!
I felt it was fairly trivial but i can be quite dense on Friday
mornings.
I really like the generalisation.
Many thanks,
baptiste
On 3 Apr 2009, at 12:11, Wacek Kusnierczyk wrote:
baptiste auguie wrote:
Dear list,
I often need to convert several variables from numeric or
Dear all,
Another newbie just got attracted to this mailing list.
I am a biologist currently working my way through R, had sort play around with
python earlier this year.
I have some data exhibiting periodicity ** my data consists of peaks and
valleys, with peaks arising due to the presence
On Fri, Apr 3, 2009 at 4:43 AM, baptiste auguie ba...@exeter.ac.uk wrote:
Dear all,
I'm puzzled by the following example inspired by a recent question on
R-help,
cc - textConnection(user_id website time
20 google 0930
21 yahoo 0935
20
I am not sure that ties are the only reason. If I create a few ties in
the ovarian dataset that Therneau and Lumley provide, all I get are
some warnings:
ovarian[4:5, 1] - mean(ovarian[4:5, 1])
ovarian[6:8, 1] - mean(ovarian[6:8, 1])
fit - coxph( Surv(futime, fustat) ~ age + rx, ovarian)
Hi,
I have been using maxLik to do some MLE of Geometric Brownian Motion Process
and everything has been going fine, but know I have tried to do it with jumps.
I have create a vector of jumps and then added this into my log-likelihood
equation, know I am getting a message:
NA in the
Hi
for some data I working on I am merely plotting time against temperature for
a variable named filmclip. So for example, I have volunteers who watched
various film clips and have used infared camera to monitor the temperature
on their face at every second of the clip.
The variable names I
Dear R-ers,
I'm not sure if this is a missing feature, a support request, or stupidity on my part, but nevertheless, its a question. Is it possible to add titles to colorkey legends? As far as I can tell, there is a command to do it for normal key legends, but not for colorkeys.
eg it works
baptiste auguie-2 wrote:
ddply(d, .(website), transform, count = table(user_id)) # why two new
columns?
Try this to see why:
as.data.frame(table(d$user_id))
This works more like you expect:
ddply(d, .(website), transform, count = unclass(table(user_id)))
- Tom
--
View this
Hi. I am sure there is a better way in R to do this then using a loop but I
am new to it and not sure what to do. I think it might be something about
using a function as an argument but not sure.
I have a 1 x 2000 vector TS2 which has entries from the set {x: x is in Z
and 0x8} (where Z is the
On Fri, Apr 3, 2009 at 8:43 AM, baptiste auguie ba...@exeter.ac.uk wrote:
That makes sense, so I can do something like,
count - function(x){
as.integer(unclass(table(x)))
}
count(d$user_id)
ddply(d, .(user_id), transform, count = count(user_id))
user_id website time count
1
Melissa2k9 wrote:
Hi
for some data I working on I am merely plotting time against temperature for
a variable named filmclip. So for example, I have volunteers who watched
various film clips and have used infared camera to monitor the temperature
on their face at every second of the clip.
Hi!
I'm reading a tab-seperated CVS file with:
test1 - read.table(data.txt, header=TRUE)
It's in the following format:
Date_Time qK qL vL vP ...
0 30 22 110 88 ...
...
(BTW: It seems to me R shifts the column descriptions by one.)
Anyway, I would like to Fourier-transform one column.
That makes sense, so I can do something like,
count - function(x){
as.integer(unclass(table(x)))
}
count(d$user_id)
ddply(d, .(user_id), transform, count = count(user_id))
user_id website time count
1 20 google 930 2
2 20 facebook 1000 2
3 21yahoo
I will try to make this more precise. In the lme() function, the
correlation argument allows the user to specify a within-group correlation
structure, i.e. the structure of the Lambda matrix using the mixed model
notation in Pineiro and Bates. What I want to do is specify a distinct
structure for
I've really been on a roll this week; the formula for the lines that I
presented was completely wrong.
But I'm glad I learned about mapply. I used this:
mapply(abline,
(converge$kY + tan((90-converge$kT) * pi / 180)*(-converge$kX)),
tan((90-converge$kT) * pi / 180))
Tom!
On Thu, Apr 2,
of course!
Thanks,
baptiste
On 3 Apr 2009, at 14:48, hadley wickham wrote:
On Fri, Apr 3, 2009 at 8:43 AM, baptiste auguie ba...@exeter.ac.uk
wrote:
That makes sense, so I can do something like,
count - function(x){
as.integer(unclass(table(x)))
}
count(d$user_id)
ddply(d,
Thank you for your comments. I have about 200 out of 2000 tied data
points which makes the situation more complicated! I'll have at look
at the book section you referred to. With regards to making the ylim
finite, I'm not sure how I can go about that given that I don't
understand why it isn't
Hi,
The error message is clear in that the gradient cannot be evaluated at your
starting value for the parameters.
Is your likelihood a smooth function of parameters? If so, then provide a
different starting value. If it is not smooth, then you may have to use a
method that does not depend
Hi,
canvas is a new R package implementing a graphics device that emits
javascript code conforming to the HTML 5 CanvasRenderingContext2D
interface. Available on CRAN soon, but you can get it here immediately:
http://www.rforge.net/canvas
If you have access to a beta web browser like
Hi,
The error message is clear in that the gradient cannot be evaluated at your
starting value for the parameters.
Is your likelihood a smooth function of parameters? If so, then provide a
different starting value. If it is not smooth, then you may have to use a
method that does not depend
p.si...@gmx.net wrote:
Hi!
I'm reading a tab-seperated CVS file with:
test1 - read.table(data.txt, header=TRUE)
It's in the following format:
Date_Time qK qL vL vP ...
0 30 22 110 88 ...
...
(BTW: It seems to me R shifts the column descriptions by one.)
No, look into your data
Dear R users,
I am trying to do exact matching on a large dataset (500.000 obs), about equal
size of treatment and controll group, with replacement: As for the moment I use
the Match function of the Matching library. I match on 2 covariates and all
observations in the treatment group have at
onyourmark wrote:
Hi. I am sure there is a better way in R to do this then using a loop but I
am new to it and not sure what to do. I think it might be something about
using a function as an argument but not sure.
I have a 1 x 2000 vector TS2 which has entries from the set {x: x is in Z
and
Melissa2k9 wrote:
Hi,
I have written a for loop as such:
model-lm(Normalised~Frame,data=All,subset=((Subject==1)(Filmclip==Strand)))
summary(model)
###
#To extract just the Adjusted R squared
###
Try
r-sig-ecol...@r-project.org
bests
miltinho
brazil-toronto
On Fri, Apr 3, 2009 at 6:27 AM, Capelle, Jacob jacob.cape...@wur.nl wrote:
Dear all,
For my PhD study I'm looking for relevant courses/workshops (short term)
in ecological data anlysis with R in Europe. After 2 days searching
I have a question on the function 'embed'. I ran the example
x - 1:10
embed(x, dimension=3)
This gives the output:
[,1] [,2] [,3]
[1,]321
[2,]432
[3,]543
[4,]654
[5,]765
[6,]876
[7,]987
[8,] 1098
Dear Elisabeth:
Have you tried it? I have not, but I suspect the answer is no.
What problem are you trying to solve? You might get more useful
suggestions from this list if you provide commented, minimal,
self-contained, reproducible code describing your problem and what
you've
Its lets you perform rolling summaries using apply:
apply(embed(1:10, 3), 1, mean)
[1] 2 3 4 5 6 7 8 9
Note that 2 is the mean of 1:3, 3 is the mean of 2:4, ...,
9 is the mean of 8:10.
On Fri, Apr 3, 2009 at 11:04 AM, rkevinbur...@charter.net wrote:
I have a question on the function 'embed'.
Kevin,
The documentation is quite clear.
What embedding does is that it takes a scalar time series, x[t], and
embeds it in a higher-dimensional space of dimension, dimension. The
entries in the matrix you see are the indices of the time-series.
So, for example, if dimension = 2, you embed
At 05:27 AM 4/3/2009, Capelle, Jacob wrote:
Dear all,
For my PhD study I'm looking for relevant courses/workshops (short term)
in ecological data anlysis with R in Europe. After 2 days searching I'm
convinced that google is probably not the right medium to find this
information. If anyone can
Here is the gif that didn't come through earlier
http://www.nabble.com/file/p22870832/signal.gif signal.gif
--
View this message in context:
http://www.nabble.com/Curve-fitting%2CFDA-for-biological-data-tp22868069p22870832.html
Sent from the R help mailing list archive at Nabble.com.
rkevinbur...@charter.net wrote:
I have a question on the function 'embed'. I ran the example
x - 1:10
embed(x, dimension=3)
This gives the output:
[,1] [,2] [,3]
[1,]321
[2,]432
[3,]543
[4,]654
[5,]765
[6,]87
What is your end goal? If it is to try and account for the
variability of the timeseries you may want to look at ?spectrum
If it is to model the periodicity...
Stephen Sefick
On Fri, Apr 3, 2009 at 11:30 AM, trias t.gkikopou...@dundee.ac.uk wrote:
Here is the gif that didn't come through
Hi kevin: one use ( there are probably many others ) is for use inside
a vector autoregression model where the RHS is
lags of the independent variable. So, x_t is the 3rd column, x_t-1 is the
second, x_t-2 is the third etc.
On Apr 3, 2009, rkevinbur...@charter.net wrote:
I
What a very useful package! Thanks for pointing out its existence.
Sadly ?melt is basically useless, but I did find the following quasi-vignette
by the author of the reshape package to be quite useful.
http://www.jstatsoft.org/v21/i12/paper
Cheers.
-Original Message-
From:
Feng Jingyu wrote:
I used gls and it still does not provide me different estimates of
variance for each treatment group. Did I do anything wrong?
lm3-gls(GSI~treatment,data=z,weights=varIdent(form=~treatment),method=ML)
try
weights = varIdent(form~1|treatment)
See the example in
I have decided to use this SANN approach to my problem but to keep the run time
reasonable instead of 20,000 variables I will randomly sample this space to get
the number of variables under 100. But I want to do this a number of times. Is
there someone who could help me set up WINBUGS to repeat
I am starting to use R for almost any sort of calculation that I need.
I am a biologist that works in the states, and there is often a need
to convert from standard units to metric units. Is there a package in
R for this already? If not I believe that I am going to write some of
the most often
On Apr 3, 2009, at 7:10 AM, Anders Bjorn wrote:
Hi,
I'm trying to extract a histogram over the results from a bootstrap.
However
I keep receiving the error message Error in hist.default(boot.lrtest
$ll,
breaks = scott) : 'x' must be numeric.
The bootstrap I'm running looks like:
To my earlier question about updating a dataframe, and certainty that this
has been solved several times before, Dr. Winsemius suggests (Thanks!):
I am sure this is not the most elegant method, but it will work.
new - merge(nn,uu, by = c(a,b), all.x=T)
new$y - with( new,
stephen sefick wrote:
I am starting to use R for almost any sort of calculation that I need.
I am a biologist that works in the states, and there is often a need
to convert from standard units to metric units. Is there a package in
R for this already? If not I believe that I am going to write
I am trying to check for backslashes in data, then remove them when I
find them, but am having a difficult time figuring out the best way to
do it. I know the backslash is the escape character in R, and I
should be able to use 'gsub' to accomplish this, but I all I seem to
be getting are
Feng Jingyu wrote:
Thanks a lot. The problem is solved. It took me a while to understand the
output from the R. With little calculation, I am able to match results
from R to SAS.
To conserve you sanity, don't try it. They will be different.
Dieter
--
View this message in context:
I had a similar need for conversions in optics. I put together several
functions and data on r-forge, where it does not clutter CRAN but can
still be shared conveniently with others.
baptiste
On 3 Apr 2009, at 19:27, Duncan Murdoch wrote:
stephen sefick wrote:
I am starting to use R
I have a list of data.frames
str(bins)
List of 19217
$ 100026:'data.frame': 1 obs. of 6 variables:
..$ Sku : chr 100026
..$ Bin : chr T149C
..$ Count: int 108
..$ X: int 20
..$ Y: int 149
..$ Z: chr 3
$ 100030:'data.frame': 1 obs. of 6 variables:
...
As you can
I am attempting to write a function that is flexible enough to respond to the
user providing a formula (with a data= argument) or not (similar to
plot(x,y) versus plot(y~x,data=data)). I have found a method to work with
this in a simple case but am having trouble determining how to find a
I have tried to use rect.hclust() to draw a rectangle around a set of
leaves, but am running into trouble.
The rect.hclust() is drawing two rects instead of one, and of the
wrong size:
scoreClusterObj - hclust(scoreDistanceObj, method=clustMethod)
order -
Below works but it has two backslashes in the word. maybe someone can
explain why the 4 and 2 works but 2 1 doesn't ? thanks.
gsub(,,Hello\\World,perl=TRUE)
On Apr 3, 2009, Andrew Conway agc...@nyu.edu wrote:
I am trying to check for backslashes in data, then remove them
On 4/3/2009 2:33 PM, Andrew Conway wrote:
I am trying to check for backslashes in data, then remove them when I
find them, but am having a difficult time figuring out the best way to
do it. I know the backslash is the escape character in R, and I
should be able to use 'gsub' to accomplish
Feng Jingyu wrote:
Hi For my purpose, I need to match variance estimates for each group from
R and SAS. They do match now.
Consider yourself a lucky man!
Dieter
--
View this message in context:
http://www.nabble.com/Fit-unequal-variance-model-in-R-tp22829549p22874048.html
Sent from
Is there in some pacakage this test? i mean that input will be only the value
of periodogram, and the function will say that are significant.
or is there only the way to calcule Fishers statistics W =Vi/V1 + · · · +
Vm, for i =1 m and test it step by step ?
thanks
--
View this message in
Hello R-ers,
I'm trying to do a weighted principal components analysis. I couldn't find any
such option with princomp or prcomp. Does anyone know of a package or way to
do this?
More specifically, the observations I'm working with are averages from
populations of varying sizes. I thus need
On Fri, 2009-04-03 at 11:45 -0700, rkevinbur...@charter.net wrote:
I have a list of data.frames
str(bins)
List of 19217
$ 100026:'data.frame': 1 obs. of 6 variables:
..$ Sku : chr 100026
..$ Bin : chr T149C
..$ Count: int 108
..$ X: int 20
..$ Y: int 149
hi,
is there a function that calculates and prints out the partial etas of
every independent variable in a linear model?
thanks for any help!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
I do not think that the form [[1:3]] is legit.
ltest - list( a, b, c, d)
ltest[[1:3]]
Error in ltest[[1:3]] : recursive indexing failed at level 2
You might try with single brackets:
ltest[1:3]
[[1]]
[1] a
[[2]]
[1] b
[[3]]
[1] c
--
David Winsemius
On Apr 3, 2009, at 2:45 PM,
On Fri, Apr 3, 2009 at 1:45 PM, rkevinbur...@charter.net wrote:
I have a list of data.frames
str(bins)
List of 19217
$ 100026:'data.frame': 1 obs. of 6 variables:
..$ Sku : chr 100026
..$ Bin : chr T149C
..$ Count: int 108
..$ X : int 20
..$ Y : int 149
..$ Z : chr
Hi R-help:
I'm just an old guy and new to this list... But have been using R for
years now.
I want to make a map of counties in the US with shaded colors that
depend on the level of variable Y that I want to map.
I have the US county and state fips codes and the Y variable.
How do I do this?
David Winsemius wrote:
I do not think that the form [[1:3]] is legit.
sure it is.
ltest - list( a, b, c, d)
ltest[[1:3]]
Error in ltest[[1:3]] : recursive indexing failed at level 2
read the error message: *recursive* indexing failed. that's because
ltest[[1]] has only one element
Hello!
I need some help with the linear discriminant analysis in R.
I have some plant samples (divided into several groups) on which I
measured a few quantitative characteristics. Now, I need to infer some
classification rules usable for identifying new samples.
I have used the function lda from
Hello,
I need to multiply the number of counts in the y-axis of a lattice histogram
by a constant factor, such that the plot would represent a different type of
variable plotted in the y-axis, not counts. Can this be done?
Thank you,
Judith
__
Try:
library(lattice)
histogram( ~ height | voice.part,
data = singer, type = c,
scales = list(y =
list(at = seq(0, 20, 5),
labels = seq(0, 200, 50
HTH,
--sundar
On Fri, Apr 3, 2009 at 2:01 PM, Judith Flores jur...@yahoo.com wrote:
Hello,
Hi all,
I created a plot function which used par(mfcol=c(2,1)) so that I could
have two plots together using just one command.
For exampe:
plot.foo - function(data){
par(mfcol=c(2,1))
hist(data)
plot(data)
}
Later I wanted to show 4 of these foo objects in the same picture. So
I used
see the row.w argument of the function dudi.pca in the ade4 package.
Cheers.
Alan Cohen wrote:
Hello R-ers,
I'm trying to do a weighted principal components analysis. I couldn't find any
such option with princomp or prcomp. Does anyone know of a package or way to
do this?
More
I'm not sure what you are doing when you Normalize. Would you explain?
To see if the slope is significant, look at the model summary, in your
example:
summary(model)
Charles Annis, P.E.
charles.an...@statisticalengineering.com
phone: 561-352-9699
eFax: 614-455-3265
reauire(MASS) ; ?predict.lda should enlighten you. Glancing at VR4
might be a bit more illuminating...
HTH
Emmanuel Charpentier
Le vendredi 03 avril 2009 à 22:29 +0200, Pavel Kúr a écrit :
Hello!
I need some help with the linear discriminant analysis
Le vendredi 03 avril 2009 à 14:17 -0400, stephen sefick a écrit :
I am starting to use R for almost any sort of calculation that I need.
I am a biologist that works in the states, and there is often a need
to convert from standard units to metric units.
rant
US/Imperial units are *not*
On 4/04/2009, at 10:37 AM, Emmanuel Charpentier wrote:
Le vendredi 03 avril 2009 à 14:17 -0400, stephen sefick a écrit :
I am starting to use R for almost any sort of calculation that I
need.
I am a biologist that works in the states, and there is often a need
to convert from standard
On 03/04/2009 5:37 PM, Emmanuel Charpentier wrote:
Le vendredi 03 avril 2009 à 14:17 -0400, stephen sefick a écrit :
I am starting to use R for almost any sort of calculation that I need.
I am a biologist that works in the states, and there is often a need
to convert from standard units to
For science yes. For pleasure I'll still take a pint instead of 570ml!
Murray
- Original Message -
From: Rolf Turner r.tur...@auckland.ac.nz
To: Emmanuel Charpentier charp...@bacbuc.dyndns.org
Cc: r-h...@stat.math.ethz.ch
Sent: Friday, April 03, 2009 6:18 PM
Subject: Re: [R] [OT ?]
Hello R community,
I have cross-posted with r-sig-geo as this issue could fall under either
interest group I believe.
I just came accross the alphahull package and am very pleased I may not
need to use CGAL anymore for this purpose. However, I am having a
problem computing alpha shapes
G'day Murray,
On Fri, 3 Apr 2009 20:01:30 -0400
Murray Cooper myrm...@earthlink.net wrote:
For science yes. For pleasure I'll still take a pint instead of 570ml!
And you might experience a disappointment then if you order it in the
US where the pint is apparently 450ml; at least, I won several
97 matches
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