On Tue, Sep 15, 2009 at 9:48 AM, ivo welch ivo...@gmail.com wrote:
dear R wizards: here is the strange question for the day. It seems to me
that nrow() is very slow. Let me explain what I mean:
ds= data.frame( NA, x=rnorm(1) ) ## a sample data set
system.time( { for (i in 1:1)
PS: Any speed suggestions are appreciated. This is experimenting time for
me.
You might want to check out the plyr package - it incorporates
everything I know about making these sorts of operations fast. The
next version will do even better.
Hadley
--
http://had.co.nz/
Here is a simpler mockup which shows the issue:
x = data.frame(rbind(c(1,2,3),c(1,2,3)))
xnames = c(a, b, c)
names(x) = xnames
for(i in 1:length(x))
{
# Create a varying string expression
expr = paste(y = x$, xnames[i], [1], sep=)
# evaluate expression
eval(parse(text=print(expr)))
Well, the title says all for this one.
Not really. What do you mean by crash? What do you mean by used together?
Hadley
--
http://had.co.nz/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
On Mon, Sep 21, 2009 at 3:04 PM, A Singh aditi.si...@bristol.ac.uk wrote:
Dear R users,
I am trying to read in a file with 105 columns, and when trying to attach
it, get an error as follows:
vc1-read.table(P:\\R\\Everything-I.txt, header=T, sep= , dec=.,
na.strings=NA, strip.white=T)
From: hadley wickham h.wick...@gmail.com
Don't use attach?
Obvously good advice but why?
Philosophically, it's better to be explicit than implicit, and the
extremely non-local effects of attach can make debugging difficult.
Hadley
--
http://had.co.nz
wtf - factor(x, levels(c(levels(wtf), NA), exclude=NULL)
xtabs (~ wtf, exclude=NULL, na.action=na.pass)
Also see addNA.
Hadley
--
http://had.co.nz/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
On Fri, Sep 25, 2009 at 7:55 AM, baptiste auguie
baptiste.aug...@googlemail.com wrote:
Thank you Paul, I was convinced I tried this option but I obviously didn't!
In ?packGrob, the user is warned that packing grobs can be slow. In
order to quantify this, I made the following comparison of 3
or with l_ply (plyr package)
l_ply(data.frame(a=1:3, b=2:4), function(x) print(deparse(substitute(x
The best way to do this is to supply both the object you want to
iterate over, and its names. Unfortunately it's slightly difficult to
create a data structure of the correct form to do this
many thanks for your answer and for the enormous work you put into plyr, a
really powerful package.
For now, I will solve my problem with a variable label attribute, I usually
attach to columns in data frames. I asked the list, because I thought, I am
overlooking something trivial, since
On Mon, Sep 28, 2009 at 9:01 AM, Gary Lewis gary.m.le...@gmail.com wrote:
Hi - Many organizations now make their data available as XML via a
REST web service architecture. Is there any R package or facility to
access this type of data directly (eg, to make the HTTP GET request
and have the
Hi David,
That's the behaviour I'd expect - the plot is 5 x 13000. What were
you expecting?
Hadley
On Fri, Feb 26, 2010 at 8:06 AM, David Hajage dhajag...@gmail.com wrote:
Hello,
I think there is a bug in coord_equal when x s a factor :
ggplot(diamonds, aes(clarity, fill=cut)) +
One of the things about R which many (and that certainly includes
me) have to find out the hard way is that you have to *learn*
what to expect! You can't just import it from prior experience in
other contexts. So, by the time you have learned that a matrix
is such that all its elements must
Suppose X is a dataframe or a matrix. What would you expect to get from
X[1]? What about as.vector(X), or as.numeric(X)?
The point is that a dataframe is a list, and a matrix isn't. If users don't
understand that, then they'll be confused somewhere. Making matrices more
list-like in one
The basic reason because apply works with matrices - it first turns
the input into a matrix, processes each column and then returns a
matrix. See colwise in the plyr package for a function that works
column wise on a data frame, returning a data frame.
Hadley
On Sun, Mar 7, 2010 at 11:07 AM,
Hi Mark,
If efficiency is a concern you might want to read Computing Thousands
of Test Statistics Simultaneously in R by Holger Schwender and Tina
Müller, http://stat-computing.org/newsletter/issues/scgn-18-1.pdf.
If you just want to do it, see the examples in
please
On Sun, Mar 7, 2010 at 2:08 PM, hadley wickham h.wick...@gmail.com wrote:
Hi Mark,
If efficiency is a concern you might want to read Computing Thousands
of Test Statistics Simultaneously in R by Holger Schwender and Tina
Müller, http://stat-computing.org/newsletter/issues/scgn-18-1
For Q2 you can use opts(legend.position = c(0.9, 0.9)).
For Q3, you can also use scale_y_sqrt().
Hadley
On Wed, Mar 10, 2010 at 2:05 PM, Tim Howard tghow...@gw.dec.state.ny.us wrote:
To answer two of my own questions to get them into the archives (I am slowly
getting the hang of ggplot):
By Q2 I was trying to refer to the Y-axis labels. For the polar plot, the
Y-axis labels reside left of the panel. I was looking for a way to get the
Y-axis labels to radiate out from the center so it would be clear which line
each label refers to. I still can't find any reference to moving
Run that function hourly with plyr
output.hourly - dlply(df.i1,tshour,cor.dat)
Why not
output.hourly - ddply(df.i1,tshour,cor.dat)
? Generally you want to work with data frames in R, if at all possible.
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics
to the
square-roots of the number of observations in the groups.
I find this option often very useful.
Thanks for any insight into how to achieve this with geom_boxplot.
Joh
On Wednesday 10 March 2010 16:12:49 hadley wickham wrote:
What is varwidth?
Hadley
On Wed, Mar 10, 2010 at 1:55 PM
ggplot2
ggplot2 is a plotting system for R, based on the grammar of graphics,
which tries to take the good parts of base and lattice graphics and
avoid bad parts. It takes care of many of the fiddly details
that make plotting a hassle
Based on a discussion found on the R mailing list but dating back to 2008, I
have compared the log-likelihoods of the glm model and of the glmer model as
follows:
lrt - function (obj1, obj2){
L0 - logLik(obj1)
L1 - logLik(obj2)
L01 - as.vector(- 2 * (L0 - L1))
df - attr(L1, df) -
Does anyone have any experience retrieving latitutde and longitude for
an address from the Google Maps API?
This thread from r-sig-geo may be of interest:
https://stat.ethz.ch/pipermail/r-sig-geo/2010-March/thread.html#7788
In particularly, note that what you are doing is against the google
Hi all,
Does any one know of any encryption/decryption algorithms in R? I'm
not looking for anything robust - I want some way of printing output
to the screen that the user can't read immediately, but can decrypt a
little later. The main thing I don't want to the user to see is a
number, so
On Fri, Mar 19, 2010 at 12:35 PM, Barry Rowlingson
b.rowling...@lancaster.ac.uk wrote:
On Fri, Mar 19, 2010 at 5:10 PM, Barry Rowlingson
b.rowling...@lancaster.ac.uk wrote:
On Fri, Mar 19, 2010 at 5:00 PM, Hadley Wickham had...@rice.edu wrote:
Hi all,
Does any one know of any encryption
On Wed, Mar 24, 2010 at 8:52 AM, mgierdal mgier...@gmail.com wrote:
I have a dataFrame variable:
L1 L2 L3 ... v1 v2 ...
1
2
3
4
...
I want to process subsets of it as defined by combinations of L1-L2-L3. I do
it successfully using nested loops:
for (i in valuesOfL1 {
for
See here:
http://learnr.wordpress.com/2010/03/23/ggplot2-changing-the-default-order-of-legend-labels-and-stacking-of-data/
Hadley
On Wed, Mar 24, 2010 at 2:40 PM, Ryan Garner
ryan.steven.gar...@gmail.com wrote:
How do I reverse the order of the legend in a bar plot to match order of the
# Set up the ratio variables
system.time({
temp - cbind(data, do.call(cbind, lapply(names(data)[3:4], function(.x)
{
unlist(by(data, data$group, function(.y) .y[,.x] /
max(.y[,.x])))
})))
colnames(temp)[5:6] - paste(colnames(data)[3:4], 'ind.to.max', sep =
exp1^(a[case] * l * 10)
would be better written out of the loop as
b - exp1^(a * l * 10)
And even better as
b - exp(a * l * 10)
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
On Wed, Mar 31, 2010 at 8:20 AM, Tony B tony.bre...@googlemail.com wrote:
Dear all,
Lets say I have the following:
x - c(Eve: Going to try something new today..., Adam: Hey @Eve, how are
you finding R? #rstats, Eve: @Adam, It's awesome, so much better at
statistics that #Excel ever was!
On Thu, Apr 1, 2010 at 3:05 AM, Peter Dalgaard pda...@gmail.com wrote:
Jeff Brown wrote:
Sorry for spamming. I swear I had worked on that problem a long time before
posting.
But I just figured it out: I have to change the values, which are
represented as integers, not strings. So the
Incidentally, there is nothing new or radical in this; indeed, John Tukey,
Leo Breiman, George Box, and others wrote eloquently about this decades ago.
And Breiman's random forest modeling procedure explicitly abandoned efforts
to build simply interpretable models (from which one might infer
Also I have seen 5,000 page listings in SAS.
Is this a pro or a con?
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
__
R-help@r-project.org mailing list
Because of the way you've constructed the plot with qplot, you need to use:
myPlot + geom_point(
data=medians,
aes(x=med,shape=cut, y=0),
size=2.5,
)
Hadley
On Wed, Apr 7, 2010 at 5:11 AM, Johannes Graumann
johannes_graum...@web.de wrote:
Hi,
Please consider the example below. How can I
df - data.frame(a = a\b)
write.table(df, test.csv, sep = ,, row = F)
Is there any to load test.csv into R correctly? I've tried the following:
read.table(test.csv, sep = ,)
[1] V1
0 rows (or 0-length row.names)
Warning message:
In read.table(test.csv, sep = ,) :
incomplete final line found
url - http://dl.dropbox.com/u/41902/22240.csv;
read.csv(url)[, 1]
[1] oppose NAoppose support
read.csv(url, header = F)[, 1]
[1] url
[2] http://maplight.org/us-congress/bill/109-hr-5825/387248;
[3] http://maplight.org/us-congress/bill/110-hr-3546/378743;
[4]
df - data.frame(a = a\b, v = 4, z = this is Z)
write.csv(df, test.csv, row.names = FALSE, quote = FALSE)
read.csv(test.csv, quote = )
Unfortunately my real example is more like:
df - data.frame(a = a\b, v = 4, z = this is: A, B, C)
so quote = F won't work.
Can write.table and read.table
Can write.table and read.table really be so asymmetric?
write() is a wrapper for cat() and read() is a wrapper for scan() so the
question should really be can cat() and scan() be so asymmetric. Looking at
their help pages, I would say that at least some degree of asymmetry is
plausible.
On Thu, Apr 8, 2010 at 8:20 AM, jim holtman jholt...@gmail.com wrote:
You were using read.csv and not read.table. The following seems to work
with using a separator that will probably not appear in the text:
df - data.frame(a = a\b, v = 4, z = this is: A, B, C)
write.table(df, test.csv,
Remove the comma and count.fields gives 11 for all rows.
From your other post(s) on escaped quotes, I assume that
this won't solve your problem with the existing files. (:
Right - but assuming I'm not crazy, that should cause an error in
read.csv, right? It shouldn't just parse the file
Other then rebuilding the plots, is there any way either (1) to combine
existing ggplot2 plots or (2) to extract a layer from an existing plot
so that it can be added to another?
Not really, although you can always pull apart the plot components.
Can you give an example of what you are trying
bar.err (agricolae)
plotCI (gplots)
xYplot (Hmisc)
error.bars (psych)
dispersion (plotrix)
plotCI (plotrix)
Not to mention: http://biostat.mc.vanderbilt.edu/wiki/Main/DynamitePlots
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
On Thu, Jun 30, 2011 at 7:31 AM, Ashim Kapoor ashimkap...@gmail.com wrote:
Dear R helpers,
I have molten data which is : -
t3
Year variable value
1 2005 ICICI.Bank 27488370
2 2006 ICICI.Bank 43166850
3 2007 ICICI.Bank 59515300
4 2008
# stringr
Strings are not glamorous, high-profile components of R, but they do
play a big role in many data cleaning and preparations tasks. R
provides a solid set of string operations, but because they have grown
organically over time, they can be inconsistent and a little hard to
learn.
On Thu, Jul 7, 2011 at 5:24 PM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
Here's another approach using the plyr package:
library(plyr)
df - data.frame(gp = factor(rep(1:3, each = 4)), x = rnorm(12), y =
rnorm(12))
mylst - split(df, df$gp)
mycoefs - ldply(mylst, function(d) coef(lm(y ~
Maybe something like this?
withNames - function(dframe, lineNames, plotName, colors){
one_day - subset(dframe, data == '1941-06-16')
one_day$lineNames - lineNames
ggplot(dframe, aes(date, value, group = factor, color = factor)) +
geom_line(size = 1) +
facet_grid(Facet~., scales =
You should only have one scale_ call for each scale type. Here, you have
three scale_colour_ calls, the first selecting a grey scale, the second
defining a single break with its label (and thus implicitly subsetting on
that single break value), and a second which defines a different
Also, if we make days a list, the class attributes are kept when looping
over the list, ie.
days- list( as.Date( c(2000-01-01, 2000-01-02) ) )
Do you realise that that's a list with length one?
I suspect you want
days - as.list( as.Date( c(2000-01-01, 2000-01-02) ) )
for (day in days) {
Thank you, this is very helpful. One final question regarding this method:
suppose a function prints multiple plots, i.e. multiple pages to a PDF. Is
it possible to record all of these plots at once? The code below only
records the final plot. I would like to record all of them, without
This is called a squarified pie chart or a waffle chart (if you want
to keep the food metaphor going):
http://eagereyes.org/communication/Engaging-readers-with-square-pie-waffle-charts.html
Hadley
On Thu, Jul 21, 2011 at 10:29 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Use plyr::rbind.fill? That does match up columns by name.
Hadley
On Thu, Jul 28, 2011 at 5:23 PM, Stavros Macrakis macra...@alum.mit.edu wrote:
I have a file of data where each line is a series of name-value pairs, but
where the names are not necessarily the same from line to line, e.g.
struc.test - breakpoints(y~x1+x2+x3+x3+x4, data=D)
*I get an error message:*
Erreur dans chol2inv(qr.R(fm$qr)) :
l'?l?ment (5, 5) est nul, donc l'inverse ne peut ?tre calcul?
(sorry for the french version, I don't know how to get the message
english translation in R).
My first
And I think Uwe is missing from that list!
Hadley
On Fri, Jul 29, 2011 at 3:34 PM, Paul Menzel
paulepan...@users.sourceforge.net wrote:
Dear R webmasters,
my browser defaults to the charset UTF-8 and since [1] seems to be
encoded in ISO-8859-1 the umlauts are not displayed correctly. It
# plyr
plyr is a set of tools for a common set of problems: you need to
__split__ up a big data structure into homogeneous pieces, __apply__ a
function to each piece and then __combine__ all the results back
together. For example, you might want to:
* fit the same model each patient subsets of
if give non-data-frames.
-s
On Thu, Jul 28, 2011 at 19:30, Hadley Wickham had...@rice.edu wrote:
Use plyr::rbind.fill? That does match up columns by name.
Hadley
On Thu, Jul 28, 2011 at 5:23 PM, Stavros Macrakis macra...@alum.mit.edu
wrote:
I have a file of data where each line
You will get the warning that last last column is not going right but
otherwise this returns what you asked for:
sapply(1:length(mydf), function(i) mydf[[i]]* as.numeric(names(mydf)[i]) )
This suits my purposes well with a couple slight modifications:
## I made this into a data.frame so I
The following example does what you want using ddply:
library(plyr)
edfPerGroup = ddply(df, .(Group), summarise, edf = edf(Value), Value =
Value)
Or slightly more succinctly:
ddply(df, .(Group), mutate, edf = edf(Value))
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
No. First, please use path.expand(~) for this, and it does not
necessarily mean the home directory (and in principle it might not expand at
all). In practice I think it will always be *a* home directory, but on
Windows there may be more than one (and watch out for local/roaming profile
No, defaults are evaluated in the evaluation frame of the function. That's
why you can use local variables in them, e.g. the way rgamma uses 1/rate as
a default for scale.
Oops, yes, I was getting confused with promises - non-missing
arguments are promises evaluated in the parent frame.
But
On Mon, Mar 14, 2011 at 9:41 AM, Strategische Analyse CSD Hasselt
csd...@fedpolhasselt.be wrote:
Hello,
we want to plot a proportional symbol map with ggplot. Symbols' area should
have the same proportions as the scaled variable.
Hereby an example we found on
Its useful for being able to set defaults for arguments that do not
have defaults. That cannot break existing programs.
Until the next program decides do co change those defaults and either
can't or does and you end up with incompatible assumptions. It also
make the code with the added
On Thu, Mar 17, 2011 at 7:25 AM, Richard D. Morey r.d.mo...@rug.nl wrote:
I would like to assign an value to an element of a list contained in an
environment. The list will contain vectors and matrices. Here's a simple
example:
# create toy environment
testEnv = new.env(parent = emptyenv())
2) I don't want to fit data with linear model of zero intercept.
3) I dont know if I understand correctly. Im 100% sure the model for my data
should have zero intercept.
The only coordinate which Im 100% sure is correct. If I had measured quality
Y of a same sample X0 number of times I would
I don't doubt that R may be the most popular in terms of discussion group
traffic, but you should be aware that the traffic for SAS comprises two
separate lists that used to be mirrored, but are no longer linked
Usenet -- news://comp.soft-sys.sas (what you counted)
listserve -- SAS-L
On Thu, Mar 24, 2011 at 8:29 AM, Michael Bach pha...@gmail.com wrote:
Dear R users,
Given this data:
x - seq(1,100,1)
dx - as.POSIXct(x*900, origin=2007-06-01 00:00:00)
dfx - data.frame(dx)
Now to play around for example:
subset(dfx, dx as.POSIXct(2007-06-01 16:00:00))
Ok. Now for
On Thu, Mar 24, 2011 at 2:24 PM, Kevin Burnham kburn...@gmail.com wrote:
I have a data file with indicates pretest scores for a linguistics
experiment. The data are in long form so for each of 33 subjects there are
400 rows, one for each item on the test, and there is a column called
filelist = list.files(pattern = K*cd.txt) # the file names are K1cd.txt
.to K200cd.txt
It's very easy:
names(filelist) - basename(filelist)
data_list - ldply(filelist, read.table, header=T, comment=;, fill=T)
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
If all you need is loess, I suspect it would be cheaper to re-write it
in C# than to get a considered legal opinion on the matter.
Hadley
On Thu, Apr 7, 2011 at 2:45 AM, Stanislav Bek
stanislav.pavel@gmail.com wrote:
Hi,
is it possible to use some statistic computing by R in proprietary
Does anyone with specific windrose experience know how to adjust the
graphic such that the data and the percent intervals are evenly spaced?
Hopefully I am making sense here
How about giving us a reproducible example?
Code is better than mere description;
code + description is best.
# plyr
plyr is a set of tools for a common set of problems: you need to
__split__ up a big data structure into homogeneous pieces, __apply__ a
function to each piece and then __combine__ all the results back
together. For example, you might want to:
* fit the same model each patient subsets of
Then, can we have the ERROR message, please?
Otherwise the only explanation I can guess is that a mirror grabs the
contents of a repository exactly in the second the repository is updated and
that is unlikely, particularly if more than one mirror is involved.
Isn't one possible explanation
On Wed, Apr 13, 2011 at 5:18 AM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
Here's one approach:
strings - c(
A5.Brands.bought...Dulux,
A5.Brands.bought...Haymes,
A5.Brands.bought...Solver,
A5.Brands.bought...Taubmans.or.Bristol,
A5.Brands.bought...Wattyl,
A5.Brands.bought...Other)
Even so, this would depend on what your publisher/printer
requires in what you submit. It would be important to obtain
from them a full and exact specification of what they require
for colour printing in files submitted to them for printing.
No one else has mentioned this, but the publisher
Am I missing something obvious on how to draw multi-line plots in base graphics?
In ggplot2, I can do:
data(Oxboys, package = nlme)
library(ggplot2)
qplot(age, height, data = Oxboys, geom = line, group = Subject)
But in base graphics, the best I can come up with is this:
with(Oxboys,
On Wed, Apr 13, 2011 at 2:58 PM, Ben Bolker bbol...@gmail.com wrote:
Hadley Wickham hadley at rice.edu writes:
Am I missing something obvious on how to draw multi-line plots in
base graphics?
In ggplot2, I can do:
data(Oxboys, package = nlme)
library(ggplot2)
qplot(age, height, data
I was trying strsplit(string,\.\.\.) as per the suggestion in Venables
and Ripleys book to (use '\.' to match '.'), which is in the Regular
expressions section.
I noticed that in the suggestions sent to me people used:
strsplit(test,\\.\\.\\.)
Could anyone please explain why I should have
Yes, it's fixed and a new version of plyr has been pushed up to cran -
hopefully will be available for download soon. In the meantime, I
think you can fix it by running library(stats) before
library(ggplot2).
Hadley
On Sun, Apr 17, 2011 at 3:51 PM, Bryan Hanson han...@depauw.edu wrote:
Is
On Wed, Apr 20, 2011 at 6:36 PM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
Perhaps you're looking for subset()? I'm not sure I understand the
problem completely, but is
do.call(rbind, lapply(database, function(df) subset(df, Symbol == 'IBM')))
or
library(plyr)
ldply(lapply(database,
If you need plyr for other tasks you ought to use a different
class for your date data (or wait until plyr can deal with
POSIXlt objects).
How do you get POSIXlt objects into a data frame?
df - data.frame(x = as.POSIXlt(as.Date(c(2008-01-01
str(df)
'data.frame': 1 obs. of 1 variable:
On Wed, Apr 27, 2011 at 4:58 AM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
You could try something like
df - data.frame( expand.grid( Week = 1:52, Year = 2002:2011 ))
expand.grid already returns a data frame... You might want
KEEP.OUT.ATTRS = F though. Even it feels like you are yelling
This has the side effect of ignoring errors
and even hiding the error messages. If you
are concerned about multiple calls to on.exit()
in one function you could define a new function
like
withOptions - function(optionList, expr) {
oldOpts - options(optionList)
On Wed, Apr 27, 2011 at 3:55 PM, Justin Haynes jto...@gmail.com wrote:
I am trying to extract the shape and scale parameters of a wind speed
distribution for different sites. I can do this in a clunky way, but
I was hoping to find a way using data.table or plyr. However, when I
try I am met
Put together a list and we can see what might make sense. If we did
take this on it would be good to think about providing a reasonable
mechanism for addressing the small flaw in this function as it is
defined here.
In devtools, I have:
#' Evaluate code in specified locale.
with_locale -
Using paste(Site,Prof) when calling ave() is ugly, in that it
forces you to consider implementation details that you expect
ave() to take care of (how does paste convert various types
to strings?). It also courts errors since paste(A B, C)
and paste(A, B C) give the same result but
z - ddply(past, c(GEO_CNTRY_NAME,PROD_SEG_NAME),
function(x) summary(lm(VAL~fy,x))$r.squared)
But when ave is not exactly doing what I need. Above code runs under a
minute for my data set where as ave runs over 8 mins.
It's hard to know without a reproducible example, but I doubt that
ddply
# Method 2: Use the plyr package
library('plyr')
bdf - ldply(mlply(files, read.csv, header = TRUE), rbind)
Or just
bdf - ldply(files, read.csv, header = TRUE)
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
Because in coding, I often end up with big chunks looking like this:
((mydataframeName$myvariableName 2
!is.na(mydataframeName$myvariableName))
(mydataframeName$myotherVariableName == male
!is.na(mydataframeName$myotherVariableName)))
Which is much less
Hi all,
Does anyone have an R implementation of the queplot (K. M. Goldberg
and B. Iglewicz. Bivariate extensions of the boxplot. Technometrics,
34(3):pp. 307–320, 1992)? I'm struggling with the estimation of the
asymmetry parameters.
Hadley
--
Assistant Professor / Dobelman Family Junior
.
Best Wishes,
Boris
On Wed, Sep 21, 2011 at 3:11 PM, Hadley Wickham had...@rice.edu wrote:
Hi all,
Does anyone have an R implementation of the queplot (K. M. Goldberg
and B. Iglewicz. Bivariate extensions of the boxplot. Technometrics,
34(3):pp. 307–320, 1992)? I'm struggling
apply(mt, 1, function(x) x[!is.nan(x)] )
[[1]]
[1] 1 3
[[2]]
[1] 4 5 6
You need to be a little careful with apply:
mt2 - matrix(c(1,4,2,5,3,6),2,3)
apply(mt2, 1, function(x) x[!is.nan(x)] )
[,1] [,2]
[1,]14
[2,]25
[3,]36
Depending on the input you will get a
On Mon, Oct 3, 2011 at 12:24 PM, Thomas Adams thomas.ad...@noaa.gov wrote:
I'm interested in creating a graphic -like- this:
c - ggplot(mtcars, aes(qsec, wt))
c + geom_point() + stat_smooth(fill=blue, colour=darkblue, size=2, alpha
= 0.2)
but I need to show 2 sets of bands (with different
You need to set the labels...
Hadley
On Sat, Sep 24, 2011 at 3:49 AM, Casper Ti. Vector
caspervec...@gmail.com wrote:
Is there any way to use expression() in legend labels with ggplot2?
It seems that things like
scale_shape_manual(value = c(
x = expression(italic(x)),
y =
# Function to compute quantiles and return a data frame
g - function(d) {
qq - as.data.frame(as.list(quantile(d$y, c(.05, .25, .50, .75, .95
names(qq) - paste('Q', c(5, 25, 50, 75, 95), sep = '')
qq }
You could cut out the melt step by making this return a data frame:
g -
On Mon, Oct 10, 2011 at 4:14 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
I could be waay off base here, but my concern about presplitting the data is
that you will have your data, and a second copy of our data that is something
like a list where each element contains the portion of the
On Fri, Feb 17, 2012 at 7:51 PM, Benilton Carvalho
beniltoncarva...@gmail.com wrote:
Hi everyone,
For reasons beyond the scope of this message, I'd like to append a
NULL element to the end of a list.
tmp0 - list(a=1, b=NULL, c=3)
append(tmp0, c(d=4)) ## works as expected
append(tmp0,
# devtools
The aim of `devtools` is to make your life as a package developer
easier by providing R functions that simplify many common tasks.
Devtools is opinionated about how to do package development, and
requires that you use `roxygen2` for documentation and `testthat` for
testing. Future
One difficulty in getting the help pages to look beautiful is that the
original input is so inconsistent, and package authors (naturally) get upset
when CRAN starts rejecting packages because of errors that used to be
ignored. The current output is definitely a compromise aimed at making most
I believe it was 2008.
Hadley
On Mon, Mar 26, 2012 at 11:46 AM, Marina Doucerain
marinadoucer...@gmail.com wrote:
Hello,
I'm wondering what was the year (or year range) of collection for the data
included in the 'diamonds' dataset in ggplot2.
This information would be very helpful in
Why not pass around a reference class?
Hadley
On Thu, Apr 5, 2012 at 3:20 PM, John C Nash nas...@uottawa.ca wrote:
In trying to streamline various optimization functions, I would like to have
a scratch pad of working data that is shared across a number of functions.
These can be called from
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