On Thu, Jul 2, 2009 at 8:15 AM, James Martinjust.strut...@gmail.com wrote:
Hadley, Sunil, and list,
This is not quite doing what I wanted it to do (as far as I can tell). I
perhaps did not explain it thoroughly. It seems to be sampling one value
for each day leaving ~200 observations. I need
On Thu, Jun 18, 2009 at 12:08 PM, Dirk Eddelbuettele...@debian.org wrote:
On 18 June 2009 at 09:36, Bert Gunter wrote:
| -- or Chapter 4 in S PROGRAMMING? (you'll need to determine if it's reader
| friendly)
+1
It helped me a lot too back in the day. But I am wondering if there are good
On Thu, Jul 2, 2009 at 3:34 PM, Sebastien
Bihorelsebastien.biho...@cognigencorp.com wrote:
Dear R-users,
I would like to know how expressions could be passed as arguments to do.call
functions. As illustrated in the short example below, concatenating lists
objects and an expression creates an
On Wed, Jul 1, 2009 at 2:10 PM, Sunil
Suchindransunilsuchind...@gmail.com wrote:
#Highlight the text below (without the header)
# read the data in from clipboard
df - do.call(data.frame, scan(clipboard, what=list(id=0,
date=,loctype=0 ,haptype=0)))
# split the data by date, sample 1
On Tue, Jun 30, 2009 at 2:12 PM, Barry
Rowlingsonb.rowling...@lancaster.ac.uk wrote:
On Tue, Jun 30, 2009 at 8:05 PM, Mark Knechtmarkkne...@gmail.com wrote:
You could wrap it in a function of your own making, right?
AddNewDev = function() {dev.new();AddNewDev=dev.cur()}
histPlot=AddNewDev()
graphing package. That code is giving me
the following error:
qplot(reorder(model,delta),delta,data=growthm.bic)
Error in UseMethod(reorder) : no applicable method for reorder
Cheers,
Chris
On 6/28/09 8:21 PM, hadley wickham wrote:
Hi Chris,
Try this:
qplot(reorder(model, delta), delta
Also consider ddply in the plyr package (although that's an over kill if
your only having two loops)
Maybe, but it sure is much simpler:
library(plyr)
ddply(data, c(industry,year), summarise, avg = mean(X1))
Hadley
--
http://had.co.nz/
__
Hi Chris,
Try this:
qplot(reorder(model, delta), delta, data = growthm.bic)
Hadley
On Sun, Jun 28, 2009 at 9:53 AM, Christopher
Desjardinscddesjard...@gmail.com wrote:
Hi,
I have 45 models that I have named: 1, 2, 3, ... , 45 and I am trying to
plot them in order of ascending BIC values. I
On Fri, Jun 26, 2009 at 10:27 PM, Osman Al-Radiosman.al.r...@gmail.com wrote:
Dear Richard and David,
Thanks for this reference. I looked into vcd and mosaic plot, it is a nice
plot for investigating associations between two or more variables. However,
I just need to plot the frequency of a
Have a look at ddply from the plyr package, http://had.co.nz/plyr.
It's made for exactly this type of operation.
Hadley
On Wed, Jun 24, 2009 at 10:34 PM, Stephan Lindnerlindn...@umich.edu wrote:
Dear all,
I have a code where I subset a data frame to match entries within
levels of an factor
You might also want to look at the plyr package,
http://had.co.nz/plyr. In particular, ddply + transform makes these
tasks very easy.
library(plyr)
ddply(mtcars, cyl, transform, pos = seq_along(cyl), mpg_avg = mean(mpg))
Hadley
On Wed, Jun 24, 2009 at 11:48 AM, David
plyr is a set of tools for a common set of problems: you need to break
down a big data structure into manageable pieces, operate on each
piece and then put all the pieces back together. For example, you
might want to:
* fit the same model to subsets of a data frame
* quickly calculate
Hi Mark,
Have a look at colwise (and numcolwise and catcolwise) in the plyr package.
Hadley
On Tue, Jun 23, 2009 at 4:23 PM, Mark Namtb...@gmail.com wrote:
Hi R-helpers,
I have a dataframe with 60columns and I would like to convert several
columns to factor, others to numeric, and yet
I have been using R for a while. Recently, I have begun converting my
package into S4 classes. I was previously using Rdoc for documentation.
Now, I am looking to use the best tool for S4 documentation. It seems that
the best choices for me are Roxygen and Sweave (I am fine with tex).
Are
Hi all,
Do you know of any good resources for learning how S3 works? I've
some how become familiar with it by reading many small pieces, but now
that I'm teaching it to students I'm wondering if there are any good
resources that describe it completely, especially in a reader-friendly
way. So
To: Hadley Wickham
Cc: r-help
Subject: Re: [R] Learning S3
There is a section on Object Orientation in MASS (I have 2nd ed).
On Thu, Jun 18, 2009 at 12:06 PM, Hadley Wickhamhad...@rice.edu wrote:
Hi all,
Do you know of any good resources for learning how S3 works? I've
some how become familiar
In revising my book Regression Modeling Strategies for a second edition, I
am seeking a dataset for exemplifying multiple regression using least
squares. Ideally the dataset would have 5-40 variables and 40-1
independent observations, and would generate significant interest for a wide
Hi all,
This is a little off-topic, but it is on the general topic of getting
data in R. I'm looking for a excel macro / vba script that will
export all spreadsheets in a directory (with one file per tab) into
csv. Does anyone have anything like this?
Thanks,
Hadley
--
http://had.co.nz/
Hi all,
Is there a cross-platform way to do this? On the mac, I cando this by
saving an eps file, and then using pbcopy. Is it possible on other
platforms?
Hadley
--
http://had.co.nz/
__
R-help@r-project.org mailing list
On Mon, Jun 8, 2009 at 10:29 AM, Herbert
Jägleherbert.jae...@uni-tuebingen.de wrote:
Hi,
i do have a dataframe representing data from a repeated experiment. PID is a
subject identifier, Time are timepoints in an experiment which was repeated
twice. For each subject and all three timepoints
On Mon, Jun 8, 2009 at 8:56 PM, Mao Jianfengjianfeng@gmail.com wrote:
Dear Ruser's
I ask for helps on how to substitute missing values (NAs) by mean of the
group it is belonging to.
my dummy dataframe is:
df
group traits
1 BSPy01-10 NA
2 BSPy01-10 7.3
3 BSPy01-10
On Sat, Jun 6, 2009 at 5:02 PM, Adam D. I. Kramera...@ilovebacon.org wrote:
Dear Colleagues,
Occasionally I deal with computer-generated (i.e., websurvey) data
files that haven't quite worked correctly. When I try to read the data into
R, I get something like this:
Error in
On Mon, Jun 1, 2009 at 2:18 PM, stephen sefick ssef...@gmail.com wrote:
library(ggplot2)
melt.updn - (structure(list(date = structure(c(11808, 11869, 11961, 11992,
12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996, 13057,
13149, 11808, 11869, 11961, 11992, 12084, 12173, 12265, 12418,
Is it really necessary to further advertise this company which already
spams R-help subscribers?
Hadley
On Thu, Jun 4, 2009 at 10:41 PM, Ajay ohri ohri2...@gmail.com wrote:
Dear All,
Slightly off -non technical topic ( but hey it is Friday)
Following last week's interview with REvolution
You might have an out-of-date version of the plyr package - try
install.packages(plyr)
Hadley
On Mon, Jun 1, 2009 at 10:20 AM, Matt Frost mwfr...@gmail.com wrote:
I'm trying to plot a time series in ggplot, but a date column in my
data frame is causing errors. Rather than provide my own data,
Hi Paul,
Unfortunately that's not something that's currently possible with
ggplot2, but I am thinking about how to make it possible.
Hadley
On Sat, May 16, 2009 at 7:48 AM, Paul Emberson em...@calidasoft.co.uk wrote:
Hi Stephen,
The problem is that the label on the graph doesn't get rendered
On Thu, May 14, 2009 at 2:14 PM, Garritt Page page2...@gmail.com wrote:
Hello,I am using xyplot to try and create a conditional plot. Below is a
toy example of the type of data I am working with
slevel - rep(rep(c(0.5,0.9), each=2, times=2), times=2)
tlevel - rep(rep(c(0.5,0.9), each=4),
On Thu, May 14, 2009 at 12:16 PM, Lori Simpson
lori.simp...@dc-energy.com wrote:
I am writing a custom function that uses an R-function from the
reshape package: cast. However, my question could be applicable to
any R function.
Normally one writes the arguments directly into a function,
On Thu, May 14, 2009 at 6:21 PM, Ping-Hsun Hsieh hsi...@ohsu.edu wrote:
Hi All,
I have a 1000x100 matrix.
The calculation I would like to do is actually very simple: for each row,
calculate the frequency of a given pattern. For example, a toy dataset is as
follows.
Col1 Col2
This does it more or less your way:
ds - split(df, df$Name)
ds - lapply(ds, function(x){x$Index - seq_along(x[,1]); x})
df2 - unsplit(ds, df$Name)
tapply(df2$X1, df2[,c(Name, Index)], function(x) x)
athough there may exist much easier ways ...
Here's one way with the plyr and reshape
On Sun, May 10, 2009 at 10:32 AM, Zeljko Vrba zv...@ifi.uio.no wrote:
Searching the mail archives I found that using legend.position as in
p.ring.3 + opts(legend.position=top)
is a known bug. I tried doing
p.ring.3 + opts(legend.position=c(0.8, 0.2))
which works, but the legend background
On Wed, May 6, 2009 at 8:12 PM, jim holtman jholt...@gmail.com wrote:
Ths should do it:
do.call(rbind, lapply(split(x, x$ID), tail, 1))
ID Type N
45900 45900 I 7
46550 46550 I 7
49270 49270 E 3
Or with plyr:
library(plyr)
ddply(x, id, tail, 1)
plyr encapsulates the
Take a look at plyr and reshape packages (http://had.co.nz/), I have a hunch
that they would have saved me a lot of headache had I found out about them
earlier :)
As the author of these two packages, I'm admittedly biased, but I
think R is unparalleled for data preparation, manipulation, and
Hi Robert,
I'm organising one - sign up to the mailing list,
http://groups.google.com/group/houston-r. I'm hoping to organise our
first meeting this summer.
Hadley
On Wed, May 6, 2009 at 10:15 AM, Robert Sanford wob...@gmail.com wrote:
I'm looking for a Users Group in or near Houston, TX.
If you do write your own, the hardest part will be picking the nice tick
marks. They should be approximately evenly spaced, but at nice round values
of the original variable: that's hard to do in general. R has the pretty()
function for the linear scale, and doesn't do too badly on log
On Thu, Apr 30, 2009 at 2:03 PM, MUHC-Research
villa...@dms.umontreal.ca wrote:
Dear R-users,
I recently began using the ggplot2 package and I am still in the process of
getting used to it.
My goal would be to plot on the same grid a number of curves derived from
two distinct datasets. The
Hi David,
I think the revolution blog is fantastic and a great service to the R
community. Thanks for all your hard work!
Hadley
On Fri, May 1, 2009 at 4:54 PM, David M Smith
da...@revolution-computing.com wrote:
I write about R every weekday at http://blog.revolution-computing.com
. In case
On Fri, May 1, 2009 at 12:22 PM, MUHC-Research
villa...@dms.umontreal.ca wrote:
Dear R-users,
I would have another question about the ggplot() function in the ggplot2
package.
All the examples I've read so far in the documentation make use of a single
neatly formatted data.frame. However,
Is situation anything better with ggplot2? It seems rather easy to get e.g.
line plots with error bars, provided that one feeds the data to some
modeling/regression function and passes the result over for plotting.. but
what
if I have generated my own error bar data? This is almost trivial
My issue is self-evident: using this method resulted in a 30 fold
increase in time. My question is why? If I time the individual
components separately, nothing is unusual. My hunch is the
interaction between the model.matrix and nsga2 methods.
Any ideas on how to speed this process up,
On Fri, May 1, 2009 at 2:38 PM, Zeljko Vrba zv...@ifi.uio.no wrote:
On Fri, May 01, 2009 at 01:06:34PM -0500, hadley wickham wrote:
It should be trivial with ggplot2 too, but it's hard to provide
concrete advice without a concrete problem.
Elementary problem:
qplot(wg, v.realtime, data
It's hard to check without a reproducible example, but the following
code should give you a 3d array of lat x long x time:
library(reshape)
df$lat - round_any(df$LATITUDE, 5)
df$long - round_any(df$LONGITUDE, 5)
df$value - df$TIME
cast(df, lat ~ long ~ time, mean)
On Thu, Apr 30, 2009 at
]
library(ggplot2)
qplot(year,value, data=data,label=countries, geom=c(line,text),
group=countries, col=countries)
But I would like to have the text labels show only once - e.g. at 1990
- and also control the size of the text. In my crude qplot, setting
size=2 e.g. changes not only the
I want to (1) create a deep copy of pop,
I have already said *I* do not know how to create a deep copy in R.
Creating a deep copy is easy, because all copies are deep copies.
You need to try very hard to create a reference in R.
Hadley
--
http://had.co.nz/
in the original were changed; the sort of behavior that
might be seen in a spreadsheet that had a copy by reference.
On Apr 26, 2009, at 11:28 AM, hadley wickham wrote:
I want to (1) create a deep copy of pop,
I have already said *I* do not know how to create a deep copy in R.
Creating a deep
Have a look at the plyr package and associated documentation -
http://had.co.nz/plyr
Hadley
On Sun, Apr 26, 2009 at 12:42 PM, mau...@alice.it wrote:
After a year my R programming style is still very C like.
I am still writing a lot of for loops and finding it difficult to recognize
where,
In statistics, a bumps chart is more commonly called a parallel
coordinates plot.
Hadley
On Sun, Apr 26, 2009 at 5:45 PM, Andreas Christoffersen
achristoffer...@gmail.com wrote:
Hi there,
I would like to make a 'bumps chart' like the ones described e.g.
here:
On Fri, Apr 24, 2009 at 5:50 AM, Duncan Murdoch murd...@stats.uwo.ca wrote:
Toby wrote:
I'm trying to figure out how I can get a generalized 2D
list/array/matrix/whatever
working. Seems I can't figure out how to make the variables the right
type. I
always seem to get some sort of error...
Hi Steve,
The general answer is yes, but the specific will depend on your
problem. Could you provide a small reproducible example to illustrate
your problem?
Hadley
On Fri, Apr 24, 2009 at 1:19 PM, sjaffe sja...@riskspan.com wrote:
Perhaps this is a common question but I haven't been able to
On Fri, Apr 24, 2009 at 3:12 PM, sjaffe sja...@riskspan.com wrote:
small example:
a-c(1.1, 2.1, 9.1)
b-cut(a,0:10)
c-data.frame(b,b)
d-table(c)
dim(d)
##result: c(10, 10)
But only 9 of the 100 cells are non-zero.
If there were 10 columns, the table have 10 dimensions each of length 10,
On Thu, Apr 23, 2009 at 5:11 PM, ozan bakis ozanba...@gmail.com wrote:
Dear R Users,
I have the following data frame:
v1 - c(rep(10,3),rep(11,2))
v2 - sample(5:10, 5, replace = T)
v3 - c(0,1,2,0,2)
df - data.frame(v1,v2,v3)
df
v1 v2 v3
1 10 9 0
2 10 5 1
3 10 6 2
4 11 7 0
5
Have you read the posting guide and the FAQs? If you do not get a reply
within two days, you may want to look at both and think about reformulating
your query. Oh, and while you are at it, look through the archives, a lot of
questions have already been asked and answered before.
As I say
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
Am I doing something wrong, here? If not, which are the real AIC and logLik
values for the different models?
I don't think it's reasonable to expect that the log-likelihood
computed by different functions be should comparable. Are the
constant terms included or dropped?
Hadley
--
On Fri, Apr 17, 2009 at 2:07 PM, Paul Warren Simonin
paul.simo...@uvm.edu wrote:
Thank you all for your advice.
I have received some good tips, but it was suggested I write back with a
small simulated data set to better illustrate my needs. So, currently my
data frame looks something like:
)
{
sum(x[!is.na(x)] == 2)
}
environment: 0x03d6c930
$freq_1
function (x)
{
sum(x[!is.na(x)] == 1)
}
environment: 0x03d6c930
I would like to use this list of functions with cast function (in package
reshape by Hadley Wickham) :
cast(melt(df, id = c(id, z), measure = c(x, y
On Fri, Apr 17, 2009 at 9:59 AM, Paul Warren Simonin
paul.simo...@uvm.edu wrote:
Hello!
Thanks for reading this request for assistance. I have a question regarding
creating a histogram-like figure from data that are not currently in the
correct format for the hist command.
Specifically, my
On Fri, Apr 17, 2009 at 12:19 PM, jim holtman jholt...@gmail.com wrote:
try this:
matrixx-function(A){
+ B=matrix(NaN,nrow=(A+1),ncol=4)
+ k - 1
+ for (i in 3:A){
+ for (j in i:A) {
+ B[k,] - c(NaN, i-2, i-1, j)
+ k - k + 1
+ }
+ }
Look at the output of pal.cr((0:40)/40)
Hadley
On Fri, Apr 17, 2009 at 2:42 PM, Etienne B. Racine etienn...@gmail.com wrote:
I try to use ColorRamp as ColorRampPalette (i.e. with the same gradient), but
it seems there is a nuance that I've missed.
pal.crp-colorRampPalette( c(blue, white,
-help-boun...@r-project.org] On Behalf Of hadley wickham
Sent: Wednesday, April 15, 2009 10:55 AM
To: r-help
Subject: [R] Extending a vector to length n
In general, how can I increase a vector of length m ( n) to length n
by padding it with m - n missing values, without losing attributes
In general, how can I increase a vector of length m ( n) to length n
by padding it with m - n missing values, without losing attributes?
The two approaches I've tried, using length- and adding missings with
c, do not work in general:
a - as.Date(2008-01-01)
c(a, NA)
[1] 2008-01-01 NA
length(a)
plyr is a set of tools for a common set of problems: you need to break
down a big data structure into manageable pieces, operate on each
piece and then put all the pieces back together. For example, you
might want to:
* fit the same model to subsets of a data frame
* quickly calculate
On Mon, Apr 13, 2009 at 4:15 AM, Peter Dalgaard
p.dalga...@biostat.ku.dk wrote:
Stavros Macrakis wrote:
It would of course be nice if the existing difftime class could be fit
into this, as it is currently pretty much a second-class citizen. For
example, c of two time differences is currently
I'm generating some images in R to put into a document that I'm producing
using Latex. This document in Latex is following a predefined model, which
does not accept compilation with pdflatex, so I have to compile with latex
- dvi - pdf. Because of that, I have to generate the images in R with
pnorm(37:39,lower.tail=FALSE)
[1] 5.725571e-300 0.00e+00 0.00e+00
This is just a limitation of double precision floating-point arithmetic
...
curve(pnorm(x,lower.tail=FALSE),from=30,to=40,log=y)
.Machine$double.xmin
But note
curve(pnorm(x,lower.tail=FALSE,
Have a look at ?gpar - it will tell you about lineheight.
Hadley
On Tue, Apr 7, 2009 at 3:28 AM, Mark Heckmann mark.heckm...@gmx.de wrote:
I am trying to change the inter-line spacing in grid.text(), but I just
don't find how to do it.
pushViewport(viewport())
grid.text(The inter-line
On Tue, Apr 7, 2009 at 8:44 AM, ryan.shef...@malbecpartners.com wrote:
I am trying to use the cast function from the reshape package, where the
formula is not passed in directly, but as the result of the as.formula()
function.
Using reshape v. 0.7.2
I am able to properly melt() by data
On Tue, Apr 7, 2009 at 4:41 PM, Jorge Ivan Velez
jorgeivanve...@gmail.com wrote:
Hi Eik,
You're absolutely right. My bad.
Here is the correction of the code I sent:
apply(mydata[,-1], 2, tapply, mydata[,1], function(x) sum(x)/length(x))
Or more simply:
apply(mydata[,-1], 2, tapply,
Hi Laura,
You might find the map_data function from the ggplot2 package helpful:
library(ggplot2)
library(maps)
head(map_data(state, iowa))
It formats the output of the map command into a self-documenting data frame.
Hadley
On Mon, Apr 6, 2009 at 7:00 AM, Laura Chihara lchih...@carleton.edu
On Mon, Apr 6, 2009 at 8:49 AM, Daniel Brewer daniel.bre...@icr.ac.uk wrote:
Hello,
What is the best way to turn a list into a data.frame?
I have a list with something like:
$`3845`
[1] 04010 04012 04360
$`1029`
[1] 04110 04115
And I would like to get a data frame like the following:
On Mon, Apr 6, 2009 at 9:34 AM, Stavros Macrakis macra...@alum.mit.edu wrote:
There are various ways to do this in R.
# sample data
dd - data.frame(a=1:10,b=sample(3,10,replace=T),c=sample(3,10,replace=T))
Using the standard built-in functions, you can use:
*** aggregate ***
On Mon, Apr 6, 2009 at 10:40 AM, baptiste auguie ba...@exeter.ac.uk wrote:
Here's one attempt with plyr, hopefully Hadley will give you a better
solution ( I could not get cast() to do it either)
test -
data.frame(a=c(A,A,A,A,B,B,B),b=c(1,1,2,2,1,1,1),c=sample(1:7))
On Mon, Apr 6, 2009 at 5:31 PM, Jun Shen jun.shen...@gmail.com wrote:
This is a good example to compare different approaches. My understanding is
aggregate() can apply one function to multiple columns
summarize() can apply multiple functions to one column
I am not sure if ddply() can actually
On Sat, Apr 4, 2009 at 12:09 PM, ds dhsha...@acad.umass.edu wrote:
I have a data frame something like:
name wrist
nLevel emot
1 4094 3.34 1
frustrated
2 4094 3.94
On Sat, Apr 4, 2009 at 12:28 PM, jim holtman jholt...@gmail.com wrote:
Does this do what you want:
x - read.table(textConnection(name wrist nLevel emot
+ 1 4094 3.34 1 frustrated
+ 2 4094 3.94
On Fri, Apr 3, 2009 at 4:43 AM, baptiste auguie ba...@exeter.ac.uk wrote:
Dear all,
I'm puzzled by the following example inspired by a recent question on
R-help,
cc - textConnection(user_id website time
20 google 0930
21 yahoo 0935
20
On Fri, Apr 3, 2009 at 8:43 AM, baptiste auguie ba...@exeter.ac.uk wrote:
That makes sense, so I can do something like,
count - function(x){
as.integer(unclass(table(x)))
}
count(d$user_id)
ddply(d, .(user_id), transform, count = count(user_id))
user_id website time count
1
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
On Thu, Apr 2, 2009 at 3:37 PM, Rowe, Brian Lee Yung (Portfolio
Analytics) b_r...@ml.com wrote:
Is this what you want:
d1[which(id != 4),]
Or just
d1[id != 4, ]
Hadley
--
http://had.co.nz/
__
R-help@r-project.org mailing list
X1 X2
1 11 0
2 11 0
3 11 0
4 11 1
5 12 0
6 12 0
7 12 0
8 13 0
9 13 1
10 13 1
and I want to select all rows pertaining to factor levels of X1 for
which exists at least one 1 for X2. To be clear, I want rows 1:4
(since there exists at least one observation for
I tried messing with the line df$FixTime[which.min(df$FixInx)] changing it
to df[which.min(df$FixInx)] or adding new lines with the additional columns
that I want to include, but nothing seemed to work. I'll admit I only have a
mild understanding of what is going on with the function .fun. :-)
On Wed, Apr 1, 2009 at 11:00 AM, hadley wickham h.wick...@gmail.com wrote:
I tried messing with the line df$FixTime[which.min(df$FixInx)] changing it
to df[which.min(df$FixInx)] or adding new lines with the additional columns
that I want to include, but nothing seemed to work. I'll admit I only
Earlier I posted a question about memory usage, and the community's input was
very helpful. However, I'm now extending my dataset (which I use when
running a regression using lm). As a result, I am continuing to run into
problems with memory usage, and I believe I need to shift to
col2rgb(#0079, TRUE)
[,1]
red 0
green0
blue 0
alpha 121
col2rgb(#0080, TRUE)
[,1]
red255
green 255
blue 255
alpha0
col2rgb(#0081, TRUE)
[,1]
red 0
green0
blue 0
alpha 129
Any ideas?
Thanks,
Hadley
--
http://had.co.nz/
On Tue, Mar 31, 2009 at 11:31 AM, baptiste auguie ba...@exeter.ac.uk wrote:
Not exactly the output you asked for, but perhaps you can consider,
library(doBy)
summaryBy(x3~x2+x1,data=x,FUN=mean)
x2 x1 x3.mean
1 1 A 1.5
2 1 B 2.0
3 1 C 3.5
4 2 A 4.0
5 2 B
On Tue, Mar 31, 2009 at 11:12 AM, Steve Murray smurray...@hotmail.com wrote:
Dear R Users,
I'm trying to use the reshape package to 'melt' my gridded data into column
format. I've done this before on individual files, but this time I'm trying
to do it on a directory of files (with variable
On Tue, Mar 31, 2009 at 5:01 PM, Marianne Promberger
mprom...@psych.upenn.edu wrote:
Hi,
I'm having problems with qplot and the order of numeric factor levels.
Factors with numeric levels show up in the order in which they appear
in the data, not in the order of the levels (as far as I
On Mon, Mar 30, 2009 at 10:33 AM, Mike Lawrence mike.lawre...@dal.ca wrote:
To repent for my sins, I'll also suggest that Hadley Wickham's plyr
package (http://had.co.nz/plyr/) is also useful/parsimonious in this
context:
a - ldply(cust1_files,read.table)
You might also want to do
On Mon, Mar 30, 2009 at 2:58 PM, Mike Lawrence mike.lawre...@dal.ca wrote:
I discovered Hadley Wickham's plyr package last week and have found
it very useful in circumstances like this:
library(plyr)
firstfixtime = ddply(
.data = data
, .variables = c('Sub','Tr','IA')
,
Or use frequency polygons, if you want to stay with the
interpretability of a histogram.
Hadley
On Wed, Mar 25, 2009 at 12:07 PM, Greg Snow greg.s...@imail.org wrote:
Personally I find those types of plots difficult to interpret. Much easier
to create, view, and interpret is to simply plot
On Sun, Mar 22, 2009 at 5:06 PM, Blanchette, Marco m...@stowers.org wrote:
Dear all,
I am processing a very long and complicated list using lapply through a
custom function and I would like to generate some sort of progress report.
For instance, print a dot on the screen every time 1000
This came up on R-sig-geo two days ago and this is what I said:
I have the following code in ggplot2 for turning a SpatialPolygon into
a regular data frame of coordinates. You'll need to load ggplot2, and
then run fortify(yoursp).
fortify.SpatialPolygonsDataFrame - function(shape, region =
On Sat, Mar 21, 2009 at 2:03 PM, joker77 vijumo...@gmail.com wrote:
Hi, I noted a discrepancy between R and openepi when I ran a fisher test with
the same matrix. In R:
a=matrix(c(1,2,6,17), nrow=2)
a
[,1] [,2]
[1,] 1 6
[2,] 2 17
fisher.test(a, conf.int=T)
On Fri, Mar 20, 2009 at 9:07 AM, Etches Jacob jetc...@iwh.on.ca wrote:
I am trying to specify a legend title to be other than the variable name,
but I find that the legend splits because scale_shape() takes effect but
scale_colour() does not. Can someone spot my error? Here's some toy code
On Thu, Mar 19, 2009 at 8:40 PM, jim holtman jholt...@gmail.com wrote:
Try this technique. I use it with large data objects since it is
sometime faster, and uses less memory, by using indices:
x - read.table(textConnection( v1 v2 n1 n2
1 a a1 1 21
2 a a1 2 22
3 a a1 3 23
4 a
It would be pretty easy to use the output from the R parser (which is never
wrong, is it?), and dump some markup out of it. For example the showTree
function in codetools dumps an R expression as Lisp, this is not too far
from generating html, or any other markup.
As this sounds like fun,
On Mon, Mar 16, 2009 at 7:21 PM, eitan lavi lavi.ei...@gmail.com wrote:
Hello
I'm having trouble using lars and glmnet functions to predict on a new data
set with different nrow then the original :
for instance:
=
log.1 = glm(temp.data$TL~(.),temp.data,family =
Thanks for the reply - some of the sets/palettes in the RColorBrewer are
ideal, but the problem with the problem i have is that they only go up to 12
colours, and i need 15 colours - so i assume the only thing i can do is
create my own palette, but i'm having limited success in trying to work
On Thu, Mar 12, 2009 at 5:37 PM, Christopher David Desjardins
cddesjard...@gmail.com wrote:
I get the following error when I run qplot()
qplot(grade, read,data = hhm.long.m, geom = c(point, smooth))
Error in smooth.construct.cr.smooth.spec(object, data, knots) :
x has insufficient unique
Have a look at the annotations section of
http://had.co.nz/ggplot2/book/toolbox.pdf
Hadley
On Wed, Mar 11, 2009 at 8:44 AM, levyofi levy...@post.tau.ac.il wrote:
Hello,
I really like the interface and flexibility of the ggplots package. However,
I cannot find how to add text to a plot (like
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