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 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
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
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.
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$ID), tail, 1))
ID Type N
45900 45900 I 7
46550 46550 I 7
49270 49270 E 3
Or with plyr:
library(plyr)
ddply(x, id, tail, 1)
plyr encapsulates the
On Sun, May 10, 2009 at 10:32 AM, Zeljko Vrba zv...@ifi.uio.no wrote:
Searching the mail archives I found that using legend.position as in
p.ring.3 + opts(legend.position=top)
is a known bug. I tried doing
p.ring.3 + opts(legend.position=c(0.8, 0.2))
which works, but the legend background
This does it more or less your way:
ds - split(df, df$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 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
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),
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
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,
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
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 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
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
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,
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
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 Felipe,
That's should be the default. How is it different to what you expect?
Hadley
On 11/16/07, Felipe Carrillo [EMAIL PROTECTED] wrote:
Hi all:
I wonder if ggplot2 can create histograms with
frequency along the Y axis
Felipe D. Carrillo
Fishery Biologist
US Fish Wildlife
Hi Michael,
To use stat_smooth as is, you need a smoothing function that basically
works like lm. So you have two options: to make an interface to
HoltWinters (or find another exponential smooth that already has that
interface) or write your own stat object, which might look something
like this:
On 11/21/07, Punit Anand [EMAIL PROTECTED] wrote:
Hello everyone,
Since the fields in variables column are unique with respect to ID and
fiscal year; any function like
sum,min,max,mean etc will lead to the desired result
You should probably check that, as the warning only occurs when
Hi Juan,
Assuming that your data frame is named df, something like the
following should work:
library(ggplot)
qplot(X, Y, data = df, colour = Z, label = Z, geom = text) +
scale_colour_continuous(low=orange, high = blue)
You can find out more about ggplot2 at http://had.co.nz/ggplot2.
Hadley
Above all there are lots of packages. As the software editor of the
Journal of Statistical Software I suggested we should review R
packages. No one has shown any enthusiasm for this suggestion, but I
think it would help. Any volunteers?
There are two common types of review. When reviewing
However, I don't know what exactly glht does, and the help file is
extremely terse. It offers the following options (in contrMat()):
contrMat(n, type=c(Dunnett, Tukey, Sequen, AVE,
Changepoint, Williams, Marcus,
McDermott), base = 1)
The
This is a strange argument. A good package will get a good review, which
may help it to become better. A review of a weak package can point out how
it can be fixed. Reviews will not become stale, just because packages are
frequently updated by their authors (like some that could be
On 11/23/07, thegeologician [EMAIL PROTECTED] wrote:
Hi everyone!
I'm digging into ggplot for some while now, I must say it's great! But - as
some others have posted before, and hadley knows very well himself - the
documentation is lacking some bits...
So I have to pose that question
On 11/23/07, hadley wickham [EMAIL PROTECTED] wrote:
Above all there are lots of packages. As the software editor of the
Journal of Statistical Software I suggested we should review R
packages. No one has shown any enthusiasm for this suggestion, but I
think it would help. Any
What do you mean here? Surely all packages authors aim to provide reliable
and effective software. If they know that they are offering something
unstable, they should say so clearly. In fact, they should wait until it is
stable. Most R users are not researchers, but users.
Now I'm
On 11/26/07, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Others have already answered your direct question but consider that
what you may want without realizing it is object-oriented programming.
I agree with Gabor, you're not actually looking for a global state,
but a mutable object
Hi:I recently started using the melt function and
found that it is very useful and powerful,however I
can't seem to find the way to change the default
column header value to my custom column name.
melted data are data frames. You can names[..]-myvalue for renaming, but
make
sure you
A nice model might be http://www.workingwithrails.com/.
It also suggests other possible measures of authority:
* member of R core
* packages written by/contributed to
* R conference attended/presented at
* contributer to wiki
Hadley
On 11/30/07, Doran, Harold [EMAIL PROTECTED] wrote:
Hi Thierry,
Have a look at http://had.co.nz/ggplot2/scale_identity. (The way you
are currently doing it only works because you are not using any kind
of grouping/facetting, and I'm thinking about adding an explicit
message/warning for this case)
Hadley
On 12/3/07, ONKELINX, Thierry [EMAIL
On 12/2/07, Jim Lemon [EMAIL PROTECTED] wrote:
[EMAIL PROTECTED] wrote:
Hi Jim,
Thanks for getting back to me so quickly.
I did look at color.legend, but that seems to plot colored blocks for
the observations (in this case the mean) and not for the color.scale
(which represents
On 12/3/07, Bert Gunter [EMAIL PROTECTED] wrote:
... but the best option is not to do this kind of technicolor extravaganza
at all!
Yes, good point! And of a course of a scatterplot of mean vs variance
will best reveal the relationship between the two variables.
Hadley
--
http://had.co.nz/
On Dec 4, 2007 10:34 AM, Stéphane CRUVEILLER [EMAIL PROTECTED] wrote:
Hi,
I tried this method but it seems that there is something wrong with my
data frame:
when I type in:
qplot(x=as.factor(Categorie),y=Total,data=mydata)
It displays a graph with 2 points in each category...
but if
Hmm, I would have recommended
colorRampPalette(c('dark red','white','dark blue'),
space = Lab)
where the 'space = Lab' part also makes sure that a
perceptually-based space rather than RGB is used.
I think the functions colorRamp() and (even more)
colorRampPalette()
,
Stéphane.
hadley wickham wrote:
On Dec 4, 2007 10:34 AM, Stéphane CRUVEILLER
[EMAIL PROTECTED] wrote:
Hi,
I tried this method but it seems that there is something wrong with my
data frame:
when I type in:
qplot(x=as.factor(Categorie),y=Total,data=mydata
On Dec 7, 2007 4:28 AM, Roger Levy [EMAIL PROTECTED] wrote:
I'm interested in writing a function that constructs a new plot on the
current graphics device if no plot exists there yet, but adds lines to
the existing plot if a plot is already there. How can I do this? It
seems to me that the
On 12/6/07, 정 태훈 [EMAIL PROTECTED] wrote:
Hi, all;
I got a reply for my previous several postings saying that I was
spamming the r-help mailing list.
I am very sorry to all subscribers if I did that.
But I've been reposting my message to the mailing list several times
because I didn't know
Could someone recommend a good book on regular expressions with focus on
applications/use as it might relate to R. I remember there was a mention of
such a reference book recently, but I could not locate that message on the
archive.
Mastering regular expressions by Jeffrey Friedl
I don't think anyone has mentioned the references given on the help page
?regexp: they are a great deal more reliable than some of the third-party
write-ups that have been mentioned.
They are good technical references, but I don't think they're very
helpful for someone trying to learn regular
Hi list,
I have a question regarding post-hoc extraction of data from a two-way
categorical anova.
I have a categorical anova of this form:
width ~ steepness + patchiness (4 steepness levels, 4 patchiness levels)
This simple setup answers if for the widths I collected across different
Hi Jiho,
The key to solving this problem is to use aes_string instead of aes.
Instead of the complicated munging that aes does to get the names of
the variables, aes_string works directly with strings, so that:
aes_string(x = mpg, y = wt) == aes(x = mpg, y = wt)
So your function would look
Is the function current.grobTree() implemented in ggplot2? According to
the draft ggplot2 book, on page 43, we can get a list of all grobs with
current.grobTree(). But when I try that, I get 'Error: could not find
function current.grobTree'.
Ooops - the book is now out of date. Paul Murrell
PS: If you care to explain: why do all parameters in the code below
have a . before the name, except precisely limits? I know it has
to do with proto, but could not find out why this one was different.
Not really - I started off with the convention that variables should
start with a . to
great! I knew you would have thought this through. That's perfect. As
always there's the trade-off between writing code and documenting the
code already written. In this case the trade-off turned toward the
code part I guess.
Autodetection of strings by aes would be even greater but that
Hi Seth,
An alternative would be to use ggplot2, http://had.co.nz/ggplot2:
model - function(a,b,c,X1,X2) {
(exp(a + b*X1 + c*X2)) / (1 + exp(a + b*X1 + c*X2))
}
g - expand.grid(X1 = seq(0.40, 0.8,0.01), X2 = seq(0.03,0.99,0.03))
a - -37.61
b - 34.88
c - 28.44
g$z- model(a, b, c, g$X1,g$X2)
Hi John,
One alternative would be to use ggplot2 with a polar coordinate system:
library(ggplot2)
qplot(mpg, wt, data=mtcars) + coord_polar()
qplot(mpg, wt, data=mtcars, ylim=c(3,4)) + coord_polar()
etc. You can see more examples of polar coordinates at
On 12/16/07, Bob Green [EMAIL PROTECTED] wrote:
Hello,
Below is the code for a basic bar graph. I was seeking advice
regarding the following:
(a) For each time period there are values from 16 people. How I can
change the colour value so that each person has a different colour,
which
This has nothing to do really with the question that Troels asked,
but the exposition quoted from the AA paper is unnecessarily
confusing.
The phrase ``Because X0 and X1 have identical marginal
distributions ...''
throws the reader off the track. The
I have a question regarding colors in bar plots. I want to stack a
total of 18 cost values in each bar. Basically, it is six cost types and
each cost type has three components- direct, indirect, and induced
costs. I would like to use both solid color bars and bars with the
slanted lines
On 12/18/07, Bert Jacobs [EMAIL PROTECTED] wrote:
Hi,
I'm having a bit of problems in creating a new dataframe.
Below you'll find a description of the current dataframe and of the
dataframe that needs to be created.
Can someone help me out on this one?
library(reshape)
dfm - melt(df, id =
On Dec 18, 2007 2:06 PM, [EMAIL PROTECTED] wrote:
I am plotting fishing vessel positions and want these points to be
relative in size to the catch at that point. Is this possible? I am just
begining to use R and my search of the help section didnt help in this
area. Heres what Im using so
And the url is:
http://cran.r-project.org/src/contrib/Views/ExperimentalDesign.html
Hadley
On 12/21/07, Ulrike Grömping [EMAIL PROTECTED] wrote:
Dear UseRs,
the new Task View ExperimentalDesign (Title: Design of Experiments (DoE)
Analysis of Experimental Data) has just been uploaded to
On Dec 30, 2007 10:47 AM, Peter Dalgaard [EMAIL PROTECTED] wrote:
Gabor Grothendieck wrote:
Read the warning in ?ifelse
Yep.
And, yes, it is annoying that ifelse() strips attributes, including
class, but it is one of those things that have been in the S languages
forever, and nobody really
Hi Pedro,
Sorry for missing this email - the good news is that in the next
version of ggplot (hopefully coming out in the next few days), you'll
be able to do:
qplot(x, y, data=plotdata) + scale_x_reverse()
qplot(x, y, data=plotdata) + scale_y_reverse()
to get reversed scales
Hadley
On Oct
On Jan 5, 2008 7:45 PM, Milton Cezar Ribeiro [EMAIL PROTECTED] wrote:
Dear all,
I have a dataset which I need to estimate the regression model and plot the
estimated curve two other curves with low and high confidence interval
(CI=95%). How can I do that?
The easiest way is to use the
On Jan 5, 2008 8:32 AM, Keith Jones [EMAIL PROTECTED] wrote:
Hi,
Maybe I have not been looking in the right spot, but, I have not been
able to fine a command to automatically calculate the running
cumulative sum of a vector. Is there such a command?
Try
help.search(cumulative sum)
Hadley
I followed the instructions at
http://cran.r-project.org/bin/linux/ubuntu/README.html, but I'm
getting the following error:
~: sudo apt-get install r-base
Reading package lists... Done
Building dependency tree... Done
Some packages could not be installed. This may mean that you have
requested an
On 1/7/08, Emmanuel Charpentier [EMAIL PROTECTED] wrote:
hadley wickham a écrit :
I followed the instructions at
http://cran.r-project.org/bin/linux/ubuntu/README.html, but I'm
getting the following error:
~: sudo apt-get install r-base
Reading package lists... Done
Building
errors in the data usually because I know the data. I find errors
because I can say things like
library(Hmisc)
datadensity(mydata) # show all raw data in small rug plots
hist.data.frame(mydata) # postage-stamp size histograms of all
variables in dataset
latex(describe(mydata))
On Jan 10, 2008 8:36 AM, Petr PIKAL [EMAIL PROTECTED] wrote:
Dear all
I want to display 4 dimensional space by some suitable way. I searched
CRAN and found miscellaneous 3 dim graphics packages which I maybe can
modify but anyway I am open to any hint how to efficiently display data
like:
On 1/14/08, Jim Price [EMAIL PROTECTED] wrote:
A suggestion for a family of such functions:
ceilGenerator - function(num)
function(x) num * ceiling(x / num)
ceil10 - ceilGenerator(10)
ceil20 - ceilGenerator(20)
ceil10(1:10 * 4)
ceil20(1:10 * 4)
libraray(reshape)
On Jan 14, 2008 7:30 PM, Erin Steiner [EMAIL PROTECTED] wrote:
#After spending the entire day working on this question, I have
decided to reach out for support:
#I am trying to overlay a densityplot from one data set over a
histogram of another, if I were to plot the two individually, they
Currently ggplot doesn't support any interactivity. What sort of
interaction were you looking for with the periodogram? It might be
possible to do something with rggobi.
Hadley
On Jan 15, 2008 10:38 AM, stephen sefick [EMAIL PROTECTED] wrote:
Any Ideas to get an interactive periodogram?
--
Hi Felix,
I'm a bit stumped too - it's been a while since I've played around
with viewports in ggplot. I have a feeling that this might be due to
an old bug in grid that I thought had been fixed - you can't
downViewport to a viewport underneath a frame grob. I've cc'd Paul in
the hope that he
On Jan 15, 2008 12:27 PM, Dieter Menne [EMAIL PROTECTED] wrote:
hadley wickham h.wickham at gmail.com writes:
qplot(vals, ..density.., data = data.frame.A, geom=histogram, facets
= factor.1 ~ factor.2, binwidth = 1) + geom_density(data=data.frame.B)
uhh.. how should I understand y
I'm new to R and am evaluating it to see whether it would be an appropriate
tool to create a dashboard (a graphical statistical summary page). Could
someone tell me if it is possible to display data as dials or meters. e.g.
the four dials here:
Previously I used SAS for 23 years and now R/S-Plus for 17. SAS is
effective for large datasets (in my work 500,000 subjects) but except
for that, R is far superior to SAS for data management and manipulation.
Just four of the reasons are that you can
- merge data frames multiple ways
Happy families are all alike; every unhappy family is unhappy in its own way.
Leo Tolstoy
and every messy data is messy in its own way - it's easy to define the
characteristics of a clean dataset (rows are observations, columns are
variables, columns contain values of consistent types). If you
On Jan 17, 2008 9:53 AM, Thompson, David (MNR)
[EMAIL PROTECTED] wrote:
Hello Hadley,
I am trying to reproduce the following with ggplot:
a - seq(0, 360, 5)*pi/180 ; a
ac - sin(a + (45*pi/180)) + 1 ; ac
plot(a, ac, type='b', xaxt = n)
axis(1, at=seq(0,6,1),
Typical plots with vertical bars are not histograms. Consider barplot
or plot(*, type = h) for such bar plots. . But no worry, I've mixed
them up myself a number of times.
Or you can use ggplot2, which will do the right thing regardless of
whether you have continuous or categorical data:
Do you have an example graphic that shows what you're trying to
create? I can't figure out if you want something like a square pie
chart (aka waffle chart), a stacked barchart, a levelplot, or
something else.
Hadley
On Jan 18, 2008 6:06 AM, Marta Rufino [EMAIL PROTECTED] wrote:
Dear R users,
On Jan 18, 2008 1:19 PM, Jeffrey J. Hallman [EMAIL PROTECTED] wrote:
Frank E Harrell Jr [EMAIL PROTECTED] writes:
Rob Robinson wrote:
I wonder if those who complain about SAS as a programming environment have
discovered SAS/IML which provides a programming environment akin to Matlab
No one else mentioned this, but if those 99s represent missings, you
should be using NA not a special numeric value.
Hadley
On Jan 22, 2008 5:40 PM, Dimitri Liakhovitski [EMAIL PROTECTED] wrote:
Thanks a lot, everyone!
Dimitri
On 1/22/08, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Slight
On Jan 22, 2008 8:39 PM, dxc13 [EMAIL PROTECTED] wrote:
useR's,
I want to create a movie of a sin function (from 0 to pi/2) using
levelplot() in the lattice package. I basically want to create 20 or so
plots of the sin function starting with an amplitude of 0 and ending at
amplitude 1. By
On Jan 23, 2008 3:29 AM, Pilar Loren [EMAIL PROTECTED] wrote:
Hi!, I need help, I have a problem with geom_line function because I have
this error:
df
LENGTH LAT
091639 10.002 42.26282
091640 30.808 42.26834
091641 21.591 42.31689
091642 22.030 41.53246
091643 22.744
I can reach my goal in ggplot2, although the relative heights of the
bar's pieces don't seem quite right (it does generate a warning):
library(ggplot2)
x-factor(1)
y-factor( c(Male,Male,Female) )
mydata - data.frame(x,y)
rm(x,y)
mydata
Ooops! There was a bug in stat_bin that occurred
On Jan 25, 2008 2:04 PM, Domenico Vistocco [EMAIL PROTECTED] wrote:
You could use ggplot2:
library(ggplot2)
x=rnorm(100)
y=rnorm(100)
df=melt(data.frame(x,y))
#for vertical histogram
ggplot(data=df,aes(x=value))+geom_histogram()+facet_grid(.~variable)
#for horizontal histogram
I am using ggplot2 at the moment and I must say it is definitely better
then ggplot - good work.
My problem is that I am using facet_grid() in the following way:
p - ggplot(ssq, aes(x=year, y=-log(ssq)))
p + geom_point() + facet_grid(me*gi~cs*rz)
and it works nicely, except that I
And subsetting a factor retains the original factor levels. To drop
unused levels, just use factor(f[index]) or f[index, drop=TRUE]. The
opposite behaviour can be even more annoying/dangerous because it leads
to empty cells dropping out of tables and bars disappearing from barplots.
Of course
ggopt(strip.text = function(variable, value) paste(variable, value, sep==
))
That's exactly what I was looking for - thanks.
One thing that I should mention is that this is likely to change at
some point in the future. Eventually it will become:
+ facet_grid(strip.text = function(variable,
On Jan 28, 2008 1:25 PM, Greg Snow [EMAIL PROTECTED] wrote:
I had heard the same thing about Florence Nightingale, but it seems that this
is a confusion of different graphs.
Nightingale developed a graph based on a circle, but all the angles were
equal and the different values were encoded
On Jan 29, 2008 9:12 AM, Markus Didion [EMAIL PROTECTED] wrote:
Dear all,
I would like to create a stacked area chart to show the development of
species biomass over time. Since it is intended for publication I need
to prepare in black and white. I have tried to modify the stackedPlot
On Jan 30, 2008 2:15 AM, Markus Didion [EMAIL PROTECTED] wrote:
Thank you Greg for your comments.
Here is a (very) simple example of what my data might look like:
res - matrix(sample(20:25, 41*12, replace=T), nrow=12, ncol=41,
dimnames=list(c(paste(sp,1:12)), c(seq(0, 200, by=5
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
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
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 =
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
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
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)
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