Have a look at:
"Computing Thousands of Test Statistics Simultaneously in R" by Holger
Schwender and Tina Müller, in
http://stat-computing.org/newsletter/issues/scgn-18-1.pdf
Hadley
On Mon, Sep 13, 2010 at 4:26 PM, Alexey Ush wrote:
> Hello,
>
> I have a question regarding how to speed up the t
Hi Uwe,
The problem is most likely because the original poster doesn't have
the latest version of plyr. I correctly declare this dependency in
the DESCRIPTION
(http://cran.r-project.org/web/packages/reshape2/index.html), but
unfortunately R doesn't seem to use this information at run time,
genera
On Fri, Sep 10, 2010 at 10:23 AM, Henrik Bengtsson
wrote:
> Don't underestimate the importance of the choice of the algorithm you
> use. That often makes a huge difference. Also, vectorization is key
> in R, and when you use that you're really up there among the top
> performing languages. Her
>>> I think this is a really bad idea. data.frames are not meant to be
>>> used in this way. Why not use a list of lists?
>>
>> It can be very convenient, but I suspect the original poster is
>> confused about the different between vectors and lists.
>
> I wouldn't be surprised if someone were conf
>>> I'm having trouble parsing this. What exactly do you want to do?
>> 1 - Put a list as an element of a data.frame. That's quite convenient for my
>> pricing function.
>
> I think this is a really bad idea. data.frames are not meant to be
> used in this way. Why not use a list of lists?
It can
Reshape2 is a reboot of the reshape package. It's been over five years
since the first release of the package, and in that time I've learned
a tremendous amount about R programming, and how to work with data in
R. Reshape2 uses that knowledge to make a new package for reshaping
data that is much mo
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 a data f
> daply(data.test, .(municipality, employed), function(d){mean(d$age)} )
> employed
> municipality no yes
> A 41.58759 44.67463
> B 55.57407 43.82545
> C 43.59330 NA
>
> The .drop argument has a different meaning in daply. Some R functio
Why don't you read the answers to your stackoverflow question?
http://stackoverflow.com/questions/3665885/adding-a-list-of-vectors-to-a-data-frame-in-r/3667753
Hadley
On Wed, Sep 8, 2010 at 1:17 AM, raje...@cse.iitm.ac.in
wrote:
>
> Hi,
>
> I have a preallocated dataframe to which I have to add
Have a look at match and merge.
Hadley
On Wednesday, September 8, 2010, Michael Haenlein
wrote:
> Dear all,
>
> I'm working with two data frames.
>
> The first frame (agg_data) consists of two columns. agg_data[,1] is a unique
> ID for each row and agg_data[,2] contains a continuous variable.
>
>
> One common way around this is to pre-allocate memory and then to
> populate the object using a loop, but a somewhat easier solution here
> turns out to be ldply() in the plyr package. The following is the same
> idea as do.call(rbind, l), only faster:
>
>> system.time(u3 <- ldply(l, rbind))
> u
# testthat
Testing your code is normally painful and boring. `testthat` tries to
make testing as fun as possible, so that you get a visceral
satisfaction from writing tests. Testing should be fun, not a drag, so
you do it all the time. To make that happen, `testthat`:
* Provides functions that ma
> first ddply result did I see that some sort of misregistration had occurred;
> Better with:
>
> res <-ddply(egraw2, .(category), .fun=function(df) {
> sapply(df,
> function(x) {mnx <- mean(x, na.rm=TRUE);
> sapply(x, function(z) if
>> That doesn't justify the use of a _histogram_ - and regardless of
>
> The usage highlights meaningful characteristics of the data.
> What better justification for any method of analysis and display is
> there?
That you're displaying something that is mathematically well founded
and meaningful
> I have counts ranging over 4-6 orders of magnitude with peaks
> occurring at various 'magic' values. Using a log scale for the
> y-axis enables the smaller peaks, which would otherwise
> be almost invisible bumps along the x-axis, to be seen
That doesn't justify the use of a _histogram_ - and
It's not just that counts might be zero, but also that the base of
each bar starts at zero. I really don't see how logging the y/axis of
a histogram makes sense.
Hadley
On Sunday, August 29, 2010, Joshua Wiley wrote:
> Hi Derek,
>
> Here is an option using the package ggplot2:
>
> library(ggplot
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. Additionally, they
On Wed, Aug 25, 2010 at 6:05 AM, abotaha wrote:
>
> Woow, it is amazing,
> thank you very much.
> yes i forget to attach the dates, however, the dates in my case is every 16
> days.
> so how i can use 16 day interval instead of month in by option.
Here's one way using the lubridate package:
libr
On Tue, Aug 24, 2010 at 11:25 AM, Martin Morgan wrote:
> On 08/24/2010 07:27 AM, Doran, Harold wrote:
>> There is the stringMatch function in the MiscPsycho package.
>>
>>> stringMatch('Hadley', 'Hadley Wickham', normalize = 'no')
>>
On Mon, Aug 23, 2010 at 1:02 PM, Alison Macalady wrote:
> Hi,
>
> I have a 5-paneled figure that i made using the facet function in qplot
> (ggplot). I've managed to arrange the panels into two rows/three columns,
> but for the sake of easy visual comparisons between panels in my particular
> dat
Hi all,
all.equal is generally very useful when you want to find the
differences between two objects. It breaks down however, when you
have two long strings to compare:
> all.equal(a, b)
[1] "1 string mismatch"
Does any one know of any good text diffing tools implemented in R?
Thanks,
Hadley
I should note that I realise this function is pretty trivial to write
(see below), I just want to avoid reinventing the wheel.
recyclable <- function(...) {
lengths <- vapply(list(...), length, 1)
all(max(lengths) %% lengths == 0)
}
Hadley
On Mon, Aug 23, 2010 at 10:33 AM, Hadley W
Hi all,
Is there a function to determine whether a set of vectors is cleanly
recyclable? i.e. is there a common function for detecting the
error/warnings that underlie the following two function calls?
> 1:3 + 1:2
[1] 2 4 4
Warning message:
In 1:3 + 1:2 :
longer object length is not a multiple
You may find a close reading of ?merge helpful, particularly this
sentence: "If there is more than one match, all possible
matches contribute one row each" (so check that you don't have
multiple matches).
Hadley
On Sat, Aug 21, 2010 at 10:45 AM, Cecilia Carmo wrote:
> Hi everyone,
>
> I have be
On Wed, Aug 11, 2010 at 10:14 PM, Brian Tsai wrote:
> Hi all,
>
> I'm interested in doing a dot plot where *both* the size and color (more
> specifically, shade of grey) change with the associated value.
>
> I've found examples online for ggplot2 where you can scale the size of the
> dot with a va
> When ggplot2 verifies the widths before stacking (the default position for
> histograms), it computes the widths from the minimum and maximum values for
> each bin. However, because the width of the bins (0.28) is much smaller
> than the scale of the edges (6.8e+09), there is some underflow and
On Mon, Aug 9, 2010 at 4:30 PM, Matthew Dowle wrote:
>
>
> Another option for consideration :
>
> library(data.table)
> mydt = as.data.table(mydf)
>
> mydt[,as.list(coef(lm(y~x1+x2+x3))),by=fac]
> fac X.Intercept. x1 x2 x3
> [1,] 0 -0.16247059 1.130220 2.988769 -19.14719
>> That's exactly what dlply does - so you should never have to do that
>> yourself.
>
> I'm unclear what you are saying. Are you saying that the plyr function
> _should_ have examined the objects in that list and determined that there
> were 4 rows and properly labeled the rows to indicate which l
> There is one further improvement to consider. When I tried using dlply to
> tackle a problem on which I had been bashing my head for the last three days
> and it gave just the results I had been looking for, I also noticed that the
> dlply function returns the grouping variable levels in an attri
On Mon, Aug 9, 2010 at 9:29 AM, David Winsemius wrote:
> If you look at the output (as I did) you should see that despite whatever
> expectations you have developed regarding plyr, that it did not produce a
> grouping variable:
>
>> ldply(dl, function(x) coef(summary(x)) )
> fac Estimate Std
With sweave, you need to explicitly print() the output of ggplot2 and
lattice plots.
Hadley
On Mon, Aug 9, 2010 at 6:32 AM, W Eryk Wolski wrote:
> qplot does (?) what I was looking for!
> At least it plots what I want to plot in the interactive modus.
> However, it seems not to work with Sweave.
On Sat, Aug 7, 2010 at 2:54 AM, Michael Bedward
wrote:
> On 7 August 2010 06:26, Hadley Wickham wrote:
>
>> library(ggplot2)
>> qplot(x, y, fill = z, data = df, geom = "tile")
>
> Hi Hadley,
>
> I read the original question as being about irregularly spac
On Fri, Aug 6, 2010 at 9:24 AM, W Eryk Wolski wrote:
> Hi,
>
> Would like to make an image
> however the values in z are not on an uniform grid.
>
> Have a dataset with
> length(x) == length(y) == length(z)
> x[1],y[1] gives the position of z[1]
>
> and would like to encode value of z by a color.
> Well, here's one way that "might" work (explanation below):
>
> The ideas is to turn each row into a character vector and then work with the
> two character vectors.
>
>> bigs <- do.call(paste,TheBigOne)
>> ix <- which(bigs %in% setdiff(bigs,do.call(paste,TheLittleOne)))
>> TheBigOne[ix,]
>
> Ho
Here's one way, using a function from the plyr package:
TheLittleOne<-data.frame(cbind(c(2,3),c(2,3)))
TheBigOne<-data.frame(cbind(c(1,1,2),c(1,1,2)))
keys <- plyr:::join.keys(TheBigOne, TheLittleOne)
!(keys$x %in% keys$y)
TheBigOne[!(keys$x %in% keys$y), ]
Hadley
On Thu, Jul 29, 2010 at 1:38
plyr is a set of tools for a common set of problems: you need to break
down a big data structure into manageable pieces, operate on each
piece and then put all the pieces back together. For example, you
might want to:
* fit the same model to subsets of a data frame
* quickly calculate summary
On Sat, Jul 24, 2010 at 2:23 AM, Jeff Newmiller
wrote:
> Fahim Md wrote:
>>
>> Is there any function/way to merge/unite the following data
>>
>> GENEID col1 col2 col3 col4
>> G234064 1 0 0 0
>> G2340
What is your null hypothesis? What is your alternate hypothesis? What
is the test statistic? Why do you want a p-value?
Hadley
On Thu, Jul 22, 2010 at 5:40 PM, jd6688 wrote:
>
> Here is my dataframe with 1000 rows:
>
> employee_id weigth p-value
>
> 100 150
> 1
Did you look at the examples in sample?
# sample()'s surprise -- example
x <- 1:10
sample(x[x > 8]) # length 2
sample(x[x > 9]) # oops -- length 10!
sample(x[x > 10]) # length 0
## For R >= 2.11.0 only
resample <- function(x, ...) x[sample.int(length(x), ...)]
resample(x[x > 8]) #
> ddply(ma, .(variable), summarise, mean = mean(value), sd = sd(value),
> skewness = skewness(value), median = median(value),
> mean.gt.med = mean.gt.med(value))
In principle, you should be able to do:
ddply(ma, .(variable), colwise(each(mean, sd, skewness, median, mean.gt.med)))
but
> The problem is in data.frame[ and any NA in a logical vector will return a
> row of NA's. This can be avoid by wrapping which() around the logical vector
> which seems entirely wasteful or using subset().
The basic philosophy that causes this behaviour is sensible in my
opinion: missing values m
On Thu, Jul 15, 2010 at 11:08 PM, Dennis Murphy wrote:
> Hi:
>
> I sincerely hope there's an easier way, but one method to get this is as
> follows,
> with d as the data frame name of your test data:
>
> d <- d[order(with(d, Age, School, rev(Grade))), ]
> d$Count <- do.call(c, mapply(seq, 1, as.ve
On Wed, Jul 14, 2010 at 1:32 AM, Ian Bentley wrote:
> I've got a couple of more changes that I want to make to my plot, and I
> can't figure things out. Thanks for all the help.
>
> I'm using this R script
>
> library(ggplot2)
> library(lattice)
> # Generate 50 data sets of size 100 and assign th
> So distributing code to other people is preferably done using R packages,
> which gives you this option.
However (as far as I am aware), note that this option is checked at
package build time, not at load time.
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statist
For a quick fix, you probably need to reinstall plyr.
Hadley
On Wed, Jul 14, 2010 at 11:03 PM, stephen sefick wrote:
> This is the first time that I have tried to update packages with a
> tinkered around with .Rprofile. I start R with R --vanilla and it
> does not load my .Rprofile, but when I i
>> For some reason package writers seem to prefer maximally uninformative
>> names for their packages. To take some examples of recently announced
>> packages, can anyone guess what packages 'FDTH', 'rtv', or 'lavaan'
>> do? Why the aversion to informative names along the lines of
>> 'Freq_dist_a
On Wed, Jul 14, 2010 at 7:39 AM, thmsfuller...@gmail.com
wrote:
> Hi All,
>
> The last line if the following code returns the error right below this
> paragraph. Essentially, I use the operator %:% to retrieve a variable
> in a nested frame. Then I want to use the same operator (with '<-') to
> ch
strings <- replicate(1e5, paste(sample(letters, 100, rep = T), collapse = ""))
system.time(strings[-1] == strings[-1e5])
# user system elapsed
# 0.016 0.000 0.017
So it takes ~1/100 of a second to do ~100,000 string comparisons. You
need to provide a reproducible example that illustrates
Hi Ian,
Have a look at the examples in http://had.co.nz/ggplot2/geom_tile.html
for some ideas on how to do this with ggplot2.
Hadley
On Sat, Jul 10, 2010 at 8:10 PM, Ian Bentley wrote:
> Hi all,
>
> Thanks for the really great help I've received on this board in the past.
>
> I have a very part
== ?
Hadley
On Sun, Jul 11, 2010 at 2:08 PM, Ralf B wrote:
> What is the fastest way to compare two strings in R?
>
> Ralf
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http
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 (l
And the profr package for an alternative display.
Hadley
On Tuesday, July 6, 2010, Uwe Ligges wrote:
> or just see
>
> ?Rprof
>
> and
>
> ?Rprofmem
>
>
> Uwe Ligges
>
>
> On 06.07.2010 01:21, Jim Callahan wrote:
>
> Message: 21
> Date: Mon, 5 Jul 2010 02:26:29 -0400
> From: Ralf B
> To: "r-help@r
Or wait a couple of days for the next release of ggplot2...
Hadley
On Mon, Jul 5, 2010 at 11:28 AM, Sebastian Wurster
wrote:
> Thank you for this nice patch!
> To incorporate it you have to open the ggplot2 file in "path to your R
> packages\ggplot2\R", search for the first line of code and repl
Hi Mark,
Try this to get you started:
table(roe1 > median(roe1), roe0 > median(roe0))
Hadley
On Sun, Jul 4, 2010 at 6:29 AM, Mark Carter wrote:
> I'm not very good at statistics, but I know enough to be dangerous. I'm
> completely new to R, having just discovered it yesterday. Now that the
>
This is possible in ggplot2, but it's an not appropriate use of a bar
chart - because length is used to convey value, chopping the bottoms
of the bars of will give a misleading impression of the data.
Instead, use a dot plot:
data$Q <- unlist(lapply(data$Q, function(x) paste(strwrap(x, 20),
collap
> Sure. The code uses objects() to find the exported objects in the
> package, so I guess the offending object will be there. You can check
> for yourself by loading the package and calling objects() on the package
> environment.
So I guess my question then is how do data sets and namespaces
int
> ?formula in R 2.9.2 says in para 2:
> "The %in% operator indicates that the terms on its left are nested
> within those on the right. For example a + b %in% a expands to the
> formula a + a:b. "
Ooops, missed that. So b %in% a = a:b, and that's what's meant by
"different coding".
Hadley
--
A
Hi all,
In preparation for teaching a class next week, I've been reviewing R's
standard modelling algebra. I've used it for a long time and have a
pretty good intuitive feel for how it works, but would like to
understand more of the technical details. The best (online) reference
I've found so far
Here's another version that's a bit easier to read:
na.roughfix2 <- function (object, ...) {
res <- lapply(object, roughfix)
structure(res, class = "data.frame", row.names = seq_len(nrow(object)))
}
roughfix <- function(x) {
missing <- is.na(x)
if (!any(missing)) return(x)
if (is.numer
On Tue, Jun 29, 2010 at 12:22 PM, Dimitri Liakhovitski
wrote:
> Hello, everyone!
> I have a very simple task - I have a data frame (see MyData below) and
> I need to stack the data (see result below).
> I wrote the syntax below - it's very basic and it does what I need.
> But I am sure what I am t
On Tue, Jun 29, 2010 at 8:02 AM, Matthew Dowle wrote:
>
>> dt = data.table(d,key="grp1,grp2")
>> system.time(ans1 <- dt[ , list(mean(x),mean(y)) , by=list(grp1,grp2)])
> user system elapsed
> 3.89 0.00 3.91 # your 7.064 is 12.23 for me though, so this
> 3.9 should be faster for y
library(plyr)
n<-10
grp1<-sample(1:750, n, replace=T)
grp2<-sample(1:750, n, replace=T)
d<-data.frame(x=rnorm(n), y=rnorm(n), grp1=grp1, grp2=grp2)
system.time({
d$avx1 <- ave(d$x, list(d$grp1, d$grp2))
d$avy1 <- ave(d$y, list(d$grp1, d$grp2))
})
# user system elapsed
# 39.300 0.279
1) Create a table with two columns: payor and payor.group.
2) Merge that table with your original data
Hadley
On Mon, Jun 28, 2010 at 10:46 AM, GL wrote:
>
> I'm guessing there's a more efficient way to do the following using the index
> features of R. Appreciate any thoughts
>
> for (i in
Hi Simon,
Here are two ways to do that with ggplot:
qplot(test2, data = test_df, geom = "freqpoly", colour = test,
binwidth = 30, drop = F)
qplot(test2, data = test_df, geom = "bar", fill = test, binwidth = 30)
binwidth is in days. If you want to bin by other intervals (like
months), I'd recomm
> I'm having the same problem as Stephan (see below), but what I'm trying to
> jitter is not a numeric vector, but a factor. How do I proceed? (Naively
> jittering a factor makes it numeric, no longer factor, so I don't get the
> custom ordering which conveniently comes with using a factor. I'm not
> Finally, FWIW, 1 is not considered "very large" these days; maybe
> 10,000,000,000 might be...
It's off topic, but I rather like Mike Driscoll's definition of big
data: it's too big to fit on a single machine and must be stored on
many (http://www.slideshare.net/dataspora/s-4455027). A smal
> If anybody has quick fix, that would be helpful.
Write your own function that wraps xtable...
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
__
R-help@r-project.org mailing
y
> similar SQL set. The select clause carries more coloumns in the failing data
> set.
>
> ottar
>
> On 20 June 2010 18:28, Hadley Wickham wrote:
>>
>> Hi Ottar,
>>
>> It's impossible to tell what the problem is without a reproducible
>> example (http:/
six months, particularly with values that
> have no time portion. You have promised a fix before, but l haven't seen it,
> so I convert to Date to work around the bug.
>
> "Hadley Wickham" wrote:
>
>>Hi Ottar,
>>
>>It's impossible to tell what th
Hi Ottar,
It's impossible to tell what the problem is without a reproducible
example (http://gist.github.com/270442)
Hadley
On Sun, Jun 20, 2010 at 4:38 PM, Ottar Kvindesland
wrote:
> I have a problem that puzzles me a bit today. When loading off data from a
> database and plotting using ggplot
> I've given thought in the past to the question of estimating the R
> user base, and came to the conclusion that it is impossible to get
> an estimate of the number of users that one could trust (or even
> put anything like a margin of error to).
I find it hard to believe that it should be harder
> I agree with all your points. What I have so far is nowhere near the big
> picture, but it's a start. When you install some software it asks if you
> mind it reporting usage stats back to its home site. I know that sort of
> thing has been discussed before on R-help. I'd love to see that added so
> The glitches are the cases where you would have a bundle of lines belonging
> to a specific cluster, but had spaces between them (because the place of one
> of the lines was saved for another line that in the meantime moved to
> another cluster).
I think that display looked just fine!
> I just
> My current solution is to use a constant jitter (based on "seq") on all the
> k number of clusters, but that causes glitches in the produced image (run my
> code to see).
What are the glitches? It looks pretty good to me. (I'm not sure if
the colour does anything apart from make it pretty thou
Try multcompView
Hadley
On Sat, Jun 12, 2010 at 8:42 AM, Tal Galili wrote:
> Hello dear R-help mailing list,
>
> A friend of mine teaches a regression and experimental design course and
> asked me the following question.
>
> She is trying to find a way to display the "homogeneous groups" (after
On Fri, Jun 11, 2010 at 1:32 PM, Ian Bentley wrote:
> I'm an R newbie, and I'm just trying to use some of it's graphing
> capabilities, but I'm a bit stuck - basically in massaging the already
> available data into a format R likes.
>
> I have a simulation environment which produces logs, which re
Cool! Thanks Karsten. If you send me a github pull request I'll incorporate it.
Hadley
On Thursday, June 10, 2010, Karsten Loesing wrote:
> Hi everyone,
>
> here's the same patch as a new branch on GitHub.
>
> http://github.com/kloesing/ggplot2/commit/a25e4fbfa4017ed1
>
> Best,
> --Karsten
>
>
On Thu, Jun 3, 2010 at 4:06 PM, Wu Gong wrote:
>
> Hope it helps.
>
> text <- "var1 var2
> 9G/G09 abd89C/T90
> 10A/T9 32C/C
> 90G/G A/A"
>
> x <- read.table(textConnection(text), header = T)
Or with the stringr package:
library(stringr)
str_match(x$var1, "(.)/(.)")
Hadley
--
Hi Arnaud,
I've added this case to the set of test cases in plyr and it will be
fixed in the next version.
Hadley
On Tue, Jun 1, 2010 at 2:33 PM, arnaud Gaboury wrote:
> Maybe not the cleanest way, but I create a fake data frame with one row so
> ddply() is happy!!
>> if (nrow(futures)==0) futu
Hi Karsten,
There's no easy way to do this because behind the scenes geom_ribbon
uses grid.polygon.
Hadley
On Sun, May 30, 2010 at 7:26 AM, Karsten Loesing
wrote:
> Hi everyone,
>
> it looks like geom_ribbon removes missing values and plots a single
> ribbon over the whole interval of x values.
Use aaply from the plyr package.
Hadley
On Thu, May 27, 2010 at 6:24 AM, Johannes Graumann
wrote:
> Hi,
>
> Why is the result of below "apply" call rotated with respect to the input
> and how to remedy this?
>
> Thanks, Joh
>
> .ZScore <- function(input){
> #cat(input,"\n")
> z <- (input - mea
On Tue, May 25, 2010 at 8:39 AM, Mohan L wrote:
>
>
> On Tue, May 25, 2010 at 6:59 PM, Hadley Wickham wrote:
>>
>> > I trying to get a new data frame for 1 bedroom using cast. But I am not
>> > able
>> > to get the below data for 1 Bedroom using cost.
&
> I trying to get a new data frame for 1 bedroom using cast. But I am not able
> to get the below data for 1 Bedroom using cost.
>
> State Jan Feb
> xxx 2 0
> yyy 2 2
> zzz 1 0
What do those numbers represent?
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Depar
> precip.1 <- subset(DF, precipitation!="NA")
> b <- ddply(precip.1$precipitation, .(precip.1$gauge_name), cumsum)
> DF.precip <- precip.1
> DF.precip$precipitation <- b$.data
I suspect what you want here is
ddply(precip.1, "gauge_name", transform, precipitation = cumsum(precipitation))
Hadley
Hi Giovanni,
Have a look at the classifly package for an alternative approach that
works for all classification algorithms. If you provided a small
reproducible example, I could step through it for you.
Hadley
On Sat, May 15, 2010 at 6:19 AM, Giovanni Azua wrote:
> Hello,
>
> I have a labelled
Hi Ken,
Could you please provide a small reproducible example? There are
some hints on how to do so at http://gist.github.com/270442.
Hadley
On Wed, May 12, 2010 at 11:22 AM, Ken Minns wrote:
> I am using the package segmented to fit a simple breakpoint regression to a
> large number of sets
You might want to look at the answers to that question on stackoverflow.com:
http://stackoverflow.com/questions/2628621
Hadley
On Sat, May 8, 2010 at 1:59 AM, Ruihong Huang
wrote:
> Hi,
>
> In my memory, "<<-" means assigning via a pointer or alias. But this is not
> officially defined in "R Lan
> As pointed out to me offline, data.table should be added to the list
> of relevant packages as well. Its primary advantage is for large data
> sets as it is very fast. Its interface does take some getting used to
> but its most recent version on CRAN does have several vignettes which
> should e
Hi Giovanni,
The basic idea is:
classiclimits <- aes(x=x[1:100],ymax = classiccis[1:100,e,p,
ymin=classiccis[1:100,e,p,2], colour = "classic")
ownlimits <- aes(x=x[1:100]+0.4,ymax = owncis[1:100,e,p,1],
ymin=owncis[1:100,e,p,2], colour = "own")
rbootlimits <- aes(x=x[1:100]+0.8,ymax = rbootcis
t with widths equal to the durations
> of the recessions. Once I create a free-standing plot, I'd like to be able
> to use it in various other contexts, including adding it to other existing
> plots. The alternative is to reconstruct the plot as a layer and add it to
> the other plots, but t
> The problem is that I want HCount and HProbCount to use custom
> gradients. i.e. a colour for 0-10, next shade for 10-30, next for 30-70
> etc.
Use cut to create factor with those levels, and then scale_fill_manual
to match values to colours.
> Due to some magic done on the data, one uses inter
Yes.
Hadley
PS. If you provide a small reproducible example you are bound to get
more useful answers ;)
On Tue, Apr 27, 2010 at 8:45 AM, arnaud chozo wrote:
> Hi all,
>
> I'd want to plot a table of numbers such that the values are represented by
> gray level. Is there an easy way to do that u
> Many people seem to be reluctant to define functions,
> even thought I think it is a pretty small step from
> writing scripts to writing functions.
I'm not so sure - I find most students struggle to grasp that next
level of abstraction. Generalising from a specific task to a general
function is
On Sat, Apr 24, 2010 at 12:54 PM, Peter Ehlers wrote:
> Well, this has seriously gotten off the original topic.
>
> While Hadley makes some sense, it is nevertheless
> sometimes the case (surely so for David, I would surmise)
> that one is putting together a response to an R-help
> query when a ne
> Perhaps, true in some respects. I am still chiseling out work using
> primitive editing tools. But it still takes several minutes to load the
> objects I am working on into memory and then several minutes each to build
> new models. The models still reside in memory, since I do not know any
> met
>> If clearing out your workspace destroys *any* work, then something is
>> seriously wrong with your workflow.
>>
>
> Yes, of course. Lets all post viruses to run on each others'
> machines. That will teach those users who don't run antivirus and
> backup software between each posting to r-help.
>> rm(list=ls())
>
> PLEASE, DON'T DO THAT. Or rather you can do it in your workspace but
> don't post it. It's not fair to a person who may not read your code line by
> line before pasting it into their workspace and having it wiped out. Do you
> expect us to completely clear out our workspace
Use geom_segment and calculate the end points yourself.
Hadley
On Fri, Apr 23, 2010 at 8:38 AM, arnaud chozo wrote:
> Hi all,
>
> I'd want to plot a segment from a line specified by slope and intercept.
> I want to plot this line between two limits, x1 and x2, without imposing
> these limits to t
Hi Liam,
Unfortunately this currently isn't supported. It's on my to do list:
http://github.com/hadley/ggplot2/issues/issue/94
Hadley
On Tue, Apr 20, 2010 at 7:59 PM, Liam Blanckenberg
wrote:
> Hi all,
>
> I have a question about setting arbitrary breaks/labels when using GGPLOT
> and date/tim
On Mon, Apr 19, 2010 at 5:13 AM, Paul Rigor (ucla) wrote:
> Hi all,
> I'm an R novice.
>
> I have data that's already formatted as "molten" that reshape should be able
> to work with. For example, the following was read in with
> read.csv(filename,sep=" ", header=FALSE)
>
> V1 V
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