> To answer one of your other questions: ggplot (and lattice) is/are
> very powerful, but base graphics are (a) easier to get your head around
> and (b) easier to adjust if you don't like the defaults. Changing things
> just a little bit in ggplot can be difficult (as an example, the answer to
>
On Sun, Oct 18, 2009 at 10:29 AM, John Kane wrote:
> Thanks Stefan, the annotate approach works beautifully. I had not got that
> far in Hadley's book apparently :(
>
> I'm not convinced though that the explaination
>
>> you shouldn't use aes in this case since nampost,
>> temprange, ... are not
> What are the values of
>
> length((Ht_cm[type=='SD'][from_treeline=='above'])[1])
I suspect the error is in the subsetting - the following seems more plausible:
Ht_cm[type=='SD' && from_treeline=='above']
Hadley
--
http://had.co.nz/
__
R-help@r-p
Hi John,
Could you please provide a small reproducible example?
Thanks,
Hadley
Sent from my iPhone
On 26/10/2009, at 6:50 PM, Jonathan Bleyhl > wrote:
I'm trying to plot values based on a date and then overlay a
histogram also
by date. The problem is that both data sets don't have exact
Hi Bryan,
Thanks for the reproducible example. The problem is actually in your
code, not mine ;) You probably want: y = min(res, na.rm = TRUE) - 0.1
* diff(range(res, na.rm = TRUE))
Hadley
(drop = TRUE solves a difference problem - it controls whether or not
to remove bins with zero count)
On
> Do you have write permission in C:\Program Files\R\R-2.9.2\library? It
> could be that the installer just tried to create the QRMlib subdir, and
> failed, and that's why it doesn't exist.
One possible reason for failure is that your virus checker prevented
the R installer from creating a new di
> I can reproduce it with for example
> x=c(-9.23, -9.56, -1.40)
>
> But adding a single positive number, even .001, fixes it, while
> adding a similar negative number introduces a new error message, so it
> really looks like a bug in ggplot2 when all the values are negative.
>
> Report it to t
On Wed, Oct 28, 2009 at 8:19 PM, Bill Gillespie wrote:
> I currently use lattice functions to produce multiple pages of plots using
> the "layout" argument to specify the number of rows and columns of panels,
> e.g.,
>
> xyplot(price ~ carat | clarity, diamonds, layout = c(2, 2))
>
> This results
If you package "depends" on another package, it will be automatically installed.
Hadley
On Mon, Nov 2, 2009 at 12:56 PM, Jonathan Greenberg
wrote:
> R-helpers:
>
> I'm working on an r-package that I want to make as easy-to-use as possible
> for a novice R-user, which includes automatically ins
Hi Paul,
You might want to try the gray colour scale - scale_fill_grey()
Unfortunately grid (the underlying graphics library that ggplot2 uses)
does not currently support patterns.
Hadley
On Wed, Nov 4, 2009 at 4:17 AM, Paul Chatfield wrote:
>
> Am trying to produce a graph which prints out w
> If readShapePoly() (deprecated - use readShapeSpatial() instead) says that
> the data are not polygons, then they are not. If you want to fill
> administrative boundaries polygons, you need polygons, not lines. The source
> you are using is based on OpenStreetMaps, so more likely to be lines, and
Hi all,
Is there a tool in base R to extract matched expressions from a
regular expression? i.e. given the regular expression "(.*?) (.*?)
([ehtr]{5})" is there a way to extract the character vector c("one",
"two", "three") from the string "one two three" ?
Thanks,
Hadley
--
http://had.co.nz/
an wrote:
> Is this what you want:
>
>> x <- ' one two three '
>> y <-
>> sub(".*?([^[:space:]]+)[[:space:]]+([^[:space:]]+)[[:space:]]+([ehrt]{5}).*",
> + "\\1 \\2 \\3", x, perl=TRUE)
>> unlist(strspl
Have a look at ?split
Hadley
On Mon, Nov 9, 2009 at 10:41 AM, agm. wrote:
>
> Hello,
>
> I'm trying to run a loop that will subset my data into specific sets by
> regions and by race/ethnicity. I'm trying to do this fairly compactly, and
> I cannot get this to work.
>
> A "simple" version of th
If you don't want to preserve factor levels when subsetting use
characters. There are very few other differences in behavior.
Hadley
On Tuesday, November 10, 2009, baptiste auguie
wrote:
> Dear list,
>
> subset has a 'drop' argument that I had often mistaken for the one in
> [.factor which remov
>> (x.n <- cast(x.m, id ~ var, function(.dat){
> + if (length(.dat) == 0) return(0) # test for no data; return
> zero if that is the case
> + mean(.dat)
> + }))
Or fill = 0.
Hadley
--
http://had.co.nz/
__
R-help@r-project.org mailing list
h
> See the above example. Is there a way to make 'drop=FALSE' as global
> default, so that when I say 'tmp[,1]', R will treat it as
> 'tmp[,1,drop=FALSE]'?
The following code won't change the defaults, but it would at least
let you know when you're making the mistake:
trace_all <- function(fs, tra
>> aggregate(dat$value, list(dat$x, dat$y), mean)
> Group.1 Group.2 x
> 1 1 1 15
> 2 1 2 30
>> newdat <-aggregate(dat$value, list(dat$x, dat$y), mean)
>> names(newdat) <- c("x","y",bquote("mean(value)") )
That bquote isn't necessary because you're already working with str
Hi all,
Is there a method for escaping strings to be used regular expressions?
i.e. if I have a user supplied string that I'd like to use as a fixed
component is there a method that will turn (e.g.) ".$^" into
"\\.\\$\\^" ?
Thanks,
Hadley
--
http://had.co.nz/
Test.$^", "Test"), fixed = TRUE)
>
> On Fri, Nov 13, 2009 at 11:33 AM, Hadley Wickham wrote:
>> Hi all,
>>
>> Is there a method for escaping strings to be used regular expressions?
>> i.e. if I have a user supplied string that I'd like to use
On Fri, Nov 13, 2009 at 9:53 AM, Peng Yu wrote:
> library(some_library_name)
Try
help(package = some_library_name)
Hadley
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PLEASE do read the pos
> Yes I tried all the basic ones like box plot, pie chart, etc but the data
> representation isnt that clear.
Given that you have neither provided your data, nor explained what you
are trying to uncover from it, what sort of advice do you expect to
get?
Hadley
--
http://had.co.nz/
of
> duties performed by the Secretary from HR dept and 4th column is time
> required to perform the duty
>
> so there are many such posts and dept with varied duties and times resp
>
>
> Regards
>
> Our Thoughts have the Power to Change our Destiny.
> Sunita
>
>
>
Hi Michael,
> I'm having trouble figuring out how to format Date variables when used as
> axis labels in graphs.
> The particular case here is an attempt to re-create Nightingale's coxcomb
> graph with ggplot2,
> where I'd like the months to be labeled as "Mar 1885", "Apr 1885", using a
> date for
You've asked the same question on stackoverflow.com and received the
same answer. This is rude because it duplicates effort. If you
urgently need a response to a question, perhaps you should consider
paying for it.
Hadley
On Sun, Nov 22, 2009 at 12:04 PM, masterinex wrote:
>
> so under which c
Hi David,
The variable names of mma are sp, dentro, variable and value - area is
not a variable in that data frame.
Hadley
On Sun, Nov 22, 2009 at 9:48 PM, David Douterlungne
wrote:
>
>
>
> Dear R Users,
>
>
>
>
> Reshape seems to be very useful for data-manipulation, but I am struggling
> wit
>> mywords<- c("harry","met","sally","subway10","1800Movies","12345", "not
>> correct 123")
>> all.letters <- grep("^[[:alpha:]]*$", mywords)
>> all.numbers <- grep("^[[:digit:]]*$", mywords) # numbers
>> mixed <- grep("^[[:digit:][:alpha:]]*$", mywords)
mywords<- c("harry","met","sally","subway
> I don't really understand what you want and the example solution throws away
> quite a lot of data, so consider this alternative:
>
> data.out2 <- read.table(textConnection("id rater.1 n.1 rater.2 n.2
> rater.3 n.3 rater.4 n.4
> 11 11 0.118 79 NA NA NA NA NA N
Because of the combinatorial nature of ggplot2, it is simply not
possible to provide an example that illustrates every single
combination of options. There are already over 600 example graphics
in the package - if you can't find one that exactly meets your need,
you need to buy the book and learn
On Thu, Dec 3, 2009 at 3:52 PM, John Filben wrote:
> Can R support data manipulation programming that is available in the SAS
> datastep? Specifically, can R support the following:
> - Read multiple dataset one record at a time and compare values from
> each; then base on if-then logic
> If you're after text, then it's probably a matter of locating the element
> that encloses the data you want-- perhaps by using getNodeSet along with an
> XPath[1] that specifies the element you are interest with. The text can
> then be recovered using the xmlValue() function.
And rather than tr
Hi Thomas,
I suspect you want geom_bar(stat = "identity", width = 1), but it's
hard to be sure without a reproducible example.
Hadley
On Thu, Dec 3, 2009 at 8:18 PM, Thomas S. Dye wrote:
> Aloha all,
>
> I love using ggplot. It took a while to get used to the grammar of
> graphics, but it is
> I'm not sure what the other lines(but the last) have to do anything, but are
> you looking for something like this:
>
> do.call(rbind, sapply(paste(1:10, 1:10), strsplit, split=' '))
strsplit is already vectorised wrt its first argument, so all you need is:
do.call(rbind, strsplit(paste(1:10,
t; |
> | "" | 90 | 0 | "13 - 24" |
> | "" | 67.5 | 0 | "13 - 24" |
> | "" | 45 | 0 | "13 - 24" |
> | "" | 22.5 | 0 | "13 - 24" |
> | "" | 360 | 1 | "13 - 24
ce
> all my graphs with ggplot2. If there is a bug that swallows the 16th bar,
> though, then I'll make my wind rose with another package and wait patiently
> until ggplot2 plots the full compass.
> Thanks again for a terrific software package.
> All the best,
> Tom
> Begi
> The idea of plotting a wind rose must be fairly common. I wonder if it
> would make sense to have a switch that would wrap data around the ends of a
> continuous scale?
Probably - but it requires a lot of work, because ggplot2 doesn't
currently support circular scales, which is what you really
You mean:
dat <- data.frame(x = rnorm(100))
dat1 <- data.frame(
x = c(0,0),
y = c(1,0),
Label = c("Point1", "Point2")
)
ggplot(dat, aes(x)) +
geom_histogram(aes(fill = ..count..)) +
geom_point(aes(x, y, colour = Label), data = dat1, size = 4) +
scale_fill_gradient("Count", low = "gre
Hi Mark,
Why are you using factors? I think for this case you might find
characters are faster and more space efficient.
Alternatively, you can have a look at the plyr package which uses some
tricks to keep memory usage down.
Hadley
On Tue, Dec 8, 2009 at 9:46 PM, Mark Kimpel wrote:
> Charles
On Wed, Dec 9, 2009 at 9:43 PM, Rolf Turner wrote:
>
> I am working with a somewhat complicated structure in which
> I need to deal with a function that takes ``basic'' arguments
> and also depends on a number of parameters which change depending
> on circumstances.
>
> I thought that a sexy way o
On Wed, Dec 9, 2009 at 4:48 PM, David Reiss wrote:
> Ideally I would like to be able to use the function f (in my example)
> as-is, without having to designate the environment as an argument, or
> to otherwise have to use "e$x" in the function body.
e <- new.env()
e$x <- 3
f <- function(xx) x <<-
For the case below, you don't need to know anything about how R
manages memory, but you do need to understand basic concepts
algorithmic complexity. You might find "The Algorithm Design Manual",
http://www.amazon.com/dp/1848000693, a good start.
Hadley
On Thu, Dec 10, 2009 at 10:26 AM, Peng Yu
Hi Sunita,
To get the bars, you want:
ggplot(mydata, aes(x = factor(jobno), y = recruits)) + geom_bar()
and to add the cumulative sum, first add it to the data:
mydata$cum_ recruits <- cumsum(mydata$recruits)
and then add another layer:
+ geom_line(aes(y = cum_recruits, group = 1))
Hadley
O
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
> Is there a version of apply that returns a list without NULL's?
>
> I try to remove NULL elements in the following example, but neither
> for loops work. Would you please let me know what the correct way is?
Try this function:
compact <- function(x) Filter(Negate(is.null), x)
compact(x)
Hadley
Hi Dave,
I have a few drills available from http://had.co.nz/stat405 - see the
right hand column, about half way down. They seem similar in spirit
to what you're thinking of. You might want to look at the "Little
Schemer" for a similar approach with a different programming language.
However, I'm
> A very common situation is that the users don't know all the possible
> return types of 'some_third_party_function()'. If the users don't know
> all the return types, he/she can not make sure the return type of
> function(x) {...} be always the same. How do you deal with this case?
It's not that
Hi Luc,
You want:
legend.title=theme_text(size=20, hjust = 0)
So the legend title is left aligned, not centred.
Hadley
On Fri, Dec 11, 2009 at 9:26 AM, MUHC_Research
wrote:
>
> Dear R-users,
>
> I am preparing graphs for an upcoming article using the different functions
> of the ggplot2 pack
> I don't understand what these addresses mean. Would you please help me
> understand it?
Did you try reading the documentation?
When an object is traced any copying of the object by the C
function ‘duplicate’ or by arithmetic or mathematical operations
produces a message to standa
> I hope I have missed a better way to do this in R. Otherwise, I
> believe what I'm after is some kind of C or C++ macro expansion,
> because the number of loops should not be hard coded.
Why not generate the list of integers that sum to 17, and then mix
with 0s as appropriate?
Hadley
--
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
On Wed, May 6, 2009 at 8:12 PM, jim holtman wrote:
> Ths should do it:
>
>> do.call(rbind, lapply(split(x, x$ID), tail, 1))
> ID Type N
> 45900 45900 I 7
> 46550 46550 I 7
> 49270 49270 E 3
Or with plyr:
library(plyr)
ddply(x, "id", tail, 1)
plyr encapsulates the common split-
On Sun, May 10, 2009 at 10:32 AM, Zeljko Vrba wrote:
> Searching the mail archives I found that using legend.position as in
> p.ring.3 + opts(legend.position="top")
>
> is a known bug. I tried doing
> p.ring.3 + opts(legend.position=c(0.8, 0.2))
>
> which works, but the legend background is trans
> This does it more or less your way:
>
> ds <- split(df, df$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 a
On Thu, May 14, 2009 at 12:16 PM, Lori Simpson
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, e.g.:
>
> result=cast(tabl
On Thu, May 14, 2009 at 6:21 PM, Ping-Hsun Hsieh 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 Col3 Co
On Thu, May 14, 2009 at 2:14 PM, Garritt Page 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), times=2)
>
>
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 wrote:
> Hi Stephen,
>
> The problem is that the label on the graph doesn't get rendered with a
> superscrip
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 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, I'll
> just re
.
All proceeds go to the GGobi Foundation to support graphics research.
Find out more, and book your tickets online at
http://lookingatdata.com
Regards,
Hadley Wickham
Dianne Cook
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https://stat.ethz.ch/mailman/listinfo
>> Let's see if I understand this. Do I iterate through
>> x <- factor(x, levels(c(levels(x), NA), exclude=NULL)
>> for each of the few hundred variables (x) in my data frame?
>
>
> Yes, for all being factors.
Wouldn't addNA() be the preferred method?
To do it for all variables is pretty simp
On Mon, Jun 1, 2009 at 2:18 PM, stephen sefick 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,
> 12600, 12631,
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 wrote:
> Dear All,
>
> Slightly off -non technical topic ( but hey it is Friday)
>
> Following last week's interview with REvolution Computing which m
On Sat, Jun 6, 2009 at 5:02 PM, Adam D. I. Kramer 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 scan(file, what, nmax,
On Mon, Jun 8, 2009 at 10:29 AM, Herbert
Jägle 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 there are 2 sets of four
> va
On Mon, Jun 8, 2009 at 8:56 PM, Mao Jianfeng 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 7.3
> 4
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
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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 far
ck
> Sent: Thursday, June 18, 2009 9:17 AM
> 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 Wickham wrote:
>> Hi all,
>>
>> Do you k
> 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).
>
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
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 Na 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 others to dates. R
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
Hugh-Jones wrote:
> That
On Sun, Mar 22, 2009 at 5:06 PM, Blanchette, Marco 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 item have been
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 wrote:
> Personally I find those types of plots difficult to interpret. Much easier
> to create, view, and interpret is to simply plot the lines from densi
On Mon, Mar 30, 2009 at 10:33 AM, Mike Lawrence 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
names(cust1_files) <-
On Mon, Mar 30, 2009 at 2:58 PM, Mike Lawrence 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')
> , .fun <- fu
> 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
On Tue, Mar 31, 2009 at 11:31 AM, baptiste auguie 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 5.
On Tue, Mar 31, 2009 at 11:12 AM, Steve Murray 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 file names) - th
On Tue, Mar 31, 2009 at 5:01 PM, Marianne Promberger
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 understand
> factors!)
>
> 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 wrote:
>> I tried messing with the line df$FixTime[which.min(df$FixInx)] changing it
>> to df[which.min(df$FixInx)] or adding new lines with the additional columns
>> that I want to include, but nothing seemed to work. I'
> Earlier I posted a question about memory usage, and the community's input was
> very helpful. However, I'm now extending my dataset (which I use when
> running a regression using lm). As a result, I am continuing to run into
> problems with memory usage, and I believe I need to shift to impl
On Thu, Apr 2, 2009 at 3:37 PM, Rowe, Brian Lee Yung (Portfolio
Analytics) wrote:
> Is this what you want:
>> d1[which(id != 4),]
Or just
d1[id != 4, ]
Hadley
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> X1 X2
> 1 11 0
> 2 11 0
> 3 11 0
> 4 11 1
> 5 12 0
> 6 12 0
> 7 12 0
> 8 13 0
> 9 13 1
> 10 13 1
>
>
> and I want to select all rows pertaining to factor levels of X1 for
> which exists at least one "1" for X2. To be clear, I want rows 1:4
> (since there exists at least one o
On Fri, Apr 3, 2009 at 4:43 AM, baptiste auguie wrote:
> Dear all,
>
> I'm puzzled by the following example inspired by a recent question on
> R-help,
>
>
> cc <- textConnection("user_id website time
> 20 google 0930
> 21 yahoo 0935
> 20 faceboo
On Fri, Apr 3, 2009 at 8:43 AM, baptiste auguie wrote:
> That makes sense, so I can do something like,
>
> count <- function(x){
> as.integer(unclass(table(x)))
> }
>
> count(d$user_id)
>
> ddply(d, .(user_id), transform, count = count(user_id))
>
>> user_id website time count
>> 1 2
On Fri, Apr 3, 2009 at 1:45 PM, 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 "3"
> $
On Sat, Apr 4, 2009 at 12:09 PM, ds wrote:
>
> I have a data frame something like:
> name wrist
> nLevel emot
> 1 4094 3.34 1
> frustrated
> 2 4094 3.94 1
> frustra
On Sat, Apr 4, 2009 at 12:28 PM, jim holtman 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
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 wrote:
>
> I would
On Mon, Apr 6, 2009 at 8:49 AM, Daniel Brewer 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"
>
> $`1029`
> [1] "04110" "04115"
>
> And I would like to get a data frame like the following:
>
On Mon, Apr 6, 2009 at 9:34 AM, Stavros Macrakis wrote:
> There are various ways to do this in R.
>
> # sample data
> dd <- data.frame(a=1:10,b=sample(3,10,replace=T),c=sample(3,10,replace=T))
>
> Using the standard built-in functions, you can use:
>
> *** aggregate ***
>
> aggregate(dd,list(b=dd$
On Mon, Apr 6, 2009 at 10:40 AM, baptiste auguie 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))
> ddply(test,.(a,b),.fun=
On Mon, Apr 6, 2009 at 5:31 PM, Jun Shen 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 apply multiple f
Have a look at ?gpar - it will tell you about lineheight.
Hadley
On Tue, Apr 7, 2009 at 3:28 AM, Mark Heckmann 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 spacing\n is too big
On Tue, Apr 7, 2009 at 8:44 AM, 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 with:
>
>> molten <- mel
On Tue, Apr 7, 2009 at 4:41 PM, Jorge Ivan Velez
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, mydata[,1], mean)
Hadley
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