Ok, we decided to have a shot at modifying gregexpr. Let's see how it
works out. If anybody is interested in discussing this please contact
me. R-help doesn't seem like the right place for further discussion.
Is there a default place for discussing things like that?
Thanks everybody for your
Bill, Michael,
good to see I'm not the only one who sees potential for improvements
in the regexpr domain. Adding a subpattern argument is certainly a
step in the right direction and would make my life much easier.
However, in my application I need to know not only the position of one
group but
On Wed, Sep 29, 2010 at 1:58 PM, Michael Bedward
michael.bedw...@gmail.com wrote:
How is your C coding ? Bill ? Anyone else ? I could have a got at
writing some prototype code to test in the next few days, though if
someone else with decent C skills is itching to do it please speak up.
We
On Tue, Sep 28, 2010 at 9:46 AM, Michael Bedward
michael.bedw...@gmail.com wrote:
What Titus wants to do is akin to retrieving capturing groups from a
Matcher object in Java.
Precisely. Here's the description:
Dear list!
gregexpr(a+(b+), abcdaabbc)
[[1]]
[1] 1 5
attr(,match.length)
[1] 2 4
What I want is the offsets of the matches for the group (b+), i.e. 2
and 7, not the offsets of the complete matches. Is there a way in R
to get that?
I know about gsubgn and strapply, but they only give me the
]])
# now determine where 'b' starts
justA[[1]][indx] + attr(justA[[1]], 'match.length')[indx]
[1] 2 7 17
On Mon, Sep 27, 2010 at 11:48 AM, Titus von der Malsburg
malsb...@gmail.com wrote:
Dear list!
gregexpr(a+(b+), abcdaabbc)
[[1]]
[1] 1 5
attr(,match.length)
[1] 2 4
What I want
On Mon, Sep 27, 2010 at 7:16 PM, Henrique Dallazuanna www...@gmail.com wrote:
You've tried:
gregexpr(b+, abcdaabbc)
But this would match the third occurrence of b+ in abcdaabbcbb. But
in this example I'm only interested in b+ if it's preceded by a+.
Titus
On Mon, Sep 27, 2010 at 7:29 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Try this zero width negative look behind expression:
gregexpr((?!a+)(b+), abcdaabbc, perl = TRUE)
[[1]]
[1] 2 7
attr(,match.length)
[1] 1 2
Thanks Gabor, but this gives me the same result as
gregexpr(b+,
Hi list,
I run R on Linux and OSX. On both systems I use R version 2.9.2 (2009-08-24)
and reshape version: 0.8.2 (2008-11-04). When I do a melt with
na.rm=T on a data frame I get different results on these systems:
library(reshape)
x - read.table(textConnection(char trial wn
p E10I13D0 4
r
On Sat, Feb 6, 2010 at 8:23 PM, hadley wickham h.wick...@gmail.com wrote:
The latest version of reshape is 0.8.3 - perhaps upgrading will fix
your problem.
Thanks for your response, Hadley!
I just did the upgrade on the Linux system. On OSX I was already at
0.8.3. Now, I get the same result
Ok, I studied the source code of melt.data.frame. With na.rm=T melt
operates normally except that it deletes rows from the molten
data.frame that have NAs in the value column. NAs in the id.vars are
not touched. This could be clearer in the documentation especially as
it seems that earlier
Mean and variance of Poisson distributed data are specified by \rho.
How can I estimate \rho for a set of measurements in R?
Many thanks!
Titus
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
On Fri, Jan 15, 2010 at 09:19:23AM -0500, David Winsemius wrote:
On Jan 15, 2010, at 5:59 AM, Titus von der Malsburg wrote:
Mean and variance of Poisson distributed data are specified by \rho.
How can I estimate \rho for a set of measurements in R?
rho - mean(x)
Yeah, thanks :-) I
On Fri, Jan 15, 2010 at 08:33:52AM -0700, Peter Ehlers wrote:
Why would anyone complain? You're free to call it 'applesauce'
if that suits you.
Good idea, I will do this from now on! ;-)
What do you mean by 'general way to fit a distribution'?
Maximum likelihood might be one way.
Somebody
On Wed, Dec 9, 2009 at 12:11 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Here are a couple of solutions. The first uses by and the second sqldf:
Brilliant! Now I have a whole collection of solutions. I did a simple
performance comparison with a data frame that has 7929 lines.
The
http://www.rseek.org/
It is particularly useful to search the mailing list archives of
r-help with rSeek. No matter what kind of problem you have, somebody
has had it before and asked on r-help.
Titus
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R-help@r-project.org mailing list
Hi, I have a data frame and want to merge adjacent rows if some condition is
met. There's an obvious solution using a loop but it is prohibitively slow
because my data frame is large. Is there an efficient canonical solution for
that?
head(d)
rt dur tid mood roi x
55 5523 200 4 subj
On Tue, Dec 8, 2009 at 4:50 PM, Gray Calhoun gray.calh...@gmail.com wrote:
I think there might be a problem with this approach if roi, tid, rt,
and mood are the same for nonconsecutive rows.
True, but I can use the index of my reshape solution. Aggregate was
the crucial ingredient. Thanks
On Tue, Dec 8, 2009 at 5:19 PM, Nikhil Kaza nikhil.l...@gmail.com wrote:
I suppose that is true, but the example data seem to suggest that it is
sorted by rt.
I was not very clear on that. Sorry.
d$count - 1
a - with(d, aggregate(subset(d, select=c(dur, x, count),
When loading mclust, it shows a license agreement message. This
message shows up in my document when I use Sweave. I did the
following:
echo=FALSE,include=FALSE=
library(mclust)
@
Is this a problem with mclust, Sweave or with me? How can it be fixed?
Thanks for any suggestions!
Titus
Hi list!
An operation that I often need is splicing two vectors:
splice(1:3, 4:6)
[1] 1 4 2 5 3 6
For numeric vectors I use this hack:
splice - function(x, y) {
xy - cbind(x, y)
xy - t(xy)
dim(xy) - length(x) * 2
return(xy)
}
So far, so good (?). But I also need
On Tue, Jun 09, 2009 at 11:23:36AM +0200, ONKELINX, Thierry wrote:
For factors, you better convert them first back to character strings.
splice - function(x, y) {
x - levels(x)[x]
y - levels(y)[y]
factor(as.vector(rbind(x, y)))
}
Thank you very much, Thierry!
I
Have a look at the merge function.
Merge two data frames by common columns or row names, or do other
versions of database _join_ operations.
Titus
On Tue, Jun 09, 2009 at 05:48:05AM -0700, Jason Rupert wrote:
I've got two data.frames and, when certain keys match, I would like to
An example:
schoolZone1_df - data.frame(address=101, schoolZone=Sherman)
schoolZone2_df - data.frame(address=108, schoolZone=Baker)
schoolZone_df - rbind(schoolZone1_df, schoolZone2_df)
schoolZone_df
address schoolZone
1 101Sherman
2 108 Baker
neighborhoodInfo1_df -
You can use barchart in package lattice. Here's a rough sketch:
library(lattice)
dataA - rep(1:4, c(3,2,2,4))
dataB - rep(1:4, c(5,4,3,2))
da - data.frame(table(dataA))
db - data.frame(table(dataB))
da$cond - a
db$cond - b
colnames(da)[1] - data
On Tue, Jun 09, 2009 at 04:39:29PM +0200, Titus von der Malsburg wrote:
is there a way to plot both of them in one plot, so that the bars for
value 1 (dataA: 3, dataB: 5) would appear side by side, followed by
the bars for value 2 etc.?
Oh, I see you want something different. I should've
On Tue, Jun 09, 2009 at 11:04:03AM -0400, Stavros Macrakis wrote:
This may seem like a minor point, but I think it is worthwhile using
descriptive names for functions.
Makes sense. I thought I've seen this use somewhere else (probably in
Lisp?). What better name do you suggest for this
Hi list,
I have a function that detects saccadic eye movements in a time series
of eye positions sampled at a rate of 250Hz. This function needs
three vectors: x-coordinate, y-coordinate, trial-id. This information
is usually contained in a data frame that also has some other fields.
The names
Hi Zeljko,
thanks for your suggestion!
On Tue, May 12, 2009 at 12:26:48PM +0200, Zeljko Vrba wrote:
Why not simply rearrange your data frames to have standardized column names
(see names() function), and write functions that operate on the standardized
format?
Actually that's what I'm
Hi list! I have a large matrix which I'd like to partition into blocks
and for each block I'd like to compute the mean. Following a example
where each letter marks a block of the partition:
a a a d g g
a a a d g g
a a a d g g
b b b e h h
b b b e h h
c c c f i i
looking for a way to get the means more or less at
once because the matrix is pretty large and doing it block-wise is too
slow.
Thanks again!
Titus
On 2/16/09, Titus von der Malsburg malsb...@gmail.com wrote:
Hi list! I have a large matrix which I'd like to partition into blocks
On Mon, Feb 16, 2009 at 11:08:24AM -0800, Christian Langkamp wrote:
I could of course log the whole data
set, but then explaining that transformation within a presentation is
generally not a pleasant exercise.
You don't have to explain it. Just calculate the hist of the log and
label the axis
Hi list! I have a data frame called fix and a list of index vectors
called rois:
head(rois, 3)
[[1]]
[1] 2 1
[[2]]
[1] 3
[[3]]
[1] 6 7 28 26 27 24 25
The part that's causing the issue is the following line:
lapply(rois, function(roi) fix$x[roi] - 100)
So for every index
I have a data frame called s3. This data frame has a column called
saccade which has two levels 1 and -1.
head(s3$saccade, 100)
[1] NA NA NA NA -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
[26] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1
[51]
Hi Tomek, have a look at R News, Volume 3/3, December 2003. There you
find an article about different algorithms that are available in R.
Titus
On Sat, Jan 31, 2009 at 01:36:29AM +0100, Tomek Wlodarski wrote:
now I see that cmdscale is not the best option for my problem
So I am wondering if
I have a data frame that is the result of a cast (reshape) operation. I
deleted the variable column and tried to melt the resulting data frame.
Depending on which method I use to delete the column I get different
error messages when melting:
head(tinfos)
vpn group trial_no item
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