Dear All,
I have a large data frame ( 270 lines and 14 columns), and I would like to
extract the information in a particular way illustrated below:
Given a data frame "df":
> col1=sample(c(0,1),10, rep=T)
> names = factor(c(rep("A",5),rep("B",5)))
> df = data.frame(names,col1)
> df
names
gers does
> t4 <- system.time(res <- which(as.integer(x) == match("A", levels(x
> print(t4/t1);
> usersystem elapsed
> 0.417 0.000 0.3636364
>
> So, the latter seems to be the fastest way to identify those elements.
>
> My $.02
>
> /Hen
l example
> that shows what you have and what you want?
>
> Is ?split what you are after?
>
> Emmanuel Levy wrote:
>>
>> Dear Peter and Henrik,
>>
>> Thanks for your replies - this helps speed up a bit, but I thought
>> there would be something much faster.
>>
are doing. Can you make a small example
> that shows what you have and what you want?
>
> Is ?split what you are after?
>
> Emmanuel Levy wrote:
>>
>> Dear Peter and Henrik,
>>
>> Thanks for your replies - this helps speed up a bit, but I thought
>> t
Dear list members,
I encountered this problem and the solution pointed out in a previous
thread did not work for me.
(e.g. install.packages("RCurl", repos = "http://www.omegahat.org/R";)
I work with Ubuntu Hardy, and installed R 2.6.2 via apt-get.
I really need RCurl in order to use biomaRt ...
oblem should disappear. It relates to encoding of strings.
>
> D.
>
> Emmanuel Levy wrote:
>> Dear list members,
>>
>> I encountered this problem and the solution pointed out in a previous
>> thread did not work for me.
>> (e.g. install.packages("
Dear All,
I'm sure this is not the first time this question comes up but I
couldn't find the keywords that would point me out to it - so
apologies if this is a re-post.
Basically I've got thousands of points, each depending on three variables:
x, y, and z.
if I do a plot(x,y, col=z), I get somet
in
> the base distribution, which will do exactly what you requested.
>
>
> Bert Gunter
> Genentech Nonclinical Statistics
>
>
>
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
> Behalf Of Emmanuel Levy
> Sent: Wednesday
0 - 0.17.
I haven't looked yet at the locfit package as it is not installed, but
I will check it out!
Thanks for helping!
Emmanuel
On 20/03/2008, David Winsemius <[EMAIL PROTECTED]> wrote:
> "Emmanuel Levy" <[EMAIL PROTECTED]> wrote in
> news:[EMAIL PROTECTED]
Dear list members,
I am using Mclust in order to deconvolute a distribution that I
believe is a sum of two gaussians.
First I can make a model:
> my.data.model = Mclust(my.data, modelNames=c("E"), warn=T, G=1:3)
But then, when I try to plot the result, I get the following error:
> mclust1Dplot(
this would be great; is it possible to
somehow force the parameters (e.g variance) to be
greater than a particular threshold?
Thanks,
Emmanuel
2008/10/20 Emmanuel Levy <[EMAIL PROTECTED]>:
> Dear list members,
>
> I am using Mclust in order to deconvolute a distribution that I
&
,0.15),type="n",xlab=" ",ylab=" ",axes=F, ylim=c(0,0.4) )
axis(side=1)
for (i in 1:2) {
ni <- v$parameters$pro[i]*dnorm(x0,
mean=as.numeric(v$parameters$mean[i]),sd=1)
lines(x0,ni,col=1)
nt <- nt+ni
}
lines(x0,nt,lwd=3)
segments(my.data,0,my.data,0.02)
Best,
Dear All,
I have a distribution of values and I would like to assess the
uni/bimodality of the distribution.
I managed to decompose it into two normal distribs using Mclust, and
the BIC criteria is best for two parameters.
However, the problem is that the BIC criteria is not a P-value, which
I wo
Hi Duncan,
I'm really stupid --- yes of course!!
Thanks for pointing me out the (now) obvious.
All the best,
E
2008/10/21 Duncan Murdoch <[EMAIL PROTECTED]>:
> On 10/21/2008 2:56 PM, Emmanuel Levy wrote:
>>
>> Dear All,
>>
>> I have a distribution of
Dear All,
I hope the title speaks by itself.
I believe that there should be a solution when I see what Mclust is
able to do. However, this problem is quite
particular in that d3 is not known and does not necessarily correspond
to a common distribution (e.g. normal, exponential ...).
However it mu
Dear All,
I have a long string and need to search for regular expressions in
there. However it becomes horribly slow as the string length
increases.
Below is an example: when "i" increases by 5, the time spent increases
by more! (my string is 11,000,000 letters long!)
I also noticed that
- the s
Hi Chuck,
Thanks a lot for your suggestion.
> You can find all such matches (not just the disjoint ones that gregexpr
> finds) using something like this:
>
>twomatch <-function(x,y) intersect(x+1,y)
>match.list <-
>list(
>which( vec %in% c(3
Dear All,
I just read an announcement saying that Mathematica is launching a
version working with Nvidia GPUs. It is claimed that it'd make it
~10-100x faster!
http://www.physorg.com/news146247669.html
I was wondering if you are aware of any development going into this
direction with R?
Thanks f
Dear Brian, Mose, Peter and Stefan,
Thanks a lot for your replies - the issues are now clearer to me. (and
I apologize for not using the appropriate list).
Best wishes,
Emmanuel
2008/11/19 Peter Dalgaard <[EMAIL PROTECTED]>:
> Stefan Evert wrote:
>>
>> On 19 Nov 2008, at 07:56, Prof Brian Ri
Dear all,
I have been using normalize.loess and I get the following error
message when my matrix contains NA values:
> my.mat = matrix(nrow=100, ncol=4, runif(400) )
> my.mat[1,1]=NA
> my.mat.n = normalize.loess(my.mat, verbose=TRUE)
Done with 1 vs 2 in iteration 1
Done with 1 vs 3 in iteration 1
Dear All,
I cannot find a solution to the following problem although I imagine
that it is a classic, hence my email.
I have a vector V of X values comprised between 1 and N.
I would like to get random samples of X values also comprised between
1 and N, but the important point is:
* I would like
Dear All,(my apologies if it got posted twice, it seems it didn't
get through)
I cannot find a solution to the following problem although I suppose
this is a classic.
I have a vector V of X=length(V) values comprised between 1 and N.
I would like to get random samples of X values also compri
lp me solve it.
Many thanks!
Emmanuel
PS: I apologize that I sent a second post. This one did not appear in
my "R-help" label so I assumed it wasn't sent for some reason.
2009/8/12 Ted Harding :
> On 12-Aug-09 22:05:24, Emmanuel Levy wrote:
>> Dear All,
>&
with this problem? Or even better of a package?
Thanks for your help,
Emmanuel
2009/8/12 Nordlund, Dan (DSHS/RDA) :
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
>> Behalf Of Emmanuel Levy
>> Sent: Wedne
> But if the 1st order differences are the same, then doesn't it follow that
> the 2nd, 3rd, ... order differences must be the same between the original and
> the new "random" vector. What am I missing?
You are missing nothing sorry, I wrote something wrong. What I would
like to be preserved is
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