I believe you could also set your subscription to NOMAIL and then read the
posts from the R-help archive. This would also allow you to post to R-help
since you are still subscribed.
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
> -Original Message-
> From: r-hel
t; [1] 0
> > C = t(A)%*%B%*%A
> > (max(abs(C - t(C
> [1] 3.552714e-15
>
> Any help on the matter would be very much appreciated.
>
>
Welcome to the world of floating-point calculation on finite precision
computers. You need to read R FAQ 7.31. Your maximum difference
site sampdate TDS Cond SO4 p4 p5 p6 p7
BC-0.5 1996-06-02 530 NA 194 NA NA NA NA
if you have other lines for site/sampdate
BC-0.5 / 1996-06-02
then you would use the quant values to fill in for Cond and the other
paramenters. Then your xyplot code should work. Someone else will need
There is a friedman.test() function. Any reason you want to do it by hand?
If so, you can do:
#Simulated data matrix
x<-matrix(rnorm(9),3,3,byrow=T)
x
#Rank matrix
r<-matrix(rank(x),dim(x))
HTH,
Daniel
JohnnyJames wrote:
>
> My data loo
;-factor(column)
contrasts(row)<-contr.treatment(levels(row))
contrasts(column)<-contr.treatment(levels(column))
# Works for Terps
fit.terp<-glm(counts ~ row + column + row*column,
family=poisson(link="log"))
summary(fit.terp)
HTH,
Daniel Malter
University of Maryland, College Park
in
your data (100) when it should be of length (1), i.e., the sum of the
former. If you define log.lik as the appropriate sum, you should be
successful.
HTH.
Daniel
djbanana wrote:
>
> I am trying to run this code and obtain the MLEs for my parameters.
> However I am getting this error a
we can copy paste to the R-prompt). Moreover, chances are that when you try
to reproduce the error with simulated code, you will figure out what your
mistake is.
Best,
Daniel
shish matt wrote:
>
> I have a spatial weight file in csv that I want as listw object in R.
> The file has the
,nrow=5, ncol=10)
repeat{
c1 <- sample(0:10, 4, replace=TRUE)
if(sum(c1) <= 10) break
}
n[,1] <- c(c1,10-sum(c1))
n
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
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lar results.
>
> Puzzled,
>
> Rich
>
Rich,
I don't see a 'data=' parameter in your call to lm(). How does lm() know where
to find the variables referenced in the model parameter?
If that is not the problem, then we need to see str() output for the data frame
th
t there was 'nothing to model.' If that were the
case, you would have gotten non-significant parameter estimates, not NA's. I
would guess that there is something problematic with the how the data frame is
structured relative to what lm() is expecting. So, I would not give up l
f.
>
You are correct, R can handle files that size with no problem (given sufficient
memory). You say you can't provide a working example. Well, it will be
difficult for anyone to give you a working solution. We don't have access to
the file you are trying to read so we can'
47.513 in the string above just over 47 and one-half minutes? In the
calculation above, the .513 minutes is being treated as 513 seconds (I know
that is what the OP asked for). Shouldn't the calculation actually be
something like this
c(1, 1/60) %*% strapply(x, "\\d+.?\\d+", as.numer
ring that
actually has two consecutive backslashes, which print() will display as four
consecutive backslashes. If you are talking about a variable, tmp, that
actually has two backslashes in it, then it will display like this
> tmp
[1] "C:"
> print(tmp)
[1] "C:\\
#The code of rank 1 in the previous post should have read
#rank1<-apply(iterator1,1,function(x) x+base1)
#corrected code below
siegel.tukey=function(x,y,id.col=TRUE,adjust.median=F,rnd=-1,alternative="two.sided",mu=0,paired=FALSE,exact=FALSE,correct=TRUE,conf.int=FALSE,conf.level=0.95){
if(id.col=
because R itself stops working. I
downloaded R-2.14.2patched and experienced the same problem.Any suggestions
about how to debug this problem?
Thanks,
Dan
Daniel Nordlund
Bothell, WA USA
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https://stat.
<- rpois(50,theta4)
>
> b5 <- rigamma(50,1,1)
> theta5 <- rgamma(50,0.5,(1/b5))
> sim5 <- rpois(50,theta5)
>
>
>
> par(mfrow=c(1,5))
> boxplot(sim1)
> boxplot(sim2)
> boxplot(sim3)
> boxplot(sim4)
> boxplot(sim5);
>
> Thanks,
> Raphael
&
months<-gsub('[[:alpha:]]|[[:punct:]]|0','',x[grep("/M|/m",x)]) #months
months
Convert the resulting character vectors into numeric vectors by
as.numeric(as.character(years)) , for example.
HTH,
Daniel
--
View this message in conte
This just saved me a lot of time.
Thank you!
Daniel
--
View this message in context:
http://r.789695.n4.nabble.com/Remove-leading-and-trailing-white-spaces-tp907851p4489725.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r
t the variables I've created
within R into an SPSS-usable file.
Thanks for your help!
-Daniel
[[alternative HTML version deleted]]
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PLEASE do read the posting
statistical consulting. However, you might
want to read up on Zipf's Law (for example, the link below which gives an
answer to your question), and then come back and ask an R related programming
question.
http://en.wikipedia.org/wiki/Zipf%27s_law
Dan
Daniel Nordlund
Bothell, WA
Hi,
I hope this is not too trivial, but I've had this recurring problem
and I think there is super easy solution, just not sure what it is.
Please see short example below. I would like to get the frequency
(counts) of all the variables in a single column (that is easy), but I
would also like to r
gt;
>
You don't provide context, so I don't know what results you can't reproduce.
But you need to reread the help page for the kruskal.test and look at the
examples give. If x and y are the results for two independent groups, then the
call to the kruskal.test should be e
, greatly appreciated.
>
> Thanks,
>
> Phil
>
Not my area of expertise, but if you go to your favorite CRAN mirror and look
at the task views, you will find info on spatial analysis. Also, Googling 'R
GIS' brings up a "ton" of hits.
Good luck,
Dan
Dani
; ',
na.strings='-99', row.names='OBSNO')
Something may have become corrupted on your system. Try restarting R and
re-running the code. If that doesn't solve the problem, then you may need to
show us exactly the code you are executing, and provide releva
gt; in
> my R books - no luck so far. Maybe I'm missunderstanding the entire
> ISOdatetime function?
>
Julia,
are the times from your GPS really GMT times, or are they your local time (it
looks like you are in Germany)? It looks lik
reciate if you can explain it in the simplest way you can. i
> am a pure molecular biologist and trying to learn this new monster called R.
> :-)
>
> thanks,
>
> daniel
[[alternative HTML version deleted]]
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E help?
thanks,
daniel
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, m
can someone tell me if there is an easier way to do this in R - create a
design matrix? thanks.
> design <- model.matrix(~
-1+factor(c(1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5)))
# Creates an appropriate design matrix
can someone tell me
.
>
> Thanks a lot
>
> Alex
>
maybe look at ?as.vector. But it is really hard to know without knowing what
you expect your "vector to look like or what transforms you will be doing.
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
__
ile = stderr()).Internal(printDeferredWarnings())}
invisible(structure(msg, class = "try-error", condition = e))})
19: try(gsub("\\s+", " ", paste(capture.output(print(args(cem))), collapse
= "")), silent = TRUE)
Possible actions:
1: abort
ebook$width[i])
if(codebook$skip[i] > 0) read.col <- c(read.col,-codebook$skip[i])
}
## recode type values to R classes
codebook$Rtype <- ifelse(codebook$type %in% c('int','float'),'numeric',
'character')
## now read in the data
fwfdata <- re
n the codebook page
http://www.cdc.gov/nchs/nsfg/nsfg_2006_2010_puf.htm#codebooks
hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
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PLEASE do read the posting
> HTH,
> -steve
>
> --
> Steve Lianoglou
> Graduate Student: Computational Systems Biology
> | Memorial Sloan-Kettering Cancer Center
> | Weill Medical College of Cornell University
> Contact Info: http://cbio.mskcc.org/~lianos/contact
Daniel Nordlund
Bothell, WA U
)
> > rep(data.recieved[x,1], times=data.recieved[x,2] )) ) )
> > [1] No No No No No No No No No No Yes Yes Yes
> > Levels: Yes No
> >
> >
> >
> >>>
> >>> par(mfrow=c(1,2));
> >>> plot(data.recieved$**kindergarten_a
, C2=C2), "c:/tmp/Venn_2set_simple.tiff")
If that is not what you want, then you will need to provide an example of what
you want your output to be.
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
__
R-help@r-project.org mailing
The previously posted code contains bugs. The code below should work:
x: a vector of data
y: Group indicator (if id.col=TRUE); data of the second group (if
id.col=FALSE). If y is the group indicator it MUST take 0 or 1 to indicate
the groups, and x must contain the data for both groups.
id.col:
uot;,NA,NA,2)
d<-data.frame(x1,y1,x2,y2,x3,y3)
d
# d looks like:
x1 y1 x2 y2 x3 y3
1xyxyxy
213
3 21
432
>From this, I want to create the table or data frame
x y
1 3
2 1
3 2
I would appreciate your help.
Daniel
if(N) )
>
> Mike
>
I don't know what the OP is really trying to accomplish yet, and I am not
motivated (yet) to try to figure it out. However, all this "flooring" and
"ceiling) and "rounding" is not necessary for generating uniform random
integers.
ou need to use
write.table() directly.
That being said, what was the result of the output of the very first command
that you said you used?
write.csv(practice, file.choose(new=T), quote=F, row.names=F)
I used the built-in data frame, cars, instead of your file practice and it
worked just fine in producing a comma separated value file.
Dan
Daniel Nordlund
Bothell, WA USA
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Does anyone know a package that will conduct Vuong and Clarke tests to
compare non-nested rare-events logit ("relogit") models via Zelig? I know
the pscl package can conduct a Vuong test, however, it doesn't seem to work
with relogit. I haven't been able to find a package that can implement the
Cla
I have doubled buffered animations that I show in class.
They used to work but now flash.
The default windows() option is buffered = TRUE.
Just in case, I tried using windows( buffered = TRUE)
but this made no difference.
I am not sure when the change occurred.
An older R2.11 version in one cla
arch for: emacs ess windows
>
> 2nd hit, 4th bullet from the end.
>
> --
For emacs-ess on windows here is a recent post from Vincent Goulet that
provided me with a very simple install on both WinXP and Vista.
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/125362.html
Hope
so that the last element would be the newly added one while the previous
elements all remain the same?
Thanks!
Daniel
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PLEASE do read the posting guide http://www.R-project.
e for
sum(rbinom(100, 1, p=.15))
So, you can randomly arrange a certain percentage of ones in a sequence, but
you should not expect to see that exact percentage in any subset of the
sequence.
One other example, if you flip a fair coin 100 times, do you expect to get
exactly 50 heads?
Hop
0.059190.55646 0.1060.916
p53plus 0.229770.55646 0.4130.682
Time:p53plus 0.118870.02524 4.709 4.62e-05 ***
But I do not think that is doing what I want.
Many thanks
--
******
Daniel Brewer, Ph.D.
Inst
Hi everyone!
I have a question about data processing efficiency.
My data are as follows: I have a data set on quarterly institutional
ownership of equities; some of them have had recent IPOs, some have not
(I have a binary flag set). The total dataset size is 700k+ rows.
My goal is this: For
would a density plot do? try
plot(density(x))
if you are specifically after the histogram tops rather than a density
estimate, then get the hist object with plot=F, then look at the counts
attribute:
histobj = hist(x, freq="TRUE", breaks=1000, plot=F)
plot(histobj$counts)
hope this helps.
o
oosen said the following:
Maybe you should provide a minimal, working code with data, so that we all
can give it a try.
In the mean time: take a look at the Rprof function to see where your code
can be improved.
Good luck
Bart
Daniel Folkinshteyn-2 wrote:
Hi everyone!
I have a question
i know this is an R mailing list :) but... i'll recommend you try python
with the beautifulsoup module - makes html processing a cinch.
another thing to note is that wunderground provides very handy RSS feeds
for every location, so rather than parsing the html page (with it's
associated bundle
looks like you don't have permission to write a file to C:\
try writing to some other directory where you have write access
(e.g., your user's home dir, or your "my documents", or something like
that).
on 06/05/2008 11:57 PM Megh Dal said the following:
Hi,
I got following error in write.tab
the '00' entries may be in a numeric column, so it gets typecast to a
number, and of course 00 == 0, numerically speaking, so they get
'condensed'.
to be sure you read everything "as is", specify "colClasses='character'. :
data<-read.table("data.txt",sep='\t', header=T, colClasses='character')
according to the helpfile, comment only takes one character, so you'll
have to do some 'magic' :)
i'd suggest to first run mydata through sed, and replace one of the
comment chars with another, then run read.table with the one comment
char that remains.
sed -e 's/^\^/!/' mydata.txt > mydata2
should work - don't even have to put them in quotes, if your field
separator is not space. why don't you just try it and see what comes out? :)
on 06/06/2008 08:43 AM stephen sefick said the following:
if I wanted to use a name for a column with two words say Dick Cheney and
George Bush
can I p
try this:
FullData <- merge(ETC, SURVEY, by.x = "ord", by.y = "uid", all.x = T,
all.y = F)
on 06/06/2008 07:30 AM Michael Pearmain said the following:
Hi All,
Newbie question for you all but i have been looking at the archieves and the
help dtuff to get a rough idea of what i want to do
I wo
names(f)[which.max(f)]
on 06/06/2008 09:14 AM Muhammad Azam said the following:
Dear R users
I have a very basic question. I tried but could not find the required result.
using
dat <- pima
f <- table(dat[,9])
f
0 1
500 268
i want to find that class say "0" having maximum frequency i.e
than by.y
and by.x?
I think when i was playing around i tried the all. command in that setup
as well
Mike
On Fri, Jun 6, 2008 at 2:07 PM, Daniel Folkinshteyn <[EMAIL PROTECTED]
<mailto:[EMAIL PROTECTED]>> wrote:
try this:
FullData <- merge(ETC, SURVEY, by.x = &
Anybody have any thoughts on this? Please? :)
on 06/05/2008 02:09 PM Daniel Folkinshteyn said the following:
Hi everyone!
I have a question about data processing efficiency.
My data are as follows: I have a data set on quarterly institutional
ownership of equities; some of them have had
i did! what did i miss?
on 06/06/2008 11:45 AM Gabor Grothendieck said the following:
Try reading the posting guide before posting.
On Fri, Jun 6, 2008 at 11:12 AM, Daniel Folkinshteyn <[EMAIL PROTECTED]> wrote:
Anybody have any thoughts on this? Please? :)
on 06/05/2008 02:09 PM
well, where are you getting the filename in the first place? are you
looping over a list of filenames that comes from somewhere?
generally, for concatenating strings, look at function 'paste':
write.table(myoutput, paste(myfilename,"_out.txt", sep=''),sep="\t")
on 06/06/2008 11:51 AM DAVID ARTE
008 at 12:03 PM, Daniel Folkinshteyn <[EMAIL PROTECTED]> wrote:
i did! what did i miss?
on 06/06/2008 11:45 AM Gabor Grothendieck said the following:
Try reading the posting guide before posting.
On Fri, Jun 6, 2008 at 11:12 AM, Daniel Folkinshteyn <[EMAIL PROTECTED]>
wrote:
Anybod
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")
Daniel Folkinshteyn wrote:
Anybody have any thoughts on this? Please? :)
on 06/05/2008 02:09 PM Daniel Folkinshteyn said the following:
Hi everyone!
I have a que
ci = rainbow(7)[c(4:7, 1:3)]
on 06/06/2008 01:02 PM avilella said the following:
Hi,
I want to reorder the colors given by rainbow(7) so that the last half
move to the first 4.
For example:
ci=rainbow(7)
ci
[1] "#FFFF" "#FFDB00FF" "#49FF00FF" "#00FF92FF" "#0092"
"#4900"
[7] "#FF
just in case, uploaded it to the server, you can get the zip file i
mentioned here:
http://astro.temple.edu/~dfolkins/helplistfiles.zip
on 06/06/2008 01:25 PM Daniel Folkinshteyn said the following:
i thought since the function code (which i provided in full) was pretty
short, it would be
== end function code===
on 06/06/2008 01:35 PM Gabor Grothendieck said the following:
I think the posting guide may not be clear enough and have suggested that
it be clarified. Hopefully this better communicates what is required and why
in a shorter amount of space:
https://stat.ethz.c
d not far from optimal.
If you pick the possibly too small route, then increasing
the size in largish junks is much better than adding
a row at a time.
Pat
Daniel Folkinshteyn wrote:
thanks for the tip! i'll try that and see how big of a difference that
makes... if i am not sure what exactl
works for me:
> sub('1.00', '1', '1.00E-20')
[1] "1E-20"
remember, according to ?sub, it's sub(pattern, repl, string)
try doing it step by step. first, see what yr1bp$TreeTag[1501] is.
then, if it's the right data item, see what the output of sub("1.00",
"1", yr1bp$TreeTag[1501]) is.
that'll l
another vote for ubuntu here - works for me, and quite trouble-free. add
the r-project repositories, and you're sure to always have the latest,
too. (if you don't care for the latest R, you can of course also just
get R from the distro's repos as well)
on 06/06/2008 05:22 PM Abhijit Dasgupta s
Burns said the following:
That is going to be situation dependent, but if you
have a reasonable upper bound, then that will be
much easier and not far from optimal.
If you pick the possibly too small route, then increasing
the size in largish junks is much better than adding
a row at a time.
Pat
D
t those columns, convert them to a matrix, do all the matching,
and then based on some sort of row index retrieve all of the associated
columns.
-Don
At 2:09 PM -0400 6/5/08, Daniel Folkinshteyn wrote:
Hi everyone!
I have a question about data processing efficiency.
My data are as follows: I
on 06/06/2008 06:55 PM hadley wickham said the following:
Why not try profiling? The profr package provides an alternative
display that I find more helpful than the default tools:
install.packages("profr")
library(profr)
p <- profr(fcn_create_nonissuing_match_by_quarterssinceissue(...))
plot(p)
install.packages("profr")
library(profr)
p <- profr(fcn_create_nonissuing_match_by_quarterssinceissue(...))
plot(p)
That should at least help you see where the slow bits are.
Hadley
so profiling reveals that '[.data.frame' and '[[.data.frame' and '[' are
the biggest timesuckers...
i suppose
[1]][1:n])==FALSE))
But again, if the text is large, I cannot assign it to x. I'd be grateful
for any suggestions.
Cheers,
Daniel
-
cuncta stricte discussurus
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Hi R-users
I think about some web service with stock charts and I plan use python as
web framework with R as mathematical and chart engine. I use rpy to connect
R with python. It works good.
Is it good idea to use R as chart engine? Maybe it's better (faster) to use
c/c++ plugin?
D
ge but I haven't tested this yet.
R engine is very universal and this is the strong of R choice in this kind
of web service... then begin write new (faster?) c/c++ plugin. What you
think?
daniel cegielka
-Original Message-
From: jim holtman [mailto:[EMAIL PROTECTED]
Sent: Monday,
Does anyone know how to insert a jpeg image into a widget.
This is a portion of the code I have used to create the widget. I would like
to insert an image in the bottom corner.
Cheers
tt<-tktoplevel()
fontHeading2 <- tkfont.create(family="arial",size=16, slant="italic")
fontHeading3 <- tkfont
value etc
are set to NA.
2) Suppress the Error and Warning messages
What is the best way to do this?
Thanks
--
**
Daniel Brewer, Ph.D.
Institute of Cancer Research
Molecular Carcinogenesis
Email: [EMAIL PROTECTED
know whether your problem is solved.
Cheers,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im
Auftrag von Monna Nygård
Gesendet: Tuesday, June 17, 2008 5:04 AM
An: r-help@r
> Terry Therneau
> (author of coxph)
>
>
--
******
Daniel Brewer, Ph.D.
Institute of Cancer Research
Molecular Carcinogenesis
Email: [EMAIL PROTECTED]
**
The Institute of
source('yourscript.R')
on 06/19/2008 03:11 PM [EMAIL PROTECTED] said the following:
Dear R-Users,
I've written a number of functions in a .R/script file. I would like to
call those functions from another script file. How can I execute all the
code in a script file so that the functions are av
ors
> y=x+e ##create Ys
>
> ###plot
> plot(y~x,pch=NA) ##plot Ys against Xs but suppress all symbols (i.e.
> plot invisibly)
> text(y~x,labels=round(x),pch=NULL) ##use values of X (rounded to its
integer
> value) as symbols for the X-Y plot
>
> ###End of example
>
Maybe this help you
http://www.rforge.net/Cairo/index.html
http://www.rosuda.org/R/GDD/
daniel cegielka
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of milton ruser
Sent: Sunday, June 22, 2008 5:56 PM
To: Josep Lluís Figueras
Cc: r-help@r-project.org
Hi,
random=~1|B/C
C is nested in B
##Example
data=Oats
regress=lme(yield~nitro*Variety,data=Oats,random=~1|Block/Variety)
##i.e. variety is nested in block
summary(regress)
##End of example
Best,
Daniel
-
cuncta stricte discussurus
on 06/23/2008 03:40 PM Thomas Frööjd said the following:
1. Shift the mean and std on the reference dataset to the mean
and std of my clinic birth weight data.
to shift the mean by any distance, just add or subtract that distance
from each observation (e.g., to move mean from m1 to m2, t
just cbind the cols in the appropriate order:
m.2 = cbind( m.1[,1:5], yourthreecolumns, m.1[,6:ncol(m.1)] )
on 06/24/2008 07:02 AM Daren Tan said the following:
Instead of prepend or append new columns to a matrix, how to insert them to a matrix ? For example, I would like to insert 3 new column
I don't understand this. Why not just get hist() to plot on the
density scale,
thereby making its output commensurate with the output of density()?
The hist() function will plot on the density scale if you ask it
to. Set freq=FALSE
(or prob=TRUE) in the call to hist.
ehrm
If I analyze a client's data using an R script I created then I can
charge the client a $20,000 consulting fee, but, if I let the client
push the button to execute the R script and charge him 10 cents for the
privilege then I can be sued for violating the GPL? Or are my
I think you cannot be su
no need for a for loop - we can vectorize this:
> dt <- data.frame(a = c(1, 2, 3), b = c(3, 2, 2), c = c(1, 3, 5))
> dt
a b c
1 1 3 1
2 2 2 3
3 3 2 5
> dt[,paste("test", 1:2, sep="")] = rep(1:2, each=3)
> dt
a b c test1 test2
1 1 3 1 1 2
2 2 2 3 1 2
3 3 2 5 1 2
on 06
this is probably a cludge, and there may be a "neater" way to do this,
but... here's one:
> a = 0:1
> for (i in 1:9){ a= merge(unname(a), 0:1) }
> a = t(a)
after the for loop, 'a' will contain a 1024 row by 10 col dataframe.
putting it through a transpose, gives you the 10 rows by 1024 cols ma
not sure why it doesn't work, but try the following:
first, plot to a regular window, then run:
> dev.copy(device=png, file="yourfilename.png")
> dev.off()
see if that produces a file you want.
another note: what do you mean you can't just "copy and paste the graph"
in ubuntu? doesn't pressing
this should do what you want:
> myexstrings = c("*AAA.AA","BBB BB","*.CCC.","**dd- d")
> a = gsub("^\\W*","", myexstrings,perl=T)
> b = gsub("\\W.*", "", a, perl=T)
> b
[1] "AAA" "BBB" "CCC" "dd"
first one, removes any non-word characters from the beginning (as you
already figured out)
second o
try this:
firstgenes = lapply(geneset, function(x){return(x[1,1])})
firstgenes = do.call(rbind(firstgenes))
on 06/27/2008 10:33 AM Rajasekaramya said the following:
Hi,
I have a problem in assessing the list element.
i have list called geneset it contains the following elements
oh, unlist - very nice function, thanks :)
on 06/27/2008 11:23 AM Jorge Ivan Velez said the following:
Hi Ramya,
Try something like this:
as.character(unlist(lapply(geneset,function(x) x[1])))
HTH,
Jorge
On Fri, Jun 27, 2008 at 10:33 AM, Rajasekaramya <[EMAIL PROTECTED]>
wrote:
Hi,
I h
if there's nothing specific for it, you could probably do it with merge?
on 06/27/2008 02:41 PM Agustin Lobo said the following:
Hi!
Given a vector (or a factor within a df),i.e. v1 <- c(1,1,1,2,3,4,1,10,3)
and a dictionary
cbind(c(1,2,3),c(1001,1002,1003))
is there a function (on the same lin
togram but since the audience is not
very statistically experienced I would prefer to do it this way.
Anyone have an idea?
Thanks again for your help.
Thomas Fröjd
On Wed, Jun 25, 2008 at 6:16 PM, Daniel Folkinshteyn <[EMAIL PROTECTED]> wrote:
I don't understand this. Why not
ta in it.
Can anyone give me some hints on what Im missing here? I would be very
grateful for that.
Thanks.
Daniel Pires
Daniel Pires
Faculdade de Ciências da Universidade de Lisboa
Dep. Biologia Animal
Ed. C2, piso 2
Campo Grande 1749-016 Lisboa
Portu
Assuming that you have installed and loaded the mclust library, type ?Mclust
in the R-prompt. An example is provided there with the popular iris dataset.
It seems to be as simple as Mclust(yourdata), where "yourdata" contains the
dataset (data columns of your dataset) on which you want to perform c
(importance[2,1,])
####End of example
Best,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im
Auftrag von Tanya Yatsunenko
Gesendet: Monday, June 30, 2008 10:09 PM
An: r-help@
o the
bootstrapped data, you would have to store the data in an array as well.
Cheers,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im
Auftrag von Tanya Yatsunenko
Gesendet: Tuesday,
Hello,
I am trying to extract t and pvalues from a 1000 ttests using the by-function
but everythinhg I tried did not work. Unfortunately googling "by" is not very
helpful. Any help will be very appreciated.
Cheers,
Danile Stall
*creating a data set
library(MASS)
dataset <- mvrnorm(160, mu, Sig
Hi
I just discovered the answer thanks to a previous thread from Peter Daalgart
The command is:
sapply(test, "[[", "statistic")
Cheers, Daniel
- Original Message -
To: r-help@r-project.org
Subject: extracting values from a "by" function
Date: Sun, 6 Jul 2
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