Did you try the example described on the ff man page?
On Monday, April 2, 2012, Bond, Stephen wrote:
> Thomas,
>
> I tried biglm and it does not work see
>
>
> http://r.789695.n4.nabble.com/unable-to-get-bigglm-working-ATTN-Thomas-Lumley-td2276524.html#a2278381
>
> . There are other posts from pe
you want to check the ff man page (?ff), there is an example described
there with biglm. b
On 30 March 2012 21:05, Bond, Stephen wrote:
> Greetings useRs,
>
> Can anyone provide an example how to use ff to feed a very large data
> frame to glm?
> The data.frame cannot be loaded in R using conven
You probably want:
sql<-"UPDATE testtable SET vals=21 WHERE countries='NewZealand'"
dbGetQuery(con, sql)
instead...
b
On 27 March 2012 14:18, Thomas Adams wrote:
> All:
>
> I am using RSqlite and want to be able to update individual values in a
> record, such as with this simple example:
>
>
t;, gsub('^\"|\"$', "'",
> readLines('../teste.csv', sep = ',', quote = "'", header = TRUE)
>
> On Mon, Mar 26, 2012 at 8:09 PM, Benilton Carvalho
> wrote:
> > I need to read in csv files, created by 3rd party, with
I need to read in csv files, created by 3rd party, with fields
containing single quotes (as shown below).
"header1","header2","header3","header4"
"field1r1","field2r1","field3r1","field4r1"
"field1r2","field2r2","field3r2PartA), field3r2PartB Very" Long","field4r2"
"field1r3","field2r3","field3r3"
Thanks Henrik!!! Hope to pay your beer soon. :) b
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-containe
Hi,
I have a long list of lists from which I want to efficiently extract
and rbind elements. So I'm using the approach below:
f <- function(i){
out <- replicate(5, list(matrix(rnorm(80), nc=20)))
names(out) <- letters[1:5]
out
}
set.seed(1)
lst <- lapply(1:1.5e6, f)
(t0 <- system.tim
Hi Chuck, thank you *very* much! That really helped! b
On 9 March 2012 17:15, wrote:
> Benilton Carvalho writes:
>
>> Hi,
>>
>> what is the proper of of "passing a missing value" so I can extract
>> the entire i-th row of a matrix (in a list of lists
Hi,
what is the proper of of "passing a missing value" so I can extract
the entire i-th row of a matrix (in a list of lists) without
pre-computing the number of cols?
For example, if I know that the matrices have 2 columns, I can do the following:
set.seed(1)
x0 <- lapply(1:10, function(i) repli
Thanks guys... I'm already embarrassed given how simple the solutions are. b
On Saturday, 18 February 2012, Hadley Wickham wrote:
> On Fri, Feb 17, 2012 at 7:51 PM, Benilton Carvalho
> > wrote:
> > Hi everyone,
> >
> > For reasons beyond the scope of this messag
Hi Mark,
thank you very much! This is perfect! Naive from my part not thinking of c()...
Cheers, b
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PLEASE do read the posting guide http://www.R-project.org/posting-guide
Thank you very much, David.
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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.
Hi everyone,
For reasons beyond the scope of this message, I'd like to append a
NULL element to the end of a list.
tmp0 <- list(a=1, b=NULL, c=3)
append(tmp0, c(d=4)) ## works as expected
append(tmp0, c(d=NULL)) ## list with a/b/c only
Given that I could use
tmp0$a <- NULL
to remove 'a', I see
Have you removed the *.so and *.o files prior to "R CMD SHLIB hello.c"?
b
On Tue, May 11, 2010 at 11:39 PM, Elizabeth Lawson
wrote:
> I recently bought a new macbook pro 10.6.3 and I am trying to compile some C
> code I have. I reinstalled R and Xcode on the Mac but I keep running into
> the sa
or possibly use smoothScatter() to produce the scatter plots
On Fri, Apr 30, 2010 at 12:06 PM, ONKELINX, Thierry
wrote:
> Dear Nevil,
>
> Converting your pdf to png will be the most efficient way to reduce the
> file size with scatter plots.
>
> You can either export directly to pdf or first
maybe
for (i in 1:10) assign(paste("matrix", i, sep="_"), matrix(nrow=i, ncol=i))
suffices?
On Thu, Apr 22, 2010 at 5:08 PM, karena wrote:
>
> I need to create 10 matrices. say matrix 1-10.
>
> matrix_1 is 1 by 1
> matrix_2 is 2 by 2
> matrix_3 is 3 by 3
> .
> .
> .
> matrix_10 is 10 by 1
save(list=paste("unislopes", master.i, sep=""),
file=paste("unislopes",master.i,".Rdata",sep=""))
On Mon, Apr 12, 2010 at 12:06 PM, David Young wrote:
> I'd like to use a string to refer to an R object with the end
> objective of going through a loop and saving various files of the same
> name wi
maybe
inc.freqy <- sapply(unit.dist, function(x) rowSums( test <= x))
suffices?
b
On Mon, Apr 5, 2010 at 5:31 PM, pinusan wrote:
>
> Dear R users,
>
> I would like to transform the following "for loop" from R-code to C-code
> because it takes really long time to have inc.freqy table.
> Unfortu
FAQ 7.31: Why doesn't R think these numbers are equal?
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
b
On Wed, Mar 31, 2010 at 10:10 AM, Gianluca Baio wrote:
> Dear list,
> I have observed a weird behaviour from R --- apologies if I am missing
and in addition to bill's suggestion, you may want to consider not
growing a list.
Instead of 'v <- list()', use:
v <- vector("list", nrow(A))
b
> Will your code work as you wish if you
> replace the "for(v[[1]] in x[[1]]) { ... }"
> with the following?
> for(i in x[[1]]) {
> v[[2]] <- i
can you also post an example of A and an example of the expected result?
On Tue, Mar 23, 2010 at 3:36 PM, Pj253 wrote:
>
> Thanks for your reply Ben!
>
> I don't think I want v to be identical to x... I guess I haven't put the
> question in the right context. What I'm actually trying to do is...
if this was to work, wouldn't the object 'v' be identical to 'x'?...
so, why not use 'x' itself?
b
On Tue, Mar 23, 2010 at 2:53 PM, Pj253 wrote:
>
> I have a list of vectors, x, with x[[1]]=1:5, say.
>
> And I need to go through each element of each vector in a for loop.
> Something like:
>
> for
To access elements of a list (object returned by split), you need to use "[[".
Therefore,
summary(temp[[1]])
is what you meant to use (or even summ = lapply(temp, summary) - which
will give you the summaries for every subject).
About using PDFs, I'd recommend you to take a look at Sweave (
http
how about
y = X[rep(1:nrow(X), 6), ]
?
On Mon, Mar 15, 2010 at 5:37 PM, wrote:
> What is an easy way to stack a matrix multiple times? E.g. I have a 6x6
> matrix that I need to stack vertically 154 times to get a 6*154 by 6
> matrix. I would rather not rbind(X,X,...,X) matrices. --Joe
>
If you could describe exactly what is it that you're trying to
accomplish, we could be of better help (the reason I say this is
because the way you're trying to implement things is a bit
inefficient).
Anyways, you can't use two indices with a list.
One approach would be to nest lists, and you'd g
dd = as.POSIXlt(c("2007-02-21 05:19:00", "2007-02-20 14:21:53"),
format="%Y-%m-%d %H:%M:%S")
dd[1]-dd[2]
b
On Tue, Feb 23, 2010 at 2:48 PM, karine heerah wrote:
>
> Hi,
>
>
>
> I have date and time in a format like this: " 2007-02-21 05:19:00".
>
> Do you which function i can use to derterminat
i'd use:
test[["Y"]] <- NULL
b
On Tue, Feb 23, 2010 at 4:13 PM, adam naples wrote:
> try
>
> test <- subset(test, select = -c(Y))
>
> The key is the minus sign before c() in the select argument.
> You can put in as many columns as you like.
> -a
>
> On Feb 23, 2010, at 11:02 AM, David Winsemiu
gsub("one|two", "something else", y)
?
On Mon, Feb 22, 2010 at 4:25 PM, Marianne Promberger
wrote:
> Dear list,
>
> I have two vectors:
>
> x <- c("one","two")
> y <- paste(rep(x,2),"blah")
>
> I want to replace all occurrences of each element of x in y with
> something else, so that y looks lik
assigning rownames (after the object is created) triggers a copy of
the object... if you assign the rownames at creation time, no extra
copies...
b
On Mon, Feb 22, 2010 at 9:46 AM, Larson, TR wrote:
> Hi,
>
> On R 2.10.1 for Windows, when I do the following to duplicate the structure
> of a larg
x[nrow(x):1,]
b
On Thu, Feb 18, 2010 at 11:45 AM, Amelia Livington
wrote:
> Dear 'R' friends
>
> I have a sort of stupid question to ask.
>
> I have a matrix say of the order 4 X 3 as
>
> 83 98 90
> 21 83 84
> 70 39 56
> 65 29 38
>
> Is there any command in R which will r
you mean something along the lines of
filter(x, rep(1/4, 4))
(which you can combine with na.omit)
?
On Wed, Feb 17, 2010 at 5:18 PM, Mohsen Jafarikia wrote:
> Hello All:
>
> If I do have:
>
> x = (2, 4, 5, 5, 6, 4, 5, 2, 1)
> y = (9, 11.5, 12.5, 13, 14, 19, 20, 21, 22)
>
> I wanted to find a s
say you read the quantity.csv file into a variable called
'quantity'... similarly, 'equity_price.csv' to equity.
sweep(equity, 2, quantity, "*")
b
On Tue, Feb 16, 2010 at 11:20 AM, Sarah Sanchez
wrote:
> Dear Madam / R helpers,
>
> Unfortunately the solution you have suggested is not working in
http://cran.r-project.org/web/views/HighPerformanceComputing.html
2010/2/15 :
> Dear all,
>
> I'm sitting here just in front of my new p...@work and wonder about the
> following question:
> * How can I adress multiple CPUs (cores) out of R to speed up the
> simulations I run?
> *
lation and for that I need to calculate LN(New rate / old
>> rate for each of the instrument separately).
>>
>> Here is my actual code -
>>
>> ONS = read.csv('Instrument.csv')
>> n = length(ONS)
>> Y = NULL
>> B = array()
>>
>> for
sorry, meant to type:
B = ONS^2
cheers,
benilton
On Mon, Feb 15, 2010 at 12:28 PM, Benilton Carvalho
wrote:
> maybe you just want
>
> Y = ONS^2
>
> ?
>
> b
>
> On Mon, Feb 15, 2010 at 12:22 PM, Madhavi Bhave
> wrote:
>> Dear R Helpers
>>
>> (Th
maybe you just want
Y = ONS^2
?
b
On Mon, Feb 15, 2010 at 12:22 PM, Madhavi Bhave wrote:
> Dear R Helpers
>
> (There is a small correction in my earlier mail. In the 'instrument.csv'
> file, I had mentioned only three columns. Actually there are 7 columns. I
> regret the error. Rest contents
On Thu, Feb 11, 2010 at 4:18 PM, Philipp Rappold
wrote:
> Dear all,
>
> I have two probably very easy questions:
>
> (1) Is there a way to access certain variables by their string-based name
> representation?
>
> Example:
> numbers <- c("one", "two", "three")
> varname <- "numbers"
> print(varname
Hi Matt,
what's your sessionInfo()? Can you try installing bigmemory as follows:
install.packages("bigmemory", repos="http://R-Forge.R-project.org";)
it'll get you the latest version, in which I cannot reproduce the
problem you're reporting (ie, after gc(), I get all the RAM back)
b
On Sat, Fe
u can use other aproaches to fill in the ff dataframe
b
On Fri, Feb 5, 2010 at 11:43 PM, Benilton Carvalho
wrote:
> Hi Matt,
>
> you're correct: length(ffObject) must be smaller than 2^31-1... at
> least until R has a 64bit integer type, it seems...
>
> in the meantime
Hi Matt,
you're correct: length(ffObject) must be smaller than 2^31-1... at
least until R has a 64bit integer type, it seems...
in the meantime, use the bigmemory package. ;-)
b
On Fri, Feb 5, 2010 at 4:18 PM, Matthew Keller wrote:
> Hello all,
>
> I hate to add to the daily queries regarding
On Thu, Jan 28, 2010 at 11:21 AM, Duncan Murdoch wrote:
> On 28/01/2010 5:30 AM, Benilton Carvalho wrote:
>>
>> Hi Duncan,
>>
>> On Tue, Jan 26, 2010 at 9:09 PM, Duncan Murdoch
>> wrote:
>>>
>>> On 26/01/2010 3:25 PM, Blanford, Glenn wrote:
nge soon and we
were to have 64bit integer type (say, when under a 64 bit OS),
wouldn't this allow us to have objects whose length exceeded the
2^31-1 limit?
Benilton Carvalho
> The double type in R can hold exact integer values up to around 2^52. So for
> example calculations like
check aggregate() (the examples are quite helpful)
b
On Mon, Jan 25, 2010 at 4:07 PM, wrote:
> Dear R community,
> I'm trying to develop a fast way of summing specific rows of a large data
> frame.
> Here is an example of the kind of data frames I'm dealing with:
>
>> refls
> H K L M/I
adapted from the help files of rowsum
x <- matrix(runif(100), ncol=5)
group <- sample(1:8, 20, TRUE)
xsum <- rowsum(x, group)
sweep(xsum, 1, table(group), "/")
or
aggregate(x, list(group), mean)[-1]
b
2010/1/15 Joel Fürstenberg-Hägg :
>
> Hi all,
>
>
>
> I want to calculate the row wise mean o
theFiles <- list.files(inputdir, full=T, pattern="\\.[eE][xX][tT]$")
for (file in theFiles){
...
}
On Fri, Jan 15, 2010 at 11:43 AM, Albert Vilella wrote:
> Hi all,
>
> I'm trying to process all files with a certain extension "*.ext" in a
> directory like this:
>
>> R --slave --args /my/dir < di
you need to be more clear on your question...
what is it (exactly) that you want? Is it the following?
y1 ~ x1 + ... + xm
y2 ~ x1 + ... + xm
...
yn ~ x1 + ... + xm
?
if so:
lm(cbind(y1, y2, ..., yn) ~ x1+x2+...+xm)
b
On Dec 24, 2009, at 3:33 PM, Hao Cen wrote:
> Hi,
>
> I have multiple res
replace data[1:i,] by data[1:i,drop=FALSE].
b
On Dec 24, 2009, at 12:46 AM, Francesco Napolitano wrote:
> Hi all,
>
> I'm trying to learn R after years of Matlab's experience. Here is an
> issue I couldn't solve today.
>
> Consider the following piece of code (written by memory):
>
> for(i in
instead of densityplot(...) use
print(densityplot(...))
b
On Dec 16, 2009, at 3:19 PM, c...@autistici.org wrote:
> Hi,
> i have a script how i launch lattice to make a densityplot.
> in the script:
>
> jpeg(file="XXX.jpg")
> densityplot(~f_diametro+m_diametro+n_diametro, plot.points="rug",
> a
use tracemem() to figure out... and read its documentation in detail.
b
On Dec 15, 2009, at 1:03 AM, Peng Yu wrote:
> I'm wondering if lazy copy is available in R or not. For example, in
> the following code, I'm wondering if the memory for y is allocated in
> the 2nd line or the 3rd line. Is th
oh.. and i just saw the bonus part... just replace lapply() by sapply().
b
On Dec 11, 2009, at 8:33 PM, Jennifer Young wrote:
> Good evening
>
> I often have as output from simulations a list of various values, vectors
> and matrices.
> Supposing that I then run said simulation several times, I
lapply(output, "[[", "vec")
b
On Dec 11, 2009, at 8:33 PM, Jennifer Young wrote:
> Good evening
>
> I often have as output from simulations a list of various values, vectors
> and matrices.
> Supposing that I then run said simulation several times, I often want to
> extract a particular result
matrix(t(G), nc=1)
b
On Dec 4, 2009, at 1:19 PM, Jose Narillos de Santos wrote:
> Hi all,
>
> Imagine I have a matrix G
>
> with N rows
>
> and M columns
>
> So L=NxM is the number of different cells in my matrix.
>
> I want to create a column vector F whose size will be F(L,1)
>
> So the
isn't it simpler just to pass two vectors, say v1 and v2, in which one contains
the object names and the other has the associated variances? (btw, "data" isn't
a good function name)
myData <- function(v1, v2){
Vec <- matrix(v1)
varF <- v2
}
I may have misunderstood your question, but IMHO a
take a look at geshi. b
On Nov 20, 2009, at 4:26 PM, Tal Galili wrote:
My question if in the Subject, but if to extend: I am specifically
curious
about WordPress blogs. But any solution will give me a lead.
Thanks,
Tal
--
Contact me: tal.gal...@gm
I can't think of anything that is already built in.
But you can always:
read.as.matrix <- function(...) as.matrix(read.delim(...))
and now you get one step only ;-)
b
On Nov 20, 2009, at 4:01 PM, Peng Yu wrote:
On Sat, Nov 21, 2009 at 11:55 AM, Steve Lianoglou
wrote:
read.delim gives me a
readFASTA in the BioConductor Biostrings package.
b
On Nov 18, 2009, at 8:14 AM, Tal Galili wrote:
Hello dear R help group,
I would like to download the tRNA data on:
http://gtrnadb.ucsc.edu/download.html
And then import it into R.
Can anyone direct me as to how to do so?
Thanks,
Tal
-
it doesn't skip...
think about why:
seq(.30,.5,.01) * 100 - 29 == 1:21
isn't always TRUE.
b
On Nov 17, 2009, at 9:17 PM, Bruno Giovannetti wrote:
Hello,
Sometimes the looping (using "for") seems to skip some iterations.
An example:
arg <- matrix(NA,length(seq(.30,.5,.01)),1)
for (i in se
align the else with the curly brackets
if (yes){
be happy
}else{
complain
}
b
On Nov 13, 2009, at 3:33 PM, anna_l wrote:
Hello, I am getting an error with the following code:
if( P2 > P1)
+ {
+ P<-P2
+ }
else
Erro: unexpected 'else' in "else"
{
+ P<-P1
+ }
I checked the syntax so I don´
ready for the next release.
If you're trying to run RMA on your data, I can think of ways of
working around this problem.
Cheers,
b
On Nov 7, 2009, at 5:46 PM, Benilton Carvalho wrote:
ok, i'll take a look at this and get back to you during the week. b
On Nov 7, 2009, at 1:19 PM
Hi Peng,
in a very simplistic manner, what happens is that the Operating System
"thinks" it is too dangerous to let the R process to use so much
memory. So, to protect the whole system, it kills R, before the system
becomes unstable.
I've been looking at the problem you observed last week
ok, i'll take a look at this and get back to you during the week. b
On Nov 7, 2009, at 1:19 PM, Peng Yu wrote:
Most of the 8GB was available, when I run the code, because R was the
only computation session running.
On Sat, Nov 7, 2009 at 7:51 AM, Benilton Carvalho
wrote:
you ha
ave enough resources when trying
to read.
best,
b
On Nov 7, 2009, at 10:12 AM, Peng Yu wrote:
On Fri, Nov 6, 2009 at 8:19 PM, Benilton Carvalho
wrote:
this is converging to bioc.
let me know what your sessionInfo() is and what type of CEL files
you're
trying to read, additionall
it appears that what you really want is to use:
task[[i]]
instead of task[i]
b
On Nov 1, 2009, at 11:04 PM, dadrivr wrote:
I would like to preface this by saying that I am new to R, so I
would ask
that you be patient and thorough, so that I'm not completely
clueless. I am
trying to con
or
unlist(l)
and possibly you want a unique() on that...
b
On Oct 28, 2009, at 5:15 PM, Jorge Ivan Velez wrote:
Hi Peng,
Here is a suggestion:
unique(do.call(c, l))
# [1] 1 3 4 6 7
Best regards,
Jorge
On Wed, Oct 28, 2009 at 3:07 PM, Peng Yu <> wrote:
Suppose that I have a list of ve
bigmemory and biglm packages may be of your interest.
b
On Oct 28, 2009, at 8:50 AM, Georg Ehret wrote:
Dear R community,
I have a fairly large file with variables in rows. Every variable
(thousands) needs to be regressed on a reference variable. The file
is too
big to load into R (or R ge
colnames(x) <- NULL
On Oct 27, 2009, at 8:09 PM, Peng Yu wrote:
I only see how to assign values to colnames() in help. Is there a way
to remove colnames?
colnames(x) <- value
__
R-help@r-project.org mailing list
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Maybe
http://cran.r-project.org/doc/manuals/R-exts.pdf
?
b
On Oct 23, 2009, at 2:10 AM, Peng Yu wrote:
I found the following document on making R packages. But it is old.
I'm wondering if there is more current ones and hopefully more
complete ones.
http://biosun1.harvard.edu/courses/individ
Dear Lasse,
This won't answer your specific questions and I apologize for that.
AFAIK, pdf() produces uncompressed PDFs only. But you could use tools
like pdftk to compress your PDFs. About the PNGs, you can always set
the 'res' argument to improve resolution, but it won't beat the PDFs.
aves = aggregate(df1$score, by=list(col1=df1$col1, col2=df1$col2), mean)
results = merge(df1, aves)
b
On Oct 21, 2009, at 9:03 AM, Tony Breyal wrote:
Dear all,
Lets say I have the following data frame:
set.seed(1)
col1 <- c(rep('happy',9), rep('sad', 9))
col2 <- rep(c(rep('alpha', 3), rep('
# cheap
[1] 50
b <- a
area(b)
[1] 50
length(b) <- 20
area(a)
[1] 50
area(b)
expensive
[1] 100
Martin
Thanks,
-steve
On Oct 14, 2009, at 2:57 PM, Benilton Carvalho wrote:
If you change 'area' to an environment, you may be able to get
something close
If you change 'area' to an environment, you may be able to get
something close to what you want.
For example:
setClass("Square",
representation(
length='numeric',
width='numeric',
area='environment'
align the 'else if' and 'else' with the closing curly brackets.
if (condA){
doStuff()
} else if (condB){
doOtherStuff()
} else {
doWhatever()
}
b
On Oct 3, 2009, at 12:54 PM, Chen Gu wrote:
Hello,
I am doing a simple if else statement in R. But it always comes out
error
such as 'une
mb,
while the
other took ~4GB for reading the matrix.
I will try the scan() and see if it helps.
Thanks!
Mike
-Original Message-
From: Benilton Carvalho [mailto:bcarv...@jhsph.edu]
Sent: Wednesday, September 23, 2009 4:56 PM
To: Ping-Hsun Hsieh
Cc: r-help@r-project.org
Subject: Re: [R] r
rvation:
The one with the options does save memory a lot. It took ~150mb,
while the other took ~4GB for reading the matrix.
I will try the scan() and see if it helps.
Thanks!
Mike
-Original Message-
From: Benilton Carvalho [mailto:bcarv...@jhsph.edu]
Sent: Wednesday, September 23, 2009
use the 'colClasses' argument and you can also set 'nrows'.
b
On Sep 23, 2009, at 8:24 PM, Ping-Hsun Hsieh wrote:
Hi,
I am trying to read a tab-delimited file into R (Ver. 2.8). The
machine I am using is 64bit Linux with 16 GB.
The file is basically a matrix(~600x70) and as large as
if you want to sample integers in [1, 33K] without replacement:
theSample = sample(33000, 18000)
b
On Sep 23, 2009, at 7:29 PM, phoebe kong wrote:
Hi all,
I would like to generate ~18K random number from range 1 to ~33K.
I was thinking to use round(runif(18000,1,33000)), however the some
and note that if, instead of zip files, you were using gzip files, you
could:
conn <- gzfile("file.gz", "rt")
theData <- read.table(conn)
close(conn)
b
On Sep 22, 2009, at 11:21 PM, Gabor Grothendieck wrote:
Linux is a type of UNIX so follow the instructions I gave for UNIX.
On Tue, Sep 22
library(Matrix)
a = sparseMatrix(i=c(20, 30, 1), j=rep(1, 3), x=c(2.2, 3.3,
4.4))
b = sparseMatrix(i=c(3, 30), j=rep(1, 2), x=c(0.1, 0.1), dims=dim(a))
theSum = a+b
summary(theSum)
hth,
b
On Sep 8, 2009, at 10:19 AM, Henrique Dallazuanna wrote:
Try this:
abMerge <- merge(a, b, by
have you tried:
fits <- lm(a~b)
fstat <- sapply(summary(fits), function(x) x[["fstatistic"]][["value"]])
it takes 3secs for 100K columns on my machine (running on batt)
b
On Aug 23, 2009, at 9:55 PM, big permie wrote:
Dear R users,
I have a matrix a and a classification vector b such that
it's 'scipen' you want to look at.. b
On Aug 22, 2009, at 11:16 PM, David Winsemius wrote:
On Aug 22, 2009, at 6:13 PM, Dajiang J. Liu wrote:
Dear all,I want to convert a long integer to a string, and for
example,
1
I used as.character(10) e.g, and it gives me back 1e+???.
Wha
result <- Reduce("+", unlist(z, recursive=FALSE))
b
On Aug 22, 2009, at 2:03 PM, kathie wrote:
Dear Gabor Grothendieck,
thank you for your comments.
Ive already tried that. but I've got this error message.
Reduce("+",z)
Error in f(init, x[[i]]) : non-numeric argument to binary operato
you may also consider installing from the command line...
R CMD INSTALL ~/Desktop/Camino-downloads/DiagnosisMed_0.2.2.1.tar.gz
b
On Aug 21, 2009, at 11:27 PM, Juliet Hannah wrote:
Hi Janet,
Were you able to install the package? I just installed it without
problems. I don't
think there should
set.seed(1)
x <- matrix(runif(1000), 100)
system.time(tmp1 <- exp(rowMeans(log(x
system.time(tmp2 <- apply(x, 1, function(y) prod(y)^(1/length(y
all.equal(tmp1, tmp2)
## tmp1 is more robust, btw
On Aug 20, 2009, at 3:22 PM, Edward Chen wrote:
Is there a function or an easier way to comp
Hi,
Here's one toy example that shows what I believe to be a problem with
plot.lm. This was brought up by another user (Marcos Tadeu Andrade
Cordeiro). I took a look at the source and the problem appears to be
related to the fact that you need to reorder the data by the group
means and th
what do you get with:
capabilities()[["jpeg"]]
?
b
On Aug 18, 2009, at 7:27 PM, Edward Chen wrote:
My script:
raw = read.table("c:\\Documents and Settings\\protein\\My Documents\
\My
Fragments\\file1.txt", header = TRUE)
#Normal Average Signal VS Pool
jpeg("Normal Average Signal vs Pool_t
you could stick everything in a 1-liner, but that would make it less
readable:
myf <- function(x){
tmp <- as.character(x)
c(tmp[1], paste(tmp[1], tmp[-1], sep=""))
}
df2 <- as.data.frame(sapply(df, myf))
b
On Aug 12, 2009, at 3:39 AM, milton ruser wrote:
Hi Jill,
Completely not elega
what are exactly "some operations"? if you could provide a
reproducible code, it would make it easier to understand what you're
trying to achieve.
for example, if you were to get the means, you could do something like:
theMeans <- rapply(test, mean)
cheers,
b
On Aug 11, 2009, at 8:59 PM,
?assign
b
On Apr 2, 2008, at 4:26 PM, Georg Ehret wrote:
Dear R community,I wish to include a variable (e.g. slice of "a"
below)
in another variable's name. My objective would be to get a variable-
name
"data_A" and so on. How can I do this?
a<-LETTERS[1:25]
a
[1] "A" "B" "C" "D" "E" "F
Brain indeed.
On Mar 19, 2008, at 11:21 AM, sun wrote:
Thanks Prof. Brain,
Following your suggestion I found the reason of my problem in the
document
''R Installation and Administration' and the links it points to'.
After I updated to R2.6.2, the RTools's default compiler chaged to
gcc4
get("x1")
On Mar 19, 2008, at 1:07 AM, Erin Hodgess wrote:
Dear R People:
I have a function x1
Next, I have a character vector x which has one element, "x1".
How would I retrieve the original function x1 from the character
vector, please?
thanks,
Erin
--
Erin Hodgess
Associate Professo
On Mar 18, 2008, at 11:44 PM, Rolf Turner wrote:
On 19/03/2008, at 4:39 PM, Erin Hodgess wrote:
do.call(class,list=ls())
sapply(ls(),function(x){class(get(x))})
or, in case you want to save some typing:
eapply(globalenv(), class)
b__
R-help@r-
or the suggestive :)
?"["
b
On Mar 13, 2008, at 2:58 PM, Henrique Dallazuanna wrote:
On 13/03/2008, John Kane <[EMAIL PROTECTED]> wrote:
Can you point me to some documentation that discusses
these usages. I have seen them before but I have never
actually figured out how to use them.?
See ?
gene.pair.tf.lst[sapply(gene.pair.tf.lst, "[[", "sig.cor")]
b
On Mar 12, 2008, at 5:24 PM, Mark W Kimpel wrote:
I have a very long list that I'd like to subset based on a logical
value
within each element. Example below. I'd like to get just those list
elements for further study whose $sig.
imprecise are our eyes :)
check where 4336.543 actually is and note how the range of the data
confused you.
plot(a~b)
abline(model)
abline(h=4336.543, v=10)
b
On Mar 11, 2008, at 11:27 AM, Aad Termorshuizen wrote:
# PROBLEM WITH ABLINE
# I have a question about a seemingly imprecise abl
and you might want to check ?prop.table
> prop.table(a, 2)
[,1] [,2]
[1,] 0.0 0.333
[2,] 0.5 0.333
[3,] 0.5 0.333
or even ?sweep (which will be useful for more complex situations)
> sweep(a, 2, colSums(a), "/")
[,1] [,2]
[1,] 0.0 0.333
[2,] 0.5 0.333
no, it won't.
you're doing the right math on the "valid" subset... but you're not
returning the zeros where needed therefore, the whole thing will
get recycled to match the dimensions.
b
On Mar 6, 2008, at 2:03 PM, Henrique Dallazuanna wrote:
I think this should work:
array(A[abs(B)
apparently you forgot the "commented, minimal, self-contained,
reproducible code" part...
L = 10
M = 20
N = 30
P = 40
set.seed(1)
A = array(rnorm(L*M*N*P), dim=c(L, M, N, P))
B = array(rnorm(L*M*N), dim=c(L, M, N))
B[sample(100, 10)] = 0
C = array(0, dim=c(L, M, N, P))
for (i in 1:L) {
for (j i
or, in other terms:
> set.seed(1)
> x <- runif(1e7)
> x.add <- tracemem(x)
> y <- x
> y.add <- tracemem(y)
> identical(x.add, y.add)
[1] TRUE
ie, both objects have the same memory address - therefore, not a copy:
> x.add
[1] "<0x200>"
> y.add
[1] "<0x200>"
now, observe what happens whe
apparently you want to check the genefilter package...
it defines functions like:
rowttests
colttests
rowFtests
colFtests
rowVars
rowSds
moreover, a quick look at Biobase is recommended...
that would save you lots of time as you wouldn't have to reinvent the
wheel.
b
On Mar 3, 2008, at
target = c("u", "d")
apply(expand.grid(rep(list(target), 6)), 1, paste, sep="", collapse="")
b
On Mar 1, 2008, at 9:13 AM, Megh Dal wrote:
Hi all,
Suppose I have to letters 'u' and 'd'. Now I want to find all
combinations like that :
uu
ud
du
.
dd
This t
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