On Wed, 16 Nov 2011, Abraham Mathew wrote:
I'm running R in Ubuntu 10.10 and am trying to install the RODBC package.
However, I get the following error message:
ERROR: configuration failed for package ?RODBC?
* removing ?/home/amathew/R/i686-pc-linux-gnu-library/2.13/RODBC?
The downloaded pack
On 17/11/11 18:54, Sachin Abeywardana wrote:
Hi all,
I was wondering if it was possible to get R to throw an error if a
certain condition was reached. For example:
if dim(A)[2]<=1
error("Need a matrix not a vector")
?stop
cheers,
Rolf Turner
__
?source
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Enginee
I am looking for online courses to learn Spatial Statistics using R.
Statistics.com is offering an online course in December on the same topic
but that schedule doesn't suit mine. Are there any other similar modes for
learning spatial statistics using R??? Can someone please advice???
Thank you.
Dear all,
I have the following data, which has \\n in place of \n. I introduced \n's
in the csv file so that I could use it in barchart in lattice. When I did
that and read it into R using read.csv, it read it as \\n. My question is
how do I introduce "\n" in the middle of a long string of quoted
Hi all,
I was wondering if it was possible to get R to throw an error if a
certain condition was reached. For example:
if dim(A)[2]<=1
error("Need a matrix not a vector")
thanks,
Sachin
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Michael,
Thanks a lot for the suggestions! Those are good ideas,
Xu
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hi ,I am writing a soft package based on R.
But when I try to read a configure file showed as below.
How can I read the parameter in this text file.
How can I read the parameter into each variable in this file ?
configinfo<-scan(file(configfile),ok=TRUE,n=-1)
scan seems need every line have same c
Hi All,
I am trying to convert the following piece of matlab code to R:
XX1 = sum(w(:,ones(1,N1)).*X1.*X1,1); #square the elements of X1,
weight it and repeat this vector N1 times
XX2 = sum(w(:,ones(1,N2)).*X2.*X2,1); #square the elements of X2,
weigh and repeat this vector N2 t
How can I specify 'starting values' for a cubic spline ns()? I need
starting values for all pars when fitting using the nonlinear least
squares optimiser nls(). The y variable (length) is a function of 2
variables: alpha and age. The functional form for the age-length
relationship is specified as a
Hey everyone,
I am new to R, and I'm making a scatter plot graph where i have a bunch of
plots/points that fall into 9 unique categories. I want each category to
have a unique colour, however, with the coding I have (below), the colour
black is repeated for two of my plot types. Does anyone know
Thanks Michael...!.
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Hi
I need to add an year to and date field in the dataframe.
Please help me
X Date
1 2008-01-01
2 2008-02-01
3 2003-03-01
Thanks in advance
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Hello,
Two days ago I posted a patch to R source code to colorize its output
when running it in a terminal emulator (Linux or other Unix with
support to ANSI escape codes).
I converted the patch into an R package, and people interested in
testing it should do the following:
1) Download and insta
I'm running R in Ubuntu 10.10 and am trying to install the RODBC package.
However, I get the following error message:
ERROR: configuration failed for package RODBC
* removing /home/amathew/R/i686-pc-linux-gnu-library/2.13/RODBC
The downloaded packages are in
/tmp/RtmpekzPOQ/downloaded_packag
Thanks to everyone who replied to my post, I finally got it to work. I am
however not sure how well it worked since it run so quickly, but seems like
I have a 2000 x 2000 data set. My followup questions would be, how do I get
only pairs with say a certain pearson correlation value additionally it
s
Hmmm...sorry -- the only thing I can suggest is maybe striking some
sort of deal that you change when it gets however many months out of
date: if you look here
(http://cran.r-project.org/src/contrib/Archive/MASS/), you can see the
last time each version of MASS was updated and by seeing what you
h
Perhaps, if R FAQ 7.31 isn't a problem, this would work.
(df.1$AffyIds)[match(df.2$rMF, df.1$rMF)]
Michael
On Wed, Nov 16, 2011 at 1:11 PM, Rob Griffin wrote:
> As another potential route could I put something in to the original code
> that makes df.2 (maindata2) which picks one of the AffyIds
Your code produces likelihood estimates of zero and divides them.
Hence NaN. You should decide for yourself whether this is a data
problem or an algorithmic problem.
In general try using options(error = recover) to debug when you get in
a situation like this: it's super helpful.
I'll happily give
Try this
qqInteractive <- function(..., IDENTIFY = TRUE){
qqplot(...) -> X
if(IDENTIFY) return(identify(X))
invisisble(X)
}
The trick is that identify wants coordinates of the point in the
scatter plot which are not the inputs to qqplot() but rather a
transformation thereof.
Micha
Thanks Michael,
I just started on the following code (below), and realized I should as as
this might exist.
basically what I'd like is for the function to return (basically) what you
just suggested, plus the names of the two variables (I suppose pasted
together would be good).
I hope that is
What exactly do you mean "returns" them? More generally I suppose,
what do you have in mind to do with this?
You could do something like this:
BigCorrelation <- function(X){
return(which(abs(cor(X)) > 0.9, arr.ind = T))
}
but it hardly seems worth its own function call.
On Thu, Nov 17, 20
hi,
If i have a list of things, like this
var.names <- c("a", "b", "c", "d", "e", "f")
how can i get this:
"a, b", "c, d", "e, f"
thanks ahead of time.
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Hello,
I would like to find out if a function already exists that returns only
pairwise correlations above/below a certain threshold (e.g, -.90, .90)
Thank you.
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On the face of it this looks like a job for ldply() in the plyr package
which specialises in taking things apart and putting them back together.
ldply() applies a function for each element of a list and then combine
results into a data frame
On 17 November 2011 04:53, Sarah Goslee wrote:
> O
On Wed, Nov 16, 2011 at 9:39 PM, Kevin Burton wrote:
> Say I have the following data:
>
>> s <- list()
>> s[["A"]] <- list(name="first", series=ts(rnorm(50), frequency=10,
> start=c(2000,1)), category="top")
>> s[["B"]] <- list(name="second", series=ts(rnorm(60), frequency=10,
> start=c(2000,2)),
Say I have the following data:
> s <- list()
> s[["A"]] <- list(name="first", series=ts(rnorm(50), frequency=10,
start=c(2000,1)), category="top")
> s[["B"]] <- list(name="second", series=ts(rnorm(60), frequency=10,
start=c(2000,2)), category="next")
If I use unlist since this is a list of lists
To add to what David suggests, and since you're new to R, something like
this:
plot(x,y, yaxt='n')
yticks <- pretty(y)
axis(2, at=yticks, labels=sprintf("%1.2f",yticks))
See the help page for par
?par
and look for the entry for 'xaxt' to see what the 'yaxt' arg to plot does.
--
Don MacQueen
One option would be, after you have converted one of your date/time
variables (V2 I would assume) to POSIXct class, you can use the interp()
function in the akima package for linear interpolation between adjacent
time points.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East
I would suggest starting by taking a look at the overlay() function in the
sp package.
(also suggest follow up questions to to R-sig-geo)
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 11/16/11 11:30 AM, "Peter Maclean
Sarah Goslee gmail.com> writes:
Thank you, Sara!
Your information is's been very useful
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and p
R. Michael Weylandt gmail.com>
gmail.com> writes:
..
Thanks Michael
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented,
Hi,
On Wed, Nov 16, 2011 at 7:11 PM, JulioSergio wrote:
> I read a table as follows:
>
>> F1 <- read.table("Rtext3.txt")
>> F1
> Price Floor Area Rooms Age Cent.heat
> a 52.00 111 830 5 6.2 no
> b 54.75 128 710 5 7.5 no
> c 57.50 101 1000 5 4.2 no
> d 57.
Don't use attach. I know it's tempting, but it's really not worth it in the
long run. (Cf. Genesis 3, ff.)
This works:
F1["a", "Price"]
Incidentally the reason yours didn't work is because the row names are part of
F1, not Price so attach() doesn't carry them along.
In short, row names good
I read a table as follows:
> F1 <- read.table("Rtext3.txt")
> F1
Price Floor Area Rooms Age Cent.heat
a 52.00 111 830 5 6.2no
b 54.75 128 710 5 7.5no
c 57.50 101 1000 5 4.2no
d 57.50 131 690 6 8.8no
e 59.7593 900 5 1.9 yes
On 17/11/11 05:37, Scott Raynaud wrote:
That might be an option if it weren't my most important predictor. I'm
thinking my best bet is to use MLWin for the estimation since it will properly
set fixed effects
to 0. All my other sample size simulation programs use SAS PROC IML which I
don't
Seems logical to me -- and, Juan, sorry for messing up earlier -- read
too hastily.
Michael
On Wed, Nov 16, 2011 at 5:55 PM, Peter Ehlers wrote:
> I must be missing something. What's wrong with
>
> t(apply(mat, 1, sample))
>
> ?
>
> Peter Ehlers
>
> On 2011-11-16 12:12, Gavin Simpson wrote:
>>
It's going to depend on how the data missing-ness is represented in the table.
If you have a csv like
1,2,3
1,,3
4,5,6
You automatically get an NA after using read.csv() -- this can be
omitted with the na.omit() command if desired.
If missingness is somehow encoded other than an empty spot, it
unlist(..., recursive = F)
Michael
On Wed, Nov 16, 2011 at 6:20 PM, wrote:
>
> I would like to make the following faster:
>
> df <- NULL
> for(i in 1:length(s))
> {
> df <- rbind(df, cbind(names(s[i]), time(s[[i]]$series),
> as.vector(s[[i]]$series), s[[i]]$c
I would like to make the following faster:
df <- NULL
for(i in 1:length(s))
{
df <- rbind(df, cbind(names(s[i]), time(s[[i]]$series),
as.vector(s[[i]]$series), s[[i]]$category))
}
names(df) <- c("name", "time", "value", "category")
Thanks so much for the prompt response. I just realized I never thanked
you. I apologize for the delay in responding.
On Tue, Aug 9, 2011 at 3:25 PM, Eik Vettorazzi <
e.vettora...@uke.uni-hamburg.de> wrote:
> Hi Devon,
> check your rprofile.site file and comment the following lines
>
> options(ID
Sorry I didn't respond earlier. No, Rq does not take a vector tau.
Frank
David Winsemius wrote:
>
> I suppose this constitutes thread drift, but your simple example, Frank,
> made wonder if Rq() accepts a vector argument for tau. I seem to remember
> that Koencker's rq() does.. Normally I would
I must be missing something. What's wrong with
t(apply(mat, 1, sample))
?
Peter Ehlers
On 2011-11-16 12:12, Gavin Simpson wrote:
On Wed, 2011-11-16 at 14:29 -0500, R. Michael Weylandt wrote:
Suppose your matrix is called X.
? sample
X[sample(nrow(X)),]
That will shuffle the rows at rando
(Assuming I get your meaning) Sounds like a homework exercise ...
The perhaps simplest way to do this is directly with a loop and
indexing. Did you think about this at all, as this seem fairly obvious
to me.
As an exercise for the reader, it can also be (very simply) done by
vectorizing the calcu
??console ## This you SHOULD have tried yourself!
?loadRconsole
?windows
It looks to me like you would use e.g. windows() to position and size
the graphics display and then would write a temp file and load it with
loadRconsole() to reposition the Gui window as desired.
a) This sounds clumsy
b)
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of muzz56
> Sent: Wednesday, November 16, 2011 12:28 PM
> To: r-help@r-project.org
> Subject: Re: [R] Pairwise correlation
>
> Thanks Peter. I tried this after reading in the csv (r
Say I have two matrixes: one is 8x28 and other 8x8
I'd like to multiply, for example, first and second column from the 8x8 and
plant them in first column of 8x28.
Then take first and third of 8x8 and plant into second column of 8x28.etc.
Any ideas?
Thanks!
[[alternative HTML ver
I'm just curious if Professor Koenker responded to this (e.g. in an
individual email or otherwise), as I would be interested in the answer.
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Thanks Peter. I tried this after reading in the csv (read.csv) and
converted the data to matrix (as.matrix). But when I tried the correlation,
I keeping getting the error (x must be numeric) yet when I view the data,
its numeric.
On 16 November 2011 14:32, plangfelder [via R] <
ml-node+s789695n407
I am very sorry for multiple mails.
Hi R users,
I try to read a data file (tab delimited format) in which some of the
entries in a particular field are missing. Is it possible to fill the
unavailable data with 'UnAvailable' string while performing read.table()
Something like
df = read.table(DataF
Hi R users,
I try to read a data file (tab delimited format) in which some of the
entries in a particular field are missing. Is it possible to fill the
unavailable data with 'UnAvailable' string while performing read.table()
Something like
df = read.table(DataFile, header=FALSE, fill_missing_entr
Hi all,
I know that I can run an one way ANOVA on the absolute residual values
to check the assumption of equal variances:
model<- lm(y~x)
summary(lm(abs(model$resid~x)))
What if I have two factors? As far as I know I have to check this
assumption on very factor/level combination. So if I have 3
Hi R users,
I try to read a data file (tab delimited format) in which some of the
entries in a particular field are missing. Is it possible to fill the
unavailable data with 'UnAvailable' string while performing read.table()
Something like
df = read.table(DataFile, header=FALSE, fill_missing_ent
Try this: (it would have been easier if you had used 'dput' on your data)
x <- read.table('/temp/example.txt', skip = 1, as.is = TRUE)
# convert to POSIXct
x$beg <- as.POSIXct(paste(x$V4, x$V5))
x$end <- as.POSIXct(paste(x$V6, x$V7))
# determine breaks over midnight
x$over <- format(x$beg, "%d")
To expand on what Sarah and Michael said:
if you have a 3d array:
> x<-array(1:4,c(2,2,4))
> x
, , 1
[,1] [,2]
[1,]13
[2,]24
, , 2
[,1] [,2]
[1,]13
[2,]24
, , 3
[,1] [,2]
[1,]13
[2,]24
, , 4
[,1] [,2]
[1,]13
[2,]
Hi,
On Wed, Nov 16, 2011 at 3:13 PM, wrote:
>
> I have the following scenario:
>
>> m <- matrix(1:4, ncol=2)
>> m
> [,1] [,2]
> [1,] 1 3
> [2,] 2 4
>> apply(m, 2, sum)
> [1] 3 7
>> apply(m, 1, sum)
> [1] 4 6
>
> So I can apply to rows *or* columns. According to the documentation
It's the same as what you began with -- and that's because you broke
it down by columns and rows and took the sum of everything that
resulted.
I.e.,
sum(m[1,1])
sum(m[2,1])
sum(m[1,2])
sum(m[2,2])
and put them back together.
Michael
On Wed, Nov 16, 2011 at 3:13 PM, wrote:
>
> I have the fol
On Wed, 2011-11-16 at 14:29 -0500, R. Michael Weylandt wrote:
> Suppose your matrix is called X.
>
> ? sample
> X[sample(nrow(X)),]
That will shuffle the rows at random, not permute within the rows.
Here is an alternative, first using one of my packages (permute -
shameful promotion ;-) !:
mat
I have the following scenario:
> m <- matrix(1:4, ncol=2)
> m
[,1] [,2]
[1,]13
[2,]24
> apply(m, 2, sum)
[1] 3 7
> apply(m, 1, sum)
[1] 4 6
So I can apply to rows *or* columns. According to the documentation
(?apply)
MARGIN a vector giving the subscripts which the functio
I have csv data that extend beyond the area I want for an existing map. I want
using the boundaries of the polygon shape file as a cookie cutter so that I can
overlay the csv data on map without including anything outside the map
boundaries and create a dbf file or shapefile of the clipped data
Hello all,
I have a data frame that looks like this:
http://r.789695.n4.nabble.com/file/n4077622/Capture.png
I would like to know if it's possible to split a single row into two rows
when the time frame between "beg" and "end" overlaps midnight. I want to
compare the frequency of each activity
And you got a reply on the ggplot2 list, which is why you're asked not
to cross-post.
For those who are wondering, geom_boxplot() in the ggplot2 package
will by default plot outside points along the same line as the boxplot
whiskers at their actual values. The gentleman jittered the original
point
OK, I'm using William Browne's MLPowSim to create an R script which will
simulate samples for estimation of sample size in mixed models. I have subjects
nested in hospitals with hospitals treated as random and all of my covariates
at level 1. My outcome is death, so it's binary and I'll have a
?cor
X = matrix(rnorm(400),ncol = 4)
cor(X)
Michael
On Wed, Nov 16, 2011 at 11:37 AM, muzz56 wrote:
> Dear All,
> I am not familiar with R yet I want to use it to perform some task, hence my
> posting here. I hope someone can help.
> I have a set of data, genes (rows) and samples (columns). I w
On Wed, Nov 16, 2011 at 8:37 AM, muzz56 wrote:
> Dear All,
> I am not familiar with R yet I want to use it to perform some task, hence my
> posting here. I hope someone can help.
> I have a set of data, genes (rows) and samples (columns). I want to do a
> Pearson correlation on all the possible pa
Suppose your matrix is called X.
? sample
X[sample(nrow(X)),]
Michael
On Wed, Nov 16, 2011 at 11:45 AM, Juan Antonio Balbuena wrote:
> Hello
> This is probably a basic question but I am quite new to R.
>
> I need to permute elements within rows of a binary matrix, such as
>
> [,1] [,2] [,3]
> -Original Message-
> From: Dennis Murphy [mailto:djmu...@gmail.com]
> Sent: Wednesday, November 16, 2011 11:22 AM
> To: Andy Bunn
> Cc: r-help@r-project.org
> Subject: Re: [R] equal spacing of the polygons in levelplot key
> (lattice)
>
> OK, how about this instead?
>
> # library('latti
Hello,
I generate box plots from my data like this:
qplot(x=xxx,y=column,data=data,geom="boxplot") + xlab("xxx") + ylab(ylabel) +
theme_bw() + scale_y_log10() + geom_jitter(alpha=I(1/10))
The problem is that I see lot of points above the maximum at the same level as
some outliers. It looks ver
Hello
This is probably a basic question but I am quite new to R.
I need to permute elements within rows of a binary matrix, such as
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]000010000 0
[2,]001100011
Dear All,
I am not familiar with R yet I want to use it to perform some task, hence my
posting here. I hope someone can help.
I have a set of data, genes (rows) and samples (columns). I want to do a
Pearson correlation on all the possible pairwise combinations of all the
genes (2000). Does anyone h
OK, how about this instead?
# library('lattice')
levs <- as.vector(quantile(volcano,c(0,0.1,0.5,0.9,0.99,1)))
levq <- seq(min(levs), max(levs), length = 6)
levelplot(volcano, at = levs,
colorkey = list(at = levq,
labels = list(at = levq,
On 16.11.2011 17:37, Scott Raynaud wrote:
That might be an option if it weren't my most important predictor. I'm
thinking my best bet is to use MLWin for the estimation since it will properly
set fixed effects
to 0. All my other sample size simulation programs use SAS PROC IML which I
do
> -Original Message-
> From: Dennis Murphy [mailto:djmu...@gmail.com]
> Sent: Tuesday, November 15, 2011 8:54 PM
> To: Andy Bunn
> Cc: r-help@r-project.org
> Subject: Re: [R] equal spacing of the polygons in levelplot key
> (lattice)
>
> Hi:
>
> Does this work?
Thanks Dennis.
This alm
Hi:
Try dict[[a]] rather than dict$a, and read the section on lists in the
Introduction to R manual to learn how to properly index them.
HTH,
Dennis
On Wed, Nov 16, 2011 at 7:16 AM, Antje Gruner wrote:
> Dear list,
>
> I'd like to access the elements of a list within another list with the
> hel
Dear All,
Could anybody please advice me the way to check the goodness of fit of the
linear mixed-effects model and the suitable function in R to do so, and thank
you in advance for your assistance.
Fir
[[alternative HTML version deleted]]
Three things:
1) This isn't reproducible without your data file. Work out a minimal
reproducible example -- I bet you'll find your answer along the way --
and supply it.
2) What do the warnings say: they are usually pretty good at directing
you to trouble.
3) Don't use attach. Seriously -- just
Perfect, thanks! Naturally, now I need to resize the console so it doesn't
cover my new plots. I'd like to resize it on the fly (from within the
function) then reset the size to its previous size.
So, curConsoleDims() and resize.console() are a made up functions, but they
demonstrate what I am try
As another potential route could I put something in to the original code
that makes df.2 (maindata2) which picks one of the AffyIds at random for the
duplicated FlybaseCG values (shown below)
maindata2<-aggregate(maindata[,c(161,172,168,255,254,258,264,265,263,271)],
by = maindata[,167, drop =
Happy to help, but could you provide a dput() example of your data
div1 -- I'm not sure if your table elements are coded as factors or
doubles and that can make quite a difference.
Anyways, once you confirm they are doubles, something like this will do it:
apply(D, 1, function(x) all(diff(x)>0))
Try the add = TRUE argument to contour.
Michael
On Wed, Nov 16, 2011 at 12:35 PM, Sramkova, Anna (IEE)
wrote:
> Hi all,
>
> I would like to plot one data set as a 2d histogram and another one as a
> contour. I can do it separately with the
> "hist2d" and "contour" functions, but I wonder if the
This is a little ugly, but I think it should work pretty robustly:
T <- table(unlist(df[,-1])) # Take a close look at this to see how it
works -- it's the key to the whole thing and basically creates
something we will roughly use like a hash-table (if that term is
familiar to you) (or more accurat
??focus ## admittedly, not the first keyword that comes to mind
?bringToTop
-- Bert
On Wed, Nov 16, 2011 at 9:07 AM, Ben quant wrote:
> Hello,
>
> After I plot something how do I reactivate the console (and not the plot
> window) so I don't have to click on the console each time to go to the n
Hi all,
I would like to plot one data set as a 2d histogram and another one as a
contour. I can do it separately with the
"hist2d" and "contour" functions, but I wonder if there is a way to combine
these two plots into a single one (the ranges of the two plots are the same).
Any suggestions?
T
Can you provide a dput() example of your data and put it in a workable
form -- I have no idea what that long list of numbers nor the 1 / 0 on
the end represent. I think if you do this with a real time series
class like xts you can create a new object with the interpolated
times; fill in the values
Hello,
After I plot something how do I reactivate the console (and not the plot
window) so I don't have to click on the console each time to go to the next
command?
Example that does not work:
fun = function(x){ plot(x); dev.set(dev.prev())}
fun(1:4)
...and another that does not work:
fun = fun
Put pdf() before your loop and dev.off() after -- as it stands now, on
each iteration you re-open the pdf and thereby wipe everything that
was previously on it.
For the axis, look at the axis() command if you want something
detailed or the xaxt, xlab arguments to boxplot if its relatively
simple.
On Nov 16, 2011, at 10:16 AM, Antje Gruner wrote:
Dear list,
I'd like to access the elements of a list within another list with the
help of a variable.
dict <- list( "24" = c(1,2,3,6,12,24,48,72,96,120,144,168,720),
"168" = c(1,12,24,48,72,96,120,144,168,336,504,672),
"8760"
dict[[a]]
? `[[`
Michael
On Wed, Nov 16, 2011 at 10:16 AM, Antje Gruner wrote:
> Dear list,
>
> I'd like to access the elements of a list within another list with the
> help of a variable.
>
> dict <- list( "24" = c(1,2,3,6,12,24,48,72,96,120,144,168,720),
> "168" = c(1,12,24,48,72,96
R FAQ 7.31 strikes again.
:-)
Michael
tmp[3105] == 35.52
[1] FALSE
tmp[3105] - 35.52
[1] -7.105427e-15
all.equal(tmp[3105], 35.52)
TRUE
On Wed, Nov 16, 2011 at 10:42 AM, oscar wrote:
> Hello.
>
> I recently came across the extremely strange problem (see below) of R
> telling me that elements
That might be an option if it weren't my most important predictor. I'm
thinking my best bet is to use MLWin for the estimation since it will properly
set fixed effects
to 0. All my other sample size simulation programs use SAS PROC IML which I
don't have/can't afford. I like R since it's fre
Dear Michael,
Thank you very much.
Ram
From: R. Michael Weylandt
Sent: Wednesday, November 16, 2011 1:48 PM
Subject: Re: [R] Conversion of symmetry matrix into a vector ONE MORE QUESTION
Try the as.dist() function -- it won't change any values, only giv
Are you looking for something like nlm()? It does minimization in any
number of dimensions using a Newton-type algorithm;
it can be manually given gradients / Hessians, or alternatively you can
give it neither and let them approximated empirically.
Taylor
On Wed, Nov 16, 2011 at 11:00 AM, Edward
Dear all,
I am aiming to calculate variograms using variogram() from gstat.
The problem is that some of my data-sets are very large (> 40 points).
Running the command takes some hours of time and does not give any
error-message.
Nevertheless the result seem not to be appropriate - the first fe
Hi Michael,
Thanks for your response. Using the binary seems to solve partially. I am
able to install (I think!) RCurl but not able to load the library. Below is
the info you required and the error while loading RCurl.
*> sessionInfo()*
R version 2.13.2 (2011-09-30)
Platform: x86_64-apple-darwin9.
Thanks for the answer. I found that smth like below works
subset <- c(1,0,0)
series <- c(1,1,1,0,1,0,0,1,1,0,0,0,0,1)
subset1 <- paste(subset, collapse='')
series1 <- paste(series, collapse='')
xx <- gregexpr(subset1, series1, fixed=TRUE)
as.numeric(xx[[1]]) ## yields the first element numbers i
HI,
I have subsetted a list of exons and put them in a vector exon_list
> exon_list=levels(test5$V1)
I want to loop using that vector to create a big pdf which will contain
all my barplots and doing this I got only the last element in my
exon_list plotted rather than all of them, I guess th
Dear Community,
I want to identify outliers in my data. I don't know how to use identify
command in the plots obtained.
I've gone through help files and use mahalanobis example for my purpose:
NormalMultivarianteComparefunc <- function(x) {
Sx <- cov(x)
D2 <- mahalanobis(
I am scraping data from a web page using XML (excellent package BTW - that's
scraping data the easy way!).
So far, I've got the code:
tables <- readHTMLTable(theurl)
rhf <- tables$tabResHistFull
div1 <- rhf[which(rhf$V1=="Div ps"),]
div1
which is giving me the result:
V1 V2 V3 V4
Le mercredi 16 novembre 2011 à 06:12 -0800, arunkumar a écrit :
> Hi
>
> I have a data frame and i need to change the date format in it.
> my dataframe
>
> X Date
> 1 1/1/2009
> 2 2/1/2009
> 3 3/1/2009
>
> I need to change it to 2009-01-01
See ?as.Date. In your case, I think you should use
Hello.
I recently came across the extremely strange problem (see below) of R
telling me that elements of a list do not exist...
I have tried running this small script on 2 computers (Mac OSX 10.7 &
Windows 7) running the latest releases or R (2.13.2 on Mac, 2.14.0 on Win).
I have tried it both i
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