Luca Braglia gmail.com> writes:
> g++ -shared -o pROC.so RcppExports.o delong.o perfsAll.o >
> Rcpp:::LdFlags() > > -L/usr/lib/R/lib -lR
> g++: error: >: File o directory non esistente
This can happen when you have a ~/.Rprofile that messes up the invocation
of Rscript in the src/Makevars (which
I have a time series with 1100 rows of data and am getting contradicting
results from these tests:1) My ACF and PACF plots have lot of lags and upon
looking visually, it looks as if the series in not-stationary.2) But when I
do the ADF test with with lag=30, I reject the null and conclude that seri
Hi,
After loading my data in R I get the error:
Error in .jcall("com/github/egonw/rrdf/RJenaHelper",
"Lcom/hp/hpl/jena/rdf/model/Model;", :
java.lang.OutOfMemoryError: GC overhead limit exceeded
How can I solve it?
[[alternative HTML version deleted]]
Is this what you're going for?
factor(values, levels=mylevels[[1]], labels=mylevels[[2]])
[1] a b e e j
Levels: a b c d e f g h i j
On 2014-03-28 16:38, Jonathan Greenberg wrote:
R-helpers:
Hopefully this is an easy one. Given a lookup table:
mylevels <- data.frame(ID=1:10,code=letters[1:1
Thank you very much!
I used indeed rgl.
The problem was apperently that I used text3d() in the wrong way.
When I use it this way:
plot3d(x, y, z,xlab="PC1", ylab="PC2", zlab="PC3", size=5)
text3d(x=x,y=y,z=z,texts=names, col=4)
It workes the way I wanted.
So thanks again and have a nice w
Here're a couple alternatives if you want to use the index instead of
the variable name:
# Reproducible data frame
a1 <- 1:15
a2 <- letters[1:15]
a3 <- LETTERS[1:15]
a4 <- 15:1
a5 <- letters[15:1]
df <- data.frame(a1, a2, a3, a4, a5)
df
# a1 a2 a3 a4 a5
# 1 1 a A 15 o
# 2 2 b B 14
Hi R,
I am a new Chinese user of R, I like this language very much.In the
recent months,I spend over 4 hours every day leanring R,but during the learning
I come across some problems I can not understand.For R is still not very
popular in China,I ask some net friends but nobody can give m
On 29/03/14 01:34, peter dalgaard wrote:
On 28 Mar 2014, at 02:37 , Rolf Turner wrote:
So there you are. Feel enlightened?
Somewhat, actually, but not to such an extent as to have reached nirvana.
"Promises" blow me away.
Here's the most useful part of the post: to get what you want,
On Mar 28, 2014, at 3:38 PM, Jonathan Greenberg wrote:
> R-helpers:
>
> Hopefully this is an easy one. Given a lookup table:
>
> mylevels <- data.frame(ID=1:10,code=letters[1:10])
>
> And a set of values (note these do not completely cover the mylevels range):
>
> values <- c(1,2,5,5,10)
>
Note that your example may be misleadingly simple, so I made it a bit
more complicated.
The key is ?match
> mylevels <- data.frame(ID=10:1,code=letters[1:10])
> values <- c(1,2,5,5,10)
> with(mylevels,code[match(values,ID)])
[1] j i f f a
Levels: a b c d e f g h i j
## Note that you may have to
R-helpers:
Hopefully this is an easy one. Given a lookup table:
mylevels <- data.frame(ID=1:10,code=letters[1:10])
And a set of values (note these do not completely cover the mylevels range):
values <- c(1,2,5,5,10)
How do I convert values to a factor object, using the mylevels to
define the
Hello,
Maybe there are other ways but the following works.
tst2 <- matrix(nrow = dim(test)[1], ncol = dim(test)[2])
tst2[] <- test
tst2 <- cbind(dimnames(test)[[1]], tst2)
colnames(tst2) <- c("ID", dimnames(test)[[2]])
tst2 <- as.data.frame(tst2)
tst2
Hope this helps,
Rui Barradas
Em 28-03-
Hi R Users,
I was trying to select a sample with columns (measured data) by stratum. I was
able to select rows by stratum. But, I wanted to repeat this procedures 1000
and want to take an average from the 1000 times. I think it is different than
bootstraping since I wanted to select (rondomly
Daer Massimiliano Tripoli,
Re:
> Dear all,
>
>
> I wrote a function :
>
> idrecfun <- function(x)
> {
>
> idi <- 1
> esn <- x[order(x)]
> if !(all(esn==x)) stop("x not ordered")
> esn <- as.character(esn)
> for (i in 2:length(esn)) if(esn[i]==esn[i-1])
> idi[i] <- idi[i-1] else idi[i
Thanks first. Your solutions works nearly perfect, i only have one problem
left.
The result looks like perfect, my problem is now, that i want to convert the
solution into a data.frame.
If i try
test<-xtabs(df2$Value~df2$ID + df2$Group)
test
df2$Group
df2$ID 1 2 3 4 5
10 1
This is a bit more direct. It works by forcing R to treat test
as a matrix rather than a table:
> tst2 <- data.frame(ID=dimnames(test)$ID,
as.data.frame.matrix(test),
check.names=FALSE)
> tst2
ID 1 2 3 4 5
10 10 10 20 30 40 50
11 11 0 0 60 70 0
12 12 0 0 0 80 0
> rownames(tst2) <-
Evidently different sized dashes were used in my data set. Using gsub or some
other method, is there a way to use a consistent dash? With the different
dash types it is difficult to build histograms, tables, barplots and perform
other analysis.Â
Thanks again for your help and insights.Â
Hello,
> install.packages("pROC")
Installing package into
‘/home/l/R/x86_64-pc-linux-gnu-library/3.0’
(as ‘lib’ is unspecified)
provo con l'URL
'http://cran.mirror.garr.it/mirrors/CRAN/src/contrib/pROC_1.7.1.tar.gz'
Warning in download.file(url, destfile, method, mode = "wb", ...)
:
connesso a '
Dear all,
I wrote a function :
idrecfun <- function(x)
{
idi <- 1
esn <- x[order(x)]
if !(all(esn==x)) stop("x not ordered")
esn <- as.character(esn)
for (i in 2:length(esn)) if(esn[i]==esn[i-1])
idi[i] <- idi[i-1] else idi[i] <- idi[i-1] + 1
idi
}
for whichever I send th
Hello,
I want to draw 3D plot. The coordinates should be inticated with a
red point and additional I want to label them with a name.
I tried this:
plot3d(x, y, z,xlab="PC1", ylab="PC2", zlab="PC3",main="Country
score resemblance (Stoxx600 rated by
Vigeo)",text3d(x=x,y=y,z=z,texts=name
I agree with the others that you should consult with a statistician,
but here are some additional things to consider:
The usual equivalence test can be done much simpler (both computation
and conception) by just calculating a confidence interval on the
difference and seeing if the entire interval
Have you tried running it using lmer() in lme4 instead, see if that helps?
Patrick
2014-03-27 6:21 GMT-06:00 Laura Thomas :
> Hi All,
>
> I am using R for the purpose of multilevel modelling for the first time. I am
> trying to examine individuals interpersonal changes in the dependent variable
No, I don't think that will make any difference.
1) Post this to the r-sig-mixed-models list rather than here, as you
are likely to get a much better answer there.
2) Did you realize that treatment is a linear term in the fixed
effects portion, not a factor? If you don't understand the question,
Hi,
I would like to un-compress a gz or a bz2 file to current working directory.
I am able to open the connection in a variety of ways, but than cannot
access the *.*nc file inside.
Is there a way of accessing the netcdf wiathout uncompressing?
(I am using windows8, R 3.0.3)
An example of what I
Dear Richard M. Heiberger,
Re:
> Apple has been popping an offer for a free upgrade to Mavericks.
> (I currently have OS X Lion10.7.5 on an 8GB MacBook Air).
> Other than this offer from Apple, I have no dissatisfaction with the Mac
> for my use pattern.
>
> There has been much comment on this
On 28/03/2014 6:51 AM, palad...@trustindata.de wrote:
Hello,
I want to draw 3D plot. The coordinates should be inticated with a
red point and additional I want to label them with a name.
I tried this:
plot3d(x, y, z,xlab="PC1", ylab="PC2", zlab="PC3",main="Country
score resemblance (Stoxx
On 28 Mar 2014, at 02:37 , Rolf Turner wrote:
>>
>> So there you are. Feel enlightened?
>
> Somewhat, actually, but not to such an extent as to have reached nirvana.
> "Promises" blow me away.
>
>>
>> Here's the most useful part of the post: to get what you want, use
>>
>> do.call(plot,
On 28/03/2014, 2:04 AM, Philippe Grosjean wrote:
On 28 Mar 2014, at 00:57, Duncan Murdoch wrote:
On 27/03/2014, 7:38 PM, Thomas Lumley wrote:
You get what you wanted from
do.call(plot,list(x=quote(x),y=quote(y)))
By the time do.call() gets the arguments it doesn't know how y was
originally
If you don't need a 'running average', there are other 'smoothers' you can
look at. Try 'supsmu':
> set.seed(1)
> x = c(0,1, 3,3.4, 5, 10, 11.23)
> y = x**2 + rnorm(length(x))
> rbind(x, y)
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
x 0.000 1.00 3.00 3.4
Sorry, that should be
temp.tables <- temp.tables[-(31:32),]
Rui Barradas
Em 28-03-2014 10:58, Rui Barradas escreveu:
Hello,
Maybe the following will help.
library(XML)
oneurl="http://www.tutiempo.net/en/Climate/FUKUSHIMA/11-2012/475950.htm";
temp.tables=readHTMLTable(oneurl, which = 3)
st
Hello,
Maybe the following will help.
library(XML)
oneurl="http://www.tutiempo.net/en/Climate/FUKUSHIMA/11-2012/475950.htm";
temp.tables=readHTMLTable(oneurl, which = 3)
str(temp.tables)
temp.tables <- temp.tables[-31,]
temp.tables[,1:10] <- lapply(temp.tables[,1:10], function(x)
as.numeric
Try xtabs()
>> df2 = data.frame(ID = c(10,10,10,10,10,11,11,12),Group =
c(1,2,3,4,5,3,4,4),Value = c(10,20,30,40,50,60,70,80))
>> xtabs(df2$Value~df2$ID + df2$Group)
I think this is exactly what you want.
On Fri, Mar 28, 2014 at 4:51 PM, Mat wrote:
> Hello togehter,
>
> i have a litte problem.
Hello togehter,
i have a litte problem. I have an output data.frame which look like this
one:
ID
1 10
2 11
3 12
Now I have another data.frame with more than one line for each ID:
IDGroupValue
1 10110
2 10220
3 103
Thanks Jim,
this partially helps (I wasn't aware of the functions you used, which
are good to know).
I have two main doubts about your solution though.
First of all, in my application x is large ~25 elements and I have
to repeat this for ~1000 different y vectors; moreover the range of x
is muc
Hello,
You are right, the "which" option avoid the selection in my example.
Also, I think ?"[[" might be helpful for you.
Regards.
Pascal
On Fri, Mar 28, 2014 at 3:02 PM, Bill wrote:
> Hi. I found that this works:
> tt=readHTMLTable(url,header=TRUE, as.data.frame=TRUE,which=c(3))
> tt[1,]
> I
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