Hello,
You probably should reconfigure R using:
> R CMD javareconf
before installing the package you want.
And when you install a package, don't forget the quote.
HTH
Pascal
Le 02/08/2012 14:45, am a écrit :
Hi,
I'm working with R on Cassandra and need to install RJDBC on my Ubuntu
virtual
Hi,
I am glad it is solved.
#ir3 <- data.frame(rbind(ir1[1:4],ir2))
should be "cbind" in your previous code.
Even if that was corrected,
n<-
neuralnet(Output~Sepal.Length+Sepal.Width+Petal.Length+Petal.Width,data=ir3,err.fct="sse",hidden=c(3),linear.output=FALSE)
Error in neurons[[i]] %*% we
Hello,
I guess the error might be due to the struture of your dataset.
Try this:
data(iris)
iris1<-iris
iris1[iris1$Species=="setosa","Output"]<- 1
iris1[iris1$Species=="versicolor","Output"]<- 0
#Here, if you set "virginica" to 2, still it runs, with a warning message "--
response is not bin
HI,
I tested the code for another set of data with added complexity. Seems to
be working fine.
list1<-list(data.frame(x=1:6,y=c(4,5,3,2,1,8),bw=c(4,18,12,3,4,9)))
list2<-list(data.frame(x=c(1,2,18,16,15),y=c(4,5,9,2,1),bw=c(4,18,22,3,4)))
list3<-list(data.frame(x=c(4,6,9),y=c(8,24,12),bw=c(14,31,3
Hi,
I'm working with R on Cassandra and need to install RJDBC on my Ubuntu
virtual machine. The traditional commands (install.packages(RJDBC, dep=TRUE)
and library(RJDBC)) don't work: they throw an error that they can't find the
package. I tried installing it using the Rkward GUI, but can't write
On Wed, Aug 1, 2012 at 10:28 PM, Bert Gunter wrote:
> On Wed, Aug 1, 2012 at 6:45 PM, Michael Weylandt
> wrote:
>> Isn't this what package dependencies are for?
>
> No. It's what package imports are for (preferably).
> As always, the OP should RTFM -- in this case the one to which you
> refer on
On Wed, Aug 1, 2012 at 6:45 PM, Michael Weylandt
wrote:
> Isn't this what package dependencies are for?
No. It's what package imports are for (preferably).
As always, the OP should RTFM -- in this case the one to which you
refer on the next line, especially the NAMESPACES section.
-- Bert
>
> S
If I understand you correctly you want to exclude columns where all successes
equal trials, all successes equal 0, or successes are a mixture of trials and 0
with no in between values. You did not make it clear if the number of trials
can vary, but in your example they do not. Given that all thr
I don't see anything obviously wrong, but I seem to remember that SPSS uses
equal prior probabilities as the default whereas lda() uses "the class
proportions for the training set." That might explain the difference.
--
David L Carlson
Associate Professo
I am sorry, but I don't know what exactly is the problem. What's wrong
with this:
library(igraph)
kar <- nexus.get("karate")
star3 <- graph.formula(A -- B:C:D)
subiso <- graph.get.subisomorphisms.vf2(kar, star3)
Then you can query the symbolic ids:
V(kar)$name[ subiso[[1]]+1 ] # This is actual
Thank you very much for your great help! I add "as.is=TRUE" in the sqlFetch
code, it works!
2012/8/1 Prof Brian Ripley
> On 01/08/2012 10:11, delvin008 delta wrote:
>
>> Hi, all:
>>This is the first time I use mail-list for help and I'm a student
>> from Chinese. So forgive my poor Engli
HI,
You can also try this:
d<-1:25
A<-sample(combn(20:30,2))
B<-sample(combn(20:30,2))
lapply(d,function(x) matrix(c(1,A[x],B[x],1),2,2))
[[1]]
[,1] [,2]
[1,]1 23
[2,] 271
[[2]]
[,1] [,2]
[1,]1 21
[2,] 211
[[3]]
[,1] [,2]
[1,]1 29
[2,] 231
[[4]
well, the article from this high impact factor journal mentioned about each
step for building a nomogram including interpretation without going too much
detail into statistical concept.
All I want to know is what the R code does to come up with the graph, and
not raising attention or get someone t
Thanks,
but the problem is that igraph doesn't even give the correct id mappings.
it only gives the mapping of the structure(without labels)
For above eg:
igraph only sees the one edge structure(of graph 2) without labels or names
and gives all possible edges in the main graph(1) to be isomorphic
I think all you need to do is add /directory/ to the library search path -- see
?.libPaths
Michael
On Aug 1, 2012, at 9:30 PM, Xuan Zhao wrote:
> Hi Michael,
> Thank you so much for your help!
> But my package is not under a directory that can be searched automatically,
> namely, when load my
Seems to work fine under 2.15.1
R version 2.15.1 (2012-06-22) -- "Roasted Marshmallows"
Copyright (C) 2012 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-pc-mingw32/x64 (64-bit)
R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistrib
Putting context back in.
On Jul 31, 2012, at 9:01 AM, B787s wrote:
Dear R-Help,
I am using 'rms' package to draw nomogram. I wonder how is the
"Points"
determined for each predictor in the model? Is it by the coefficient
estimate (beta) relative to the highest effect in the model or?
It
Thanks, it is helpful.
I knew there were several modeling capacities built into this package "rms".
I would just like to have a general ideal how the points for each predictor
determined. I read a paper by Lasonos et al 2008. It mentioned it was by the
size of the effect.
--
View this message
Isn't this what package dependencies are for?
See the description of the DESCRIPTION file in Writing R Extensions
Michael
On Aug 1, 2012, at 5:27 PM, xuan zhao wrote:
> Hi,
> I have built two R packages. One of them (PKG1) needs to use the functions
> of the other package (PKG2).
> So I need
Hello R help,
I have this data frame M2[16,5] with NAs, a simple example would be:
set.seed(1234)
M2<-expand.grid(ID=182:183, year=2012, month=1:3, day=1:3,
KEEP.OUT.ATTRS=FALSE)
M2 <- M2[with(M2, order(ID, year, month, day)),] #sort the data
M2$value <- sample(c(NA, rnorm(100)), nrow(M2),
Yuo can use source function.
Looking for ?source for an example.
Best,
Roberto
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Hi,
I have built two R packages. One of them (PKG1) needs to use the functions
of the other package (PKG2).
So I need to link these two packages together, so that the functions of PKG2
can be available to PKG1. And when I load one package using
'library("PKG1")', PKG2 can be loaded at the same.
An
Hi,
I am trying to follow the same logic starting from the initial dataset with
NaN.
Z16<-read.table(text="
Summary G Y R T
Accts 582 644 NaN 1226
AcctCov 230 165 NaN 395
Cov% 40 2
Hi all,
I obtained a strage result with LDA (MASS) function in R with NIR data.
I tried both CV (leave one out cross validation) and splitting my data in
odd (training) and even (prediction) sets.
In all the cases the minimum error was near to 0.
Due to the strange result, I tried with SPSS IBM so
Thanks for specifying the solution.
The deal is that we are faced of vectors.
Moreover I need to specify them as function of own-lagged
with expansion=1,2, … ; lag=1, 2, …
set.seed(1)
E <- cbind(as.vector(rnorm(10)),as.vector(rnorm(10)),as.vector(rnorm(10)))
#or more vectors; here 3 for illustra
This isn't the year 2038 problem and workaround #2 won't help because
R uses the tm structure, not time_t.
Here's a non-xts example of the issue, which seems to be in as.POSIXct.POSIXlt:
x <- "2038-12-31-23-59-59.9"
(y <- strptime(x, "%Y-%m-%d-%H-%M-%OS"))
unclass(y)
(z <- as.POSIXct(y))
I th
I'm not sure that's it: R uses doubles for its POSIXct values so those
aren't at play here. (And even if they were, 64-bit R wouldn't help:
that's only pointer size)
One can trace the problem out of xts down to
ISOdatetime(2035:2040, 12, 31, 23, 59, 59.999)
which is a base R function, but I'm st
I'm a little confused on your data structure -- can you use dput() as
described here [1] to give a small reproducible example?
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
Michael
On Wed, Aug 1, 2012 at 10:38 AM, gail wrote:
> Thank you!!! But I realise
Can you make an R-only example using dput()? See this page for more details:
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
Best,
Michael
On Wed, Aug 1, 2012 at 8:51 AM, PaulJr wrote:
> Good morning everyone,
>
> I have attached an Excel file that contains
On Aug 1, 2012, at 10:15 AM, gaiarrido wrote:
Hi,
I´ve got some problems with the labels of the x-axis
I,ve got two factors with two categories each: sex (males and
females) and
area (central and peninsulae),
but because of the lenngth in the graphic just appeared two of the
four
labels.
On Aug 1, 2012, at 8:50 AM, John linux-user wrote:
Hi everyone,
I try to add many vectors (L1,L2,L3) to multiple list objects
(a.list,
b.list) in a workspace. Somethings like below, but it is not
working. Any suggestions will be appreciated. Best, John
lf=ls(pattern=".lst")
for
On Aug 1, 2012, at 8:11 AM, John linux-user wrote:
Hi everyone,
I try to add many vectors (L1,L2,L3) to many list objects
(a.list, b.list) in a workspace. Somethings like below, but it
is not working. Any suggestions will be appreciated. Best, John
lf=ls(pattern=".lst")
for (x
I believe that foreach() only goes along the "shorter" of the
iteration vectors.
E.g.,
foreach(.combine = c, i = 1:10, j = 1:15) %do% print(i + j)
stops when i == 10 (and j = 10 as well). To make it more explicit
foreach(.combine = c, i = 1:10, j = 11:25) %do% print(i + j)
stops when j = 20 be
I'm afraid I don't quite understand what you're getting at: can you
say what you're trying to do big-picture wise.
This advice might help making a reproducible example:
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
Best,
Michael
PS -- Just a hunch -- but
Plz provide a reproducible example (using dput).
Anyway, I suspect your data to be 'factor' and not numeric.
y=rbinom(100,1,0.5)
x=runif(100)
plot(x,y) #OK
plot(x,as.factor(y))
See ?factor and ?as.numeric.
Regards,
Eloi
On 12-08-01 11:41 AM, Georgiana May wrote:
> Hello,
> Anyone know why the
Hi -- these sorts of graphical parameters are documented in ?par, but
it's admittedly somewhat terrifying to try to read (Oh well, the cost
of comprehensiveness) -- you're looking for either las or (less
likely, but might work if you can draw the labels manually) str, both
documented on that page.
On Wed, Aug 1, 2012 at 11:40 AM, matthewbk wrote:
> Hi
>
> I am using the nls function in R however it is causing the following error:
> REAL() can only be applied to a 'numeric', not a 'logical'
>
> Using traceback shows this occurs after the .Call(R_nls_iter, m, ctrl,
> trace) call. However I do
Perhaps y is a factor. E.g.,
> plot(x=c(3,5), y=factor(c(0,1)))
> plot(x=c(3,5), y=factor(c("Pig","Goat")))
both treat y as as.numeric(y), which is c(1,2) in the first
and c(2,1) in the second.
When you import data into R, always look at it with str()
and various plots to convince yourself it is
http://en.wikipedia.org/wiki/Year_2038_problem
Workaround 1: Avoid collecting data in the future and processing it in the
present.
Workaround 2: Use a 64-bit build of R, as appropriate for such future data sets.
---
Jeff New
The following 'f' counts the number of times the sequence x
does not increase. Is this what you want?
> f <- function(x) split(x, cumsum(c(TRUE, x[-1] <= x[-length(x)])))
> f(numbers)
$`1`
[1] 1 2
$`2`
[1] 1 2 3 4 5
> f(c(1,1,2,3,4,5,1,2,3,2,3,4,5,6,4,5))
$`1`
[1] 1
$`2`
[1] 1 2 3 4 5
$`3`
[1
Hi -- this isn't really reproducible: can you follow the advice given
here and make a small reproducible example:
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
Best,
Michael
On Wed, Aug 1, 2012 at 12:51 PM, Rahul Bhalla
wrote:
> I require some help in deb
On Wed, Aug 1, 2012 at 12:54 PM, Douglas Karabasz
wrote:
> I used quantmod to pull in price data from the ticker SPY. The data has
> date and closing price. I would like to show the day of the week for each
> closing price. Is that possible? Also, I would like to add the back into
> the data f
Hello,
Try the following.
fun <- function(x){
n.diff <- cumsum(diff(c(x[1], x)) <= 0)
split(x, n.diff)
}
numbers <- c(1,2,1,2,3,4,5)
fun(numbers)
fun( c(1,1,2,3,4,5,1,2,3,2,3,4,5,6,4,5) )
Hope this helps,
Rui Barradas
Em 01-08-2012 14:29, capy_bara escreveu:
Hello,
I have a vecto
On Wed, Aug 1, 2012 at 6:30 AM, Ingezz wrote:
> Dear Peter,
>
> I have another question about WGCNA. I am using the package for
> meta-analysis to find modules preserved in several datasets. However, I am
> unsure how to handle the softpower, because each dataset has its own ideal
> scale indepenc
On Jul 31, 2012, at 9:01 AM, B787s wrote:
Dear R-Help,
I am using 'rms' package to draw nomogram. I wonder how is the
"Points"
determined for each predictor in the model? Is it by the coefficient
estimate (beta) relative to the highest effect in the model or?
It would be better if you ask
On Wed, Aug 1, 2012 at 8:43 AM, gail wrote:
> Hello. Please forgive me if this problem has already been posted (and solved)
> by someone else ... I can't find it anywhere though it seems so very basic.
> Here it is:
>
> I have a list comprised of several matrices, each of which has two columns.
>
I am not that proficient in R. I found some codes on web using those
indicator variables to sum up log likelihood. I block out bols in the
codes, but I also tried using them as start value for my estimation of
ologit. Didn't work. Thanks for your suggestion.
Jun
On Wed, Aug 1, 2012 at 5:44 PM, Be
On Wed, Aug 1, 2012 at 4:19 PM, Ramiro Barrantes
wrote:
> Hello,
>
> I come from using different programming languages (C++, Mathematica, Perl)
> but have been using R extensively for several months. I see the data frame
> as a key piece of the language and wanted to inquire people's experience
Can you give a reproducible example: it doesn't for me
x <- c(3, 5)
y <- c(0, 1)
plot(x, y)
What you might be getting confused with is
plot(y)
where the x axis is now at 1 and 2 -- this is because when only given
a single vector to plot, R plots the values against their indices,
which, as all i
Hi Ray,
2012/7/31 Ray Brownrigg :
> On 07/28/12 23:46, Thomas Steiner wrote:
>> Hi,
>> I'd like to have a low resolution word map in R.
>> The "maps" package has this option, but if I use the argument, the map
>> looses sense: Russia and Australia get empty etc
>>
>> library("maps")
>> m=map(col="
1. Is there ever a situation using the R recommended packages only (to
keep things reasonable) where one explicitly needs to create indicator
variables to do statistical modeling?
2. If so, does model.matrix() always suffice to do this?
In any situation I can imagine, one would create a factor an
Hello,
Try reading the data file with stringsAsFactors = FALSE, like this you
don't have character vectors, you have factors, which are coded as
consecutive integers 1, 2, etc.
?read.table
?read.csv
See in particular the argument above, stringsAsFactors
Hope this helps,
Rui Barradas
Em 01
You'll have to give a more realistic description to get detailed help:
otherwise, look up "combinatorial optimization" and "parallelization"
for some generic pointers.
Best,
Michael
On Wed, Aug 1, 2012 at 11:34 AM, loyolite270 wrote:
> Thanks for the reply, but the example i have given above is
Please don't post files to nabble in the future -- when Nabble removes
them, we then have a broken link in the real archives: dput() is much
preferred way to send data.
Anyways:
# mydata <- read.csv("http://r.789695.n4.nabble.com/file/n4638691/jan_2011.csv";)
#
# dput(head(mydata, 15))
structure
Hello everybody!
I have a problem, which I'm trying to solve. I have two Krig-fits: kr_one
and kr_two; both use x and y and additional either A (kr_one) or B (kr_two):
kr_one <- with(1, Krig(cbind(x,y), A))
kr_two <- with(1, Krig(cbind(x,y), B))
I can view the 3d-plane using:
image(predict.surf
Hi Namit,
I feel like you posted this (or a _very_ similar) question just a few
days ago: yes indeed you did:
https://stat.ethz.ch/pipermail/r-help/2012-July/319634.html
with a good amount of response.
I'm going to direct you in a different path: don't add these things
within R, at least, n
Hello,
I'm currently using MCRestimate package and I have a question regarding
the MCRestimate function.
Here is my code:
NestedCV.rf<-MCRestimate(eset, "Class", classificatin.fun="RF.wrap",
variableSel.fun="varSel.highest.var", poss.parameters=
list(var.numbers=c(100), mtry=c(10,50),
cross.oute
I think the vector is far more fundamental than the data frame. Most of the
time I write functions that return vectors, even if my input is a data frame.
If I need a large number of input vectors, I set up the input arguments to
include a data frame and additional named parameters with defaults
Thanks very much for your response. Sorry for writing to you directly.
Hereafter, I'll email to the R-Help list. I agree with all your comments and
suggestions. My analysis will be descriptive type; purely observational as the
studies are non-randomized studies. I have sample sizes and counts
After fitting and pruning an rpart model, it is often the case that one or
more of the original predictors is not used by any of the splits of the
final tree. It seems logical, therefore, that values for these "unused"
predictors would not be needed for prediction. But when predict() is called
on s
I used quantmod to pull in price data from the ticker SPY. The data has
date and closing price. I would like to show the day of the week for each
closing price. Is that possible? Also, I would like to add the back into
the data frame in a new column without changing the structure of the data
se
? ccf
Best,
Michael
On Wed, Aug 1, 2012 at 7:46 AM, arunkumar wrote:
> Hi
>
> I've to find the correlation of a variable and lag of other variable.
>
> My code currently looks like this
>
> y= 32 1 45 26 33 55 15 24 31 78
> x=51 69 96 67 71 89 61 42 78 24
> lagTimes=2
>
> i want to find th
I require some help in debugging this code
library(neuralnet)
ir<-read.table(file="iris_data.txt",header=TRUE,row.names=NULL)
ir1 <- data.frame(ir[1:100,2:6])
ir2 <-
data.frame(ifelse(ir1$Species=="setosa",1,ifelse(ir1$Species=="versicolor",0,"")))
colnames(ir2)<-("Output")
ir3 <- data.frame(rbi
I think there is a problem with R's tm package's weightTfIdf function.
The manual says that the idf is calculated as
idf(term ) = log (|D|/number of documents that contain the term)
In cases where the dictionary is passed in the control list as given below:
dtm =
DocumentTermMatrix(myCorpus,con
Specifically note that, unlike C, R has true multidimensional arrays
which are more than pointer jumps: the practical result of this is
that you want to use a single subsetting ( x[i,j] ) and specify them
both in one call -- x[i][j] instead gives you the "j"th element of
the "i"th element which ma
I have the following problem:
As expected I obtain
timeBasedSeq( 2037/2037)
[1] "2037-01-01"
>
However if I do the same for 2038 I get
timeBasedSeq( 2038/2038)
[1] "2038-01-01" "2039-01-01"
I get the same strange result if I use later years or any interval that
contains 2038:
timeBasedSeq( 203
I somehow solved the problem - kind of. The data set on which I ran the GAM
model contains many more variables than are needed in the model, so I
created a new data set in R and reran the GAM model on the slimmed down data
set. Same problem: The GAM can be computed, but the tensor product cannot be
Thank you Ricardogg and Arun. I don't know how I missed that. Too many
sleepless nights perhaps. Thank you.
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Hi -
I'm using the tune.nnet() to identify the optimal tuning parameters for a
dataset in which the dependent variables is binary and has very few 1s. As
a result, tune.nnet() provides parameters that give high accuracy, but by
increasing the specificity and decreasing the sensitivity radically. D
Hi everyone,
I try to add many vectors (L1,L2,L3) to multiple list objects (a.list,
b.list) in a workspace. Somethings like below, but it is not
working. Any suggestions will be appreciated. Best, John
lf=ls(pattern=".lst")
for (x in listfiles) {
dat=read.delim(x,header=F)
Hello,
Anyone know why the command:
> plot(x,y) where y is a 0,1 result
sometimes plots the y values as 1,2 rather than 0,1?
And how to prevent this?
Thank you,
Georgiana May
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
Hi everyone,
Let me have a dataframe named “mydata” and created as below,
*> n=c(5,5,5,5) #number of trils
> x1=c(2,3,1,3) ) #number of successes
> x2=c(5,5,5,5) #number of successes
> x3=c(0,0,0,0) #number of successes
> x4=c(5,0,5,0) #number of successes
> mydata=data.frame(n,x1,x2,x3,x4)
> myda
Hi,
I´ve got some problems with the labels of the x-axis
I,ve got two factors with two categories each: sex (males and females) and
area (central and peninsulae),
but because of the lenngth in the graphic just appeared two of the four
labels.
I thought the solution could be rotate it 45º degrees
Hi
I am using the nls function in R however it is causing the following error:
REAL() can only be applied to a 'numeric', not a 'logical'
Using traceback shows this occurs after the .Call(R_nls_iter, m, ctrl,
trace) call. However I do not know how to debug this further to find out
what is causing
Hi everyone,
I try to add many vectors (L1,L2,L3) to many list objects (a.list,
b.list) in a workspace. Somethings like below, but it is not working. Any
suggestions will be appreciated. Best, John
lf=ls(pattern=".lst")
for (x in listfiles) {
dat=read.delim(x,header=F)
for
Thanks for the reply, but the example i have given above is a sample
function. I would be using a function with input vector of size more than
100 and the function will compute some score based on the 100 vector
combinations.
Since i don't want to do this computation sequentially for all these
com
Thank you for the answer, but I did´n find the solution. I'm a bit lost,
cos I am a beginner in R. I hope you can give me a little bit more help.
2012/7/29 jholtman [via R]
> Try this: You had some 'indexing' problems. You were accessing the
> columns of the dataframe incorrectly. You are
Thank you!!! But I realise I've simplified my data to the point that your
solution doesn't actually work -- not your fault, mine! My list is actually
more complicated than what I presented it to be; it's not composed of
numerical matrices but of lists, each being composed of 7 columns, the first
tw
For me it owrks when i write it like:
as.Date(paste(mydata$Delivery.Date), "%m/%d/%Y")
Hope it works,
Regards,
Ricardo
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You're totally right Jeff. My mistake! to use with, we write it like this:
DF$season <- factor ( with ( *DF*, ifelse (( month == 12 | nonth == 1 |
month == 2 ), "Win",
ifelse ((month == 3 | nonth == 4 | month == 5 )
, "Spr",
ifelse
I have an object, which I pull in from a csv file here
http://r.789695.n4.nabble.com/file/n4638691/jan_2011.csv jan_2011.csv
mydata <- read.csv("jan_2011.csv", header=TRUE, sep=",")
> head(mydata)
Delivery.Date Hour.Ending Repeated.Hour.Flag Settlement.Point
Settlement.Point.Price
101/01
Hi Rui,
Thanks. It was my fault. I had both of these saved (read.table (),
data.frame(read.table()). I think I accidentally pasted the second one, and
got the error.
A.K.
- Original Message -
From: Rui Barradas
To: arun
Cc: r-help
Sent: Wednesday, August 1, 2012 9:41 AM
Subj
Hi,
just during these vacation days, I'm trying to approach with multicore
package
and I have some troubles with foreach.
ex.
foreach(.combine=c, ii=1:200, jj=1:500) %dopar% makefunction[ii,jj]
Command seem to work but if I see ii and jj value at the and of cycle, they
are both 200
instead than i
Good morning everyone,
I have attached an Excel file that contains a macro from which I call and
use R's auto.arima function to generate forecasts. The program runs
perfectly and it gets me the results; however, those results are pretty
unusual. I also tried using the auto.arima function directly
Hello. Please forgive me if this problem has already been posted (and solved)
by someone else ... I can't find it anywhere though it seems so very basic.
Here it is:
I have a list comprised of several matrices, each of which has two columns.
> list
[[1]]
[,1] [,2]
[1,]1 3
[2,]
Hello,
I have a vector with positive integer numbers, e.g.
> numbers <- c(1,2,1,2,3,4,5)
and want to split the vector whenever an element in the vector is smaller or
equal to its predecessor.
Hence I want to obtain two vectors: c(1,2) and c(1,2,3,4,5).
I tried with which(), but it is not so el
Dear Peter,
I have another question about WGCNA. I am using the package for
meta-analysis to find modules preserved in several datasets. However, I am
unsure how to handle the softpower, because each dataset has its own ideal
scale indepence value. When combining several datasets what should I do?
Hi Petr,
The following is different line of thought which is posted in different form,
maybe you have some wise input on it.
"I need to find Efficient(tracktable) deterministic algorithm for Matching
Weighted Graphs with bounded degree. Now we all know Graph matching is
non-tractable but when
HI,
The code was working perfectly fine yesterday and today, until half an hour
ago. Couldn't find any problems in the code. Still, I am getting error message.
myMatrix <- data.matrix(read.table(text="
Name Age
ANTONY 27
IMRAN 30
RAJ 22
NAHAS
Hi everyone
I'm trying to estimate both the rho parameter and the degree of freedom of
t-ev copula from the data using "fitCopula" function in the "copula"
package.
However, there is always an error and it says:
Error: length(copula@parameters) == length(param) is not TRUE
I also tried the sam
Hi Friends,
I'm new to R ,I have a data frame Z16 which is genarated from another data
frame, and I want to add “%” & “$” in row 4 and 5 respectively. when I’m
trying using below logic, I’m getting warning message. I'm using R 2.14.2
Version
Can anyone help me out on this.
Note: Initially
HI,
Try this:
#This might get you started:
startdate <- as.POSIXct("01/01/2006", format = "%d/%m/%Y")
enddate <- as.POSIXct("01/12/2006", format = "%d/%m/%Y")
date <- seq(from = startdate, to = enddate, by = "months",format = "%d/%m/%Y")
DF <-
data.frame(data=c(2.5,1.4,3.6,0.5,-1.2,6.5,7,4.5,
Hi Petr,
It been sometime since I wrote. But here are the latest developments.
When I give the binary linear opt problem to lpSolve optimizer than for small
instance it gives correct answer but when size of nodes increase let's say 16
then there are about 2000 binary variables and lpSolve just
Hi,
Try this:
dat1<- read.table(text="2000-01-05 00:00:00 1.0
2000-01-05 01:00:00 1.0
2000-01-05 05:00:00 3.6
2000-01-05 06:00:00 3.6
2000-01-05 07:00:00 2.2
2000-01-05 08:00:00 2.2
2000-01-05 09:00:00 2.2
2000-01-05 10:00:00
Here is my approximation:
# Creation of the temporal variables
DF$year <- as.numeric(format(DF$date, format = "%Y"))
DF$month <- as.numeric(format(DF$date, format = "%m"))
# For years with data from 2006 to 2008
DF_type1 <- DF [ - which (year == 2006 & month ==1 | year == 2006 & month ==
2 |
Hi
I've to find the correlation of a variable and lag of other variable.
My code currently looks like this
y= 32 1 45 26 33 55 15 24 31 78
x=51 69 96 67 71 89 61 42 78 24
lagTimes=2
i want to find the correlation between y and Lag(x,1) and Y &Lag(x,2)
i'm using loop to find the lag and then
Hi Petr and Jean,
thanks very much, problem solved! Really appreciate your help.
Jennifer
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Disclaimer: I have not followed this thread at all, but only wish to note:
1) Indicator variables are (almost?) never needed in R -- that you are
fooling with them suggests that there is probably a better approach.
2) Your bols is just least regression, no? -- If so, there are far
better ways to
Thanks Michael. Now I switched my approach after doing some google.
Following are my new codes:
###
library(foreign)
readin <- read.dta("ordfile.dta", convert.factors=FALSE)
myvars <- c("depvar", "x1", "x2", "x3")
mydta <- readin[myvars]
# remov
I have no answer to your question, but note:
1. You do not need to return a data frame at all, of course. Most
functions do not -- e.g., say, lm() .
2. See ?with and ?within for perhaps relevant functionality.
-- Bert
On Wed, Aug 1, 2012 at 2:19 PM, Ramiro Barrantes
wrote:
> Hello,
>
> I come f
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