On Aug 28, 2012, at 05:11 , David Winsemius wrote:
> That's exactly it. If a logical index returns NA, its row is included in the
> output of "[" extraction. You can correct what I consider a failing and
> others consider a feature with:
>
> df[df$Renewal==TRUE & !is.na(df$Renewal), 1:2]
>
P
On Aug 27, 2012, at 8:49 PM, Elaine Kuo wrote:
I want to predict wing length using regression commands. (lm and
predict)
The data details are as followed
Data:
Bird physiological data
1. body mass
2. body length
3. wing length
Data type:
Order A: consisting of 20 species, body mass and l
On 28-08-2012, at 03:12, Christopher Kelvin wrote:
> Hello,
> I want to estimate the exponential parameter by using optim with the
> following input, where t contains 40% of the data and q contains 60% of the
> data within an interval. In implementing the code command for optim i want it
> to
summary: I can successfully ncvar_put(...) data to a file, but when I
try to ncvar_get(...) the same data, I get
> Error in if (nc$var[[li]]$hasAddOffset) addOffset = nc$var[[li]]$addOffset
> else addOffset = 0 :
> argument is of length zero
How to fix or debug?
details:
R code @
https://
On Aug 27, 2012, at 10:04 PM, R. Michael Weylandt > wrote:
It's because read.table returns a data frame, not a matrix. You can
coerce to a matrix with as.matrix() but you might loose information
if your variables are of different classes.
Michael
On Aug 27, 2012, at 7:07 PM, Cheryl Johns
Just looked at your code, and you do call nc_sync. I am wondering if it works
if you call nc_close, reopen and then write. Perhaps the nc_sync is sufficent
when you have been in define mode in the nc_create call.
-Roy
On Aug 27, 2012, at 8:31 PM, Tom Roche wrote:
>
> summary: I can successfu
If I had to guess, you did not call nc_close or nc_sync. In netcdf, the data
is actually really written until that is called.
-Roy
On Aug 27, 2012, at 8:31 PM, Tom Roche wrote:
>
> summary: I can successfully ncvar_put(...) data to a file, but when I
> try to ncvar_get(...) the same data, I g
On Aug 27, 2012, at 5:08 PM, Mauricio Cornejo wrote:
Hi,
Would anyone have any idea as to why I would obtain completely
different results when subsetting using the subset function vs
bracket notation?
I have a data frame with 65 variables and 4382 rows. When I use
execute the following
Take a look at gsub()
Michael
On Aug 27, 2012, at 6:47 PM, Sapana Lohani wrote:
> Hi,
>
> My data frame (Students) is
>
> ID Name Fav_Place
> 101 Andrew� Phoenix AZ
> 201 John San Francisco
> 303 JulieCalifornia / New York
> 304 Monica� New York
>
> How can I replace Phoenix with Tucson & N
On Aug 27, 2012, at 1:40 PM, Sapana Lohani wrote:
Hi, my data frame is
x<-data.frame(ID=c("abc/def","abc/def/ghi","abc","mno/pqr/st/ab"))
I want to split my column ID using "/" as the place to split. How
can I do that without telling the code how many sub-columns. I could
use nchar(gsub("
It's because read.table returns a data frame, not a matrix. You can coerce to a
matrix with as.matrix() but you might loose information if your variables are
of different classes.
Michael
On Aug 27, 2012, at 7:07 PM, Cheryl Johnson wrote:
> Greetings,
>
> When I try to use the write.table c
subset(dataFrame, subset) does the equivalent of dataFrame[!is.na(subset) &
subset,].
I.e., it treats the NA's in the subset argument the same as FALSEs. Doesn't
help(subset)
mention this?
By the way, if Renewal is a logical vector, it will be identical to
Renewal==TRUE so
you may as well leav
Hello,
I want to estimate the exponential parameter by using optim with the following
input, where t contains 40% of the data and q contains 60% of the data within
an interval. In implementing the code command for optim i want it to contain
both the t and q data so i can obtain the correct estim
Hi,
I am new to R and using Rcmdr and like to automate and get into scripting...
I am using for evaluating two variables
.Table <- xtabs(~CurSWI+BckMo, data=swanalysis_run1)
.Table
fisher.test(.Table)
remove(.Table)
the output for this will be
> .Table
BckMo
CurSWI N/A No Yes
N/A
Greetings,
When I try to use the write.table command to save a matrix as a file and
then open the file with read.table, if I try to take the mean of the entire
matrix, instead each column of the matrix has its mean calculated. I have
copied and pasted an example of my code below. When I try to mak
Hi,
My data frame (Students) is
ID Name Fav_Place
101 Andrew Phoenix AZ
201 John San Francisco
303 JulieCalifornia / New York
304 Monica New York
How can I replace Phoenix with Tucson & New York with New York City in the df?
Thanks
[[alternative HTML version deleted]]
__
I have read multiple books and looked at many posts online and can't seem to
figure out how to add a 1:1 line in lattice xyplot. I've tried multiple
versions of getting this to work, including trying "panel.abline(h=0,v=1),
panel=function and others.
Second question, how do I get the legend creat
I am making the assumption that all the columns are character and not factors:
for (i in c("A", "B", "C", "D", "E")){
yourdf[[i]] <- ifelse(yourdf[[i]] == 'x'
, 'y'
, ifelse(yourdf[[i]] == 'a'
, 'b'
, yourd
Hi,
Would anyone have any idea as to why I would obtain completely different
results when subsetting using the subset function vs bracket notation?
I have a data frame with 65 variables and 4382 rows. When I use execute the
following subset command I get the correct results (125 rows)
> subset(
HI,
#In addition to,
nchar1<-nchar(gsub("[^/]","",x$ID))
max(nchar1)
#[1] 3
library(stringr)
data.frame(str_split_fixed(x$ID,"/",max(nchar1)+1))
#or strsplit(as.character(x$ID),"/")
#You can also use:
strsplit(gsub("(.*)/(.*)","\\1 \\2",gsub("(.*)/(.*)/(.*)","\\1 \\2
\\3",gsub("(.*)/(.*
done, thanks for the suggestion.
David
On 2012-08-27 21:15, Sam Steingold wrote:
* jim holtman [2012-08-27 14:55:08 -0400]:
Most likely when 'y' is converted to a dataframe (not sure what the
function 'write.matrix.csr' does since you did not say where you got
it),
sorry,
library(e1071)
'
I have 5 (A,B,C,D,E) columns in my dataframe. I want to replace all "x" with
"y" and all "a" with "b" within these 5 columns. Can I do it in one step?
Thanks
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat
Dear Michael,
Thanks very much for your explicit explanation! That makes much sense.
Best wishes,
Jeff
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-generate-a-matrix-of-Beta-or-Binomial-distribution-tp4641422p4641472.html
Sent from the R help mailing list archive at
Splitting is easy:
strsplit(as.character(x$ID), "/")
That produces a list with four elements, each of which is a character
vector. R does not have a representation for "sub-columns" so you will have
to be clearer about how you want to represent the data and what you are
planning to do with it.
-
Hi All,
I am trying to install RODBC Package in Linux version of R. PFB the error
message. Request all, if you have any solution how to overcome this error.
configure: error: "ODBC headers sql.h and sqlext.h not found"
Regards,
Madana
--
View this message in context:
http://r.789695.n4.nabbl
Hello,
I'm glad it helped.
Inline.
Em 27-08-2012 19:00, Zuofeng Shang escreveu:
Hi Rui,
Many thanks for your excellent code! That works very well. But I still
have a kind of naive question about the set.seed(1)
Consider shape1=c(1,3) and shape2=c(2,4)
Using R I got that
> set.seed(1)
> rbe
Hey everyone,
I am an R-newby... so sorry for bothering you with simple-to-solve
questions;) I have the following issue: trying to add labels to my
scatterplots (with geom_dl in ggplot2). Everything works fine, but after
checking every resource I do not find a way to change the font size of my
lab
Hello
I want to predict wing length using regression commands. (lm and predict)
The data details are as followed
Data:
Bird physiological data
1. body mass
2. body length
3. wing length
Data type:
Order A: consisting of 20 species,
body mass and length of all 20 species are measured
Perfectly sensible, and indeed what I originally wrote. But it only
works for my trivial example, not the general situation where it might
not be the first level that needs changing.
-- Bert
On Mon, Aug 27, 2012 at 1:52 PM, David Winsemius wrote:
>
> On Aug 27, 2012, at 12:18 PM, Bert Gunter wro
On Aug 27, 2012, at 12:18 PM, Bert Gunter wrote:
Well ...See below.
-- Cheers, Bert
On Mon, Aug 27, 2012 at 9:19 AM, David Winsemius > wrote:
On Aug 27, 2012, at 3:09 AM, Fridolin wrote:
What is a smart way to change an entry inside a column of a
dataframe or
matrix which is of type "fa
On 27/08/2012 1:47 PM, Stephane Chantepie wrote:
Dear all,
I have tried to plot a triangular matrix with the function persp3d(rgl).
for example
z=rbind(c(1,NA,NA,NA),c(5,3,NA,NA),c(4,2,9,NA),c(8,6,5,11))
x=1:4
y=1:4
persp3d(x,y,z,color="gray")
The two extreme points are not plotted (value=1 a
I wasn't an Rstudio user at that point, but now that I am a new user,
I am wondering "how" you were interacting with the Rstudio windows. I
have an editing window open and is has both a "Run" button and a
"Source" button. I get the behavior you describe when I hit the Source
button and the
Emil,
It's helpful to provide code that generates your (example) data.
You can use the function dput() to do that.
# I set up your matrix a bit differently than what you presented.
output <- structure(c(-2.3109, -0.2934, -0.5539, -2.1352, -2.6968,
-2.0429,
-0.3973, -0.6468, 0.4181, -0.6243, 2.9
On Mon, Aug 27, 2012 at 9:56 AM, Alok Bohara, PhD wrote:
> Hi:
>
> I am trying to understand the link between ".csv" and ".rda" files.Is
> there any easy to follow tutorial on this?
>
> (I could do some of the operations below, but I got lost in the details.)
>
> 1. Reading .rda file ?
>
> d
Typo:
> y <- as.character(x[[1]]) ## You need a character vector argument
> strsplit(y,"/") ## works
-- Bert
On Mon, Aug 27, 2012 at 12:58 PM, Bert Gunter wrote:
> (Re-)read ?strsplit. You do not have to tell the code how many columns ...
> etc.
> And the split isn't into sub columns.
>
> y <-
(Re-)read ?strsplit. You do not have to tell the code how many columns ... etc.
And the split isn't into sub columns.
y <- as.character(x[[1]] ## You need a character vector argument
strsplit(y,"/") ## works
-- Bert
On Mon, Aug 27, 2012 at 11:40 AM, Sapana Lohani wrote:
> Hi, my data frame is
>
On Mon, Aug 27, 2012 at 9:49 AM, JeffND wrote:
> Hi folks,
>
> I have a question about how to efficiently produce random numbers from Beta
> and Binomial distributions.
>
> For Beta distribution, suppose we have two shape vectors shape1 and shape2.
> I hope to generate a 1 x 2 matrix X whose i
I need some help with using recursive least squares lm.fit.recursive
{quantreg}.
I found some references online but cannot find a reproducible example as to
how to use recursive least squares.
I'd really appreciate if anyone can point me to a reproducible example.
[[alternative HTML versi
I tried to calculate the pairwise pearson correlation of the column
vector of a matrix of 20,000 columns using the gpuCor fucntion from the
package gputools, and compare the speed with the ordinary cor function.
The gpuCor is supposed to be much faster than cor. But I found that the
ordinay cor is
Hi, my data frame is
x<-data.frame(ID=c("abc/def","abc/def/ghi","abc","mno/pqr/st/ab"))
I want to split my column ID using "/" as the place to split. How can I do that
without telling the code how many sub-columns. I could
use nchar(gsub("[^/]","",x$ID)) to get how many "/" are in each row of t
Dear all,
I have tried to plot a triangular matrix with the function persp3d(rgl).
for example
z=rbind(c(1,NA,NA,NA),c(5,3,NA,NA),c(4,2,9,NA),c(8,6,5,11))
x=1:4
y=1:4
persp3d(x,y,z,color="gray")
The two extreme points are not plotted (value=1 and value=10). It seems
because the half of the matr
Dear All,
could you please confirm (or disconfimr, and then please explain:-)) my
understanding of the results generated by the following code:
hdr.den(rnorm(200,55,3))
result:
[,1] [,2]
99% 47.07170 63.26864
95% 48.83670 61.55108
50% 53.04884 57.37916
the way I am interpreti
Hi David.
I mean that I want to get the *standard error of the predicted means* (which
is a type standard deviation, if I have understand everything right), which
the se.fit switch mentioned above should require from the "predict()"
function.
But the se.fit switch doesn't seem to work for manova ob
Thank you, character indexing (?"[") is what I was looking for:
sapply(paste('var', 1:100, sep=''), function(x)
weighted.mean(data[[x]], data$weight))
Daniel
On Mon, Aug 27, 2012 at 6:26 PM, Bert Gunter wrote:
> Have you read "An Introduction to R". If not, please do so before
> osting and pay
Hey all,
I was wondering if someone could tell me a way to remove the lines in my
annoated BeadSummary Data file that have a PROBEQUALITY of "BAD" and
"Match".
They method they used in the Bead summary tutorial gives me the following
error.
Anyone have any ideas?
Thanks in advance
Will
ids <-
> * jim holtman [2012-08-27 14:55:08 -0400]:
>
> Most likely when 'y' is converted to a dataframe (not sure what the
> function 'write.matrix.csr' does since you did not say where you got
> it),
sorry,
library(e1071)
> '0' and '1' are converted to factors which probably show up as 1
> and 2 in t
I was going to point you to the same utility, which I have used over the years
on both Linux and OSX. A Google search using "pdf to text" will bring up a
variety of non-R related possibilities.
It is possible that somebody, somewhere has built an interface in R to
pdftotext, such as a wrapper f
When a sparse matrix is multiplied by a regular one, the result is
usually not sparse. However, when matrix.csr is multiplied by a regular
matrix in R, a matrix.csr is produced.
Is there a way to avoid this?
Thanks!
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.1110300
Most likely when 'y' is converted to a dataframe (not sure what the
function 'write.matrix.csr' does since you did not say where you got
it), '0' and '1' are converted to factors which probably show up as 1
and 2 in the file. Here is a quick example:
> x <- sample(c(0,1), 20,TRUE)
> y <- table(x)
Thanks Berend for your reply. However I was expecting something may be
available within R itself (or perhaps some added package.)
Thanks and regards,
On Tue, Aug 28, 2012 at 12:23 AM, Berend Hasselman wrote:
>
> On 27-08-2012, at 20:21, Christofer Bogaso wrote:
>
>> Dear all, I have got a pdf fi
> write.matrix.csr(mx, y = y, file = file)
> table(y)
0 1
5194394 23487
$ cut -d' ' -f1 f | sort | uniq -c
23487 2
5194394 1
i.e., 0 is written as 1 and 1 is written as 2.
why?
is there a way to disable this?
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X
On 27-08-2012, at 20:21, Christofer Bogaso wrote:
> Dear all, I have got a pdf file with lot of numerical data which I
> want to export to R for some analysis. Is there any way to doing that?
>
> Thanks for your time.
Possibly the program pdftotext from xpdf tools
(http://www.foolabs.com/xpdf
Thank you very much!
It worked nicely.
Best!
Giovanni
On Aug 27, 2012, at 4:18 PM, jim holtman wrote:
> try this:
>
>
> > input <- readLines(textConnection("Iteration JminError Elapsed
> > Corral Duality Gap Step1 Step2 Step3 Step4
> + 2 2 0.00e+001
Dear all, I have got a pdf file with lot of numerical data which I
want to export to R for some analysis. Is there any way to doing that?
Thanks for your time.
Thanks and regards,
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listin
res3 = Profile.LS(fhn, data=data2, times=times, pars=pars,
coefs=coefs, lambda=lambda, out.meth='nls',
control.in=control.in, control.out=control.out)
#Rather than parameter estimates, as with the single replicate
analysis, this produces
Thank you, David. I found some good stuffs for which I have been
looking.
On Sun, 2012-08-26 at 09:01 -0700, David Winsemius wrote:
> On Aug 25, 2012, at 9:00 PM, Chel Hee Lee wrote:
>
> > I will very appreciate if anyone can provide some materials to draw a
> > simplex plot of a Dirichlet dist
try this:
> x <- read.table(text = "2012-07-22 12:12:00, 21
+ 2012-07-22 12:15:00, 22
+ 2012-07-22 12:18:00, 24
+ 2012-07-22 12:39:00, 21
+ 2012-07-22 12:45:00, 25
+ 2012-07-22 12:49:00, 26
+ 2012-07-22 12:53:00, 20
+ 2012-07-22 13:00:00, 18
+ 2012-07-22 13:06:00, 22", colClasses = c("POSIXct", "
Well ...See below.
-- Cheers, Bert
On Mon, Aug 27, 2012 at 9:19 AM, David Winsemius wrote:
>
> On Aug 27, 2012, at 3:09 AM, Fridolin wrote:
>
>> What is a smart way to change an entry inside a column of a dataframe or
>> matrix which is of type "factor"?
>>
>> Here is my script incl. input data:
On Aug 27, 2012, at 7:11 AM, Daniel Caro wrote:
Hello,
This is a beginner question.
You should read the Posting Guide. (As well as following Bert's
suggestion to work through "Intro to R".
I am trying to loop through numbered
variables with "apply" to calculate weighted means. My data i
Dear R helpers,
I have a social network data including repated measures of ten alters (whom
you contact) and their attributes(gender, age, strength of tie).
I wrote variables related with alters just for wrote alter 1 and alter 2.
I'd like to change the like below. I'd change each name separetely
On Aug 27, 2012, at 3:09 AM, Fridolin wrote:
What is a smart way to change an entry inside a column of a
dataframe or
matrix which is of type "factor"?
Here is my script incl. input data:
#set working directory:
setwd("K:/R")
#read in data:
input<-read.table("Exampleinput.txt", sep="\t", h
Have you read "An Introduction to R". If not, please do so before
osting and pay particular attention to the section on indexing. If so,
re-read the sections on indexing.
For a terser exposition, ?"["
-- Bert
On Mon, Aug 27, 2012 at 7:11 AM, Daniel Caro wrote:
> Hello,
>
> This is a beginner qu
Hi folks,
I have a question about how to efficiently produce random numbers from Beta
and Binomial distributions.
For Beta distribution, suppose we have two shape vectors shape1 and shape2.
I hope to generate a 1 x 2 matrix X whose i th rwo is a sample from
reta(2,shape1[i]mshape2[i]). Of co
Can anybody, please, explain me how many parameter are estimated using
randomLCA?
For examples, model "dentistry.lca2random" estimate 1 scale (or
variance, b_j) parameter and 2 position parameters (a_cj)? Doesn't
it?
Do I need at least 4 diagnostic tests for such a model?
What happens if I
> importance(rfor.pdp11_t25.comb1,type=1)
%IncMSE
v1 -0.28956401263
v2 1.92865561147
v3 -0.63443929130
v4 1.58949137047
v5 0.03190940065
I wasn't entirely confident with interpreting these results based on the
documentation.
Could you please interpret?
[[alternative HTML vers
Hello,
If the op says he has tried strsplit, maybe a simple subsetting
afterwards would solve it.
a <- "12345_mydata"
strsplit(a, "_")[[1]][1]
Hope this helps,
Rui Barradas
Em 27-08-2012 15:22, jim holtman escreveu:
Is this what you want:
a<- "12345_mydata"
sub("_.*", "", a)
[1] "12345
Dear All,
wondering if I could get some help on the following using a box-percentile
plotting in Hmisc:
1. how could I put limits on the y axis? the following does NOT seem to work:
bpplot(b,c,d,e,f,g,h, ylim=c(0,2500))
2. I would like to color in each box plot from b to h the line represe
One way is to read in your text data as character columns with
read.table(stringsAsFactors=FALSE,...), then fiddle with the values,
and finally make factors out of the columns that you want to be factors,
specifying the levels explicitly. (Do you really want those number/number
things to be treat
Dear UseRs!
We are pleased to announce the release of our new package 'PVAClone'.
The 'PVAClone' package implements Population Viability Analysis (PVA)
methodology using data cloning. The data cloning algorithm by Lele et
al. (2007, 2010) is employed to compute maximum likelihood estimates
of the
Hello,
This is a beginner question. I am trying to loop through numbered
variables with "apply" to calculate weighted means. My data is "data",
the variables are "var1" to "var100", the weight is "weight". The
command works using
sapply(paste('data$var', 1:100, sep=''), function(x)
weighted.mean(
Hi:
I am trying to understand the link between ".csv" and ".rda" files.
Is there any easy to follow tutorial on this?
(I could do some of the operations below, but I got lost in the details.)
1. Reading .rda file ?
data <- load("profit.rda") # supposed to have four variable --y
VCU faculty and non-profit Georgia R School offer 11-14 substantial (4-10
classes each) online courses using R on: Fundamentals; Graphics; Data
Manipulation; Data Mining; Multivariate Statistics; PLS Path Modeling;
Programming; Simulation; and more. Take until May 15, 2013 to complete as
many cours
Hi
you could save yourself time to read help page for factor
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Fridolin
> Sent: Monday, August 27, 2012 12:09 PM
> To: r-help@r-project.org
> Subject: [R] Changing entries of colu
Maybe the unix time command comes in handy (provided you work with linux)?
http://en.wikipedia.org/wiki/Time_(Unix)
Best
Jan
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von William Dunlap
Gesendet: Montag, 27. August 2012 1
I found something that looks a bit odd to me:
The memory consumption of reference classes seems to increase dramatically
(roughly 10x in the example below) with serialization/deserialization
(saveRDS() / readRDS()).
Maybe creating instances of the same class has a synergy effect (references?)
w
On Aug 27, 2012, at 6:37 AM, Jean V Adams wrote:
Not sure why that is. You can always use manova() to get the
multivariate
summary statistics and then use individual models to get predictions
for
each response. For example,
mydata <- data.frame(y1=rnorm(50), y2=rnorm(50), x1=rnorm(50),
x
I didn't notice a discussion of what system or user time meant in
those two articles. (The first correctly says that system.time() is
a more convenient way to time things, as it calls proc.time() before
and after your expression and subtracts. It also does garbage
collection before the timing, so
Hi Dirk,
Thanks for the quick reply. Setting LD_LIBRARY_PATH does the trick, so
I can get by with that in my environment.
I would be happy to look into potential improvements. The ./configure
may work, but I installed littler through the debian package and thus
bypassed manual configuration. I'll
On 2012-08-27 22:25, David Winsemius wrote:
On Aug 27, 2012, at 1:53 AM, Jinsong Zhao wrote:
On 2012-08-27 9:35, David Winsemius wrote:
On Aug 26, 2012, at 5:06 PM, Jinsong Zhao wrote:
Hi there,
In my code, there is a for loop like the following:
pmatrix <- matrix(NA, nrow = 99, ncol =
On Aug 27, 2012, at 1:53 AM, Jinsong Zhao wrote:
On 2012-08-27 9:35, David Winsemius wrote:
On Aug 26, 2012, at 5:06 PM, Jinsong Zhao wrote:
Hi there,
In my code, there is a for loop like the following:
pmatrix <- matrix(NA, nrow = 99, ncol = 1)
qmatrix <- matrix(NA, nrow = 99, ncol
Is this what you want:
> a<- "12345_mydata"
> sub("_.*", "", a)
[1] "12345"
On Mon, Aug 27, 2012 at 5:29 AM, Rantony wrote:
> Hi,
>
> here im unable to run a string handle function called unpaste().
>
> for eg:- a<- "12345_mydata"
> Actually my requirement what is i need to get , only 12345.
try this:
> input <- readLines(textConnection("Iteration JminError
Elapsed Corral Duality Gap Step1 Step2 Step3 Step4
+ 2 2 0.00e+001.912976e-031 0.00e+00
1.779780e-037.214600e-051.243600e-052.246700e-05
+
../test/genmoons_data/l
Hi,
I guess this helps you.
library(reshape)
dat1<-read.table(text="
Iteration Jmin Error Elapsed Corral Duality_Gap Step1
Step2 Step3 Step4
10 32 3.133476e-03 6.075853e-03 8 4.057531e-01
1.613035e-03 3.956920e-03 3.077200e-05 4.390900e-05
20
Hi all R-users,
I’m trying to assign colors on those p-value in my table output that fall
above a certain critical value, let’s say a p-value >0.05.
My table looks like this:
Assets ADF-Level P-Value
ADF-First D P-Value ADF-Se
What is a smart way to change an entry inside a column of a dataframe or
matrix which is of type "factor"?
Here is my script incl. input data:
> #set working directory:
> setwd("K:/R")
>
> #read in data:
> input<-read.table("Exampleinput.txt", sep="\t", header=TRUE)
>
> #check data:
> input
I
Hi,
here im unable to run a string handle function called unpaste().
for eg:- a<- "12345_mydata"
Actually my requirement what is i need to get , only 12345. Means that , i
need the all letter as a word which is before of first " _ " - symbol of
"a". i tried to do with unpaste, it says functio
Hello, together
I'm trying to use user defined kernel. I know that kernlab offer user
defined kernel(custom kernel functions) in R.
I used data spam including package kernlab.
(number of variables=58 number of examples =4061)
i'm user defined kernel's form,
kp=function(d,e){
as=v*d
bs=v*e
cs=
Hi,
I'm pretty new to R and have run into a task which although I'm certain is
within R's capabilities, falls outside of mine. :-) Consider the following
data set:
2012-07-22 12:12:00, 21
2012-07-22 12:15:00, 22
2012-07-22 12:18:00, 24
2012-07-22 12:39:00, 21
2012-07-22 12:45:00, 25
2012-07-22 12
John,
I see this:
edd@max:~$ r -e'require(rJava)'
Loading required package: rJava
Loading required package: methods
Error : .onLoad failed in loadNamespace() for 'rJava', details:
call: dyn.load(file, DLLpath = DLLpath, ...)
error: unable to load shared object
'/usr/lib/R/site-library/rJava
Not sure why that is. You can always use manova() to get the multivariate
summary statistics and then use individual models to get predictions for
each response. For example,
mydata <- data.frame(y1=rnorm(50), y2=rnorm(50), x1=rnorm(50),
x2=rnorm(50), x3=rnorm(50))
myfit <- manova(cbind(y1,
Hello list,
I'm having some difficulty getting rJava to load in littler. Even
after a R CMD javareconf and a reinstall of littler, I get this:
jlaing@xenon:~$ r -e "require(rJava)"
Loading required package: rJava
Loading required package: methods
Error : .onLoad failed in loadNamespace() for 'rJav
If the problem is the same then you still have an incomplete version
of R. As you can see from the August 27th checks gsubfn works fine on
the latest version of R:
http://cran.r-project.org/web/checks/check_results_gsubfn.html
Try installing Rcmdr. I suspect it won't install either on your ver
Hello,
What would be the best set of R functions to parse and transform a file?
My file looks as shown below. I would like to plot this data and I need to
parse it into a single data frame that sorts of "transposes the data" with the
following structure:
> df <- data.frame(n=c(1,1,2,2),iter=c(
On 26.08.2012 15:33, Reza Salimi-Khorshidi wrote:
Thanks Uwe,
Am I right that in ksvm's internal cross-validation, there is no
guarantee for having *at least one* of each classes in each subset?
That is my guess, but I haven't read the code. Please read it yourself
in case you want more det
On Aug 24, 2012, at 21:51 , Daniel Nordlund wrote:
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
>> On Behalf Of Marc Schwartz
>> Sent: Friday, August 24, 2012 12:06 PM
>> To: ramoss
>> Cc: r-help@r-project.org
>> Subject: Re: [R] if the
On 2012-08-27 9:35, David Winsemius wrote:
On Aug 26, 2012, at 5:06 PM, Jinsong Zhao wrote:
Hi there,
In my code, there is a for loop like the following:
pmatrix <- matrix(NA, nrow = 99, ncol = 1)
qmatrix <- matrix(NA, nrow = 99, ncol = 3)
paf <- seq(0.01, 0.99, 0.01)
for (i in 1
Hello,
I would like to write a program that compute the principal components of
a set of data and then
1. Run the dependent variable against the principal components (lagged
value)
2. Do prediction
, following Stock and Watson (1999) "Forecasting Inflation". All data
are time series.
97 matches
Mail list logo
