I am trying to model a state space process using the MARSS package. My
model has two unobservable states and 5 observable time series
along with external covariates in the observation process only. None of the
coefficients in either of the two processes are time varying.
After running my setup and
The answer to your question is yes. You can convert a column of values to Date
using the as.Date function with the appropriate format, and then test if any
values are NA using the is.na function, and find them with the which function.
If you want something less vague then you should read the Pos
I'm very new to R. I have a data file that I have read in via read.csv. I
expect one of the "columns" to be of type date for example. However at
least one value in that column is not of date type. I know this because
another program I am trying to process the file with is erroring, yet it
doesn't
Look at the "missing" function.
Or set the default value of the arguments to NA.
On Thu, Aug 29, 2013 at 3:23 PM, newruser12345 wrote:
> Hi All,
>
> I'm very green user and have little programming background, but appreciate
> any and all help/direction. I have a spreadsheet that successfully s
Hi,
I am getting the same error with R 3.0.1
SpectrumSearch(y, sigma=3.0, threshold=1.0, background=TRUE, iterations=13,
markov=FALSE, window=3)
#Error in .Call("R_SpectrumSearchHighRes", as.vector(y), as.numeric(sigma), :
# "R_SpectrumSearchHighRes" not available for .Call() for package "Pea
Hi All,
I'm very green user and have little programming background, but appreciate
any and all help/direction. I have a spreadsheet that successfully sends
values from Excel cells to R as variables for a function, which then runs
and generates a plot. I cannot figure out how to make R recognize
Hi,
I apologize for not following the posting rules…
Here is the text from my previous post:
"I started evaluating the 'Peaks' package a couple of months ago and found it
to be quite
useful. Getting back to it last week I had to set up my R environment due to
hardware
changes again. The Peaks
Dear All,
I am trying to use the felm function in the lfe package. However it does
not seem to deal with missing values the way the lm function does. I wish
to tell it na.omit or na.action = na.omit but it does not recognize this. I
need to allow for missing values as I have different specificatio
I upgraded R from 2.12.1 to 3.0.0 (on windows XP(, and as soon as I saved the
3.0.0 workspace, was unable to access .Rdata from 2.12.1. The message in the R
console is "Error in loadNamesSpace(name): there is no package called parallel"
and a popup window that says "Fatal error: unable to resto
Thank you for your answer. But further calculations will be much more
difficult, like
(1-b)^2 * Var(V1) for all matching columns
where b is the slope from a regression V1 (from datset 1) on V1 (dataset 2)
and Var(V1) the variance from V1(from dataset2).
So what I'm looking for is somethi
Hi thereI've got two datasets of the following form (just an example, the
real dataset got a lot more columns)dataset1V1 V2 V32 6 84
3 41 9 8and
dataset 2V1 V2 V36 8 42 0 78 1 3First,
I'd like to calculate the
followin
Or you can use with:
> a <- new.env()
> with(a, b <- function(x) x )
> a$b
function(x) x
On Wed, Aug 28, 2013 at 3:45 PM, ivo welch wrote:
> duh!
>
>
>
> Ivo Welch (ivo.we...@gmail.com)
> http://www.ivo-welch.info/
> J. Fred Weston Professor of Finance
> Anderson School at UCLA, C519
>
On 08/30/2013 01:28 AM, Shane Carey wrote:
Hello all,
I have decided to go ahead with gap.boxplot. I am trying to suppress the
axis labels, both x and y labels. I tried using axis.labels=NULL but it
would not work.
Hi Shane,
To suppress the axis labels, pass an empty string:
gap.barplot(...,x
Hi,
On Thu, Aug 29, 2013 at 3:03 PM, Rolf Turner wrote:
> On 29/08/13 12:10, Ista Zahn wrote:
>>
>> On Wed, Aug 28, 2013 at 7:44 PM, Steve Lianoglou
>> wrote:
>>>
>>> Hi,
>>>
>>> On Wed, Aug 28, 2013 at 3:58 PM, Ista Zahn wrote:
Or go all the way and put
options(stringsAsFac
Hi,
You could try:
res<-colMeans(aperm(moo,c(2,1,3)))
resOld<-apply(moo,c(1,3),mean)
identical(res,resOld)
#[1] TRUE
#Speed:
set.seed(285)
moo1<- array(runif(1400*9*15),dim=c(1400,9,15))
system.time({res1<- colMeans(aperm(moo1,c(2,1,3)))})
#user system elapsed
# 0.004 0.000 0.002
syste
On 29/08/13 12:10, Ista Zahn wrote:
On Wed, Aug 28, 2013 at 7:44 PM, Steve Lianoglou
wrote:
Hi,
On Wed, Aug 28, 2013 at 3:58 PM, Ista Zahn wrote:
Or go all the way and put
options(stringsAsFactors = FALSE)
at the top your script or in your .Rprofile. This will prevent this
kind of annoyanc
For matrices, colMeans/rowMeans are quick, vectorized functions. But
say I have a higher dimensional array:
moo <- array(runif(400*9*3),dim=c(400,9,3))
And I want to get the mean along the 2nd dimension. I can, of course,
use apply:
moo1 <- apply(moo,c(1,3),mean)
But this is not a vectorized
Thanks, I'll try this as well.
Srecko
On Thu, Aug 29, 2013 at 3:26 PM, arun wrote:
>
>
> Hi Srecko,
> Try this:
> dat1<- read.table(text="
> id module event time time_on_task Categurl
> 1sys login 1373502892 80 B http://
> 2 taskadd 1373502892
Hi Srecko,
Try this:
dat1<- read.table(text="
id module event time time_on_task Categ url
1 sys login 1373502892 80 B http://
2 task add 1373502892 80 A
http://post/add?id=33&idp=67
3 task add 1373502972 23 A
Hi Arun,
this could to the work...
Thanks so much!
On Thu, Aug 29, 2013 at 3:10 PM, arun wrote:
> HI,
> It's not really clear, but you can try this:
> dat1<- read.table(text="
> id module event time time_on_task Categurl
> 1sys login 1373502892 80 B
> http://po
HI,
It's not really clear, but you can try this:
dat1<- read.table(text="
id module event time time_on_task Categ url
1 sys login 1373502892 80 B http://post/add?id=42&idp=45
2 task add 1373502892 80 A http://post/add?id=33&idp=45
3 task
Dear Gerard,
Without your data, it's not possible to reproduce your problem exactly, but
it's clear that it isn't specific to the scatterplot() function in the car
package. For example, try
plot(1:10)
title(main=bquote(paste("Hypothesis 9.4.1\nBaseline XYZ with Disease
Activity (DAS28)\nat Month
In addition to the other suggestions, try typing
help('&')
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 8/29/13 1:16 AM, "Mª Teresa Martinez Soriano"
wrote:
>Hi to everyone and sorry for my question, I would like
Hi All,
I'm using R 3.0.0. I'm trying to add the sample size of the paired data
(calculated by a function n(), which returns a value of 70, correctly).
My main title works fine except that the '70' appears far to the right on the
line as in:
at Month 18 (N=
Hi,
Try:
res<-sapply(seq_len(ncol(dat1)),function(i)
setNames(((1-coef(lm(dat1[,i]~dat2[,i]))[2])^2)*var(dat2[,i]),NULL))
res
#[1] 21.0 16.11842 18.69231
A.K.
Thank you for your answer. But further calculations will be much more
difficult, like
(1-b)^2 * Var(V1) for all matching
Hi,
Try:
dat1<- read.table(text="
V1 V2 V3
2 6 8
4 3 4
1 9 8
",sep="",header=TRUE)
dat2<- read.table(text="
V1 V2 V3
6 8 4
2 0 7
8 1 3
",sep="",header=TRUE)
res1<- as.matrix(dat1-dat2)
res1
# V1 V2 V3
#[1,] -4 -2 4
#[2,] 2 3 -3
#[3,] -7 8 5
res2<-t(t(dat1)-colMeans(dat2))
res2
#
Hi,
You could try this:
dat1<- read.table(text="
id module event time time_on_task
1 sys login 1373502892 80
2 task add 1373502892 80
3 task add 1373502972 23
4 sys login
Thanks Berend,
I don't know why I didn't try that before posting the question... but...
anyways, thanks for your help
Srecko
On Thu, Aug 29, 2013 at 11:34 AM, Berend Hasselman wrote:
>
> On 29-08-2013, at 20:15, srecko joksimovic
> wrote:
>
> > Thanks Arun,
> >
> > this is great. However, it
Hi Arun,
There is one more question... you explained me how to
use split(dat1,cumsum(dat1$action=="login")) in one of previous questions,
and that is great.
Now, if I have something like this:
id moduleevent time time_on_task
1 sys login 13735028
On 29-08-2013, at 20:15, srecko joksimovic wrote:
> Thanks Arun,
>
> this is great. However, it should be just a little bit different:
>
> # id event time time_on_task
> #1 1add 1373502892 80
> #2 2add 1373502972 23
> #3 3 delete 1373502995 901
Thanks Arun,
this is great. However, it should be just a little bit different:
# id event time time_on_task
#1 1add 1373502892 80
#2 2add 1373502972 23
#3 3 delete 1373502995 901
#4 4 view 1373503896 100
#5 5add 1373503996
Hi,
Try:
dat1<- read.table(text="
id event time
1 add 1373502892
2 add 1373502972
3 delete 1373502995
4 view 1373503896
5 add 1373503996
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat1$time_on_task<- c(NA,diff(dat1$time))
dat1
# id event time
You got the point. Thank you for pointing out the problem.
Thanks again.
David
2013/8/30 Duncan Murdoch
> On 29/08/2013 1:37 PM, Marino David wrote:
>
>> Hi all R users:
>>
>>
>>
>> I am a little bit confused about the following results. See as follows:
>>
>>
>>
>> library(mvtnorm)
>>
>>
>>
>>
On 29/08/2013 1:37 PM, Marino David wrote:
Hi all R users:
I am a little bit confused about the following results. See as follows:
library(mvtnorm)
xMean<-c(24.12,66.92,77.65,131.97,158.8)
xVar<-c(0.01,0.06,0.32,0.18,0.95)
xFloor<-floor(xMean)
# use “mvtnorm” package
p1<-dmvnorm(xF
Hi,
I have a following data set:
ideventtime (in sec)
1 add 1373502892
2 add 1373502972
3 delete 1373502995
4 view 1373503896
5 add 1373503996
...
I'd like to add new column "time on task" which is time elapsed between two
events (id2 - id1...).
Hi all R users:
I am a little bit confused about the following results. See as follows:
library(mvtnorm)
xMean<-c(24.12,66.92,77.65,131.97,158.8)
xVar<-c(0.01,0.06,0.32,0.18,0.95)
xFloor<-floor(xMean)
# use mvtnorm package
p1<-dmvnorm(xFloor,mean=xMean,sigma=diag(xVar))
p2<-dmvnor
Hello,
You should post your questions to r-help@r-project.org, the odds of
getting more and better answers are greater.
As for the question, try the following. Note that the functions now have
an extra argument.
incub <- function(x, n = 2){
x$Incubation <- 0
x$Incubation[1] <- x$Sympt
"I would also like to display a y-axis value in the upper box"
I got this part working now.
On Thu, Aug 29, 2013 at 4:28 PM, Shane Carey wrote:
> Hello all,
>
> I have decided to go ahead with gap.boxplot. I am trying to suppress the
> axis labels, both x and y labels. I tried using axis.label
Hello all,
I have decided to go ahead with gap.boxplot. I am trying to suppress the
axis labels, both x and y labels. I tried using axis.labels=NULL but it
would not work.
gap.boxplot(DATA$Conductivity~factor(DATA$UnitName_1),ylim=c(LOWER_Y_Conductivity,UPPER_Y_Conductivity_int),gap=gap_Conductiv
On Thu, Aug 29, 2013 at 5:40 AM, jim holtman wrote:
> Here is how I would do it since are reading in the entire file. This
> breaks on each "Flow Budget" section, extracts the RECHARGE values and
> puts them in a list with the name of the Flow Budget:
>
I learned more R in studying your solutio
Please use dput() to supply data. It's a lot easier for readers to just copy
and paste into R.
I have no idea of what variables are associated with the columns below.
John Kane
Kingston ON Canada
> -Original Message-
> From: mohan.radhakrish...@polarisft.com
> Sent: Thu, 29 Aug 2013 09
As said by arun, the code is not clear.
Ma Teresa, what is it that you actually want to do?
Regards,
José
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of arun
Sent: 29 August 2013 15:12
To: R help
Subject: Re: [R] Help R
HI,
You
Looks like school is starting up again.
We don't usually help with homework especially at this level. Read a text book
John Kane
Kingston ON Canada
> -Original Message-
> From: bal.chan...@gmail.com
> Sent: Thu, 29 Aug 2013 15:57:29 +0530
> To: r-help@r-project.org
> Subject: [R] Few do
HI,
Your code is not clear:
for( i in 1: nrow(D))
{
for( j in 5:ncol(D))
{
D[((D[i,j]/D[i,2])>1.5)]15999)]<-100
## "1.5)]15999)]"
}
^^^
}
D<- structure(list(X. = c(1108L, 1591L,
Do you have NA/NAN in your data set? If yes our check with an IF or
substitute them with a value that fits your need.
I hope I understood correctly your problem.
-
- László-András Zsurzsa,
Hi to everyone,
I would like to replace some values in a data.frame (D)
> str(D)
'data.frame': 116 obs. of 10 variables:
$ X. : int 1108 1591 3408 3872 5823 8099 10640 12600 14680 14698 ...
$ media : num 22 86.6 807 103.2 73 ...
$ IE.2003: num 32 92 166 237 161 ...
$ IE.2004: num
I understand you response but it does not solve the problem. I'am aware
that one can simply color every cell in an excel file by using his own
algorithm.
The question was if I can write my data to a *single* cells and use
different formatting for every piece of data.
-
Hi All,
I apologize for the opaqueness and I will try to make it clearer.
I am comparing two diagnostic tests G (gold standard) and N (new). Both
are real tests, real experiments. G is currently the gold standard because
it is the best test available, not because it is a perfect test. G is a
g
Am 29.08.2013 15:03 (UTC+1) schrieb Zsurzsa Laszlo:
> First of all thank you for the quick resposen.
>
> I know I can color and set up every cell. I will take a look again *
> CellStyle* but is it possbile for example to write an array to a single
> cell that has different colors for some data. Ba
HI,
May be this helps:
ts1<- ts(1:20)
ts2<- ts(1:25)
ts1[-(1:3)]<- ts1[-(1:3)]+ts2[1:17]
as.numeric(ts1)
# [1] 1 2 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37
A.K.
Hey everyone,
I`m an absolut beginner in R and need some help for an exercise:
I want to do ordinary calculati
Hi
can you please give the brief explanation about anova?
what is the purpose of null hypothesis in anova?
how can we find future predictive value from existing data?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
Hi
can you please give the brief explanation about anova?
what is the purpose of null hypothesis in anova?
how can we find future predictive value from existing data?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
First of all thank you for the quick resposen.
I know I can color and set up every cell. I will take a look again *
CellStyle* but is it possbile for example to write an array to a single
cell that has different colors for some data. Basically the color depends
on the data.
--
Am 29.08.2013 12:08 (UTC+1) schrieb Zsurzsa Laszlo:
> Dear R users,
>
> I have a question about the xlsx package. It's possible to create excel
> files and color cells and etc.
yes, with package xlsx you can colourize you data sheets, even the
fonts. See for example ?CellStyle .
A good demonstra
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
-- Forwarded message --
From: jim holtman
Date: Thu, Aug 29, 2013 at 8:43 AM
Subject: Re: [R] Narrowing values collected from .txt file
To: "Mo
Here is how I would do it since are reading in the entire file. This
breaks on each "Flow Budget" section, extracts the RECHARGE values and
puts them in a list with the name of the Flow Budget:
> # read entire file
> input <- readLines("C:\\Users\\jh52822\\Downloads\\MCR_Budgets.txt")
> # determ
On 13-08-29 8:23 AM, David Epstein wrote:
I have two data frames, "train" and "response". Here is my attempt to do a
linear regression. All entries of both data frames are numeric. I am
expecting the intercept value to lie between 2 and 3 (in particular,
non-zero).
lm expects the variables in t
At 15:18 28/08/2013, Donald Catanzaro wrote:
Good Day All,
I am working with a diagnostic test and comparing the new test to an old
test. Normally I would be able to calculate sensitivity and specificity
quite easily.
However, the 'gold standard' that I am comparing my new diagnostic with is
r
I have two data frames, "train" and "response". Here is my attempt to do a
linear regression. All entries of both data frames are numeric. I am
expecting the intercept value to lie between 2 and 3 (in particular,
non-zero).
Here is a record of my interaction with R:
> class(response)
[1] "data.fr
On 08/29/2013 02:19 PM, mohan.radhakrish...@polarisft.com wrote:
Hi,
...
The plots are all there but the x=axis labels are not there. The graph
labels are only '12:30', '13:30' and '14:30'
I think I need to use your code to get all the values.
Hi Mohan,
Try this:
plot(strptime(data$Time,"%H:
Dear R users,
I have a question about the xlsx package. It's possible to create excel
files and color cells and etc.
My question would be that is it possible to color only some part of the
data hold in a cell. Let's assume I've got the following data :
167,153,120,100 and I want to color to red
Hey
if (( (1==1) && (2==2) ) || (3==3)) { print( "hello world") }
-
- László-András Zsurzsa,-
- Msc. Infromatics, Technical University Munich, Germany -
- Scientif
Ok, thanks all :-)
On Thu, Aug 29, 2013 at 2:39 AM, Jim Lemon wrote:
> On 08/29/2013 02:52 AM, Shane Carey wrote:
>
>> Hi,
>>
>> Has anyone ever created scale breaks in R something like what is shown
>> here
>> in the section,
>> Use a Scale Break
>>
>> http://www.r-bloggers.com/**graphing-high
Hello,
and is && ; or is || ; and print() needs the parenthesis around its argument
if((condition1 && condition2) || (condition3 && condition4)) {print(uhvef)}
Hope this helps,
Rui Barradas
Em 29-08-2013 09:16, Mª Teresa Martinez Soriano escreveu:
Hi to everyone and sorry for my question,
Hi to everyone and sorry for my question, I would like to use IF in an example
like this:
If((condition1 and condition2) Or (condition 3 and condition4)) {print uhvef}
BUt I don´t know how to write it correctly,
Thanks in advance
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