Thank you Greg,
This is what I figured out... The problem with txtProgressBar() is that
many packages display some information during installation (even with
quiet=TRUE), especially the installation of dependencies, so that the
progress bar is not very useful. So I have tried with tkProgressBa
Hi All,
I have uploaded a new release of the R Toolbox.
R Toolbox is a collection of simple but useful functions which I developed
for myself to shorten the develoment process. Currently all functions use
base R. No other packages are needed. One exception is "t_openxlsx" cause
this module dea
Hello Everyone,
I am trying to replace the values in the 2nd column (Variable 1)
corresponding to certain dates (Date)
with NAs as shown below. Both Date and Variable1 are numeric vectors . I am
trying to use the transform function
as shown below but it doesn’t seem to work even though if I
I am having trouble installing R Commander on a Mac (OS X 10.7.5). I would
really appreciate some help!
I receive the following message after following the instructions to install
Rcmdr via Package Installer:
Warning: dependency 'rglwidget' is not available
trying URL
'http://cran.ms.unimel
Hi everybody,
I have a vector of character strings.
Each string has the same pattern and I want
to split them in pieces and get a vector made
of the first pieces of each string.
The problem is that strsplit returns a list.
All I found is
uu<- matrix(unlist(strsplit(x,";")),ncol=3,byrow=T)[,1]
Hi,
I'm not sure about the more generalized solution, but how about this for a
start.
x <- c("a;b;c", "d;e", "foo;g;h;i")
x
#[1] "a;b;c" "d;e" "foo;g;h;i"
sapply(strsplit(x, ";",fixed = TRUE), '[',1)
#[1] "a" "d" "foo"
If you want elegance then I suggest you take a look at the s
This might not be the whole story, but part of the problem is that you
want to select a _*character string*_ greater/smaller than another. That
doesn't make much sense!
I am not sure how to best compare two dates, but if you convert the Date
values into numeric, then that would work. The probl
Dear Eric
I think you are looking for sub or gsub
Without an example set of input and output I am not quite sure but you
would need to define an expression which matches your separator (;)
followed by any characters up to the end of line. If you have trouble
with that then someone here will n
Hello,
Another thing to consider is to use Variable1 = NA, not '=='.
With '==' it will probably return TRUE/FALSE/NA.
Hope this helps,
Rui Barradas
Citando Ivan Calandra :
> This might not be the whole story, but part of the problem is that
> you want to select a _*character string*_ great
Dear Judy,
I see that the rglwidget package *is* available on CRAN but also that you're
using a very old version of Mac OS X. You may have to upgrade OS X and install
the current version of R (which is not available for OS X 10.7) in order to use
the Rcmdr package.
You may also be able to get
On 20/07/2016 11:47 PM, Judy Munday wrote:
I am having trouble installing R Commander on a Mac (OS X 10.7.5). I would
really appreciate some help!
I receive the following message after following the instructions to install
Rcmdr via Package Installer:
Warning: dependency 'rglwidget' is not
Oops, missed that one!
--
Ivan Calandra, PhD
Scientific Mediator
University of Reims Champagne-Ardenne
GEGENAA - EA 3795
CREA - 2 esplanade Roland Garros
51100 Reims, France
+33(0)3 26 77 36 89
ivan.calan...@univ-reims.fr
--
https://www.researchgate.net/profile/Ivan_Calandra
https://publons.com/a
You may like to look at
?suppressMessages
P.S. sorry for posting in HTML, new laptop and it’s next on my list of things
to fix.
From: Ivan Calandra
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE a
Not sure I can translate the format of your Date column correctly, however the
command
DF1$Date <- as.Date(DF1$Date, format=’formatstr’)
Will convert the dates into a format correctly handled by R.
?strptime
Should give you an idea of what formatstr should look like.
I.e. if
date = 160721
as.
Dear All
I have created a permanent corpus and I can see the file exist:
setwd("E:/textmining/texts")
(data_mined_permanent <- PCorpus(DirSource(("E:/textmining/texts")),
readerControl = list(languages = "eng"),
dbControl =
list(
On 7/20/2016 8:26 PM, lily li wrote:
Yes, I tried to create a dataframe and merge it with the shortened
dataframe. The resulting dataframe goes with the short one and truncates
the complete date column, so it does not work.
On Wed, Jul 20, 2016 at 6:38 PM, David Winsemius
wrote:
On Jul 20,
> Ralf Goertz
> on Wed, 20 Jul 2016 16:37:53 +0200 writes:
> Am Wed, 20 Jul 2016 11:35:31 +0200
> schrieb Ralf Goertz :
>> Hi,
>>
>> after a recent update to version 3.3.1 on Opensuse Leap I have
>> problems with command lines longer than the terminal width.
Hi,
A not unusual task is performing a multiple regression in a rolling window
on a time-series.A standard piece of advice for doing in R is something
like the code that follows at the end of the email. I am currently using
an "embed" variant of that code and that piece of advice is out there
Just replacing lm with a faster version would speed it up. Try lm.fit
or even faster is fastLm in the RcppArmadillo package.
On Thu, Jul 21, 2016 at 2:02 PM, jeremiah rounds
wrote:
> Hi,
>
> A not unusual task is performing a multiple regression in a rolling window
> on a time-series.A stand
Jeremiah,
for this purpose there are the "roll" and "RcppRoll" packages. Both use
Rcpp and the former also provides rolling lm models. The latter has a
generic interface that let's you define your own function.
One thing to pay attention to, though, is the numerical reliability.
Especially o
Thanks all. roll::roll_lm was essentially what I wanted. I think maybe
I would prefer it to have options to return a few more things, but it is
the coefficients, and the remaining statistics you might want can be
calculated fast enough from there.
On Thu, Jul 21, 2016 at 12:36 PM, Achim Zeile
Hi,
I am trying to use system() to run some command in OS. such as
system("cmd 'a\tb')
however, it alway runs
cmd 'ab'
instead of
cmd 'a\tb'
How can I prevent system to interpret 'a\tb' to 'ab'?
Thanks
[[alternative HTML version deleted]]
___
You could escape the backslash.
system("cmd 'a\\tb'")
On Thu, Jul 21, 2016 at 4:00 PM, Michael Peng
wrote:
> Hi,
>
> I am trying to use system() to run some command in OS. such as
>
> system("cmd 'a\tb')
>
> however, it alway runs
> cmd 'ab'
> instead of
> cmd 'a\tb'
>
> How can I prevent
Yes. I forgot to escape the escape. Thank you so much.
2016-07-21 16:07 GMT-04:00 Sarah Goslee :
> You could escape the backslash.
>
> system("cmd 'a\\tb'")
>
>
> On Thu, Jul 21, 2016 at 4:00 PM, Michael Peng
> wrote:
> > Hi,
> >
> > I am trying to use system() to run some command in OS. such a
I would be careful about making assumptions regarding what is faster.
Performance tends to be nonintuitive.
When I ran rollapply/lm, rollapply/fastLm and roll_lm on the example
you provided rollapply/fastLm was three times faster than roll_lm. Of
course this could change with data of different di
I appreciate the timing, so much so I changed the code to show the issue.
It is a problem of scale.
roll_lm probably has a heavy start-up cost but otherwise completely
out-performs those other versions at scale. I suspect you are timing the
nearly constant time start-up cost in small data. I
Dear all,
I have the following GDP data by latitude and longitude at 0.5 degree by
0.5 degree.
temp <- dput(head(ptsDF,10))
structure(list(longitude = c(-68.25, -67.75, -67.25, -68.25,
-67.75, -67.25, -71.25, -70.75, -69.25, -68.75), latitude = c(-54.75,
-54.75, -54.75, -54.25, -54.25, -54.25, -5
Hi Jermiah: another possibly faster way would be to use a kalman filtering
framework. I forget the details but duncan and horne have a paper which
shows how a regression can be re-computed each time a new data point is
added .I
forget if they handle taking one off of the back also which is what you
Dear all,
I have two logistic regression models:
• model <- glm(Y ~ X1+X2+X3+X4, data = data, family = "binomial")
• modelInteraction <- glm(Y ~ X1+X2+X3+X4+X1*X4, data = data, family =
"binomial")
To calculate the marginal effects (MEM approach) for these models, I used the
`mfx` p
I have no idea which method produces the fastest results, but the package KFAS
has a function to do recursive regressions using the Kalman filter. One
difference is that it is not, as far as a I can telll, a moving window (so past
data are being dropped), just a recursively computed regressio
I use this code, and it works. So has to set 'all=TRUE'.
merge(df, data.frame(time=seq(as.Date("1990-01-01"),
to=as.Date("1990-12-31"), by="days")), all=TRUE)
On Thu, Jul 21, 2016 at 9:22 AM, Daniel Nordlund
wrote:
> On 7/20/2016 8:26 PM, lily li wrote:
>
>> Yes, I tried to create a dataframe
I agree that when appropriate Kalman Filter/Smoothing the higher-quality
way to go about estimating a time-varying coefficient (given that is what
they do), and I have noted that both the R package "dlm" and the function
"StructTS" handle these problems quickly. I am working on that in
parallel.
I have a question about interpolating missing values in a dataframe. The
dataframe is in the following, Column C has no data before 2009-01-05 and
after 2009-12-31, how to interpolate data for the blanks? That is to say,
interpolate linearly between these two gaps using 5.4 and 6.1? Thanks.
df
ti
Hi Florian,
As I suspected, you will have to use something other than the "arrows"
function, which most people use to draw "error bars". This is where
the "lty" argument gets used for all of the lines. You could
substitute a function like this:
foobars<-function(x0,y0,x1,y1,length=0.02,col=par("fg
Depending on your situation you may want all=TRUE, all.x=TRUE, or
all.y=TRUE.
I think the SQL people call these outer joins, left outer joins, and right
outer joins.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Jul 21, 2016 at 2:35 PM, lily li wrote:
> I use this code, and it works. So
Try approx(), as in:
df <-
data.frame(A=c(10,11,12),B=c(5,5,4),C=c(3.3,4,3),time=as.Date(c("1990-01-01","1990-02-07","1990-02-14")))
with(df, approx(x=time, y=C, xout=seq(min(time), max(time), by="days")))
Do you notice how one can copy and paste that example out of the
mail an into R to see how
Thanks, I meant if there are missing data at the beginning and end of a
dataframe, how to interpolate according to available data?
For example, the A column has missing values at the beginning and end, how
to interpolate linearly between 10 and 12 for the missing values?
df <- data.frame(A=c(NA,
> On 22 Jul 2016, at 01:34, lily li wrote:
>
> I have a question about interpolating missing values in a dataframe.
First of all, filling missing values action must be taken into account very
carefully. It must be known the nature of the data that wanted to be filled and
most of the time, to
> On 22 Jul 2016, at 01:54, lily li wrote:
>
> Thanks, I meant if there are missing data at the beginning and end of a
> dataframe, how to interpolate according to available data?
>
> For example, the A column has missing values at the beginning and end, how
> to interpolate linearly between 10
Hi R users,
I’m having some issues trying to extract texts from PDF file using tm
package.
Here are the steps that were carried out:
1. Downloaded and installed the following programs:
- Xpdf (Copied the ‘bin32’, ‘bin64’, ‘doc’ folders into ‘C:\Program
Files\Xpdf’ directory; also added C:\Progr
This may be useful:
Sven Hammarling and Craig Lucas
"Updating the QR factorization and the least squares problem"
http://eprints.ma.man.ac.uk/1192/01/covered/MIMS_ep2008_111.pdf
http://www.maths.manchester.ac.uk/~clucas/updating/
2016-07-21 20:02 GMT+02:00 jeremiah rounds :
> Hi,
>
> A not unusua
Hi R users,
I'm trying to download lubricate from this website, and then install it on
my mac.
https://github.com/hadley/lubridate
but it says windows version does not apply to mac. How to install the
package for mac? Thanks.
[[alternative HTML version deleted]]
Using the same data, if I ran
fit2
<-glm(formula=AR~Age+LumA+LumB+HER2+Basal+Normal,family=binomial,data=RacComp1)summary(fit2)exp(coef(fit2))
I obtained:
> exp(coef(fit2))(Intercept) Age LumA LumB HER2
> Basal Normal 0.24866935 1.00433781 0.10639937
Hiya,
I am trying to assign minimum values to a dataframe based on existing columns.
I can do this via loops, but surely R has much more elegant solutions...
Here is my code:
set.seed(666)
xyz<-as.data.frame(cbind(x=rep(rpois(50,10),2)+1,
y=rep(rpois(50,10),2)+1,z=runif(100, min=-3, max=40)))
Hi Jeremiah: I think I wasn't that clear. I'm not suggesting the kalman
filter to deal with time varying coefficients. As Roy pointed out, one can
use the kalman filter to do regular regression where one "sees" a new data
point as each time unit passes. It can be assumed that the coefficients do
n
You don't have to download and install from github. You can install
lubridate package easly from cran repository. If you really intend to
install from github, i advise you install devtools package first and use
install_github function.
http://www.inside-r.org/packages/cran/devtools/docs/install_gi
> On Jul 21, 2016, at 2:22 PM, Faradj Koliev wrote:
>
> Dear all,
>
> I have two logistic regression models:
>
>
> • model <- glm(Y ~ X1+X2+X3+X4, data = data, family = "binomial")
>
>
>
> • modelInteraction <- glm(Y ~ X1+X2+X3+X4+X1*X4, data = data, family =
> "binomial")
>
> To ca
Dear David Winsemius,
Thank you!
The sample make no sense, I know. The real data is too big. So, I only want to
understand how to plot marginal effects, to visualize them in a proper way.
Best,
> 22 juli 2016 kl. 08:35 skrev David Winsemius :
>
>>
>> On Jul 21, 2016, at 2:22 PM, Faradj
> jeremiah rounds
> on Thu, 21 Jul 2016 13:56:17 -0700 writes:
> I appreciate the timing, so much so I changed the code to show the issue.
> It is a problem of scale.
> roll_lm probably has a heavy start-up cost but otherwise completely
> out-performs those other vers
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