I have a data set with daily claim numbers. I group with monthly totals and
create a time series as follows
v.ts<- ts(v, start=2008,frequency=12)
How can the daily data be converted into a time series where there is no
claim number for sundays and public holidays.
Charles.
[[alternati
On Tue, 20 May 2014 02:39:17 PM Shane Carey wrote:
> Hi,
>
> I amtrying to plot a cfd with a histogram on one plot. The problem is
the
> scale (y-axis) of the plots are hugely different and as a result the
> histogram plot is hard to read. Are there any examples of plots like
this
> done in R:
>
You did not show us how "x" was created, so we cannot tell whether this
indicates a bug or not. My bet would be that "x' was created in some way that
bypassed the standard POSIXct creation code.
Note that your timezone specification is not atypical for output shorthand but
may not yield consist
On 21/05/2014 08:33, Jeff Newmiller wrote:
You did not show us how "x" was created, so we cannot tell whether this indicates a
bug or not. My bet would be that "x' was created in some way that bypassed the standard
POSIXct creation code.
Note that your timezone specification is not atypical fo
Hi all,
I ended up using twoord.plot from library(plotrix)
It worked great. Thanks for all the help.
Cheers
On Wed, May 21, 2014 at 8:15 AM, Jim Lemon wrote:
> On Tue, 20 May 2014 02:39:17 PM Shane Carey wrote:
> > Hi,
> >
> > I amtrying to plot a cfd with a histogram on one plot. The problem
Hello together,
i want to create a line graph, from a data.frame like this one:
Categorie Y10_11 Y11_12Y12_13 Y13_14
AService 1 2 3 4
BSupport 2 55 0
CTe
On Wed, 21 May 2014 01:20:52 AM Mat wrote:
> Hello together,
>
> i want to create a line graph, from a data.frame like this one:
> Categorie Y10_11 Y11_12Y12_13 Y13_14
> AService 1 2 3 4
> BSupport
Hey R Users!
I´m trying to run a waldtest comparing an restricted and unrestricted model.
But sample size is different (because of NA´s), so I get an error message:
> caus2<-plm(CSmean~ lag(CSmean,1)+ lag(numfull_FCRlong,1)+
+GDP+linpol_GDPpCapita+fort_budget+fort_percentDeb
On 07 May 2014, at 17:55 , Marc Schwartz wrote:
>> as.data.frame(sapply(mtcars[, c("mpg", "disp")], summary))
I'd use lapply() rather than sapply() here. That goes directly from a list to a
data frame, avoiding the detour via a matrix.
--
Peter Dalgaard, Professor
Center for Statistics, Cop
Dear R forum
I have a vector as
dat = c("ABC 1", "ABC 2", "ABC 3", "DEF 10", "DEF 20")
> dat
[1] "ABC 1" "ABC 2" "ABC 3" "DEF 10" "DEF 20"
I need to split the names into two parts say
p1 p2
ABC 1
ABC 2
ABC 3
DEF 10
DEF 20
Kindly guide
Katherine
[[al
Hi Katherin,
the following code does the trick:
x <- strsplit(dat," ")
df <- data.frame(p1=sapply(x, function(x) x[1]), p2 = sapply(x,
function(x) x[2]))
Hope it helps,
Luca
2014-05-21 11:29 GMT+02:00 Katherine Gobin :
> Dear R forum
>
> I have a vector as
>
> dat = c("ABC 1", "ABC 2", "ABC 3",
Hi Katherine,
#or you can use:
read.table(text=dat,sep="",header=FALSE,stringsAsFactors=FALSE)
# V1 V2
#1 ABC 1
#2 ABC 2
#3 ABC 3
#4 DEF 10
#5 DEF 20
A.K.
On Wednesday, May 21, 2014 5:38 AM, Luca Cerone wrote:
Hi Katherin,
the following code does the trick:
x <- strsplit(dat," ")
df <-
Thanks Arun,
I agree using read.table is probably easier :)
2014-05-21 12:08 GMT+02:00 arun :
>
>
> Hi Katherine,
>
> #or you can use:
> read.table(text=dat,sep="",header=FALSE,stringsAsFactors=FALSE)
> # V1 V2
> #1 ABC 1
> #2 ABC 2
> #3 ABC 3
> #4 DEF 10
> #5 DEF 20
> A.K.
>
>
> On Wednesday
Next I want to move the x axis labels up towards the axis to
make the gap smaller. The mgp= was suggested to me.
The staxlab function to be standard soon was sent to me by
the author to make staxlab work on the second x axis.
Runnable code to help demonstrate what I want to do:
Hurr
#trying for
Hi,
You may check ?isWeekday, ?isBizday, from library(timeDate)
library(timeDate)
tS <- timeSequence(as.Date("2008/1/1"), as.Date("2014/4/21"))
tWd <- tS[isWeekday(tS)]
length(tWd)
#[1] 1645
pubHolUS <- holiday(2008:2014,listHolidays("US"))
tS1 <- tS[isBizday(tS, holidays = holidayNYSE(2008:2
I believe you are right. A night of sleep has done wonders for my
understanding of the problem.
Thank you for your patience and help!
Raphael
On 5/21/14, Rolf Turner wrote:
>
>
> It sounds to me as though you are simply getting yourself flummoxed by
> the fact that the different packages produc
An easy non-R solution is to use a clipboard utility like PathCopy.
Most have the capability to configure how a windows path is copied
to the clipboard.
On 5/20/2014 6:37 AM, Knut Krueger wrote:
Is there any function to change the windows path to linux path?
especially I would like to have th
Run the examples in the ?boot help page. Notice that boot needs a function that
accepts a dataframe and an index vector.
Sent from my iPhone
> On May 21, 2014, at 1:49 AM, 张以春 wrote:
>
> Dear friends,
>
>
> I have a numeric vector composed of 320 numbers.
>
>
> Now, I want to do resample
Am 20.05.2014 19:00, schrieb David L Carlson:
Now I understand. Not really a solution, but you can instruct students to use
only forward slashes in their paths since R will convert to backslash on
Windows systems automatically. After a couple of broken commands, they should
get the hang of it.
Perhaps it should be mentioned that resampling does not produce a
correct CI for extreme quantiles, so the whole exercise may be
pointless.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certain
I tend to view this as enforced training in string escaping.
One option i have not seen mentioned is the file.choose function, which is GUI
but a little tedious for deep directories.
---
Jeff Newmiller
On Wed, May 21, 2014 at 9:25 AM, Jeff Newmiller
wrote:
> I tend to view this as enforced training in string escaping.
>
> One option i have not seen mentioned is the file.choose function, which is
> GUI but a little tedious for deep directories.
I also was thinking about that too but note that f
Dear,
I would like to stop to receive any email from R-help mailing list for this
email address. Thank you.
Mingxuan Han
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-
Then unsubscribe.
On May 21, 2014 10:00 AM, "han...@purdue.edu" wrote:
> Dear,
>
> I would like to stop to receive any email from R-help mailing list for
> this email address. Thank you.
>
>
> Mingxuan Han
>
> __
> R-help@r-project.org mailing list
> ht
Hi everyone,
How do I draw a transparent ploygon overlaying with a scatter plot?
Let's say, we call plot() to have a scatter plot, then call polygon() to
add a polygon. I was hoping the polygon can be transparent so the scatter
plot is still visible. I can't find any argument in polygon() for suc
The standard colors in R are opaque, but you can add an alpha value to make
them semi-transparent. In this example we set alpha halfway between 0 and 255
to define a semi-transparent red:
> set.seed(42)
> x <- runif(10)*10
> y <- runif(10)*10
> plot(x, y, pch=16)
> col2rgb("red", alpha=TRUE)
You can use adjustcolor() to do some of the arithmetic for you. E.g.,
the following draws red squares with 10 opacities ('alpha's) from 5%
to 95%:
plot(1:10)
square <- 0.45 * cbind(c(-1,1,1,-1),c(-1,-1,1,1))
for(i in 1:10) polygon(square+i, col=adjustcolor("red", alpha=(i-.5)/10))
Bill Du
Yes and you do that by going to
https://stat.ethz.ch/mailman/listinfo/r-help
and fill in the required information.
Sorry to loose you on this list.
Best regards
Frede
Sendt fra Samsung mobil
Oprindelig meddelelse
Fra: Steve Friedman
Dato:21/05/2014 16.03 (GMT+01:00)
Til:
... I just specify alpha values directly in the hex-code for a color:
set.seed(112352)
plot(runif(20),runif(20), pch=16)
polygon(runif(3), runif(3), col="#FF22")
polygon(runif(3), runif(3), col="#00FF0022")
polygon(runif(3), runif(3), col="#FF22")
B.
On 2014-05-21, at 12:05 PM, William
Thanks everyone who replied.
Boris, could you explain a bit more how to obtain this hex-code for a
specific color? Thanks.
Jun
On Wed, May 21, 2014 at 1:37 PM, Boris Steipe wrote:
> ... I just specify alpha values directly in the hex-code for a color:
>
> set.seed(112352)
> plot(runif(20),runi
try this:
> x <- structure(list(date = structure(c(15765, 15766, 15767, 15768,
+ 15769, 15770, 15771, 15772, 15773, 15780, 15781, 15782, 15788,
+ 15789, 15790, 15791, 15792, 15795), class = "Date"), mon = c("Mrz",
+ "Mrz", "Mrz", "Mrz", "Mrz", "Mrz", "Mrz", "Mrz", "Mrz", "Mrz",
+ "Mrz", "Mrz", "Mr
Hex-codes are ubiquitously used on the Web - they are a short-hand version of
specifying rgb triplets. The first two digits specify the red value, the second
two are for green, then blue. R accepts a fourth pair of digits to optionally
specify the transparency, the semantics of the code is thus
Hi All,
I've successfully gotten out the colMeans for 60 columns using:
col <- colMeans(x, na.rm = TRUE, dims = 1)
My next question is: is there a way of getting a mean of all the
column means (ie a mean of a mean)?
Thanks!
__
R-help@r-project.org ma
Inline.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
H. Gilbert Welch
On Wed, May 21, 2014 at 11:26 AM, Boris Steipe wrote:
> Hex-codes are ubiquitously used on the
That would be because col is a function in base R, and thus a poor
choice of names for user objects. Nonetheless, it worked when I ran
it, but you didn't provide reproducible example so who knows.
R> set.seed(1)
R> x <- data.frame(matrix(runif(150), ncol=10))
R> # col is a function, so not a good
Thanks for the explanation. And tip... this was a quick a dirty code
so didn't really think about naming something that is already a
function in R. Data was generic - just a bunch of columns with
numbers so didn't bother including that as I know that wasn't the
problem. Same goes with replying -
Hhhhmmhhhm, but is that not the same as
mean(as.matrix(x))
?
Yours sincerely / Med venlig hilsen
Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
Plant Performance & Modeling
Technology & Service Solutions
T +45 9730 5135
M +45 2547 6050
fr...@vestas.com
http://www.vestas.com
Company
Dear R forum,
I have following data.frame -
my_dat = data.frame(sr = c(0,0,0,0, 1, 1, 1, 1, 2, 2, 2, 2), bond = c("A", "B",
"B", "B", "A", "B", "B", "B", "A", "B", "B", "B"), cashflow = c(1000, 2000,
2000, 4000, 10, 200, 300, 100, 80, 40, 60, 120))
> my_dat
sr bond cashflow
1 0 A
Dear Katharina,
if I get it right, you just want to assess the significance of
'lag(numfull_FCRlong,1)'. For this purpose, you can simply look at the relevant
t-statistic in the model summary. BTW, the latter is the square root of the
(F-version of the) Wald test for the restriction you are in
Dear Jeff,
2014-05-21 9:33 GMT+02:00 Jeff Newmiller :
> You did not show us how "x" was created, so we cannot tell whether this
> indicates a bug or not. My bet would be that "x' was created in some way
> that bypassed the standard POSIXct creation code.
>
I debugged it, this morning:
The probl
Kate Ignatius writes:
> Hi All,
>
> I've successfully gotten out the colMeans for 60 columns using:
>
> col <- colMeans(x, na.rm = TRUE, dims = 1)
>
> My next question is: is there a way of getting a mean of all the
> column means (ie a mean of a mean)?
maybe
mean(x)
?
--
:: Igor Sosa Mayor
Thanks a lot for the input (and thanks for R),
but nope, both R versions are the same ... and I really do not think
that NppToR is actually causing the problem.
Since RMySQL has problems (as I understand the errors/warnings) to find
what it expects to find I thought that maybe there is a way
Hi list
This doesn't particularly concern the interpretation of a mixed model, so I
thought it best to post here. I assume it has something to do with using the
asList call for an S4 object or some form of grouping done by lme, but I might
be way off. I'd like to extract the residuals of a lmer
Self-taught R newbie (and stats newbie) here.
I'd like to run a MANOVA test to see whether self- and peer ratings differ
within teams. There are 3 peer ratings per person on a team.
My data is in a tab-delimited text file, structured as follows:
*Team* *Person**Rating t
Hello,
I would be really thankful if I could bother you with two questions.
1. I ran into some problems when I used the code "optim()" in my work. I used
it to optimize a function-fn(x,y), where x is a two-dimensional parameter. For
y I want to pass the value later when I use optim(), because
Yichun,
1. You are doing this wrong!
2. It is the wrong thing to do!
In more detail:
1. To use boot, it helps to read the help file which clearly says the
statistic must be a function of two arguments, the original dataset and
the indices saying which of the original datapoints are in the boot
Please help me to find the reason when applied this program gave me the same
result for
b1,b2,b3 for all (BS,BIAS, RMSE , and SE)
I want to compare between these beta's
tq.
## Ridge MM ##
rm(list=ls())
library(MASS)
library(mvoutlier)
library(robu
Sorry for the naive question, but is it actually possible do SEE the
approximation function y=f(x) in any of the output options of approxfun?
Cheers,
--
Helio
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://st
So,
some feedback.
Have installed MariaDB 10.0.10 on the Linux box. That speeded things up.
Changed from InnoDB/XtraDb to Aria. That speeded loading of the data up.
Have installed MariaDB on the iMac. That speeded things up more.
Tried to tune MariadDB's config. Didn't speed things up much, but f
Just a note on this earlier thread: R package collections for public APIs
seem to be now emerging and thriving in various disciplines:
- Bioinformatics the Bioconductor collection contains many API packages
- rOpenSci: R tools for open science-related APIs
- rOpenHealth: R tools for open healthc
Not necessarily. Missing values may base the means on unequal numbers of
observation:
x <- c(1,1,1,2,NA,2)
dim(x) <- c(3,2)
cM <- colMeans(x, na.rm = TRUE, dims = 1)
mean(cM)
mean(x, na.rm = TRUE)
B.
On 2014-05-21, at 3:38 PM, Frede Aakmann Tøgersen wrote:
> Hhhhmmhhhm, but is that not the
On 21/05/14 23:34, Raphael Päbst wrote:
I believe you are right. A night of sleep has done wonders for my
understanding of the problem.
Oh sleep it is a gentle thing
Beloved from pole to pole!
Thank you for your patience and help!
Patience? Moi? This must be some new use of the word "pati
On 21 May 2014, at 22:31 , Helio Camargo wrote:
> Sorry for the naive question, but is it actually possible do SEE the
> approximation function y=f(x) in any of the output options of approxfun?
Yes and no...
> a <- approxfun(x, y, rule = 2:1)
> a
function (v)
.approxfun(x, y, v, method, yleft
On Wed, 21 May 2014 03:59:26 PM Anna Carter wrote:
> Dear R forum,
>
> I have following data.frame -
>
> my_dat = data.frame(sr = c(0,0,0,0, 1, 1, 1, 1, 2, 2, 2, 2), bond =
c("A",
> "B", "B", "B", "A", "B", "B", "B", "A", "B", "B", "B"), cashflow = c(1000,
> 2000, 2000, 4000, 10, 200, 300, 100,
On the page listing all of the post topics and time last post it
says that this topic has a post at 6:05 am today, but
that post doesn't show up when I view this topic.
There must be some kind of programming error.
Hurr
--
View this message in context:
http://r.789695.n4.nabble.com/Second-axis
I'm using the sizeplot() function from the plotrix package to plot binomial
data in which there are a number of repeated values.
a portion of my data look like this:
mortality = c(1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1,
0, 0, 1, 0, 0, 1, 0, 1, 0)
ww.max = c(4.0, 5.0, 3.0,
On Wed, 21 May 2014 06:34:58 PM Liz Duermit wrote:
> I'm using the sizeplot() function from the plotrix package to plot
binomial
> data in which there are a number of repeated values.
>
> a portion of my data look like this:
>
> mortality = c(1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0,
Well, yes, and also a user error.
On Wed, May 21, 2014 at 6:50 PM, Hurr wrote:
> On the page listing all of the post topics and time last post it
> says that this topic has a post at 6:05 am today, but
> that post doesn't show up when I view this topic.
> There must be some kind of programming er
Hi Anna,
In addition, you may also try:
library(data.table)
dt1 <- data.table(my_dat, key=c('sr', 'bond'))
dt2 <- dt1[,list(cashflow_total=sum(cashflow)), by=c('sr','bond')]
A.K.
On Wednesday, May 21, 2014 6:46 PM, Jim Lemon wrote:
On Wed, 21 May 2014 03:59:26 PM Anna Carter wrote:
> Dear R
Thanks Jim, that was a great help!
On Wed, May 21, 2014 at 7:49 PM, Jim Lemon wrote:
> On Wed, 21 May 2014 06:34:58 PM Liz Duermit wrote:
> > I'm using the sizeplot() function from the plotrix package to plot
> binomial
> > data in which there are a number of repeated values.
> >
> > a portion
You can specify whatever plotting size you want for graphics devices,
so for example you could send the plot to a pdf device set up as A0 or
other poster size. Then you just have the challenge of finding a
place to print it.
The pairs2 function in the TeachingDemos package will create pairs
like
Like this:
> x <- 1:10
> y <- rnorm(10)
> f <- approxfun(x, y)
> f
function (v)
.approxfun(x, y, v, method, yleft, yright, f)
Yours sincerely / Med venlig hilsen
Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
Plant Performance & Modeling
Technology & Service Solutions
T +45 9730 5135
M +45
Am 20.05.2014 19:03, schrieb Duncan Murdoch:
I have no idea what you mean by a "callback from utils".
clipboard {utils}
> readClipboard
function (format = 1L, raw = FALSE)
.Call(C_readClipboard, format, raw)
>
I do not know whether it is an real callback or a normal call to an C
subroutine
Sorry
https://en.wikipedia.org/wiki/Callback_%28computer_programming%29
Knut
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide comme
Hi,
I would like to unsubscribe from the mailing list.
Thanks
--
--
Martina MarbÃ
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://
65 matches
Mail list logo
