Thanks Gabor for your reply. Your script will find missing days in date1
range but my intention was to find missing days in each level (df$lev)
in corresponding month. The resulting df could look like this:
lev date1
1 lev1 2007-09-03
2 lev1 2007-09-04
4 lev1 2007-09-24
5 lev1 2007-09
Try this. It creates a sequence of dates from the range of df$date1 and
then does a setdiff between that and the original dates. The result is
numeric so we create a Date structure out of it.
structure(setdiff(do.call(seq, as.list(range(df$date1))), df$date1),
class = "Date")
On Jan 2, 2008 1:
Hi,
I have a data.frame like this:
y <- rnorm(60)
lev <- gl(3,20, labels=paste("lev", 1:3, sep=""))
date1 <- as.Date(seq(ISOdate(2007,9,1), ISOdate(2007,11,5),
by=60*60*24))
date1 <- date1[-c(3,4,15,34,38,40)]
df <- data.frame(lev=lev, date1=date1, y=y)
I would like to produce a new data.frame w
On Tue, 1 Jan 2008, jim holtman wrote:
Is this what you want?
x <- c('5/5/2007', '12/31/2007')
# convert to day of year (Julian date) -- use POSIXlt
strptime(x, "%m/%d/%Y")$yday+1
[1] 125 365
I don't think that is the usual definition of JDN: see
http://en.wikipedia.org/wiki/Julian_day. A
Hi R Gurus!
There was a function in S called "usa()" which would plot the US.
I found map('usa') in R for the lower 48 states. Is there a way to
include Alaska and Hawaii as well, please?
Thanks,
Edna Bell
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https:/
Dear R-users,
I am using stepplr for L2 regularized logistic regression. Since number of
attribute is too large i discarded interaction terms. Everything is fine but
only problem i have faced that i cannot choose a good shrinkage coefficient
(lambda). If CV is the best way to estimate, can you pl
Or use complete.cases
df.complete <- df[complete.cases(df),]
Simon.
On Wed, 2008-01-02 at 13:21 +1000, Ross Darnell wrote:
> You could try
>
>
> > complete.case.df <- na.omit(df)
>
>
> Ross Darnell
>
>
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> On B
My guess is that this combination of variables produces separation in
the data: Too many (all?) of the response 1's are in at level of VAR3,
and the 0's are at the other level (or vice versa).
HTH,
Simon.
On Sat, 2007-12-29 at 18:39 -0500, Charles Willis wrote:
> Hello,
>
> I am trying to run t
You could try
> complete.case.df <- na.omit(df)
Ross Darnell
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Marko Milicic
Sent: Wednesday, 2 January 2008 11:50 AM
To: r-help@r-project.org
Subject: [R] Subsetting data frame problem
Dear R users,
?complete.cases
On Jan 1, 2008 8:50 PM, Marko Milicic <[EMAIL PROTECTED]> wrote:
> Dear R users,
>
> I'm new but already fascinated R user so please forgive for my
> ignorance. I have the problem, I read most of help pages but couldn't
> find the solution. The problem follows
>
> I have large
Dear R users,
I'm new but already fascinated R user so please forgive for my
ignorance. I have the problem, I read most of help pages but couldn't
find the solution. The problem follows
I have large data set 10,000 rows and more than 100 columns... Say
something like
var1,var2,var2,var4.
Actually this works beautifully as it stands:
breaks <- c(-Inf, -1, 1, Inf)
zz <- lapply(breaks, function(x) if(x==-Inf) quote(-infinity) else
if (x==Inf) quote(infinity) else format(x))
lbl <- mapply(function(x,y)
bquote("(" * .(x) * "," * .(y) * "]"),
Try:
das$danger <- with(das, (age > 65) * (bmi > 30))
On Jan 1, 2008 5:03 PM, Gerard Smits <[EMAIL PROTECTED]> wrote:
> Hi All,
>
> I have a small dataset named das (43 cases) in which I am trying to
> create a binary outcome (1/0) based on the following code:
>
> if (das$age>65 && das$bmi>30) {
Hi Domenico,
I was incorrectly assuming it would use a vector of equal length to
my data. frame. Thanks for the clarification.
Also, thanks for the many alternate programming approaches provided by others.
Gerard
At 02:25 PM 1/1/2008, Domenico Vistocco wrote:
>You should look for your answer
you could try the following:
>das$danger <- 0
>das$danger[das$bmi > 30 & das$age > 65] <- 1
On Jan 2, 2008 9:16 AM, Gerard Smits <[EMAIL PROTECTED]> wrote:
>
> Thanks, but I tried the single ampersand, but got a warning msg with
> the same lack of correct assignment:
>
> > if (das$age>65 & das
You should look for your answer using the help for the if statement (?"if").
The cond argument should be a scalar (otherwise only the first element
is used).
?"if"
.
cond: A length-one logical vector that is not 'NA'. Conditions of
length greater than one are accepted with a warnin
Thanks, but I tried the single ampersand, but got a warning msg with
the same lack of correct assignment:
> if (das$age>65 & das$bmi>30) {das$danger<-1} else das$danger<-0
Warning message:
In if (das$age > 65 & das$bmi > 30) { :
the condition has length > 1 and only the first element will be
You need to use '&' instead of '&&':
A shorter version of your code using ifelse:
das$danger <- with(das, ifelse(age>65 & bmi>30, 1, 0))
HTH
-Christos
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Gerard Smits
> Sent: Tuesday, January 01, 200
Is this what you want?
> x <- c('5/5/2007', '12/31/2007')
> # convert to day of year (Julian date) -- use POSIXlt
> strptime(x, "%m/%d/%Y")$yday+1
[1] 125 365
On Jan 1, 2008 4:59 PM, Nüzhet Dalfes <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Is there a package for converting day-month-year type date to
Hi All,
I have a small dataset named das (43 cases) in which I am trying to
create a binary outcome (1/0) based on the following code:
if (das$age>65 && das$bmi>30) {das$danger<-1} else das$danger<-0
I am setting a flag called 'danger' to 1 of the subject is over 65
and has a BMI > 30.
I fin
Hi,
Is there a package for converting day-month-year type date to julian
day number (JDN)? I looked around and I couldn't find any (I am pretty
new to R...)
Thanks and happy New Year to everybody!
H. Nüzhet Dalfes
Professor,
Istanbul Technical University
Eurasia Institute of Earth Scienc
On Tue, 1 Jan 2008, zhijie zhang wrote:
> Dear all,
> I have two variables, y and x. It seems that the relationship between them
> is Piecewise Linear Functions. The cutpoint is 20. That is, when x<20, there
> is a linear relationship between y and x; while x>=20, there is another
> different lin
Dear Anisah,
ENSEMBL data are at your hand under R:
> library(seqinr)
> choosebank("ensembl")
> cat(banknameSocket$details, sep = "\n")
ACNUC Data Base Content
Ensembl Release 47 Last Updated: Dec 12, 2007
76,798,685,993 bases; 3,138,133 s
Hello there,
Happy new year.
As I know, we can get proximity of new data in randomForest package. How to
get the proximity matrix of new data in party package then? Thanks.
Joseph
[[alternative HTML version deleted]]
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R-help@r-project.org
Dear Lindsay,
If you "pkgconfig" installed by MacPorts.
Maybe it likely to be caused by JAGS install.
You'll better to do
sudo make uninstall
./configure --prefix=/usr/local --mandir=/usr/local/share/man
make
sudo make install
sudo R CMD INSTALL -l /Library/Frameworks/R.framework/Resources/li
Is this what you want (same as your except the draw.key and
reference to grid is removed key=... is added in the xyplot call:
library(lattice)
x <- 1:3
colorseq < c(0.2, 0.5, 0.8)
mycolors <- gray(colorseq)
xyplot(x ~ x, col = mycolors, aspect = 1,
key = list(rect = list(col = mycolors), text =
On 1/1/08, Hofert Marius <[EMAIL PROTECTED]> wrote:
> Happy New Year to all R users!
>
> I have two short questions concerning a xyplot with a color key:
>
> 1) How do I properly place (align) the color key beside the xyplot?
> As you can see from the code listed below, the placement of the color
>
Richard Müller wrote:
> Happy New Year to all!
>
> I want to run a script from a directory on a central server on different
> machines, linux and windows. To determine which machine is calling the script
> I want to pass an argument with the call (e.g. for choosing the display
> device, for w
baptiste Auguié wrote:
> Dear all,
>
> Happy new year!
>
> I posted a very similar question a few days ago, but probably too
> cluttered. Here is a tidy, minimal version:
>
> I want to make a package, with a data.frame d and a function f given
> below. Now, the function f needs to use the
Dear all,
I have two variables, y and x. It seems that the relationship between them
is Piecewise Linear Functions. The cutpoint is 20. That is, when x<20, there
is a linear relationship between y and x; while x>=20, there is another
different linear relationship between them.
How can i specify t
Thank you very much for your help, Chuck. But I don't understand the function
"statistic" nor that his arguments make. Those arguments do not take value
at any moment, according to I understand (I have not given values to "d" nor
"ind"). It is not thus?
Can you explain me, please?
Thanks.
_Fede
_Fede_ wrote:
> Hi again.
>
> Watching this example that appears in the help page
>
> ratio <- function(d, w) sum(d$x * w)/sum(d$u * w)
> city.boot <- boot(city, ratio, R = 999, stype = "w",sim = "ordinary")
> boot.ci(city.boot, conf = c(0.90,0.95),type =
> c("norm","basic","perc","bca"))
>
> I
Hello,
You use nlrq() pretty much the same as nls(). Look at ?nls and you will
find there many examples on how to use it. The easiest way is to use
with a "self-start model". Do apropos("^ss") to get the list of
self-starting models defined, and look at their respective help pages to
see if on
Please,
I have a problem with nonlinear quantile regression.
My data shows a large variability and the quantile regression seemed perfect
to relate two given variables. I got to run the linear quantile regression
analysis and to build the graph in the R (with quantreg package). However, the
up p
Hi again.
Watching this example that appears in the help page
ratio <- function(d, w) sum(d$x * w)/sum(d$u * w)
city.boot <- boot(city, ratio, R = 999, stype = "w",sim = "ordinary")
boot.ci(city.boot, conf = c(0.90,0.95),type =
c("norm","basic","perc","bca"))
I have tried to do the following (c
Happy New Year to all!
I want to run a script from a directory on a central server on different
machines, linux and windows. To determine which machine is calling the script
I want to pass an argument with the call (e.g. for choosing the display
device, for writing path names etc.)
I tried the
I have the following procedure which worked just fine for in R 2.2.0.
Recently I upgraded to 2.6.1 and now get an error:
> ScatterOutlier(pass_500_506[1:1000,6:12], marginal_500_506[,6:12])
Error in eval(expr, envir, enclos) : object "out" not found
Note that I use the same workspace (and h
Happy New Year to all R users!
I have two short questions concerning a xyplot with a color key:
1) How do I properly place (align) the color key beside the xyplot?
As you can see from the code listed below, the placement of the color
key is not correct. I would like the upper and lower end poin
Hello there,
How to get the proximity matrix of new data in party package? Thanks.
Joseph
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PLEASE do read the posting guide h
Dear all,
Happy new year!
I posted a very similar question a few days ago, but probably too
cluttered. Here is a tidy, minimal version:
I want to make a package, with a data.frame d and a function f given
below. Now, the function f needs to use the data.frame d. I could
(and that's what I'
Thanks.
library(ROCR) was used finally. It also automatically generate a plot beside
the value of AUC.
On Dec 31, 2007 11:38 PM, Frank E Harrell Jr <[EMAIL PROTECTED]>
wrote:
> zhijie zhang wrote:
> > Dear all,
> > Some functions like 'ROC(Epi)' can be used to perform ROC analyssi,
> but it
> >
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