Hi,
I'm trying to figure out if there are any packages allowing
one to fit mixed models (or non-linear mixed models) to data
that includes censoring.
I've done some searching already on CRAN and through the mailing
list archives, but haven't discovered anything. Since I may well
have done a
Gunter
Genentech Nonclinical Statistics
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Douglas Grove
Sent: Monday, April 23, 2007 10:58 AM
To: r-help@stat.math.ethz.ch
Subject: [R] fitting mixed models to censored data?
Hi,
I'm trying to figure
() or
coxph() --- is another alternative to explore, I believe.
Hope that helps,
Bill
Nonclinical Statistics, Centocor R D
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Douglas Grove
Sent: Monday, April 23, 2007 2:29 PM
To: Bert Gunter
Cc: r-help
You need to create a new object and assign it to 'df'
so you'd do something like this:
df - sapply(factors, function (name) {
pos - match(name,df.names)
factor(df[[pos]])
})
Doug
On Thu, 11 May 2006, Sam Steingold wrote:
* jim
You really need to learn how to do some searching, as you seem to
be constantly asking questions you can answer yourself
help.search(sprintf)
On Fri, 10 Mar 2006, Michael wrote:
something like sprintf in C?
so I can do:
print(sprintf(the correct result is %3.4f\n, myresult));
---
Hi,
I'm trying to figure out if there's an automated way to get
read.table to read in my data and *not* convert the character
columns into anything, just leave them alone. What I'm referring
to as 'character columns' are columns in the data that are quoted.
For columns of alphabetic strings
reads carefully, as.is protects
a character vector from converstion to a *factor*,
but not from conversion to numeric/logical.
Doug
On Sun, 26 Feb 2006, Kjetil Brinchmann Halvorsen wrote:
Douglas Grove wrote:
Hi,
I'm trying to figure out if there's an automated way to get
read.table
So you want to create a subset of a data frame?
with components name1 name2 name3 ...
dframe[, c(name1,name2,name3,...)]
will do that
Doug
On Fri, 20 Jan 2006, Michael Reinecke wrote:
Hi! I suspect there must be an easy way to access components of a data frame
by name, i.e. the input
It's much more helpful if you show the actual command you used.
Presumably you have a data frame 'd' and you've done
hist(d), and 'hist' has complained because d is not numeric,
d is a data frame that *contains* a numeric vector.
You need to give hist() that numeric vector, which you can do
in
Help pages are useful, you should try them
e.g. ?pi or ?LETTERS
How can one discover or list all available built-in objects?
On Jun 10, 2005, at 7:23 AM, Muhammad Subianto wrote:
L3 - LETTERS[1:3]
L10 - LETTERS[1:10]
LETTERS is apparently a built-in character vector. ls() and objects
The new version of R has begun enforcing rules on regular expressions.
Your pattern is not a valid regular expression, hence it no longer works.
The meaning of '*' is with respect to a preceding character, hence it is
ill-defined without one.
On Mon, 25 Apr 2005, Ye, Bin wrote:
Hi,
I
When I think of New Zealand I think Rabbit :)
How 'bout something like the Monty Python rabbit from
the Holy Grail (nasty pointy teeth..., look at the bones!)
Doug
__
[EMAIL PROTECTED] mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
An alternate method that saves having to use order() again is
r[o] - r
Doug
On Mon, 2004-10-04 at 15:21, Wolfram Fischer wrote:
I have:
d - sample(10:100, 9)
o - order(d)
r - d[o]
How I can get d (in the original order), knowing only r and o?
Thanks - Wolfram
]
on Mon, 28 Jun 2004 10:59:26 +0200 (CEST) writes:
Torsten On Fri, 25 Jun 2004, Douglas Grove wrote:
I should have specified an additional constraint:
I'm going to need to use this repeatedly on large vectors
(length 10^6), so something efficient is needed
On Sat, 26 Jun 2004, Prof Brian Ripley wrote:
On Fri, 25 Jun 2004, Douglas Grove wrote:
I get ties in output from runif() when I generate as few as 10^5
variates and get quite a lot when I generate 10^6. Is this
expected??
It should have been.
I haven't seen any duplication
Hi,
I'm wondering if anyone can point me to a function that will
allow me to do a ranking that treats ties differently than
rank() provides for?
I'd like a method that will assign to the elements of each
tie group the largest rank.
An example:
For the vector 'v', I'd like the method to
I should have specified an additional constraint:
I'm going to need to use this repeatedly on large
vectors (length 10^6), so something efficient is
needed.
On Fri, 25 Jun 2004, Sundar Dorai-Raj wrote:
Douglas Grove wrote:
Hi,
I'm wondering if anyone can point me to a function
I get ties in output from runif() when I generate as few as 10^5
variates and get quite a lot when I generate 10^6. Is this
expected?? I haven't seen any duplication with rnorm(10^6), but
see varying amounts of duplication using rexp(), rbeta() and
rgamma(). I would have thought that there'd
You can't use this anymore. The function predict() has a method
for loess objects, but there is no longer an available function
called predict.loess. So just replace predict.loess
with predict.
On Fri, 13 Feb 2004, Thomas Jagoe wrote:
I am using R to do a loess normalisation procedure.
In
On Sat, 6 Dec 2003, Prof Brian Ripley wrote:
I think you misunderstand how R uses memory. gc() does not free up all
the memory used for the objects it frees, and repeated calls will free
more. Don't speculate about how memory management works: do your
homework!
Are you saying that
I'm sorry to insist but I still think there is something wrong with the function
kmeans. For instance, let's try the same small example:
dados-matrix(c(-1,0,2,2.5,7,9,0,3,0,6,1,4),6,2)
I will choose observations 3 and 4 for initial centers and just one iteration. The
results are
Hi,
What's the best way of dropping leading or trailing
blanks from a character string?
The only thing I can think of is using sub() to replace
blanks with null strings, but I don't know if there is
a better way (I also don't know how to represent the
trailing blank in a regular expression).
Douglas Grove [EMAIL PROTECTED] writes:
Hi,
I'm trying to understand a behaviour that I have encountered
and can't fathom.
Here's some code I will use to illustrate the behaviour:
# start with some data frame a having some named columns
a - data.frame(a=rep(1,3),c=rep
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