Dear R users,
I´m trying to optimize simultaneously two binomials inequalities used to
acceptance sampling, which are nonlinear solution, so there is no simple
direct solution.
The 'n' represents the sample size and the 'c' an acceptance number or
maximum number of defects (nonconforming) in sampl
Hi,
I have the following jackknife code which is much slower than my colleagues C
code. Yet I like R very much and wonder how R experts would optimize this.
I think that the for (i in 1:N_B) part is bad because Rprof() said sum() is
called very often but I have no idea how to optimize it.
#
Hi all,
Consider the following function:
my.func = function(x){
y=ifelse(x>-.5,0,ifelse(x< -.8,abs(x)/2,abs(x)))
print(c(x,y)) #print what was tested and what the result is
return(y)
}
curve(my.func,from=-1,1)
When I attempt to find the maximum of this function,
Hi Mike,
You function is discontinuous at -0.8, so you can
expect everything :-}! But this is not the only
problem. The algorithm for optimize never gets there.
In general there exists no universal method to find
the global maximum of a function (unless it satisfies
certain conditions). You can al
Given a univariate problem where the maximum
must be between -1 and 1, I would test with a grid
of points, then refine that if necessary.
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")
Mike Lawrence wrote:
>
alth/People/Faculty/Varadhan.html
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Mike Lawrence
Sent: Monday, October 01, 2007 1:29 AM
To: Rhelp
Subject: [R] optimize() stuck in local plateau ?
Hi all,
Consider the
p://www.jhsph.edu/agingandhealth/People/Faculty/
> Varadhan.html
>
>
>
> --
> --
>
>
>
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> project.org]
-Original Message-
From: Mike Lawrence [mailto:[EMAIL PROTECTED]
Sent: Monday, October 01, 2007 10:27 AM
To: Ravi Varadhan
Cc: 'Rhelp'
Subject: Re: [R] optimize() stuck in local plateau ?
I may be misunderstanding, but my example seems to v
Dear R users,
I´m trying to optimize simultaneously two binomials inequalities used to
acceptance sampling, which are nonlinear solution, so there is no simple
direct solution. Please, let me explain shortly the the problem and the
question as following.
The objective is to obtain the smallest v
Dear R users,
I´m trying to optimize simultaneously two binomials inequalities (used in
acceptance sampling) which are nonlinear solution, so there is no simple
direct solution. Please, let me explain shortly the the problem and the
question as following.
The objective is to obtain the smallest v
I want to find the last record for each person_id in a data frame
(from a SQL database) ordered by date. Is there a better way than
this for loop?
for (i in 2:length(history[,1])) {
if (history[i, "person_id"] == history[i - 1, "person_id"])
history[i, "order"] = history[i - 1, "order"]
I saw your post on 7/29, and I have not seen a reply, so I will
attempt a response to the question at the start of your email:
obtain the smallest value of 'n' (sample size)
satisfying both inequalities:
(1-alpha) <= pbinom(c, n, p1) && pbinom(c, n, p2) <= beta,
where alpha, p1,
Just a quick follow up to Spencer's post, you might want to look at the
AcceptanceSampling package on CRAN:
http://cran.r-project.org/web/packages/AcceptanceSampling/index.html
HTH,
Marc Schwartz
on 08/05/2008 09:00 AM Spencer Graves wrote:
I saw your post on 7/29, and I have not seen a
Thanks Mr. Graves for your support.
> I saw your post on 7/29, and I have not seen a reply,
> so I will attempt a response to the question at the start of
> your email:
>
> obtain the smallest value of 'n' (sample size) satisfying
> both inequalities:
>
> (1-alpha)
.
EToktar
> -Original Message-
> From: Marc Schwartz [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, August 05, 2008 1:12 PM
> To: Spencer Graves
> Cc: [EMAIL PROTECTED]; r-help@r-project.org
> Subject: Re: [R] optimize simultaneously two binomials
> inequalities using nlm(
Dear Andrew,
Here is one way:
# Some data
set.seed(1)
mydata<-data.frame(person_id=rep(1:10,10),x=rnorm(100))
mydata
# last register for person_id
with(mydata,tapply(x,person_id,function(x) tail(x,1)))
# test for person_id=1
mydata[mydata$person_id==1,] # see the last number in column 2
HTH,
"...from an SQL database."How? Structure of the result?
You say "ordered by date" but then you don't reference any date
variable? And your code creates an "order" column, but that would not
appear necessary for the stated purpose and you don't output the last
"order" within a "person_i
Dear Andrew:
Here is another way assuming you have an order column in your history data
as well as a person_id. Again, your variable
of interest is x:
# Some data
set.seed(1)
history<-data.frame(
person_id=rep(1:10,each=10),
record=rep(sample(10),10),
x=rnorm(100)
[,"person_id"]),,drop=FALSE]
history[order(history$date),,drop=FALSE]
}
> f2(history)
person_id date
2 Mary2
7 Alex7
9Sue9
10 Joe 10
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
On Fri, Feb 27, 2009 at 2:10 PM, William Dunlap wrote:
> Andrew, it makes it easier to help if you supply a typical
> input and expected output along with your code. I tried
> your code with the following input:
I'll be careful to avoid these mistakes. Also, I should not have used
a reserved wo
Hi Andrew,
Just remember that something was bad with the code I sent you (the same you
referred to in [1]).
This version runs with no problems:
# Some data
history_ <-
data.frame(person_id=c(1,2,2),date_=c("2009-01-01","2009-02-03","2009-02-02"),x=c(0.01,0.05,0.06))
colnames(history_) <- c("perso
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