Displaying 20 results from an estimated 7000 matches similar to: "[Bug 2339] New: openssh consumes stdin even if command isn't interactive"
2012 Jun 14
0
fixed trimmed mean for j-group
Hello...i want to find the empirical rate for type 1 error using fixed
trimmed mean. To make it easy, i'm referring to journal given by this
website
http://www.academicjournals.org/ajmcsr/PDF/pdf2011/Yusof%20et%20al.pdf.
I already run the programme and there is no error in it but i got zero for
the empirical rate of type 1 error. The empirical rate for the type 1 error
given in the journal
2012 Jul 07
0
fixed trimmed mean for group
Hello,
I haven't found errors in your code. I implemented the test in the paper
(the first, fixed symetric mean) and it also gives me zero Type I
errors, when alpha = 0.05. Try to see the value of min(pv) or to plot
the histogram of 'pv', hist(pv) and you'll see that there are no
significant p-values, at that level.
Anyway I'll continue to look at it, but my first
2017 Jun 06
0
integrating 2 lists and a data frame in R
Thank you David. Using xtabs operation simplifies the code very much, many
thanks ;)
On Tue, Jun 6, 2017 at 7:44 AM, David Winsemius <dwinsemius at comcast.net>
wrote:
>
> > On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote:
> >
> > Hi Bogdan,
> > Kinda messy, but:
> >
> > N <-
2017 Jun 06
1
integrating 2 lists and a data frame in R
Simple matrix indexing suffices without any fancier functionality.
## First convert M and N to character vectors -- which they should
have been in the first place!
M <- sort(as.character(M[,1]))
N <- sort(as.character(N[,1]))
## This could be a one-liner, but I'll split it up for clarity.
res <-matrix(NA, length(M),length(N),dimnames = list(M,N))
res[as.matrix(C[,2:1])] <-
2012 Jun 04
1
simulation of modified bartlett's test
Hi, I run this code to get the power of the test for modified bartlett's
test..but I'm not really sure that my coding is right..
#normal distribution unequal variance
asim<-5000
pv<-rep(NA,asim)
for(i in 1:asim)
{print(i)
set.seed(i)
n1<-20
n2<-20
n3<-20
mu<-0
sd1<-sqrt(25)
sd2<-sqrt(50)
sd3<-sqrt(100)
g1<-rnorm(n1,mu,sd1)
g2<-rnorm(n2,mu,sd2)
2017 Jun 06
2
integrating 2 lists and a data frame in R
> On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote:
>
> Hi Bogdan,
> Kinda messy, but:
>
> N <- data.frame(N=c("n1","n2","n3","n4"))
> M <- data.frame(M=c("m1","m2","m3","m4","m5"))
> C <-
2017 Jun 06
1
integrating 2 lists and a data frame in R
Here's another approach:
N <- data.frame(N=c("n1","n2","n3","n4"))
M <- data.frame(M=c("m1","m2","m3","m4","m5"))
C <- data.frame(n=c("n1","n2","n3"), m=c("m1","m1","m3"), I=c(100,300,400))
# Rebuild the factors using M and N
C$m <-
2004 Sep 01
0
Re: [S] [R/S] strange solution
Hi, Erin:
A cleaner way is to pass "n2" to "outer" as a "..." argument, as
in the following modification of your code:
boot1 <- function(y,method="f",p=1) {
n1 <- length(y)
n2 <- n1*p
n3 <- n2 - 1
a <- 0.5*(outer(1:n3,1:n3,function(x,y, n2.){n2. - pmax(x,y)}, n2.=n2))
return(a)
}
y1 <- c( 9, 8, 7, 3, 6)
boot1(y=y1,p=4)
2011 Sep 14
1
ruby to solve a physics question
I am trying to solve one of my graduate level physics problems with
ruby...
Here is what I have so far...
a6=0.0
for n1 in -10..10
for n2 in -10..10
for n3 in -10..10
if n1!=0 and n2!=0 and n3!=0
p=Math.sqrt(n1**2+n2**2+n3**2+n1*n2/1.414+n1*n3/1.41+n2*n3/1.414)
a6+=(1/p)**6
end
end
end
end
puts a6
What I''ve got here is a 10x10x10 face-centered
2004 Sep 02
0
Re: [S] [R/S] question re solution
> Someone else mentioned Venables and Ripley (2000) S
> Programming (Springer). Please see this or some other discussion of
the
> "..." argument.
The "Introduction to R" (from Cran website) also talks about it. See pg
49 - section 10.4 (was just reading this the other day).
Cheers
Manoj
-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
2012 Sep 27
0
problems with mle2 convergence and with writing gradient function
Dear R help,
I am trying solve an MLE convergence problem: I would like to estimate
four parameters, p1, p2, mu1, mu2, which relate to the probabilities,
P1, P2, P3, of a multinomial (trinomial) distribution. I am using the
mle2() function and feeding it a time series dataset composed of four
columns: time point, number of successes in category 1, number of
successes in category 2, and
2011 Jul 27
3
Reorganize(stack data) a dataframe inducing names
Dear Contributors,
thanks for collaboration.
I am trying to reorganize data frame, that looks like this:
n1.Index Date PX_LAST n2.Index Date.1 PX_LAST.1
n3.Index Date.2 PX_LAST.2
1 NA 04/02/07 1.34 NA 04/02/07 1.36
NA 04/02/07 1.33
2 NA 04/09/07 1.34 NA 04/09/07
2012 Oct 11
2
model selection with spg and AIC (or, convert list to fitted model object)
Dear R Help,
I have two nested negative log-likelihood functions that I am optimizing
with the spg function [BB package]. I would like to perform model
selection on these two objective functions using AIC (and possibly
anova() too). However, the spg() function returns a list and I need a
fitted model object for AIC(), ICtab() [bbmle package], or anova().
How can I perform AIC-based model
2010 Jul 05
2
Function to compute the multinomial beta function?
Dear R-users,
Is there an R function to compute the multinomial beta function? That is, the normalizing constant that arises in a Dirichlet distribution. For example, with three parameters the beta function is Beta(n1,n2,n2) = Gamma(n1)*Gamma(n2)*Gamma(n3)/Gamma(n1+n2+n3)
Thanks in advance for any assisstance.
Regards,
Greg
[[alternative HTML version deleted]]
2010 Dec 29
2
How to create an array of lists of multiple components?
Hi,
how can I create an array of lists of three components?
This approach does not work:
n1 <- 2
n2 <- 4
n3 <- 5
res <- array(rep(vector("list",3), n1*n2*n3), dim = c(n1,n2,n3))
res[1,1,1] # is not a list with three components...
The goal is that res[1,1,1] is a list with three components. Also, appending the
components didn't work. For example, I tried:
component
2005 Feb 23
1
H-F corr.: covariance matrix for interaction effect
Hi,
I'm still not quite there with my H-F (G-G) correction code. I have it
working for the main effects, but I just can't figure out how to do it
for the effect interactions. The thing I really don't know (and can't
find anything about) is how to calculate the covariance matrix for the
interaction between the two (or even n) main factors.
I've looked through some books
2005 Feb 23
1
H-F corr.: covariance matrix for interaction effect
Hi,
I'm still not quite there with my H-F (G-G) correction code. I have it
working for the main effects, but I just can't figure out how to do it
for the effect interactions. The thing I really don't know (and can't
find anything about) is how to calculate the covariance matrix for the
interaction between the two (or even n) main factors.
I've looked through some books
2017 Jun 06
0
integrating 2 lists and a data frame in R
Hi Bogdan,
Kinda messy, but:
N <- data.frame(N=c("n1","n2","n3","n4"))
M <- data.frame(M=c("m1","m2","m3","m4","m5"))
C <- data.frame(n=c("n1","n2","n3"), m=c("m1","m1","m3"), I=c(100,300,400))
2012 May 18
1
help with creating a box plot
Hi:
I am looking for some help in making two boxplots next to each other.
I have a data like this:
N1 T1 N2 T2 N3 T3 N4 T4 ... Nn Tn
7 8.2 4 5 8 10 4 5 ..... 10 11
I want to have box plot for all Normal samples (N1,N2,N3,N4,,,,Nn)
and another box plot for all tumors (T1,T2,T3,T4,...Tn).
I have data in a numeric class.
If data is represented as N1
2010 Nov 23
0
(no subject)
Dear R Help -
I am analyzing data from an ecological experiment and am having problems
with the ANOVA functions I've tried thus far. The experiment consists of a
blocked/split-plot design, with plant biomass as the response. The following
is an overview of the treatments applied (nitrogen addition, phosphorus
addition, and seeded/not seeded) and at what level (block, main-plot, and
sub-plot):