Displaying 10 results from an estimated 10 matches for "m2b".
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2011 Sep 03
2
problem in applying function in data subset (with a level) - using plyr or other alternative are also welcome
...rep(3, 3))
y <- rnorm(10, 8, 2)
# variable set 1
M1a <- sample (c(1, 2,3), 10, replace= T)
M1b <- sample (c(1, 2,3), 10, replace= T)
M1aP1 <- sample (c(1, 2,3), 10, replace= T)
M1bP2 <- sample (c(1, 2,3), 10, replace= T)
# variable set 2
M2a <- sample (c(1, 2,3), 10, replace= T)
M2b <- sample (c(1, 2,3), 10, replace= T)
M2aP1 <- sample (c(1, 2,3), 10, replace= T)
M2bP2 <- sample (c(1, 2,3), 10, replace= T)
# variable set 3
M3a <- sample (c(1, 2,3), 10, replace= T)
M3b <- sample (c(1, 2,3), 10, replace= T)
M3aP1 <- sample (c(1, 2,3), 10, replace= T)
M3bP2 <...
2012 Apr 10
1
compare two matrices
Dear Members,
I have two estimated transition matrices and I want to compare them.
In fact I want to check the hypothesis if they come from the same process.
I tried to look for some test but all I found was independence test of
contingency tables.
The following code shows that the usual chi-squared test statistic does
not follow chisq distribution.
MCRepl <- 5000
khi12 <- rep(0,MCRepl)
2011 May 04
0
Fwd: simple question
...rrected version:
Dear R experts
I have simple question, please execuse me:
#example data, the real data consists of 20000 pairs of variables
K1 <- c(1,2,1, 1, 1,1); K2 <- c(1, 1,2,2, 1,2); K3 <- c(3, 1, 3, 3, 1, 3)
M1a <- rep( K1, 100); M1b <- rep(K2, 100)
M2a <- rep(K1, 100); M2b <- rep(K1, 100)
M3a <- rep(K1, 100); M3b <- rep(K3, 100)
mydf <- data.frame(M1a, M1b, M2a, M2b, M3a, M3b)
# matrix operation
nmat <- matrix(c(paste('M', 1:3, 'a', sep = ''), paste('M', 1:3, 'b', sep =
'')), 3)
coffin <- function(x...
2010 Jan 27
1
term.formula error when updating an nls object
...directly
with nls
m2a <- nls(Response ~
Rm * Contrast^ex/(Contrast^fx + sig^fx),
data = dd,
start = list(Rm = 30, sig = 0.05, ex = 3,
fx = 3.1))
nor with fitting the model with the data transformed (though it is
not necessarily the way I would like to fit the model in this
case).
m2b <- update(m1, log(.) ~ log(.))
m3 <- update(m2b, . ~
log(Rm * Contrast^ex/(Contrast^fx + sig^fx)),
start = list(Rm = 30, sig = 0.05, ex = 3,
fx = 3.1))
sessionInfo()
R version 2.10.1 Patched (2010-01-25 r51051)
i386-apple-darwin9.8.0
locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/...
2007 Feb 13
1
lme4/lmer: P-Values from mcmc samples or chi2-tests?
...m2a: number_pollinators ~ logpatch + landscape_diversity + (1 | site)
m2: number_pollinators ~ logpatch + loghab + landscape_diversity + (1 |
site)
Df AIC BIC logLik Chisq Chi Df Pr(>Chisq)
m2a 4 84.713 91.850 -38.357
m2 5 84.834 93.755 -37.417 1.8793 1 0.1704
##
m2b=update(m2,~.-logpatch)
anova(m2,m2b,test="Chi")
Data: primula
Models:
m2b: number_pollinators ~ loghab + landscape_diversity + (1 | site)
m2: number_pollinators ~ logpatch + loghab + landscape_diversity +
m2b: (1 | site)
Df AIC BIC logLik Chisq Chi Df Pr(>Chisq)
m2b...
2006 Jan 19
1
nls profiling with algorithm="port" may violate bounds (PR#8508)
...---
library(stats4)
mfun <- function(a,b,s) {
if (a<0 || b<0 || s<0) stop("bounds violated")
-sum(dnorm(y,a*x/(1+a*b*x),sd=s,log=TRUE))
}
m2 = mle(minuslogl=mfun,
start=list(a=1,b=1,s=0.1),
method="L-BFGS-B",lower=c(0.002,0.002,0.002))
confint(m2)
m2b = mle(minuslogl=mfun,
fixed=list(b=0),start=list(a=1,s=0.1),
method="L-BFGS-B",lower=c(0.002,0.002,0.002))
## set boundary slightly above zero to avoid
## boundary cases
dev <- 2*(-logLik(m2b)+logLik(m2))
as.numeric(pchisq(dev,lower.tail=FALSE,df=1))
2014 Mar 17
5
LD50
Quiero comparar varias dosis letales 50% (LD50) usando análisis probit. He
seguido un ejemplo que viene en paquete DRC, pero no obtengo el resultado
esperado. Lo que quiero es saber si las LD50s, son diferentes y si la
diferencias son estadísticamente significativas.
Gracias de antemano.
José Arturo
e-mail. jafarfan@uady.mx <grejon@uady.mx>
e-mail alterno. jafarfan@gmail.com
2011 Jun 30
0
CCF of two time series pre-whitened using ARIMA
...reg argument.
I have some sample code:
set.seed(66)
x = arima.sim(list(order=c(1,1,0), ar=.9), n=100) + 50
y=c(x[-1:-3],x[98:100])
# Method 1
m1a=arima(diff(x),order=c(1,0,0))
m1b=arima(diff(y),order=c(1,0,0))
a=ccf(resid(m1a),resid(m1b))
# Method 2
m2a=arima(x,order=c(1,1,0),xreg=1:length(x))
m2b=arima(y,order=c(1,1,0),xreg=1:length(y))
b=ccf(resid(m2a),resid(m2b))
My question is why do the two methods generate different results for the CCF
(one having a peak correlation at lag 4 and the other at lag 3, and the
correlations are also of slightly different values)? I am assuming Method 2
is...
2006 Jan 17
0
nls profile with port/constraints
...------
library(stats4)
mfun <- function(a,b,s) {
if (a<0 || b<0 || s<0) stop("bounds violated")
-sum(dnorm(y,a*x/(1+a*b*x),sd=s,log=TRUE))
}
m2 = mle(minuslogl=mfun,
start=list(a=1,b=1,s=0.1),
method="L-BFGS-B",lower=c(0.002,0.002,0.002))
confint(m2)
m2b = mle(minuslogl=mfun,
fixed=list(b=0),start=list(a=1,s=0.1),
method="L-BFGS-B",lower=c(0.002,0.002,0.002))
## set boundary slightly above zero to avoid
## boundary cases
dev <- 2*(-logLik(m2b)+logLik(m2))
as.numeric(pchisq(dev,lower.tail=FALSE,df=1))
--
620B Bartram Hall...
2005 Dec 21
9
question about changejournal
Hi,
I''ve got a newbie question--sorry if this is covered elsewhere, I parsed
through the archives for awhile and didn''t see it.
I''d like to listen for whenever a file is renamed (e.g. foo.txt -> foo.old)
and then magically change it back. This sounds odd, but I''m working with a
stubborn application and this will actually make things work nice.
So, if I do: