search for: 0.868

Displaying 19 results from an estimated 19 matches for "0.868".

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2010 Feb 10
1
looping problem
Hi R-users,   I have this code here: library(numDeriv)   fprime <- function(z) { alp  <- 2.0165;   rho  <- 0.868;   # simplified expressions   a      <- alp-0.5   c1     <- sqrt(pi)/(gamma(alp)*(1-rho)^alp)   c2     <- sqrt(rho)/(1-rho)   t1     <- exp(-z/(1-rho))   t2     <- (z/(2*c2))^a   bes1   <- besselI(z*c2,a)   t1bes1 <- t1*bes1   c1*t1bes1*t2 }   ## Newton
2010 Jan 26
1
newton method for single nonlinear equation
Hi r-users,   I would like to solve for z values using newton iteration method.  I 'm not sure which part of the code is wrong since I'm not very good at programming but would like to learn.  There seem to be some output but what I expected is a vector of z values.  Thank you so much for any help given.   newton.inputsingle <- function(pars,n) {  runi    <- runif(974, min=0, max=1)
2009 Mar 17
2
formula question
Dear R People: Here is a small data frame and two particular formulas: > test.df y x 1 -0.9261650 1 2 1.5702700 2 3 0.1673920 3 4 0.7893085 4 5 0.3576875 5 6 -1.4620915 6 7 -0.5506215 7 8 -0.3480292 8 9 -1.2344036 9 10 0.8502660 10 > summary(lm(exp(y)~x)) Call: lm(formula = exp(y) ~ x) Residuals: Min 1Q Median 3Q Max -1.6360 -0.6435
2011 Feb 16
1
caret::train() and ctree()
Like earth can be trained simultaneously for degree and nprune, is there a way to train ctree simultaneously for mincriterion and maxdepth? Also, I notice there are separate methods ctree and ctree2, and if both options are attempted to tune with one method, the summary averages the option it doesn't support. The full log is attached, and notice these lines below for
2010 Feb 09
1
how to adjust the output
Hi R-users,   I have this code below and I understand the error message but do not know how to correct it.  My question is how do I get rid of “with absolute error < 7.5e-06” attach to value of cdf so that I can carry out the calculation.   integrand <- function(z) { alp  <- 2.0165   rho  <- 0.868   # simplified expressions   a      <- alp-0.5   c1     <-
2009 Jul 09
2
Improvement of [dpq]wilcox functions
Hi, I believe I have significantly improved [dpq]wilcox functions by implementing Harding's algorithm: Harding, E.F. (1984): An Efficient, Minimal-storage Procedure for Calculating the Mann-Whitney U, Generalized U and Similar Distributions, App. Statist., 33, 1-6 Results on my computer show (against R-2.9.1): > system.time( dwilcox( 800, 800, 80) ) user system elapsed 0.240
2006 Jul 06
1
PLS method
dear all, I am a new comer to R and statistic. Now I have a little confuse about the package pls. I have to use 5 components to form a model. There are strong relationship between some of the components, which leads to the changes of the sign of each coeficeince, of course this is unwanted when using the normal regression way. So I choose the way of PLS, which is good at solve this kind of
2006 Apr 06
1
interpreting anova summary tables - newbie
Hello, Apologies if this is the wrong list, I am a first-time poster here. I have an experiment in which an output is measured in response to 42 different categories. I am only interested which of the categories is significantly different from a reference category. Here is the summary of the results: summary(simple.fit) Call: lm(formula = as.numeric(as.vector(TNFa)) ~ Mutant.ID, data =
2007 Apr 05
2
about systemfit
Hello. I am still a newbie in R. Excuse me if I am asking something obvious. My efforts to get an answer through browsing the mailing archives failed. I want to perform an augmented Dickey-Fuller test and to obtain AIC and BIC and to be able to impose some linear restrictions on the ADF regression so as to decide the correct order of autoregression. However I could find no obvious way to impose
2010 Jan 26
1
Newton method
Hi r-users,   I hope somebody can help me with this code. I would like to solve for z values using newton iteration method.  I 'm not sure which part of the code is wrong since I'm not very good at programming but would like to learn.  There seem to be some output but what I expected is a vector of z values.  Thank you so much for any help given.   newton.inputsingle <-
2010 Feb 09
0
For and while in looping
I would like to solve a nonlinear eqn using newton method and here is the code:   library(numDeriv)   fprime <- function(z) { alp  <- 2.0165;   rho  <- 0.868;   # simplified expressions   a      <- alp-0.5   c1     <- sqrt(pi)/(gamma(alp)*(1-rho)^alp)   c2     <- sqrt(rho)/(1-rho)   t1     <- exp(-z/(1-rho))   t2     <- (z/(2*c2))^a    bes1   <- besselI(z*c2,a)  
1999 Mar 10
1
lty=2
On Wed, Mar 10, 1999 at 03:14:08PM +0000, Simon Bond wrote: > Dear all, > > I'm using R 63.2 on windows NT, when I use > > > lines(x,y, lty=2) > > it produces a dashed line between the first pair of points and then reverts > back to a solid line. It produces different colours perfectly ok, but it's > not really a solution when the plot needs to be printed
2011 Sep 26
3
survival analysis: interval censored data
hello: my data looks like: time1  time2   event  catagoria 2004    2006        1            C 2004    2005        0            C 2005    2010        1            E 2007    2009        1            C 2006    2007        0            E 2008    2010        0            C 2008    2010        1            E ... and the census interval is 1 year I have tried  this
2012 Sep 14
2
when to use "I", "as is" caret
Dear community, I've check it while working, but just to reassure myself. Let's say we have 2 models: model1 <- lm(vdep ~ log(v1) + v2 + v3 + I(v4^2) , data = mydata) model2 <- lm(vdep ~ log(v1) + v2 + v3 + v4^2, data = mydata) So in model1 you really square v4; and in model2, v4*^2 *doesn't do anything, does it? Model2 could be rewritten: model2b <- lm(vdep ~
2011 Dec 24
1
Nested model - "singularities not defined"
I am using a nested model in R and the lm output shows 47 not defined because of singularities and I have no idea why. Any help on why this is happening or how to fix this problem would be very much appreciated. Below is the output I received from R. Thanks and happy holidays! Call: lm(formula = Dist ~ Treatment/SiteL/Territory) Residuals: Min 1Q Median 3Q Max -6.646 -1.443
2011 Oct 06
1
Coefficients for lagged plm model variables not calculated
Hello, So I am afraid I am having a recurring problem that I just can't figure out. I am using the plm package to conduct a panel analysis - although I am not sure if the problem is arising as a result of the plm package or something more general. I am trying to run a fixed effects model with effects over time and individual. The model has various lags, and the problem is that these lags do
2000 Oct 03
3
prcomp compared to SPAD
Hi ! I've used the example given in the documentation for the prcomp function both in R and SPAD to compare the results obtained. Surprisingly, I do not obtain the same results for the coordinates of the principal composantes with these two softwares. using USArrests data I obtain with R : > summary(prcomp(USArrests)) Importance of components: PC1 PC2
2012 Jun 07
0
how lm behaves
I was wondering if somebody could explain why I get different results here: >treats[,2]<-as.factor(treats[,2]) >treats[,5]<-as.factor(treats[,5]) >treats[,4]<-as.factor(treats[,4]) #there are 'c' on more days than I have 'h2o2', where treats[,4] is the day. I only want 'c' that correspond to the same days that I have a 'h2o2' also.
2007 May 10
1
Follow-up about ordinal logit with mixtures: how about 'continuation ratio' strategy?
This is a follow up to the message I posted 3 days ago about how to estimate mixed ordinal logit models. I hope you don't mind that I am just pasting in the code and comments from an R file for your feedback. Actual estimates are at the end of the post. ### Subject: mixed ordinal logit via "augmented" data setup. ### I've been interested in estimating an ordinal logit model