search for: m1b

Dear R experts. I might be missing something obvious. I have been trying to fix this problem for some weeks. Please help. #data ped <- c(rep(1, 4), rep(2, 3), 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 &lt...

Hello! If I have a matrix as 1 2 2 3 and I want to change the value 2 in 0, what can I do? Thank you -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !)

...rry I had typo in previous email, this typo corrected 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', s...

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

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)

...[x[[i]],]) I want to know if yrs (a continuous variable) has a significant unique effect in the model, so I fit a simplified model with the main effect ommitted... m2<-glm(count~siteall+yrs:district,family=quasipoisson,weights=weight,data=m[x[[i]],]) then compare models using anova() anova(m1,m1b,test="F") Analysis of Deviance Table Model 1: count ~ siteall + yrs + yrs:district Model 2: count ~ siteall + yrs:district Resid. Df Resid. Dev Df Deviance F Pr(>F) 1 1936 75913 2 1936 75913 0 0 > The d.f.'s a...

...e dummy to be equal across equations # Estimate the model with `sysGmm` using different weighting matrices: identity, "optimal" and manually specified m1a <- sysGmm(g=ES_g, h=ES_h, wmatrix="ident" , weightsMatrix=NULL, vcov=vc1, crossEquConst=cec1, data=data1); summary(m1a) m1b <- sysGmm(g=ES_g, h=ES_h, wmatrix="optimal", weightsMatrix=NULL, vcov=vc1, crossEquConst=cec1, data=data1); summary(m1b) m1c <- sysGmm(g=ES_g, h=ES_h, weightsMatrix=Wmat, vcov=vc1, crossEquConst=cec1, data=data1); summary(m1c) ------------------------------ R ses...

...uations > > # Estimate the model with `sysGmm` using different weighting matrices: > identity, "optimal" and manually specified > m1a <- sysGmm(g=ES_g, h=ES_h, wmatrix="ident" , weightsMatrix=NULL, > vcov=vc1, crossEquConst=cec1, data=data1); summary(m1a) > m1b <- sysGmm(g=ES_g, h=ES_h, wmatrix="optimal", weightsMatrix=NULL, > vcov=vc1, crossEquConst=cec1, data=data1); summary(m1b) > m1c <- sysGmm(g=ES_g, h=ES_h, weightsMatrix=Wmat, > vcov=vc1, crossEquConst=cec1, data=data1); summary(m1c) > > ------------...

...a series that has already been differenced and not include an order for differencing in the function, or 2) include an xreg 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 correlatio...

...g code: > require(biglm) > options(digits=6, scipen=3, contrasts = c("contr.treatment", "contr.poly")) > dat=data.frame(y =c(rpois(50000, 10),rpois(50000, 15)), ttment=gl(2,50000)) > m1 <- glm(y~ttment, data=dat, family=poisson(link="log")) > m1big <- bigglm(y~ttment , data=dat, family=poisson(link="log")) > summary(m1) <snipped output for this email> Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 2.30305 0.00141 1629 <2e-16 *** ttment2 0.40429 0.00183 221 &lt...

...E) > m1a<-lmer(cbind(prevented,control.count)~repellant+hour+(1|volunteer)+(1|date), + family=binomial(link='probit'), method='Laplace') Error in dev.resids(y, mu, weights) : argument wt must be a numeric vector of length 1 or length 219 > # probit doesn't converge > m1b<-lmer(cbind(prevented,control.count)~repellant+hour+(1|volunteer)+(1|date), + family=binomial, method='Laplace') Error in dev.resids(y, mu, weights) : argument wt must be a numeric vector of length 1 or length 219 > # logit is overdispersed > m1<-lmer(cbind(prevented,control.cou...

...nderstand why Safari displays a raise action this way RuntimeError in Admin contactsController#set_contact_postcode {&quot;action&quot;=&gt;&quot;set_contact_postcode&quot;, &quot;id&quot;=&gt;&quot;2&quot;, &quot;value&quot;=&gt;&quot;M1B 5Y7&quot;, &quot;controller&quot;=&gt;&quot;admin_contacts&quot;} Why will it not display &quot; as " or &gt; as > ?? --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Ruby...

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