search for: lamda

...n Ch8 of Pinheiro & Bates. I am getting the following error (R version 2.1.1) *Error in qr(attr(rhs, "gradient")) : NA/NaN/Inf in foreign function call (arg 1)* Below is the function and data. /# the van genuchten moisture release function vanGen <- function(x, Vr, Vm, alpha, lamda) { if (Vr < 0) Vr <- 0 Vr + (Vm - Vr)/((1+(alpha*x)^lamda)^(1-1/lamda)) } vanGen <- deriv(~Vr + (Vm - Vr)/((1+(alpha*x)^lamda)^(1-1/lamda)), c("Vr", "Vm", "alpha", "lamda"), function(x, Vr, Vm, alpha, lamda) {} )/ the call in R /> fm1fl...

Hi r-users;   I have this values: eign_val <- c(137.810447,3.538721,2.995161,1.685670) alp    <- 1.6549 ;  lamda <- eign_val lamda_m <- min(lamda)   First I calculated manually: delta0 <- 1 delta1 <- alp*delta0*(4-lamda_m*(1/lamda[1]+1/lamda[2]+1/lamda[3]+1/lamda[4]))  delta1 delta2 <- (alp/2)*(delta1*(delta1/alp) + delta0*((1-lamda_m/lamda[1])^2+ (1-lamda_m/lamda[2])^2+(1-lamda_m/lamda[3])^2+...

...theoretical interpretation of it. I'm sending the code : ######################################################################################### f3<-function(t,r){ #calculation for t1 fb1<-function(t,r){ v1<-numeric(0) for(j in 1:r){ int1<-function(x1){ int_1<- (1/p+log(x1/lamda))^2 * j * choose(n,j) * (pweibull(x1,shape=p,scale=lamda))^(j-1) * (1 - pweibull(x1,shape=p,scale=lamda))^(n-j) * dweibull(x1,shape=p,scale=lamda) int_1 } v1[j]<-integrate(int1,lower=0,upper=t)$value } sum(v1) } #calculation for t2 fb2<-function(t,r){ v2<-numeric(0) for(j in 1:r){ int2&...

...a){x = 1:396s = sample(x,6,replace=T)bsdata = data[(s[1]):(s[1]+59),] for (j in 2:6) { a = data[(s[j]):(s[j]+59),] bsdata = rbind(bsdata,a) }return(bsdata)} #set.seed(1234)#trial<-GetBSData(OriData) ##############################the Minimisation functionOpt<-function(data, horizon, col, lamda){TbillReturn<-numeric(30/horizon)USReturn<-numeric(30/horizon)for (x in 1: (30/horizon)){ TbillReturn[x]<-prod(data[(12*horizon*(x-1)+1):(12*horizon*(x-1)+12*horizon),col])-1 USReturn[x]<-prod(data[(12*horizon*(x-1)+1):(12*horizon*(x-1)+12*horizon),2])-1}Return<-cbind(TbillReturn,USR...

Dear All, In Storey's estimator of the proportion of true nulls, the estimator depends on the tuning parameter lamda. Suppose now that an estimator of this proportion has been obtained by the qvalue package, what is the lamda that corresponds to the estimate? How to get this lamda? Thanks, -Chee [[alternative HTML version deleted]]

...perator is invalid for atomic vectors" I'll just provide the code, because I don't see where the problem is: library(lme4) options(contrasts=c("contr.helmert","contr.poly")) data = read.csv("C:/Users/Angela/Desktop/Summer_at_CDS/aging_JDM_Martine/datasets/lamda.data_07.27.09.csv", header=TRUE) attach(data) new<-data.frame(lamda, age.o_1, Item.l2_2,Subject) new Subject <- factor(Subject) Age <- factor(age.o_1,levels=c("0","1"),labels=c("young","old")) Item <- factor(Item.l2_2,levels=c("1&quot...

Hi,all: I met a problem of nls. My data: x y 60 0.8 80 6.5 100 20.5 120 45.9 I want to fit exp curve of data. My code: > nls(y ~ exp(a + b*x)+d,start=list(a=0,b=0,d=1)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates I can't find out the reason for the error. Any suggesions are welcome. Many thanks. [[alternative HTML

...l I am so glad the R can provide the efficient calculate about eigenvector and eigenvalue. However, i have some puzzle about the procedure of eigen. Fristly, what kind of procedue does the R utilize such that the eigen are obtained? For example, A=matrix(c(1,2,4,3),2,2) we can define the eigenvalue lamda, such as det | 1-lamda 4 | =0 | 2 3-lamda | then we can obtain the lamda=5 and -1 however, i am interesting in that how does the R to obatin lamda? By the way, how does the R also obtain the eigenvector? Basically, i must to know those definition in the R becau...

...with jumps. I have create a vector of jumps and then added this into my log-likelihood equation, know I am getting a message: NA in the initial gradient My codes is hear # n<-length(combinedlr) j<-c(1,2,3,4,5,6,7,8,9,10) gbmploglik<-function(param){ mu<-param[1] sigma<-param[2] lamda<-param[3] nu<-param[4] gama<-param[5] logLikVal<- - n*lamda - .5*n*log(2*pi) + sum(log(sum(for(j in 1:10)(cat((lamda^j/factorial(j))*(1/((sigma^2+j*gama^2)^.5)*exp( - (combinedlr-mu-j*nu)^2/2*(sigma^2+j*gama^2)))))))) logLikVal } rescbj<- maxLik(gbmploglik, grad = NULL, hess = NU...

...document frequency normalization H2(2) (as proposed by Amati in his PHD thesis). The formula for w(t,d) in this scheme is given by::- w(t,d) = wqf * L * P where wqf = within query frequency L = Laplace law of after effect sampling =1 / (wdfn + 1) P = wdfn * log (wdfn / lamda) + (lamda - wdfn) log(e) + 0.5 * log (2 * pi * wdfn) wdfn = wdf * (1+c * log(average length of document in database / length of document d )) (H2 Normalization ) lamda = mean of the Poisson distrubution = Collection frequency of the term / Size of the database and the b...

Hi list, I have a group of data. It looks like they follow a exponential distribution. In R, how can I esimate lamda, that is the rate in pexp, of the distribution and can I use Kolmogorov-Smirnov for hypothesis testing in such a situation? I have read the "8.2 Examing the distribution of a set of data" of "An Introduction to R" but I did not find any clues on this issue. (The data is in the...

...ry to create a vector > #GBMPJ MLE Combined Ph 1 LR > # > n<-length(combinedlrph1) > j<-c(1,2,3,4,5,6,7,8,9,10) Error in c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) : unused argument(s) (3, 4, 5, 6, 7, 8, 9, 10) > gbmploglik<-function(param){ + mu<-param[1] + sigma<-param[2] + lamda<-param[3] + nu<-param[4] + gama<-param[5] + logLikVal<- - n*lamda - .5*n*log(2*pi) + sum((for(i in 1:n)(log(sum(for(j in 1:10)(((lamda^j/factorial(j))*(1/((sigma^2+j*gama^2)^.5)*exp( - (combinedlrph1-mu-j*nu)^2/2*(sigma^2+j*gama^2)))))))))) + logLikVal + } > rescbj<- maxLik(gbmplo...

Dear R-users, I'm wondering how to obtain Wilks-lambda values when discriminant analyses have only one discriminant function (i.e. 2 categories to discriminate between). The use of manova(predictions~groups, test="Wilks") asks for multiple response and the use of anova(lm(predictions~groups), test="Wilks") simply does not consider the last term. (The alternative

Hi - I am trying to dynamically create my partition folders via attachment_fu. I have read about a hack that allows you to do this on the attachment model: :partition => lambda {|a| a.article_id} Which passes the model''s object into the lambda and uses the ''article_id'' attribute as its directory name. However, this isn''t work for me. I think the lambda

Hi, I need to generate correlated (positive as well as negative) bivariate exponential distribution with rate of 1/5 or any rate I need some guidance here. Please help. Regards, Sunny

...p <- log(z[i]) + q <- log(1 - z[j]) + + r <-(2 * i - 1) * (p + q) + #cat(p+ q, r, '\n') + k <- k + r + #p <- p * (2 * i - 1) + #cat(p, '\n') + } + + k <- - (k / len.gen) + k <- k - len.gen + + k <- (1 + 0.6/len.gen) * k + + return(list(lamda=mean.gen, ad=k)) + #cat(k, '\n') + } > > # The Exponential data from Law & Kelton "Simulation Modeling & Analysis" > # second edition, table 6.7, page 367 (with lamda 0.399) > > kelton <- c(rep(0.01, 8), rep(0.02, 2), rep(0.03, 3), rep(0.04, 6), + rep...

...Xjh bar Xjhbar=list() for (i in 1:q){ w=matrix(0,R[i],p) for (j in 1:R[i]){ w[j,]=colSums(W[[i]][,j]*X)/n11[[i]][j] } Xjhbar[[i]]=w } #calcul des TO jh TO.jh=list() for (i in 1:q){ w=Xjhbar[[i]] to=w*0 for (j in 1:R[i]){ to[j,]=w[j,]-Xbar } TO.jh[[i]]=to } #calcul des Lamda J Lamda=matrix(0,p,p) for (i in 1:q){ to=TO.jh[[i]] w=matrix(0,p,p) for (j in 1:R[i]){ w=w+(P11[[i]][j]*(to[j,]%*%t(to[j,]))) } Lamda=Lamda+w } tr1=n*sum(diag(Lamda)) # Calcul de Gamma GGamma=matrix(0,p*sum(R),p*sum(R)) PGamma=kronecker(diag(P11[[1]]),diag(p)) Ifin=p*R[1] GGamma[...

...1:q){ 80 w=matrix(0,R[i],p) 81 for (j in 1:R[i]){ 82 w[j,]=colSums(W[[i]][,j]*X)/n11[[i]][j] 83 } 84 Xjhbar[[i]]=w 85 } 86 #calculation of TO jh 87 TO.jh=list() 88 for (i in 1:q){ 89 w=Xjhbar[[i]] 90 to=w*0 91 for (j in 1:R[i]){ 92 to[j,]=w[j,]-Xbar 93 } 94 TO.jh[[i]]=to 95 } 96 #calculation of Lamda J 97 Lamda=matrix(0,p,p) 98 for (i in 1:q){ 99 to=TO.jh[[i]] 100 w=matrix(0,p,p) 101 for (j in 1:R[i]){ 102 w=w+(P11[[i]][j]*(to[j,]%*%t(to[j,]))) 103 } 104 Lamda=Lamda+w 105 } 106 tr1=n*sum(diag(Lamda)) 107 # Gamma Calculation 108 GGamma=matrix(0,p*sum(R),p*sum(R)) 109 PGamma=kronecker(diag(P1...

...[[i]][j] > 83 } > 84 Xjhbar[[i]]=w > 85 } > > 86 #calculation of TO jh > > 87 TO.jh=list() > 88 for (i in 1:q){ > 89 w=Xjhbar[[i]] > 90 to=w*0 > 91 for (j in 1:R[i]){ > 92 to[j,]=w[j,]-Xbar > 93 } > 94 TO.jh[[i]]=to > 95 } > > 96 #calculation of Lamda J > > 97 Lamda=matrix(0,p,p) > 98 for (i in 1:q){ > 99 to=TO.jh[[i]] > 100 w=matrix(0,p,p) > 101 for (j in 1:R[i]){ > 102 w=w+(P11[[i]][j]*(to[j,]%*%t(to[j,]))) > 103 } > 104 Lamda=Lamda+w > 105 } > > 106 tr1=n*sum(diag(Lamda)) > > 107 # Gamma Calculatio...

Hi I?m trying to fit a nonlinear model to a derivative of the logistic function y = a/(1+exp((b-x)/c)) (this is the parametrization for the SSlogis function with nls) The derivative calculated with D function is: > logis<- expression(a/(1+exp((b-x)/c))) > D(logis, "x") a * (exp((b - x)/c) * (1/c))/(1 + exp((b - x)/c))^2 So I enter this expression in the nls function: