search for: denom

...om the NetBSD standard library. It works on ARM, Mips, Microblaze,ppc, ppc64, and x86_64. On i386 a very strange thing happens. Here's the source: #include <stdlib.h> #define __weak_alias(sym) __attribute__ ((weak, alias (#sym))) lldiv_t lldiv(long long int num, long long int denom) __weak_alias(_lldiv); lldiv_t _lldiv(long long num, long long denom) { lldiv_t r; r.quot = num / denom; r.rem = num % denom; if (num >= 0 && r.rem < 0) { r.quot++; r.rem -= denom; } return (r); } I get...

Dear R-experts, I am running a large simulation exercise where the enough complicated integration is required. The integral is computed within a C-function called Denom by use of function qags from the gsl library. Here is a piece of R-code: denom<-try(.C("Denom",as.double(x),as.integer(n), as.integer(p), as.double(param), as.double(delta),res=as.double(results))) denomres=if (class(denom)=="try-error") NA else de...

...0.00 0.00 0.00 > iVec2 <- index(Vec2, verbose=TRUE) finalized tree user.self sys.self elapsed sort 7.78 0 7.79 tree 0.00 0 0.00 > > timefactor(any(Vec1=="a999"), any(iVec1=="a999"), 10, 100) finalized tree nom denom factor user.self 0.041 0.0036 11.38889 sys.self 0.000 0.0000 NaN elapsed 0.041 0.0036 11.38889 > timefactor(any(Vec2=="a999"), any(iVec2=="a999"), 10, 100) nom denom factor user.self 0.040 0.0036 11.11111 sys.self 0.000 0.0000 NaN elapsed 0.04...

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...ry. It works on ARM, Mips, > Microblaze,ppc, ppc64, and x86_64. On i386 a very strange thing happens. > Here's the source: > > #include <stdlib.h> > #define       __weak_alias(sym) __attribute__ ((weak, alias (#sym))) > > lldiv_t lldiv(long long int num, long long int denom) __weak_alias(_lldiv); > > lldiv_t _lldiv(long long num, long long denom) > { >        lldiv_t r; >        r.quot = num / denom; >        r.rem = num % denom; >        if (num >= 0 && r.rem < 0) { >                r.quot++; >                r.rem -= denom; &...

...cteur a[j] where j: 1...8 >this is the code in R: >aj.fun <- function(j, i, X, z, E, beta0, beta1){ + n <- length(X) + iX <- order(X) + iz <- order(z) + e1 <- -(beta)*z[ iz[1:(i - 1)] ] + numer <- E[j] - sum( X[ iX[1:(i - 1)] ] * exp(e1) ) + e2 <- -(beta)*z[ iz[i:n] ] + denom <- sum( exp(e2) ) + numer/denom + } > iX<-order(X) > iX [1] 75 37 29 60 73 20 69 55 30 70 72 38 26 35 65 61 74 50 71 57 25 54 64 76 56 [26] 58 48 67 46 63 28 62 36 49 47 66 1 42 41 19 39 43 22 51 68 33 27 9 15 11 [51] 10 59 32 40 45 44 52 16 18 34 4 53 21 23 31 7 6 13 14 12 17 24...

Dear members, I am trying to use boot () to compute the distributions of a statistic of a data set. The statistic is defined in the following code: eds<-function(x) { r<-cor(x) paren<-1-abs(r)/2 denom<-sum(sum(paren)+0.5) desvio<-sd(x) media<-mean(x) a<-desvio/media nom<-sum(a*(sum(paren)+0.5)) eds_abs<-nom/denom print(eds_abs) } I need to calculate the nonparametric bootstrap confidence in...

...*(xa-theta1[9])*(xb-theta1[10])/sqrt(theta1[9]*(1-theta1[9]))/sqrt(theta1[10]*(1-theta1[10])))/ (2*pi*theta[18]*theta[20]*sqrt(1-theta[21]^2))*exp(-2*(1-theta[21]^2))^(-1)*((xc-theta[17])^2/theta[18]^2+(xd-theta[19])^2/theta[20]^2-2*theta[21]^2*(xc-theta[17])*(xd-theta[19])/(theta[18]*theta[20])) denom=nume1+nume2+nume3+nume4+nume5 Ep1=nume1/denom Ep2=nume2/denom Ep3=nume3/denom Ep4=nume4/denom Ep5=nume5/denom elogld= sum(Ep1*(-log(2*pi*theta[2]*theta[4]*sqrt(1-theta[21]^2))-(2*(1-theta[21]^2))^(-1)*((xc-theta[1])^2/theta[2]^2+(xd-theta[3])^2/theta[4]^2-2*theta[21]^2*(xc-theta[1])*(xd-theta[3]...

Dear all, I have discovered that there are many things that I used to do in my GIS which are easily done directly in R, for example calculating interpoint distances using geoR and pick out points inside a polygon using splancs. I now wonder, is there a function to create a line object like a watercourse and then calculate the distances between many points in space and this line? I couldn't

...ate(func1, lower = 0, upper = Inf,x)$val} T <- function(x){ 0.3*dnorm(x)/(0.3*dnorm(x)+0.7*m1star(x))} func2 <- function(x,c){(T(x) <=c)*0.3*dnorm(x)} func3 <- function(x,c){(T(x) <= c)*(0.3*dnorm(x)+0.7*m1star(x))} numer <- function(c){ integrate(func2, -Inf, Inf, c)$val} denom <- function(c){ integrate(func3, lower, Inf,c)$val} The error message is as below : > numer(0.5) Error in integrate(func1, lower = 0, upper = Inf, x) : the integral is probably divergent [[alternative HTML version deleted]]

...); double mu_c = asReal(mu); double mu2_c = 1.0 - mu_c; double error_c = asReal(err); double lowest_double = 1e-15; double tmp_c; double denom; double error_temp; double error_i_c; /* Make ready vector froms input */ PROTECT(q); n_prot++; PROTECT(v); n_prot++; PROTECT(psi); n_p...

...bs(alpha[5])/(1+abs(alpha[5])) # between 0 and lstar delta <- 0.01 + exp(alpha[6]) # greater than 0.01 ########################################## # THE THREE FUNCTIONS TO INTEGRATE # FOR COMPUTING THE LOGLIKELIHOOD ########################################## denom <- 2*pi*sigep*sigeta*(sqrt(1-rho2^2)) # A CONSTANT TO BE USED # FOR DEFINING THE # THREE FUNCTIONS f1 <- function(z1) { # FIRST FUNCTION b11 <- ((z1-muep)^2)/((-2)*(1-rho2^2)*(sigep^2...

Hi All, My dependent variable is a ratio that takes a value of 0 (zero) for 95% of the observations and positive non-integer values for the other 5%. What model would be appropriate? I'm thinking of fitting a GLM with a Poisson ~. Now, becuase it takes non-integer values, using the glm function with Poisson family issues warning messages. Warning messages: 1: In dpois(y, mu, log = TRUE) :

...t;? My R code and the results are as follows. Thank you very much. :) > oneway.test(BackCalac~factor(Assay), var.equal=T,data=Control) One-way analysis of means data: BackCalac and factor(Assay) F = 56.2191, num df = 4, denom df = 20, p-value = 1.345e-10 > oneway.test(BackCalac~factor(Assay), var.equal=F,data=Control) One-way analysis of means (not assuming equal variances) data: BackCalac and factor(Assay) F = 92.8834, num df = 4.000, denom df = 9.625, p-value = 1.165e-07 -- View this message in context:...

I wonder if anyone out there has written a routine to solve the simple linear calibration problem? - fit regression of y vs x - estimate the value x0 (with 95% CI) that gives y0 Thanks for any help. Bill -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info",

...t;-rep(NA,asim) ssd1<-rep(NA,asim) ssd2<-rep(NA,asim) madN1<-rep(NA,asim) madN2<-rep(NA,asim) lo.w<-rep(NA,asim) up.w<-rep(NA,asim) lo.x<-rep(NA,asim) up.x<-rep(NA,asim) hw<-rep(NA,asim) hx<-rep(NA,asim) xbar.t<-rep(NA,asim) num<-rep(NA,asim) df2<-rep(NA,asim) denom<-rep(NA,asim) F<-rep(NA,asim) asim<-1000 for(j in 1:asim) { print(j) set.seed(j) a<-rnorm(15,0,1) b<-rnorm(15,0,1) w<-a*exp(h*a^2/2) x<-b*exp(h*b^2/2) matw<-sort(w) matx<-sort(x) trim1[j]=1/((1-2*alpha)*n)*(sum(matw[i:m])+r*(matw[k]+matw[n-k+1])) trim2[j]=1/((1-2*alp...

I have a matrix of frequency counts from 0-160. x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)) I would like to apply a function creating a new column (x[,2])containing values equal to: a) log(x[m,1]) if x[m,1] > 0; and b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero values +1) for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472 whereas x[3,2] should

...the Experimenter factor, I get a sequential anova table where factor levels are fit one at a time, ie, > anova(lmer(Response ~ Weight + Experimenter + + (1 | SUBJECT.NAME) + (1 | Date.StudyDay), + data=data.df)) Analysis of Variance Table Df Sum Sq Mean Sq Denom F value Pr(>F) Weight 1 10.8 10.8 4242.0 14.4457 0.0001463 *** Experimentercarmena 1 0.5 0.5 4242.0 0.6708 0.4128136 Experimentercnunez 1 4.8 4.8 4242.0 6.5042 0.0107969 * Experimenterlsolberg 1 0.4 0.4 4242.0 0.5910 0.4420783 Experimenter...

...test item i in class j, c is the proportion correct on the test by student k, and q is the number of items on the test in total. I have programmed this using Newton-Raphson as follows: numer <- function(R,a,c,q){ result <- sum((1-(1-R)^a)^(1/a))-c*q result } denom <- function(R,a,c,q){ result <- sum(-((1 - (1 - R)^a)^(1/a) * (log((1 - (1 - R)^a)) * (1/a^2)) + (1 - (1 - R)^a)^((1/a) - 1) * ((1/a) * ((1 - R)^a * log((1 - R)))))) result } aConst <- function(R, c, q, con){ a <- .5 # starting value for a...

...slow",data=d2) I'm expecting a value of about 0.63 to 0.64 based on the data source (simulated) and the following hand-coded MLE: negloglik = function(beta,dat) { dat$wexb = dat$wt * exp(dat$x * beta) agged = aggregate(dat$wexb,list(riskset=dat$riskset),sum) names(agged)[2] = "denom" dat = merge(dat[dat$cc==1,],agged,by="riskset") -sum(log(dat$wexb)-log(dat$denom)) } nlm(negloglik,0,hessian=T,dat=d2) Am I misunderstanding the meaning of case weights in the coxph function? The help file is pretty terse. Scott Bartell, PhD Assistant Professor Department o...