search for: tol

So i''ve inherited this solaris 10 system with a sun storedge attached. pool: tol-pool state: ONLINE scrub: none requested config: NAME STATE READ WRITE CKSUM tol-pool ONLINE 0 0 0 raidz1 ONLINE 0 0 0 c1t8d0 ONLINE 0 0 0 c1t9d0 ONLINE 0 0...

I have been told by the CRAN administrators that the following code generated an error on 64-bit Fedora Linux (gcc, clang) and on Solaris machines (sparc, x86), but runs well on all other systems): > fn <- function(x, y) ifelse(x^2 + y^2 <= 1, 1 - x^2 - y^2, 0) > tol <- 1.5e-8 > fy...

...n, under the condition that N1 % divides C1 and N2 divides C2 (optimal subgrids for an already existing % grid of size C1xC2 N) % gridsizes(N,C1,C2,1) returns [NaN;NaN] instead of an error message if the % fitting couldn't be found (used in the function N_from_d) sizes = NaN(2,1); s = sqrt(N); tol = 10?(-12); N2= ceil(s); N1 = N/N2; if(nargin 3) C1 = varargin{1}; C2 = varargin{2}; while(rem(C1,N1) > tol || rem(C2,N2) > tol || rem(N1,round(N1)) > tol) N2 = N2+1; N1 = N/N2; if(N2 > N) if(nargin 4 && varargin{3} == 1) 45 N1 = NaN; N2 = NaN; break; else error('Cannot fi...

...m Files\R\R-3.0.0\MD5, leads me to suspect there might be an error in the test itself. Thanks for the help! -------------- next part -------------- > .proctime00 <- proc.time() > > isall.equal <- function(x,y) + typeof(x) == typeof(y) && is.logical(r <- all.equal(x,y, tol=0)) && r > > report <- function(x) {print(x); stopifnot(x)} > options(error = expression(NULL)) # don't stop on error in batch > ##~~~~~~~~~~~~~~ > > ###-------- > x <- integer(0) > res <- try(as.array( x ), silent = TRUE) > if(!inherits(res, '...

...: > combined[[i]] <- performance(fit) With two iterations I get the following list, as output to a CSV: object.RMSE object.R2 object.RMSE.1object.R2.1 WA.inv 0.321543059 0.86388897 0.350494954 0.882600618 WA.cla 0.345947482 0.86388897 0.373078015 0.882600618 WA.inv.tol 0.308649941 0.874585589 0.336975196 0.89148291 WA.cla.tol 0.330038831 0.874585589 0.356895789 0.89148291 Obviously if I run thousands of iterations this continues on to the right, and the CSV files end up being too big for post-processing in most spreadsheet programs. Instead I want it to look l...

...o solve the problem This is the code I use to obtain bonpower using Sparse Matrix & alternatively the code is in the following website http://igraph.wikidot.com/r-recipes#toc6 *bonpow.sparse <- function(graph, nodes=V(graph), loops=FALSE, exponent=1, rescale=TRUE, tol=1e-07) {* * ## remove loops if requested * * ## sparse adjacency matrix d <- get.adjacency(graph, sparse=TRUE)* * ## sparse identity matrix id <- spMatrix(vcount(graph), vcount(graph), i=1:vcount(graph), j=1:vcount(graph), x=rep(1, vcount(graph)))...

I was looking for an R-package that would reduce matrices to row-echelon form, but Google was not my friend; any leads? If not, I wonder if the problem could be expressed in terms of constraint satisfaction...

Hi there, I was wondering if anyone could explain how you should set tol in the prcomp function. Using help(prcomp) explains it as "a value indicating the magnitude below which components should be omitted. (Components are omitted if their standard deviations are less than or equal to tol times the standard deviation of the first component.) With the default nul...

...show all three cubic terms. So my question: How to circumvent the singularity and estimate the coefficients for these terms in Scheffe's complete cubic model in the modeling of Chemical Blends? Thanks in advance for any help. Thank you so much Cleber Borges ####? MY_DATA: ???????? H2O TOL?????? MET????? Response 1? 1.0000000 0.0000000 0.0000000??? 0.39056923 2? 0.0000000 1.0000000 0.0000000?? -1.49611856 3? 0.0000000 0.0000000 1.0000000?? -0.33922649 4? 0.5000000 0.5000000 0.0000000?? -0.65517683 5? 0.5000000 0.0000000 0.5000000?? -1.23281158 6? 0.0000000 0.5000000 0.5000000??? 0.51...

...work well. However that is not the case if combines together, where certainly I made mistake somewhere when constructing the code. can someone help point out the mistakes for me please? Thanks! Aehan ____________________ intgfun<- function(a,b,c,mu,alpha,xi,upper, ...,subdivisions = 100, rel.tol = .Machine$double.eps^0.25, abs.tol = rel.tol, stop.on.error = TRUE, keep.xy = FALSE, aux = NULL) { x<-vector() fs<-function(x){ temp<-(-xi^(-1))*log(1-xi*(x-mu)/alpha) (((alpha^(-1))*(exp((-(1-xi)*temp)-exp(-temp))))*((a*(x-c)^0.5)+(b*(x-c)))) } return(fs) intgrt<-function(l...

Hi, I have a vector like: r <- runif(100) Now I would like to split r into 10 pieces (each with 10 elements) ? but the 'pieces' should be roughly similar with regard to mean and sd. what is an efficient way to do this in R? thanks!

...fail which ends with: > ## checking ar.yw.default() multivariate case > estd <- ar(unclass(y) , aic = FALSE, order.max = 2) ## Estimate VAR(2) > es.d <- ar(unclass(y.), aic = FALSE, order.max = 2, na.action=na.pass) > stopifnot(exprs = { + all.equal(est$ar[1,,], diag(0.8, 2), tol = 0.08)# seen 0.0038 + all.equal(est[1:6], es.[1:6], tol = 5e-3) + all.equal(estd$x.mean, es.d$x.mean, tol = 0.01) # seen 0.0023 + all.equal(estd[c(1:3,5:6)], + es.d[c(1:3,5:6)], tol = 1e-3)## seen {1,3,8}e-4 + all.equal(lapply(estd[1:6],unname), + lapply...

Two friend reported me a problem, which I can't solve: (I run R-1.0.0, Debian Linux) They hava a function "corr.matrix" (see end of mail), and when they create a 173x173 matrix with this function V <- corr.matrix(0.3, n=173) V1 <- qr.solve(V) reports: Error in qr(a, tol = tol) : NA/NaN/Inf in foreign function call (arg 1) For n < 173, qr.solve returns the correct result. Torsten ________________________________________________________________________ corr.matrix <- function(d, n=100) { mat <- NULL for(i in 1:n) { mat <- c(mat,((1-i):(n-i))...

Hello, I'm trying to run a lagsarlm (maximum likelihood estimation of a spatial lag model) in the spdep library ; but R gives following error message: Error in solve.default(inf, tol = tol.solve) : system is computationally singular: reciprocal condition number = 4.20137e-12 I get the same message when I try to run de lagsarlm with a bigger data set (4333 regions). The command I used: TestLag<-lagsarlm( mean_price ~ transcount+ C1_5_1+ C1_5_2+ C1_5_3+ C1_5_4, dat...

I was trying to find source for optimize and I ran across function (f, interval, ..., lower = min(interval), upper = max(interval), maximum = FALSE, tol = .Machine$double.eps^0.25) { if (maximum) { val <- .Internal(fmin(function(arg) -f(arg, ...), lower, upper, tol)) list(maximum = val, objective = f(val, ...)) } else { val <- .Internal(fmin(function(arg) f(arg, ...), lower, upper...

...0.0 2.8 0.0 0.0 0.0 1948 0.3 20.1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.5 0.5 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 ## To extract all the positive values x1 <- day_data1[,1] x2 <- day_data1[,2] x3 <- day_data1[,3] x4 <- day_data1[,4] x5 <- day_data1[,5] tol <- 1E-6 a1 <- x1[x1>tol] a2 <- x2[x2>tol] a3 <- x3[x3>tol] a4 <- x4[x4>tol] a5 <- x5[x5>tol] library(MASS) ## Example January Charleville 1943-2007 fitdistr(a1,dgamma, list(shape = 1, rate = 0.1), lower = 0.01) fitdistr(a2,dgamma, list(shape = 1, rate = 0.1), low...

...cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl- u1)+(1-pnorm(cu-u1)))+(0.2-p1)*(pnorm(cl-u2)+(1-pnorm(cu-u2))))} p1 <- 0.15 lam <- seq(0.01,0.99, by=0.001) x1 <- numeric(length(lam)) for (i in 1:length(lam)){ cl <- uniroot(f1, lower =-10, upper = 0, tol = 1e-10,p1=p1,lambda=lam[i])$root cu <- uniroot(f1, lower =0, upper = 10, tol = 1e-10,p1=p1,lambda=lam[i])$root x1[i]<- f2(p1=p1, cl=cl, cu=cu) } k <- 1 while(k<length(lam) && x1[k]<=0.1){ k=k+1 } k<-k-1;k lower <- uniroot(f1, lower =-...

I am fairly new to R. I find it surprising that f <- function(x,a) {x-a} uniroot(f, c(0,1), a=.5) works, but integrate(f, 0, 1, a=.5) gives an error: Error in integrate(f, 0, 1, a = 0.5) : argument 4 matches multiple formal arguments What is the best way of avoiding such surprises? Is there a way of telling integrate() that the 'a' argument is for f()? If I wrote my own function

Hi, I am looking for C/C++ codes for computing generalized inverse of a matrix. Can anyone help me in this regard? Thanks, Mahbub.

...7.19,7.19,7.19,7.19,6.79,6.79,7.19,6.79,7.19,7.19,1.407,7.19,4.42,9,1.64,1.64,6.79,1.664,1.664) > > library(MASS) > hubers(a) ## NO response! I think it is due to the infinite loop caused by the following line in the code of "hubers" (around Line 30): if ((abs(mu0 - mu1) < tol * s0) && abs(s0 - s1) < tol * s0) break where "s0" evaluates to ZERO initially (due to more than 50% of the number 7.19). I propose to change the "<" sign to "<=": if ((abs(mu0 - mu1) <= tol * s0) && abs(s0 - s1) <= tol...