search for: infs

We are trying to read a sas export file into R. (Works fine on Mac by ftping the export file from solaris box and then importing to R with read.xport, this gives rational numbers etc.) Trying to do this on an IBM Power p655 running linux, reading the same sas export file as used on MAC, in this case the results are not rational. Many of the numbers are interpreted as INF etc. Is it possible

#optim package estimate<-optim(init.par,Linn,hessian=TRUE, method=c("L-BFGS-B"),control = list(trace=1,abstol=0.001),lower=c(0,0,0,0,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf),upper=c(1,1,1,1,Inf,Inf,Inf,Inf,Inf,Inf,Inf,Inf,Inf)) #nlminb package estimate<-nlminb(init.par,Linn,gr=NULL,hessian=TRUE,control =

Hello, I have 2 functions (a and b) a = function(n) { matrix (runif(n*2,0.0,1), n) } > > > b = function (m, matrix) { > n=nrow (matrix) > p=ceiling (n/m) > lapply (1:p, function (l,n,m) { > inf = ((l-1)*m)+1 > if (l<p) sup=((l-1)*m)+m > else sup=n >

Hi, I need fit the France model : y = A{1 - exp[-b(t-T) - c(sqrt(t) - sqrt(T))]} parameters: A, b, T, c variable: t (time) resp: y I tried: time = 1:48 resp = rnorm(48, 200, 10) dados = data.frame(resp, time) attach(dados) f = function(x, A, b, T, c) A*(1-exp(-b*(x-T) - c*(sqrt(x) - sqrt(T)))) (mod1 = nls(resp~f(time, A, b, c, T), data=dados, start=c(A=500, b=152, c=100,

Instead of trying to split and parse elements from virtio-win paths, use the '*.inf' files supplied with the drivers to control how Windows drivers are installed. The following emails best explain how this works: https://www.redhat.com/archives/libguestfs/2015-October/msg00352.html https://www.redhat.com/archives/libguestfs/2015-November/msg00065.html Currently the product variant (eg.

Hi, I noticed a minor flaw in integrate() from package stats: Taking up arguments lower and upper from integrate(), if (lower == Inf) && (upper == Inf) or if (lower == -Inf) && (upper == -Inf) integrate() calculates the value for (lower==-Inf) && (upper==Inf). Rather, it should return 0. Quick fix: ### old code ### ### [snip] else {

I find this a bit puzzling ... > solve(matrix(c(5, 2, 3, 1), 2, 2), c(Inf, 3)) [1] NaN Inf > solve(matrix(c(5, 2, 3, 1), 2, 2)) %*% c(Inf, 3) [,1] [1,] -Inf [2,] Inf I would expect the answer to be c(-Inf, Inf), so why has the -Inf been replaced by NaN in solve? Cheers, Jonathan. --please do not edit the information below-- Version: platform = sparc-sun-solaris2.7 arch = sparc

The integral of any probability density from -Inf to Inf should equal 1, correct? I don't understand last result below. > integrate(function(x) dnorm(x, 0,1), -Inf, Inf) 1 with absolute error < 9.4e-05 > integrate(function(x) dnorm(x, 100,10), -Inf, Inf) 1 with absolute error < 0.00012 > integrate(function(x) dnorm(x, 500,50), -Inf, Inf) 8.410947e-11 with absolute error <

https://github.com/rwmjones/libguestfs/tree/rewrite-virtio-copy-drivers Instead of trying to split and parse elements from virtio-win paths, use the '*.inf' files supplied with the drivers to control how Windows drivers are installed. The following emails best explain how this works: https://www.redhat.com/archives/libguestfs/2015-October/msg00352.html

Hi, I think I've discovered a bug in base R. Basically, when using 'Inf' as as 'Date', is is visually displayed as 'NA', but R still treats it as 'Inf'. So it is very confusing to work with, and can easily lead to errors: # Visually displays as NA > as.Date(Inf, origin="2018-01-01") [1] NA # Visually displays as NA > str(as.Date(Inf,

...2 2 1 [2,] Inf 0 1 1 1 2 [3,] Inf Inf 0 1 2 1 [4,] Inf Inf Inf 0 1 2 [5,] Inf Inf Inf Inf 0 1 [6,] Inf Inf Inf Inf Inf 0 # and here is the error message: Error in isoMDS(dist) : an initial configuration must be supplied with NA/Infs in 'd' It appears that isoMDS doesn't like the "Inf" values in the dist matrix, although help(isoMDS) suggests it accepts them if proper initial configuration is provided: Arguments d distance structure of the form returned by dist, or a full, symmetric matrix. Data are assu...

I am working on Johansen cointegration test, using urca and var package. in the selection of var, I have got following results. >VARselect(newd, lag.max = 10,type = "none") $selection AIC(n) HQ(n) SC(n) FPE(n) 6 6 6 5 $criteria 1 2 3 4 5 6 7 8 9 AIC(n) -3.818646e+01 -3.864064e+01

Hi Gabriel, The point is that it *visually* displays as NA, but is.na() still responds as FALSE. When I (and I am sure many people) see an NA, we then use is.na(). If we see Inf displayed, we then use is.infinite(). With as.Date() this breaks down. I'm not arguing that as.Date(Inf) should be coerced to NA. I'm arguing that as.Date(Inf) should be *visually* displayed as Inf (i.e. the

Hi If I have a <- Inf + 1i then Re(a) is Inf, and Im(a) is 1, as expected. But if b <- 1 + Inf * 1i, then Im(b) = Inf , as expected, but Re(b) = NaN, which I didn't expect. Why this asymmetry? How to define an object with Re(b)=1, Im(b)=Inf? -- Robin Hankin Uncertainty Analyst Southampton Oceanography Centre European Way, Southampton SO14 3ZH, UK tel 023-8059-7743

Patch 1 moves the v2v/fake-virtio-win and v2v/fake-virt-tools directories to the recently created test-data/ hierarchy. This is just refactoring with no functional change at all. Patches 2-4 then extend the available (fake) virtio-win drivers: - Patch 2 adds all of the drivers from the virtio-win RPM. - Patch 3 adds all of the drivers from the virtio-win ISO (which are different from the

Hi, I am trying to produce an xyplot with a regression line. The data should be represented as log/log but when I fit the lmline I receive an error message - the plot is fine without the log transformation, but the then the plot is meaningless. I know it must be something simple, but I just can't see it. Hope someone can help... Thanks. xyplot(log(Pk)~log(k),data=rwpk,cex=1,

On Tue, Mar 5, 2019 at 9:54 PM Richard White <w at rwhite.no> wrote: > Hi Gabriel, > > The point is that it *visually* displays as NA, but is.na() still > responds as FALSE. > > When I (and I am sure many people) see an NA, we then use is.na(). If we > see Inf displayed, we then use is.infinite(). With as.Date() this breaks > down. > > I'm not arguing that

Hi Sebastian, It boils down to this: The previous heuristic solver could return infinite cost solutions in some rare cases (despite finite-cost solutions existing). The new solver is still heuristic, but it should always return a finite cost solution if one exists. It does this by avoiding early reduction of infinite spill cost nodes via R1 or R2. To illustrate why the early reductions can be a

Regarding patches 5-7, it seems like the mem-map code flow could be more shared. It is a bit challenging to unravel things though. I guess the only specific thing I can really point out is that PcdPciAllowFullEnumeration should be initialized in a different patch, and not within the mem-map init path. -Jordan On Tue, Nov 19, 2013 at 12:38 PM, Wei Liu <wei.liu2@citrix.com> wrote: > This

I'd like to be able to modify axlab in (C) below so that 'Inf' is replaced by the infinity symbol. y <- rnorm(40) breaks <- c(-Inf, -1, 1, Inf) x <- cut(y, breaks=breaks) plot(unclass(x), y, xaxt="n", xlab="") ## A: The following gives the axis labels "(-Inf, 1]", etc. axis(1, at=1:3, labels=expression("(-Inf,-1]", "(-1,1]",