similar to: Infinite != NaN?

On Fri, Mar 31, 2017 at 10:14 PM, Prof Brian Ripley <ripley at stats.ox.ac.uk> wrote: > From ?NA > > Numerical computations using ?NA? will normally result in ?NA?: a > possible exception is where ?NaN? is also involved, in which case > either might result. > > and ?NaN > > Computations involving ?NaN? will return ?NaN? or perhaps ?NA?: >

Full_Name: Bill Simpson Version: 0.64.1 OS: linux Submission from: (NULL) (193.62.250.209) Here is the data file: x y z 1 1 1 1 2 2 2 1 NaN 2 2 4 >data<-read.table("~/junk.dat",header=TRUE) > data x y z 1 1 1 1 2 1 2 2 3 2 1 NaN 4 2 2 4 > matrix(data$z,length(y),length(x)) [,1] [,2] [1,] 1 4 [2,] 2 3 This is not the correct matrix. It seems

Hi All, I have some data where I am doing fairly simple calculations, nothing more than adding, subtracting, multiplying and dividing. I’m running into a problem when I divide one variable by another and when they’re both 0 I get NaN. I realize that if you divide a non-zero by 0 then you get Inf, which is, of course, correct. But in my case I never get Inf, just NaN because of the structure

I calculated a large vector. Unfortunately, I have some measurement error in my data and some of the values in the vector are erroneous. I ended up wih some Infs and NaNs in the vector. I would like to filter out the Inf and NaN values and only keep the values in my vector that range from 1 to 20. Is there a way to filter out Infs and NaNs in R and end up with a clean vector? Mike

Hi, I want to recode all Inf and NaN values to NA, but I;m surprised to see the result of the following code. Could anybody enlighten me about this? > df <- data.frame(a=c(NA, NaN, Inf, 1:3)) > df[is.infinite(df) | is.nan(df)] <- NA > df a 1 NA 2 NaN 3 Inf 4 1 5 2 6 3 > Thanks! Cheers!! Albert-Jan

Hi, I want to recode all Inf and NaN values to NA, but I;m surprised to see the result of the following code. Could anybody enlighten me about this? > df <- data.frame(a=c(NA, NaN, Inf, 1:3)) > df[is.infinite(df) | is.nan(df)] <- NA > df a 1 NA 2 NaN 3 Inf 4 1 5 2 6 3 > Thanks! Cheers!! Albert-Jan

I stumbled across this and I am wondering if this is unexpected behavior or if I am missing something. > pnorm(-1.0e+307, log.p=TRUE) [1] -Inf > pnorm(-1.0e+308, log.p=TRUE) [1] NaN Warning message: In pnorm(q, mean, sd, lower.tail, log.p) : NaNs produced > pnorm(-1.0e+309, log.p=TRUE) [1] -Inf I don't know C and am not that skilled with R, so it would be hard for me to look into

I expect that atan(1i) = (0 + infinity i) and that atan(1i)/5 = (0 + infinity i)/5 = (0 + infinity i). Here's what I get in C: (0,1) = (0, 1) atan((0,1)) = (0, inf) atan((0,1))/5 = (0, inf) Note the difference between I*infinity = (0,1)*infinity = (0*infinity,1*infinity) = (NaN,infinity) and (0,infinity)/5 = (0/5,infinity/5) = (0,infinity). The former involves multiplying 0 by infinity, which

As a side note, Splus makes sin(x) NA, with a warning, for abs(x)>1.6*2^48 (about 4.51e+14) because more than half the digits are incorrect in sin(x) for such x. E.g., in R we get: > options(digits=16) > library(Rmpfr) > sin(4.6e14) [1] -0.792253849684354 > sin(mpfr(4.6e14, precBits=500)) 1 'mpfr' number of precision 500 bits [1]

This question is more out of curiosity than a complaint or suggestion, but I'm just wondering. The behavior of R on calculations that result in NaN seems a bit inconsistent. # this is expected: > 0/0 [1] NaN # but this gives a warning > sin(Inf) [1] NaN Warning message: In sin(Inf) : NaNs produced # and this again does not > exp(NaN) [1] NaN Conceptually, I like to think that R

Hi, recently I have started working with R (v. 2.7.2), and I have been using R's internal ARIMA_Like() function (from the "stats" package) to estimate some ARIMA models. In particular, I use ARIMA_Like() in a function "fn()" that I feed to the optim() method; the main goal is to find optimal ARIMA prediction models for some time series. The ARIMA_Like() function returns a

Hello, Currently, determinant(A) calculates the determinant of 'A' by factorizing A=LU and computing prod(diag(U)) [or the logarithm of the absolute value]. The factorization is done by LAPACK routine DGETRF, which gives a status code INFO, documented [1] as follows: *> INFO is INTEGER *> = 0: successful exit *> < 0: if INFO = -i, the i-th

On 2024-09-06 12:44 a.m., Richard O'Keefe wrote: > I expect that atan(1i) = (0 + infinity i) and that atan(1i)/5 = (0 + > infinity i)/5 = (0 + infinity i). > Here's what I get in C: > (0,1) = (0, 1) > atan((0,1)) = (0, inf) > atan((0,1))/5 = (0, inf) > > Note the difference between I*infinity = (0,1)*infinity = > (0*infinity,1*infinity) = (NaN,infinity) > and

Perhaps > Inf*1i [1] NaN+Infi clarifies why it is *not* a bug. (Boy, did that jog some long dusty math memories :-) ) -- Bert On Thu, Sep 5, 2024 at 2:48?PM Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > On 2024-09-05 4:23 p.m., Leo Mada via R-help wrote: > > Dear R Users, > > > > Is this desired behaviour? > > I presume it's a bug. > >

G.5.1 para 2 can be found in the C17 standard -- I actually have the final draft not the published standard. It's in earlier standards, I just didn't check earlier standards. Complex arithmetic was not in the first C standard (C89) but was in C99. The complex numbers do indeed form a field, and Z*W invokes an operation in that field when Z and W are both complex numbers. Z*R and R*Z,

On 05.03.2018 19:27, Sanjay Patel via llvm-dev wrote: > 3. fadd C, undef --> undef (where C is not NaN or Inf) > In the general constant case, the result could be anything as long as > constant operand C is not NaN or Inf. If C is the largest finite positive number, then (fadd C, X) cannot be a finite negative number. So doesn't folding (fadd C, undef) --> undef break the

Hi I am having a problem using varImpPlot in randomForest. I get the error message "Error in plot.window(xlim = xlim, ylim = ylim, log = "") : need finite 'xlim' values" When print $importance, several variables have NaN under %IncMSE. There are no NaNs in the original data. Can someone help me figure out what is happening here? Thanks! [[alternative HTML

I'm having trouble grokking complex NaN's. This first set examples using complex(re=NaN,im=NaN) give what I expect > Re(complex(re=NaN, im=NaN)) [1] NaN > Im(complex(re=NaN, im=NaN)) [1] NaN > Arg(complex(re=NaN, im=NaN)) [1] NaN > Mod(complex(re=NaN, im=NaN)) [1] NaN > abs(complex(re=NaN, im=NaN)) [1] NaN and so do the following > Re(complex(re=1,

Hi! In a current project, I am fitting loess models to subsets of data in order to use the loess predicitons for normalization (similar to what is done in many microarray analyses). While working on this I ran into a problem when I tried to predict from the loess models and the data contained NAs or NaNs. I tracked down the problem to the fact that predict.loess will not return a value at all

Dear list, I'm seeking some advice regarding a particular numerical integration I wish to perform. The integrand f takes two real arguments x and y and returns a vector of constant length N. The range of integration is [0, infty) for x and [a,b] (finite) for y. Since the integrand has values in R^N I did not find a built-in function to perform numerical quadrature, so I wrote my own after