On 16-Nov-09 11:40:29, Petr PIKAL wrote: > Hi > AFAIK, this is issue of the preference of operators. > > r-help-boun...@r-project.org napsal dne 16.11.2009 11:24:59: >
Not in this case (see below), though of course in general "-" takes precedence over "^", so, for example, in the expression -2^(1/3) the "-" is applied first, giving (-2); and then "^" is applied next, giving (-2)^(1/3). There is a work-round (see below). >> Hi, >> I want to apply ^ operator to a vector but it is applied to >> some of the elements correctly and to some others, it generates >> NaN. Why is it not able to calculate -6.108576e-05^(1/3) even >> though it exists? It only exists (in the real domain) if "^" takes precedence over "-" which (in R) it does not! >> tmp >> [1] -6.108576e-05 4.208762e-05 3.547092e-05 7.171101e-04 > -1.600269e-03 >> > tmp^(1/3) >> [1] NaN 0.03478442 0.03285672 0.08950802 NaN > > This computes (-a)^(1/3) which is not possible in real numbers. In this example, that is not accurate. "tmp" has already been defined, and contains numbers which are already stored as negative numbers, so "-" is no longer on the scene as an operator, before "^" is applied; the issue of precedence of "-" over "^" is no longer present. The NaN arises from x^(1/3) where x is negative. > You have to use as.complex(tmp)^(1/3) to get a result. > >> > -6.108576e-05^(1/3) >> [1] -0.03938341 This is not the result I get: as.complex(tmp)^(1/3) # [1] 0.01969171+0.03410703i 0.03478442+0.00000000i # [3] 0.03285672+0.00000000i 0.08950802+0.00000000i # [5] 0.05848363+0.10129662i > this is actually > -(6.108576e-05^(1/3)) > > Regards > Petr It is possible to work round the problem without using as.complex which can introduce complications -- see above, and also: x <- (-1) x^(1/3) # [1] NaN as.complex(x)^(1/3) # [1] 0.5+0.8660254i as.complex(-1)^(1/2) # [1] 0+1i which you would not want if you are working throughout in real numbers (you would want the result "-1" instead). Although, in the mathematics of complex numbers, (-1)^(1/3) has three values, one of which is -1, R only returns a single value. However, you would have to define a new operator, called say "%^%": "%^%"<-function(X,x){sign(X)*(abs(X)^x)} tmp <- c(-6.108576e-05, 4.208762e-05, 3.547092e-05, 7.171101e-04, -1.600269e-03) tmp%^%(1/3) # [1] -0.03938341 0.03478442 0.03285672 0.08950802 -0.11696726 The definition of "%^%" forces "^" to take precedence over "-", by in effect removing "-" from the scene until "^" has done its work. But, if you hope to rely on this, note that if you apply to 'tmp' any function in which the ordinary "^" will be used on a negative number, you will still have the same problem. Note: Trying to redefine "^" will not work, since invoking the result initiates an infinite recursion: "^"<- function(X,x){sign(X)*(abs(X)^x)} ## (This definition will be accepted by R) tmp%^%(1/3) # Error: evaluation nested too deeply: infinite recursion / # options(expressions=)? It's not a clean situatio, but I hope the above helps! Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <ted.hard...@manchester.ac.uk> Fax-to-email: +44 (0)870 094 0861 Date: 16-Nov-09 Time: 12:55:25 ------------------------------ XFMail ------------------------------ ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.