Re: [R] 2 x 3 Probability under the null
On 27-Oct-11 04:09:46, Jim Silverton wrote: I have a 2 x 3 matrix called snp and I want to compute the following probability: choose(sum(snp[,1]), snp[1,1]) * choose(sum(snp[,2]), snp[1,2]) * choose(sum(snp[,3]), snp[1,3])/choose(sum(snp), sum(snp[1,])) but I keep getting Infs and NaNs. Is there a function that can do this in R? -- Thanks, Jim. Since 1/0 -- Inf, and 0/0 -- NaN, it seems likely that your data lead to zero denominators. However, true diagnosis needs to see what your data really are. What is a typical 2x3 matrix that gives such results? If this guess is correct, then no possible function in R can resolve the problem! For data where the problem does not arise, then you can of course write your own function to implement your code above for an arbitrary 2x3 matrix. Hoping this helps, Ted. E-Mail: (Ted Harding) ted.hard...@wlandres.net Fax-to-email: +44 (0)870 094 0861 Date: 27-Oct-11 Time: 07:32:30 -- 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.
Re: [R] 2 x 3 Probability under the null
On 11-10-27 12:09 AM, Jim Silverton wrote: I have a 2 x 3 matrix called snp and I want to compute the following probability: choose(sum(snp[,1]), snp[1,1]) * choose(sum(snp[,2]), snp[1,2]) * choose(sum(snp[,3]), snp[1,3])/choose(sum(snp), sum(snp[1,])) but I keep getting Infs and NaNs. Is there a function that can do this in R? Work in the log scale. You're probably getting overflows. lchoose() calculates the log of choose(). Duncan Murdoch __ 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.
Re: [R] 2 x 3 Probability under the null
I have a 2 x 3 matrix called snp and I want to compute the following probability: choose(sum(snp[,1]), snp[1,1]) * choose(sum(snp[,2]), snp[1,2]) * choose(sum(snp[,3]), snp[1,3])/choose(sum(snp), sum(snp[1,])) but I keep getting Infs and NaNs. Is there a function that can do this in R? -- Thanks, Jim. [[alternative HTML version deleted]] __ 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.
