On Sun, 17 Sep 2006 15:12:30 -0500, "Sebastian P. Luque" <[EMAIL PROTECTED]> wrote:
[...] > I thought about some very contrived ways to accomplish this, involving > 'which' and 'diff', but I sense a function might already be available to > do this efficiently. I think I found a better combination of those two functions than I what I was initially playing with: R> lenh <- hist(iris$Sepal.Length, br=seq(4, 8, 0.05)) R> ok <- which(lenh$counts > 0) R> lenh$counts[ok] <- lenh$counts[ok] / diff(c(0, ok)) R> lenh$counts [1] 0.0000 0.0000 0.0000 0.0000 0.0000 0.1667 0.0000 1.5000 0.0000 0.5000 [11] 0.0000 2.0000 0.0000 1.0000 0.0000 2.5000 0.0000 3.0000 0.0000 5.0000 [21] 0.0000 4.5000 0.0000 2.0000 0.0000 0.5000 0.0000 3.0000 0.0000 3.5000 [31] 0.0000 3.0000 0.0000 4.0000 0.0000 3.5000 0.0000 1.5000 0.0000 3.0000 [41] 0.0000 3.0000 0.0000 2.0000 0.0000 4.5000 0.0000 3.5000 0.0000 2.5000 [51] 0.0000 1.0000 0.0000 4.0000 0.0000 1.5000 0.0000 2.0000 0.0000 0.5000 [61] 0.0000 0.5000 0.0000 1.5000 0.0000 0.5000 0.0000 0.5000 0.0000 0.0000 [71] 0.0000 0.2500 0.0000 2.0000 0.0000 0.0000 0.0000 0.2500 0.0000 0.0000 It makes sense, although I'm a bit nervous about floating-point issues that might make this fail in some cases. Any suggestions/comments welcome. -- Seb ______________________________________________ R-help@stat.math.ethz.ch 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.
