Hi, I think this is substantially less ugly:
A <- matrix(1:15,nrow=5,byrow=F); A a <- c(1,2,3) B <- sweep(A, 2, a, "^") apply(B, 1, prod) You could combine it into one line if you wanted, but I find it clearer as two: > apply(sweep(A, 2, a, "^"), 1, prod) [1] 47916 169344 421824 889056 1687500 Sarah On Fri, Jul 1, 2016 at 10:15 AM, Steven Yen <sye...@gmail.com> wrote: > A is a 5 x 3 matrix and a is a 3-vector. I like to exponentiate A[,1] to > a[1], A[,2] to a[2], and A[,3] to a[3], and obtain the product of the > resulting columns, as in line 3. > > I also accomplish this with lines 4 and 5. I like to have rowProducts(B) > but there is not so I came up with something ugly in line > 5--exponentiating the row sums of log. Is there a more elegant way than > than line 5 or, better yet, lines 4 and 5 together? Thanks. > > A<-matrix(1:15,nrow=5,byrow=F); A > a<-c(1,2,3) > (A[,1]^a[1])*(A[,2]^a[2])*(A[,3]^a[3]) > B<-t(t(A)^a); B > exp(rowSums(log(B))) > > Result: > > > A<-matrix(1:15,nrow=5,byrow=F); A > [,1] [,2] [,3] > [1,] 1 6 11 > [2,] 2 7 12 > [3,] 3 8 13 > [4,] 4 9 14 > [5,] 5 10 15 > > a<-c(1,2,3) > > (A[,1]^a[1])*(A[,2]^a[2])*(A[,3]^a[3]) > [1] 47916 169344 421824 889056 1687500 > > B<-t(t(A)^a); B > [,1] [,2] [,3] > [1,] 1 36 1331 > [2,] 2 49 1728 > [3,] 3 64 2197 > [4,] 4 81 2744 > [5,] 5 100 3375 > > exp(rowSums(log(B))) > [1] 47916 169344 421824 889056 1687500 > > ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
