On Sun, Jan 16, 2011 at 03:01:49PM +0800, r-help wrote: > I have a data frame with 10 columns: A:J and I want to have the output as a > data frame with 11 columns, the value of 11th column is?? > > > for each row, if any column can be divided by 13, then the 11th column has a > values of 1, otherwise, it has a value of 0. How to do that? > > > input is > > a=matrix(1:10000,1000,10) > > dimnames(a)=list(NULL,LETTERS[1:10])
For a general matrix "a", this can be done, for example x <- rowSums(a %% 13 == 0) b <- cbind(a, x) If the residues in the columns are periodic, like in the example above, then also the vector x is periodic. If this really occurs in the application, generating the new column using this fact may also be useful. The "length.out" argument of rep() can be used. y <- rep(rowSums(a[1:13, ] %% 13 == 0), length=1000) identical(x, y) # [1] TRUE Petr Savicky. ______________________________________________ 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.
