Aimin Yan wrote: > p factor have 5 levels > aa factor have 19 levels. > totally it should have 95 combinations. > but I just find there are 92 combinations. > Does anyone know how to code to find what combinations are missed?
Here is an example with fewer factor levels of one way you might do this: df <- data.frame(p = rep(c("A","B","C","D"), each=10), aa = rep(c("Yes","No"), 20)) df$aa <- replace(df$aa, df$p == "D", "No") table(df) aa p No Yes A 5 5 B 5 5 C 5 5 D 10 0 names(which(with(df, table(interaction(p, aa))) == 0)) [1] "D.Yes" > Thanks, > > Aimin > > ______________________________________________ > 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. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 ______________________________________________ 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.