[R] show unique combination of two factor

[Chuck Cleland]

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
> 
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Chuck Cleland, Ph.D.
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