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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