dplyr GPF case

[smckinney2718]

foo <- data.frame(a = letters[c(1,2)])
bar <- tbl_df(foo)
unlist(distinct(bar[, "a"][bar[, "a"]!=""]))

[hadley]

Looks like this is actually a tibble issue.

[krlmlr]

@smckinney2718: Note that bar["a"] and bar[, "a"] still return a data frame, use bar[["a"]] to return an atomic vector.

Corrupt data frames may come from elsewhere. It would be good if all verbs checked consistency of the input:

• Do all columns have the same length?
• Is the row.names attribute consistent?

I remember seeing something along these lines in the code, but I'm not sure. At the very least the check doesn't seem to be active for distinct().