z <- substitute(f(x), list(x = data.frame(y = 1)))

z
# f(list(y = 1))

str(z)
# language f(structure(list(y = 1), .Names = "y", row.names = c(NA,
-1L), class = "data.frame"))

dput(z)
# f(structure(list(y = 1), .Names = "y", row.names = c(NA, -1L), class
= "data.frame"))

Hadley

-- 
Chief Scientist, RStudio
http://had.co.nz/

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