how do i manipulate with purrr verbs v0 to return a 2x4 tibble where columns 2:4 are the nested outputs of the v(), ie names(v1) is c('x','a','b','c') instead of a 2x2 c('x','out') (which is what i get now)
v0 <- as_tibble(data.frame(x=c(15,30)))
v <- function(x) list(a=mtcars[1:x,],b=iris[1:x,],c=volcano[1:x,])
v1 <- v0%>%mutate(out=purrr::map(x,v))
obviously the elements arent assumed to be just matrices and data.frames but any class that is consistent across rows
here is one way. is there a more efficient one?
v <- function(x) list(a=mtcars[1:x,],b=iris[1:x,],c=volcano[1:x,])
v1 <- data.frame(x=c(15,30))%>%
do(cbind(x=.$x,as_tibble(transpose(purrr::map(.$x,v)))))
Maybe:
f6 <- rlang::exprs(a = map(x, ~ mtcars[1:.x, ]),
b = map(x, ~ iris[1:.x, ]),
c = map(x, ~ volcano[1:.x, ]))
mutate(v0, !!! f6)
# # A tibble: 2 x 4
# x a b c
# <dbl> <list> <list> <lis>
# 1 15.0 <data.frame [15 × 11]> <data.frame [15 × 5]> <dbl…
# 2 30.0 <data.frame [30 × 11]> <data.frame [30 × 5]> <dbl…
Unsure about efficiency.
If you want to minimize modifications of existing code and risks of typos, maybe this instead where you just have to replace function(x) list(.... with rlang::exprs(....:
f8 <- rlang::exprs(a = mtcars[1:x, ],
b = iris[1:x, ],
c = volcano[1:x, ])
f9 <- map(f8, ~ rlang::expr(map(x, function(x) !! .x)))
mutate(v0, !!! f9)