Allow spread to work with list cols

Fixes tidyverse#199

NEWS.md

@@ -31,6 +31,8 @@
   as "key<sep>value". Additionally, if sep is `NULL` missing values will be
   converted to `<NA>` (#68).
 
+* `spread()` works in the presence of list-columns (#199)
+
 * `unnest()` works with non-syntactic names (#190).
 
 * `unnest()` gains a `sep` argument. If non-null, this will rename the 

R/id.R

@@ -47,6 +47,11 @@ id_var <- function(x, drop = FALSE) {
   } else if (length(x) == 0) {
     id <- integer()
     n <- 0L
+  } else if (is.list(x)) {
+    # Sorting lists isn't supported
+    levels <- unique(x)
+    id <- match(x, levels)
+    n <- max(id)
   } else {
     levels <- sort(unique(x), na.last = TRUE)
     id <- match(x, levels)

tests/testthat/test-spread.R

@@ -145,6 +145,14 @@ test_that("complex values are preserved  (#134)", {
   expect_equal(out2$b, 3:4 + 1i)
 })
 
+test_that("can spread with nested columns", {
+  df <- tibble::data_frame(x = c("a", "a"), y = 1:2, z = list(1:2, 3:5))
+  out <- spread(df, x, y)
+
+  expect_equal(out$a, 1:2)
+  expect_equal(out$z, df$z)
+})
+
 test_that("spread gives one column when no existing non-spread vars", {
   df <- data_frame(
     key = c("a", "b", "c"),