我意识到我的标题可能有点令人困惑。我有一些JSON对unnest
有点混淆。我正试着利用这个小精灵。
样本数据
df <- structure(list(long_abbr = c("Team11", "BBS"), short_name = c("Ac ",
"BK"), division = c("", ""), name = c("AC Slaters Muscles", "Broken Bats"
), abbr = c("T1", "T1"), owners = list(structure(list(commissioner = 0L,
name = "Chris Liss", id = "300144F8-79F4-11EA-8F25-9AE405472731"), class = "data.frame", row.names = 1L),
structure(list(commissioner = 1L, name = "Mark Ortin", id = "90849EF6-7427-11EA-95AA-4EEEAC7F8CD2"), class = "data.frame", row.names = 1L)),
id = c("1", "2"), logged_in_team = c(NA_integer_, NA_integer_
)), row.names = 1:2, class = "data.frame")
)
# Unnest Owners Information
df <- df %>%
unnest(owners)
由于我有使用name
的重复列,因此出现以下错误。
Error: Column names `name` and `id` must not be duplicated.
有没有一种简单的方法可以通过在嵌套列之前使用前缀owners
(或者在我的情况下,我希望它使用包含嵌套df
的列的任何名称(的命名约定来取消对列的命名。即owners.commissioner
、owners.name
、owners.id
。我也会对使用camel case和下划线的解决方案感兴趣。例如ownersName
或owners_name
。
设置参数names_sep
:
df <- structure(
list(long_abbr = c("Team11", "BBS"),
short_name = c("Ac ", "BK"),
division = c("", ""),
name = c("AC Slaters Muscles", "Broken Bats"),
abbr = c("T1", "T1"),
owners = list(
structure(list(commissioner = 0L, name = "Chris Liss",
id = "300144F8-79F4-11EA-8F25-9AE405472731"),
class = "data.frame", row.names = 1L),
structure(list(commissioner = 1L, name = "Mark Ortin",
id = "90849EF6-7427-11EA-95AA-4EEEAC7F8CD2"),
class = "data.frame", row.names = 1L)),
id = c("1", "2"),
logged_in_team = c(NA_integer_, NA_integer_)),
row.names = 1:2, class = "data.frame"
)
tidyr::unnest(df, owners, names_sep = "_")
#> # A tibble: 2 x 10
#> long_abbr short_name division name abbr owners_commissi… owners_name
#> <chr> <chr> <chr> <chr> <chr> <int> <chr>
#> 1 Team11 "Ac " "" AC S… T1 0 Chris Liss
#> 2 BBS "BK" "" Brok… T1 1 Mark Ortin
#> # … with 3 more variables: owners_id <chr>, id <chr>, logged_in_team <int>
由reprex包(v0.3.0(于2020-04-26创建
这能解决你的问题吗?