分割リストの列を複数の行に分割[R]

Question

私は以下のようなデータフレーム/ティブルを持っています。

# # A tibble: 2 × 3 
#  from_id    created_time  text 
#   <chr>     <chr> <list> 
# 1 10113538711 2017-02-10T23:33:01+0000 <chr [3]> 
# 2 10113538711 2017-02-10T05:41:39+0000 <chr [5]> 

私はそれは次のようになりますので、行の上にテキスト列からlistアイテムを広げたいと思っています。

# # A tibble: 2 × 3 
#   from_id    created_time        text 
#   <chr>     <chr>        <chr> 
# 1 10113538711 2017-02-10T23:33:01+0000 "earlier this week we received ..." 
# 1 10113538711 2017-02-10T23:33:01+0000 "lance payne's photo struck a c..." 
# 1 10113538711 2017-02-10T23:33:01+0000 "this is his story:" 
# 2 10113538711 2017-02-10T05:41:39+0000 "i'm melting, but extreme heat ..." 
# 2 10113538711 2017-02-10T05:41:39+0000 "place the container in an area..." 
# 2 10113538711 2017-02-10T05:41:39+0000 "please share far and wide." 
# 2 10113538711 2017-02-10T05:41:39+0000 "thank you." 
# 2 10113538711 2017-02-10T05:41:39+0000 "photo © tanya-dee johnson" 

私は私の使い方では動作しませんでしたが、tidy::separate()をしようと思いました。私はそれがスプリットのいくつかの形式、または別のものであると思われますが、gather()またはmelt()が続きますが、私のRの語彙は現時点で私を失望させています。

これについてのご支援をいただければ幸いです。

私の転倒のDPUT。

> dput(df) 

structure(list(from_id = c("10113538711", "10113538711"), created_time = c("2017-02-10T23:33:01+0000", 
"2017-02-10T05:41:39+0000"), text = structure(list(c("earlier this week we received shocking photos of a turtle hatchling emerging beside a lump of coal at mackay's east point beach near hay point – the largest coal port alongside the great barrier reef.", 
"lance payne's photo struck a chord around the country.", "this is his story:" 
), c("i'm melting, but extreme heat causes significant stress particularly for all animals.", 
"place the container in an area where animals are protected from predators when drinking eg near a shrub or bush and keep your pets away from this area so that animals can drink undisturbed.", 
"please share far and wide.", "thank you.", "photo © tanya-dee johnson" 
)), class = c("get_sentences", "list"))), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -2L), .Names = c("from_id", 
"created_time", "text")) 

Answer 1

我々は速かったunnest

library(tidyverse) 
unnest(df)