如何用tidyr::unnest()展开嵌套tibble得到指定结构的数据表

现有如下嵌套结构的tibble对象test：

test
# A tibble: 2 × 4
  site  data               samp dates      
  <chr> <list>            <int> <list>     
1 ARK   <tibble [15 × 2]>    15 <date [15]>
2 BCH   <tibble [44 × 2]>    44 <date [44]>

其构造代码为：

test <- structure(list(site = c("ARK", "BCH"), data = list(structure(list(
    alcohol = c("Yes", "No", "Yes", "No", "No", "No", "Yes", 
    "No", "No", "No", "No", "No", "No", "No", "No"), opioids = c("No", 
    "No", "No", "No", "No", "No", "No", "No", "No", "No", "No", 
    "No", "No", "No", "No")), row.names = c(NA, -15L), class = c("tbl_df", 
"tbl", "data.frame")), structure(list(alcohol = c("No", "No", 
"No", "Yes", "No", "No", "No", "Yes", "No", "No", "No", "No", 
"No", "No", "No", "No", "Yes", "No", "Yes", "Yes", "No", "Yes", 
"No", "No", "No", "No", "No", "No", "No", "No", "No", "No", "No", 
"No", "No", "No", "Yes", "No", "No", "Yes", "No", "Yes", "Yes", 
NA), opioids = c("No", "No", "No", "No", "No", "No", "No", "No", 
"No", "No", "No", "No", "No", "No", "No", "No", "No", "No", "No", 
"No", "No", "No", "No", "No", "No", "No", "No", "No", "No", "No", 
"No", "No", "No", "No", "No", "No", "No", "No", "No", "No", "No", 
"No", "No", NA)), row.names = c(NA, -44L), class = c("tbl_df", 
"tbl", "data.frame"))), samp = c(15L, 44L), dates = list(structure(c(19786, 
19854.0714285714, 19922.1428571429, 19990.2142857143, 20058.2857142857, 
20126.3571428571, 20194.4285714286, 20262.5, 20330.5714285714, 
20398.6428571429, 20466.7142857143, 20534.7857142857, 20602.8571428571, 
20670.9285714286, 20739), class = "Date"), structure(c(19786, 
19808.1627906977, 19830.3255813953, 19852.488372093, 19874.6511627907, 
19896.8139534884, 19918.976744186, 19941.1395348837, 19963.3023255814, 
19985.4651162791, 20007.6279069767, 20029.7906976744, 20051.9534883721, 
20074.1162790698, 20096.2790697674, 20118.4418604651, 20140.6046511628, 
20162.7674418605, 20184.9302325581, 20207.0930232558, 20229.2558139535, 
20251.4186046512, 20273.5813953488, 20295.7441860465, 20317.9069767442, 
20340.0697674419, 20362.2325581395, 20384.3953488372, 20406.5581395349, 
20428.7209302326, 20450.8837209302, 20473.0465116279, 20495.2093023256, 
20517.3720930233, 20539.5348837209, 20561.6976744186, 20583.8604651163, 
20606.023255814, 20628.1860465116, 20650.3488372093, 20672.511627907, 
20694.6744186047, 20716.8372093023, 20739), class = "Date"))), row.names = c(NA, 
-2L), class = c("tbl_df", "tbl", "data.frame"))

需要将其展开为包含15+44=59行、5列的tibble，列分别为site、alcohol（取自data）、opioids（取自data）、samp和dates。

解决代码 #

使用tidyr::unnest()函数，同时指定需要展开的嵌套列data和dates：

library(tidyr)

# 展开嵌套列
result <- test %>% 
  unnest(cols = c(data, dates))

# 查看结果结构
str(result)

验证结果 #

执行后得到的tibble结构如下：

result
# A tibble: 59 × 5
   site  alcohol opioids  samp dates     
   <chr> <chr>   <chr>   <int> <date>    
 1 ARK   Yes     No         15 2024-01-01
 2 ARK   No      No         15 2024-03-10
 3 ARK   Yes     No         15 2024-05-18
 4 ARK   No      No         15 2024-07-27
 5 ARK   No      No         15 2024-09-14
 6 ARK   No      No         15 2024-11-22
 7 ARK   Yes     No         15 2025-01-30
 8 ARK   No      No         15 2025-04-09
 9 ARK   No      No         15 2025-06-17
10 ARK   No      No         15 2025-08-25
# ℹ 49 more rows

可以看到，结果已成功展开为59行5列，包含所需的所有字段。

内容的提问来源于stack exchange，提问作者tic-toc-choc