R 将长度不等的多级嵌套列表转换为数据帧

我有这样一个列表（biclqiues（）的输出）：

---

A这是一个基本解决方案
lappy
循环三个
biclique
元素，然后使用
[
提取列表的左、右部分。
stack
将
lappy（）
结果简化为一个漂亮的数据框：

left_df <- stack(lapply(A, `[[`, 'left'))
left_df

#    values       ind
#1   EVELYN biclique1
#2    LAURA biclique1
#3   BRENDA biclique1
#4   EVELYN biclique2
#5    LAURA biclique2
#6  THERESA biclique2
#7   EVELYN biclique3
#8    LAURA biclique3
#9  THERESA biclique3
#10  BRENDA biclique3
#11 FRANCES biclique3

left_df[left_df$values == 'EVELYN', ]

#  values       ind
#1 EVELYN biclique1
#4 EVELYN biclique2
#7 EVELYN biclique3

# and your direct question - this is base but there are simpler answers in dplyr and data.table:
unique(left_df[left_df$ind %in% left_df[left_df$values == 'EVELYN', 'ind'], 'values'])

#"EVELYN"  "LAURA"   "BRENDA"  "THERESA" "FRANCES"

left\u通过发布
dput（list\u name）
使此问题可重复。同时发布相同的预期输出。@Wendy您希望数据框具有什么结构？您可以发布一个示例吗
left_df <- stack(lapply(A, `[[`, 'left'))
left_df

#    values       ind
#1   EVELYN biclique1
#2    LAURA biclique1
#3   BRENDA biclique1
#4   EVELYN biclique2
#5    LAURA biclique2
#6  THERESA biclique2
#7   EVELYN biclique3
#8    LAURA biclique3
#9  THERESA biclique3
#10  BRENDA biclique3
#11 FRANCES biclique3

left_df[left_df$values == 'EVELYN', ]

#  values       ind
#1 EVELYN biclique1
#4 EVELYN biclique2
#7 EVELYN biclique3

# and your direct question - this is base but there are simpler answers in dplyr and data.table:
unique(left_df[left_df$ind %in% left_df[left_df$values == 'EVELYN', 'ind'], 'values'])

#"EVELYN"  "LAURA"   "BRENDA"  "THERESA" "FRANCES"

right_df <- stack(lapply(A, `[[`, 'right'))
right_df

#   values       ind
#1       1 biclique1
#2       3 biclique1
#3       5 biclique1
#4       6 biclique1
#5       8 biclique1
#6       2 biclique2
#7       3 biclique2
#8       5 biclique2
#9       6 biclique2
#10      8 biclique2
#11      3 biclique3
#12      5 biclique3
#13      6 biclique3
#14      8 biclique3