在保留索引的同时取消列表的列表 [英] unlisting a list while keeping the indices

问题描述

我有一个列表，可以包含空条目，包含一个元素的条目和包含多个元素的条目.

I have a list which can have either empty entries, entries containing one elements and entries containing multiple elements.

l1 = list(integer(0), 11L, integer(0), integer(0), 11L, 11L, c(6L, 
36L), 16L, 16L, integer(0), integer(0))

我想取消列出l1并创建一个数据框，对于空元素，索引将根本不会出现，对于多个元素，索引将出现多次:

I would like to unlist l1 and create a data frame, where for the empty elements the index will not appear at all, and for multiple elements the index will appear multiple times:

df = data_frame(entry = c(2, 5, 6, 7, 7, 8, 9), element = c(11, 11, 11, 6, 
36, 16, 16))

例如，在此条目7出现两次，因为列表中有两个元素(6和36).

so for example here entry 7 appears twice because the list has two elements there (6 and 36).

我该怎么做?

推荐答案

我们可以有两个选择.将list命名为一个，将enframe命名为tbl_df，然后将unnest命名为list元素. NULL元素将被自动删除

We can have two options. Make the list a named one, enframe it to a tbl_df and then unnest the list element. The NULL elements will be automatically removed

library(tidyverse)
l1 %>% 
     set_names(seq_along(.)) %>% 
     enframe %>%
     unnest

或者在将list命名为stack后，将其命名为2列data.frame

Or after naming the list, stack it to a 2 column data.frame

stack(setNames(l1, seq_along(l1)))[2:1]

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