如何解决使用标签作为新列的值将列表转换为数据框使用 R
require(tidyverse)
months <- c("january","february","march","october","december")
weekdays <- c("Sunday","Monday","Tuesday")
seasons <- c("Summer","Winter","Fall","autumn")
timeWords_list <- tibble::lst(months,weekdays,seasons)
我想将此列表转换为两列-data.frame： 我需要列表的所有值都在第一列中 并且标签是第二列的值
结果应该是这样的：
df <- data.frame(first_column = c("january","december","Sunday","Tuesday","Summer","autumn"),second_column = c("months","months","weekdays","seasons","seasons"))
第二列可以是字符串或因子，但应对应于列表的标签。请注意，列表可能有不同的大小，其元素可能有不同的长度。
解决方法
一个简单的选项是 stack 来自 base R
df1 <- stack(timeWords_list)
names(df1) <- c("first_column","second_column")
我认为 @akrun 的 stack 方法是迄今为止最有效的方法。这是另一个使用 unlist + rep
data.frame(
first_col = unlist(timeWords_list,use.names = FALSE),second_col = rep(names(timeWords_list),lengths(timeWords_list))
)
给出
first_col second_col
1 january months
2 february months
3 march months
4 october months
5 december months
6 Sunday weekdays
7 Monday weekdays
8 Tuesday weekdays
9 Summer seasons
10 Winter seasons
11 Fall seasons
12 autumn seasons
使用 tibble::enframe 和 tidyr::unnest
tibble::enframe(timeWords_list,name = 'first_column',value = 'second_column') %>%
tidyr::unnest(second_column)
# first_column second_column
# <chr> <chr>
# 1 months january
# 2 months february
# 3 months march
# 4 months october
# 5 months december
# 6 weekdays Sunday
# 7 weekdays Monday
# 8 weekdays Tuesday
# 9 seasons Summer
#10 seasons Winter
#11 seasons Fall
#12 seasons autumn
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