從 data.frames 列表中創建 dataframe

Question

我有一個 data.frames 列表列表，我想將其轉換為data.frame 。 結構如下：

l_of_lists <- list(
  year1 = list(
    one = data.frame(date = c("Jan-10", "Jan-22"), type = c("type 1", "type 2")),
    two = data.frame(date = c("Feb-1", "Feb-28"), type = c("type 2", "type 3")),
    three = data.frame(date = c("Mar-10", "Mar-15"), type = c("type 1", "type 4"))
    ),
  year2 = list( # dates is used here on purpose, as the names don't perfectly match
    one = data.frame(dates = c("Jan-22"), type = c("type 2"), another_col = c("entry 2")),
    two = data.frame(date = c("Feb-10", "Feb-18"), type = c("type 2", "type 3"), another_col = c("entry 2", "entry 3")),
    three = data.frame(date = c("Mar-10", "Mar-15"), type = c("type 1", "type 4"), another_col = c("entry 4", "entry 5"))
    ),
  year3 = list( # this deliberately only contains two data frames
    one = data.frame(date = c("Jan-10", "Jan-12"), type = c("type 1", "type 2")),
    two = data.frame(date = c("Feb-8", "Jan-28"), type = c("type 2", "type 3"))
  ))

數據框有兩個我試圖在上面模仿的特性：

• 列名相差 1-2 個字符（例如date與dates ）
• 某些列僅存在於某些數據幀中（例如another_col ）

我現在想將其轉換為數據框（我嘗試了對rbind和do.call 的不同調用，如此處所述，但未成功）並且希望 - 容忍地匹配列名（如果列名類似於 1- 2 個字符，我希望它們匹配並且 - 在其他列中用NA填充不存在的列。

我想要一個類似於以下的數據框

year  level       date        type  another_col                    
   1    one    "Jan-10"    "type 1"           NA
   1    one    "Jan-22"    "type 2"           NA
   1    two     "Feb-1"    "type 2"           NA
   1    two    "Feb-28"    "type 3"           NA
   1  three    "Mar-10"    "type 1"           NA
   1  three    "Mar-15"    "type 4"           NA
   2    one    "Jan-22"    "type 2"     "entry 2"
   2    two     "Feb-1"    "type 2"     "entry 2"
   2    two    "Feb-28"    "type 3"     "entry 3"
   2  three    "Mar-10"    "type 1"     "entry 4"
   2  three    "Mar-15"    "type 4"     "entry 5"
   3    one    "Jan-10"    "type 1"           NA
   3    one    "Jan-12"    "type 2"           NA
   3    two     "Feb-8"    "type 2"           NA
   3    two    "Feb-28"    "type 3"           NA

有人可以指出rbind是否是正確的路徑 - 以及我缺少什么？

Answer 1

您可以使用 purrr 和 dplyr 執行以下操作：

l_of_lists <- list(
  year1 = list(
    one = data.frame(date = c("Jan-10", "Jan-22"), type = c("type 1", "type 2")),
    two = data.frame(date = c("Feb-1", "Feb-28"), type = c("type 2", "type 3")),
    three = data.frame(date = c("Mar-10", "Mar-15"), type = c("type 1", "type 4"))
  ),
  year2 = list( # dates is used here on purpose, as the names don't perfectly match
    one = data.frame(dates = c("Jan-22"), type = c("type 2"), another_col = c("entry 2")),
    two = data.frame(date = c("Feb-10", "Feb-18"), type = c("type 2", "type 3"), another_col = c("entry 2", "entry 3")),
    three = data.frame(date = c("Mar-10", "Mar-15"), type = c("type 1", "type 4"), another_col = c("entry 4", "entry 5"))
  ),
  year3 = list( # this deliberately only contains two data frames
    one = data.frame(date = c("Jan-10", "Jan-12"), type = c("type 1", "type 2")),
    two = data.frame(date = c("Feb-8", "Jan-28"), type = c("type 2", "type 3"))
  ))

# add libraries
library(dplyr)
library(purrr)

# Map bind_rows to each list within the list
l_of_lists %>% 
  map_dfr(~bind_rows(.x, .id = "level"), .id = "year")

這將產生：

     year level   date   type  dates another_col
1  year1   one Jan-10 type 1   <NA>        <NA>
2  year1   one Jan-22 type 2   <NA>        <NA>
3  year1   two  Feb-1 type 2   <NA>        <NA>
4  year1   two Feb-28 type 3   <NA>        <NA>
5  year1 three Mar-10 type 1   <NA>        <NA>
6  year1 three Mar-15 type 4   <NA>        <NA>
7  year2   one   <NA> type 2 Jan-22     entry 2
8  year2   two Feb-10 type 2   <NA>     entry 2
9  year2   two Feb-18 type 3   <NA>     entry 3
10 year2 three Mar-10 type 1   <NA>     entry 4
11 year2 three Mar-15 type 4   <NA>     entry 5
12 year3   one Jan-10 type 1   <NA>        <NA>
13 year3   one Jan-12 type 2   <NA>        <NA>
14 year3   two  Feb-8 type 2   <NA>        <NA>
15 year3   two Jan-28 type 3   <NA>        <NA>

然后當然你可以做一些正則表達式解析只保留數字年份：

l_of_lists %>% 
  map_dfr(~bind_rows(.x, .id = "level"), .id = "year") %>% 
  mutate(year = substring(year, regexpr("\\d", year)))

如果您知道 date 和 dates 相同，則始終可以使用mutate將 then 更改為那些沒有丟失的值（即mutate(date = ifelse(.is,na(date), date, dates)) ）