R用NA為數據幀填充缺失值

Question

我目前正在嘗試使用以下列表創建數據框

location <- list("USA","Singapore","UK")
organization <- list("Microsoft","University of London","Boeing","Apple")
person <- list()
date <- list("1989","2001","2018")
Jobs <- list("CEO","Chairman","VP of sales","General Manager","Director")

當我嘗試創建數據幀時，出現（顯而易見的）錯誤，即列表的長度不相等。 我想找到一種方法，使列表具有相同的長度，或者用“ NA”填充丟失的數據框條目。 經過一些搜索后，我無法找到解決方案

Answer 1

這里是purrr （部分tidyverse ）和基礎R解決方案，假設你只是想在每個列表，以填補剩余值NA 。 我采取任何列表作為的最大長度len ，然后為每個表做rep(NA) 該列表的長度和任何列表的最大長度之間的差異。

library(tidyverse)

location <- list("USA","Singapore","UK")
organization <- list("Microsoft","University of London","Boeing","Apple")
person <- list()
date <- list("1989","2001","2018")
Jobs <- list("CEO","Chairman","VP of sales","General Manager","Director")

all_lists <- list(location, organization, person, date, Jobs)
len <- max(lengths(all_lists))

使用purrr::map_dfc ，您可以映射列表列表，根據需要添加NA ，轉換為字符向量，然后獲得所有這些向量在一個管道調用中cbind的數據幀：

map_dfc(all_lists, function(l) {
  c(l, rep(NA, len - length(l))) %>%
    as.character()
})
#> # A tibble: 5 x 5
#>   V1        V2                   V3    V4    V5             
#>   <chr>     <chr>                <chr> <chr> <chr>          
#> 1 USA       Microsoft            NA    1989  CEO            
#> 2 Singapore University of London NA    2001  Chairman       
#> 3 UK        Boeing               NA    2018  VP of sales    
#> 4 NA        Apple                NA    NA    General Manager
#> 5 NA        NA                   NA    NA    Director

在基礎R，你可以lapply跨列表列表相同的功能，然后使用Reduce到cbind所產生的名單，並將其轉換成數據幀。 采取兩個步驟而不是purrr的一個步驟：

cols <- lapply(all_lists, function(l) c(l, rep(NA, len - length(l))))
as.data.frame(Reduce(cbind, cols, init = NULL))
#>          V1                   V2 V3   V4              V5
#> 1       USA            Microsoft NA 1989             CEO
#> 2 Singapore University of London NA 2001        Chairman
#> 3        UK               Boeing NA 2018     VP of sales
#> 4        NA                Apple NA   NA General Manager
#> 5        NA                   NA NA   NA        Director

對於這兩者，您現在都可以根據需要設置名稱。

Answer 2

您可以這樣做：

data.frame(sapply(dyem_list, "length<-", max(lengths(dyem_list))))

   location         organization person date            Jobs
1       USA            Microsoft   NULL 1989             CEO
2 Singapore University of London   NULL 2001        Chairman
3        UK               Boeing   NULL 2018     VP of sales
4      NULL                Apple   NULL NULL General Manager
5      NULL                 NULL   NULL NULL        Director

其中dyem_list為以下內容：

dyem_list <- list(
  location = list("USA","Singapore","UK"),
  organization = list("Microsoft","University of London","Boeing","Apple"),
  person = list(),
  date = list("1989","2001","2018"),
  Jobs = list("CEO","Chairman","VP of sales","General Manager","Director")
)