R用数据框的NA填充缺失值

Question

我目前正在尝试使用以下列表创建数据框

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")

当我尝试创建数据框时，出现（明显的）错误，即列表的长度不相等。我想找到一种方法，要么使列表长度相同，要么用 "NA" 填充缺失的数据框条目。经过一些搜索后，我一直无法找到解决方案

Answer 1

这里是 purrr（tidyverse 的一部分）和基础 R 解决方案，假设您只想用 NA 填充每个列表中的剩余值。我将任何列表的最大长度设为 len，然后对每个列表执行 rep(NA) 以获得 that 列表的长度与最大长度之间的差异任何列表。

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，根据需要附加 NAs，转换为字符向量，然后获取所有这些向量的数据框 cbinded 在一个管道调用中：

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 base R 中，您可以 lapply 跨列表列表使用相同的函数，然后使用 Reduce 到 cbind 结果列表并将其转换为数据框。走两步而不是 purrr 的一步：

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

对于这两个，您现在可以随意设置名称。

Answer 2

你可以这样做：

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 如下：

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")
)

R用数据框的NA填充缺失值

R Filling missing values with NA for a data frame

r

list

dataframe

na