使用Tidyverse Join更新/替换Dataframe中的值

Question

使用查找表中的(正确)值更新/替换主数据集中的NA的最有效方法是什么？这是一个如此常见的操作!类似的问题似乎没有整洁的解决方案.

约束:1)请假设大量缺失值和更大的查找表比给出的示例.所以情况下明智的替换操作将是不切实际的(不case_when,if_else等等)

2)查找表没有主数据帧的所有值,只有替换的值.

Tidyverse解决方案的答案更受欢迎.类似的问题似乎没有整洁的解决方案.

library(tidyverse)

### Main Dataframe ###
df1 <- tibble(
  state_abbrev = state.abb[1:10],
  state_name = c(state.name[1:5], rep(NA, 3), state.name[9:10]),
  value = sample(500:1200, 10, replace=TRUE)
)


#> # A tibble: 10 x 3
#>    state_abbrev state_name value
#>    <chr>        <chr>      <int>
#>  1 AL           Alabama      551
#>  2 AK           Alaska       765
#>  3 AZ           Arizona      508
#>  4 AR           Arkansas     756
#>  5 CA           California   741
#>  6 CO           <NA>        1100
#>  7 CT           <NA>         719
#>  8 DE           <NA>         874
#>  9 FL           Florida      749
#> 10 GA           Georgia      580


### Lookup Dataframe ###
lookup_df <- tibble(
  state_abbrev = state.abb[6:8],
  state_name = state.name[6:8]
)

#> # A tibble: 3 x 2
#>   state_abbrev state_name 
#>   <chr>        <chr>      
#> 1 CO           Colorado   
#> 2 CT           Connecticut
#> 3 DE           Delaware

理想情况下,left_join将具有缺失值的替换选项.唉...

left_join(df1, lookup_df)
#> Joining, by = c("state_abbrev", "state_name")
#> # A tibble: 10 x 3
#>    state_abbrev state_name value
#>    <chr>        <chr>      <int>
#>  1 AL           Alabama      551
#>  2 AK           Alaska       765
#>  3 AZ           Arizona      508
#>  4 AR           Arkansas     756
#>  5 CA           California   741
#>  6 CO           <NA>        1100
#>  7 CT           <NA>         719
#>  8 DE           <NA>         874
#>  9 FL           Florida      749
#> 10 GA           Georgia      580

```

由reprex包创建于2018-07-28 (v0.2.0).

Answer 1

这是一个单行解决方案rows_update()：

df1 %>% 
  rows_update(lookup_df, by = "state_abbrev")

演示：

library(dplyr)

### Main Dataframe ###
df1 <- tibble(
  state_abbrev = state.abb[1:10],
  state_name = c(state.name[1:5], rep(NA, 3), state.name[9:10]),
  value = sample(500:1200, 10, replace=TRUE)
)

### Lookup Dataframe ###
lookup_df <- tibble(
  state_abbrev = state.abb[6:8],
  state_name = state.name[6:8]
)

df1 %>% 
  rows_update(lookup_df, by = "state_abbrev")
#> # A tibble: 10 x 3
#>    state_abbrev state_name  value
#>    <chr>        <chr>       <int>
#>  1 AL           Alabama       532
#>  2 AK           Alaska        640
#>  3 AZ           Arizona       521
#>  4 AR           Arkansas      523
#>  5 CA           California    970
#>  6 CO           Colorado      695
#>  7 CT           Connecticut   504
#>  8 DE           Delaware     1088
#>  9 FL           Florida       979
#> 10 GA           Georgia      1059

Answer 2

收集Alistaire和Nettle的建议并转变为可行的解决方案

df1 %>% 
  left_join(lookup_df, by = "state_abbrev") %>% 
  mutate(state_name = coalesce(state_name.x, state_name.y)) %>% 
  select(-state_name.x, -state_name.y)

# A tibble: 10 x 3
   state_abbrev value state_name 
   <chr>        <int> <chr>      
 1 AL             671 Alabama    
 2 AK             501 Alaska     
 3 AZ            1030 Arizona    
 4 AR             694 Arkansas   
 5 CA             881 California 
 6 CO             821 Colorado   
 7 CT             742 Connecticut
 8 DE             665 Delaware   
 9 FL             948 Florida    
10 GA             790 Georgia

OP表示更喜欢"tidyverse"解决方案.但是,更新联接已在data.table包中提供:

library(data.table)
setDT(df1)[setDT(lookup_df), on = "state_abbrev", state_name := i.state_name]
df1

    state_abbrev  state_name value
 1:           AL     Alabama  1103
 2:           AK      Alaska  1036
 3:           AZ     Arizona   811
 4:           AR    Arkansas   604
 5:           CA  California   868
 6:           CO    Colorado  1129
 7:           CT Connecticut   819
 8:           DE    Delaware  1194
 9:           FL     Florida   888
10:           GA     Georgia   501

基准

library(bench)
bm <- press(
  na_share = c(0.1, 0.5, 0.9),
  n_row = length(state.abb) * 2 * c(1, 100, 10000),
  {
    n_na <- na_share * length(state.abb)
    set.seed(1)
    na_idx <- sample(length(state.abb), n_na)
    tmp <- data.table(state_abbrev = state.abb, state_name = state.name)
    lookup_df <-tmp[na_idx] 
    tmp[na_idx, state_name := NA]
    df0 <- as_tibble(tmp[sample(length(state.abb), n_row, TRUE)])
    mark(
      dplyr = {
        df1 <- copy(df0)
        df1 <- df1 %>% 
          left_join(lookup_df, by = "state_abbrev") %>% 
          mutate(state_name = coalesce(state_name.x, state_name.y)) %>% 
          select(-state_name.x, -state_name.y)
        df1
      },
      upd_join = {
        df1 <- copy(df0)
        setDT(df1)[setDT(lookup_df), on = "state_abbrev", state_name := i.state_name]
        df1
      }
    )
  }
)
ggplot2::autoplot(bm)

data.tableupate join总是更快(注意日志时间刻度).

随着更新连接修改数据对象,每个基准测试运行都使用一个新副本.