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使用tidyverse join更新/替换数据帧中的值

dplyr r

2

Nettle · 技术社区 · 8 年前

用查找表中的(正确的)值更新/替换主数据集中的NAS最有效的方法是什么?这是一个很普通的手术!类似的问题似乎没有整齐的解决方案。

制约因素: 1)请假设有大量的缺失值和比给出的示例更大的查找表。因此,根据具体情况进行替换操作是不切实际的(不 case_when , if_else 等)

2)查找表没有主数据帧的所有值,只有替换值。

最好用tidyverse解决方案回答。类似的问题似乎没有整齐的解决方案。

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,左连接对于缺少的值有一个替换选项。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

` `

创建日期:2018-07-28 reprex package (v0.2.0)。

4 回复 | 直到 7 年前

1

5

Uwe 8 年前

和 Nettle's 建议并转化为有效的解决方案

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

运营商表示更喜欢“tidyverse”解决方案。然而, 更新连接 已经可以使用 data.table 包裹:

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)

数据表 的upate连接总是更快(请注意日志时间刻度)。

作为 更新连接 修改数据对象,每个基准运行都使用一个新的副本。

2

alistaire 8 年前

目前没有一次尝试合并多个列(可以使用 a lookup table approach ifelse(is.na(value), ..., value) there has been discussion coalesce put it in a function

library(tidyverse)

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_df <- tibble(
    state_abbrev = state.abb[6:8],
    state_name = state.name[6:8]
)

df1 %>% 
    full_join(lookup_df, by = 'state_abbrev') %>% 
    bind_cols(map_dfc(grep('.x', names(.), value = TRUE), function(x){
        set_names(
            list(coalesce(.[[x]], .[[gsub('.x', '.y', x)]])), 
            gsub('.x', '', x)
        )
    })) %>% 
    select(union(names(df1), names(lookup_df)))
#> # A tibble: 10 x 3
#>    state_abbrev state_name  value
#>    <chr>        <chr>       <int>
#>  1 AL           Alabama       877
#>  2 AK           Alaska       1048
#>  3 AZ           Arizona       973
#>  4 AR           Arkansas      860
#>  5 CA           California    938
#>  6 CO           Colorado      639
#>  7 CT           Connecticut   547
#>  8 DE           Delaware      672
#>  9 FL           Florida       667
#> 10 GA           Georgia      1142

3

1

GG LEE 7 年前

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

4

-1

sbha 8 年前

left_join(df1 %>% select(-state_name), lookup_df, by = 'state_abbrev') %>% 
  select(state_abbrev, state_name, value)

match if_else mutate

df1 %>% 
  mutate(state_name = if_else(is.na(state_name), state.name[match(state_abbrev,state.abb)], state_name))

# A tibble: 10 x 3
   state_abbrev state_name  value
   <chr>        <chr>       <int>
 1 AL           Alabama       525
 2 AK           Alaska        719
 3 AZ           Arizona      1186
 4 AR           Arkansas     1051
 5 CA           California    888
 6 CO           Colorado      615
 7 CT           Connecticut   578
 8 DE           Delaware      894
 9 FL           Florida       536
10 GA           Georgia       599