將 Dataframe 中的 NULL 值替換為 R 中的 Median of Column

Question

我有一個 dataframe 有多個NULL值。 class 列的類型是LIST而不是NUMERIC 。 是否可以用列的中值替換所有NULL值？ 我嘗試了一種手動方法，即使用as.numeric() function 將第 1 列的NULL值更改為 1，然后應用median() function。是否有更有效的方法來執行此操作？

i1 <- sapply(pivot_table_1$`Start Working`, is.null)
pivot_table_1$`Start Working`[i1] <- 0

來自dput() ：

structure(list(Day = 1:31, `Start Sleeping` = list(0, 20, 35, 
    40, 50, 0, 40, 0, 0, 40, 50, 0, 0, 40, 0, 40, 35, 45, 0, 
    0, 65, 35, 40, 40, 0, 50, 40, 0, 0, 0, 0), `Stop Sleeping` = list(
    440, 440, 440, 440, 440, 440, 440, 440, 440, 440, 440, 440, 
    440, 440, 440, 440, 440, 440, 440, 440, 440, 440, 440, 440, 
    440, 440, 440, 440, 440, 440, 440), `Start Working` = list(
    490, 490, 490, 490, 0, 0, 490, 490, 490, 490, 490, 0, 0, 
    490, 490, 490, 490, 490, 0, 0, 490, 490, 490, 490, 490, 0, 
    0, 490, 490, 490, 490), `Stop Working` = list(1005, 1005, 
    1005, 1005, NULL, NULL, 965, 965, 965, 965, 965, NULL, NULL, 
    965, 965, 965, 965, 965, NULL, NULL, 965, 965, 965, 965, 
    965, NULL, NULL, 965, 965, 965, 965), Breakfast = list(690, 
    645, 615, 540, NULL, NULL, NULL, NULL, NULL, NULL, NULL, 
    475, NULL, NULL, NULL, NULL, NULL, NULL, NULL, 475, NULL, 
    NULL, NULL, NULL, NULL, 475, NULL, NULL, NULL, NULL, NULL), 
    Dinner = list(1390, 1360, 1285, 1270, 1390, NULL, 1140, 1140, 
        1130, 1135, 1130, NULL, 1165, 1140, 1130, 1135, 1130, 
        1140, 1140, 1180, NULL, 1145, 1135, 1140, 1135, 1160, 
        1140, 1140, NULL, 1140, NULL)), row.names = c(NA, -31L
), class = c("tbl_df", "tbl", "data.frame"))

Answer 1

如果您希望將條目保留為長度為一的列表，您可以執行以下操作：

pivot_table_1[] <- lapply(pivot_table_1, function(x) {
    ifelse(lengths(x) == 1, x, list(median(unlist(x))))})

pivot_table_1
#> # A tibble: 31 x 7
#>      Day `Start Sleeping` `Stop Sleeping` `Start Working` `Stop Working`
#>    <int> <list>           <list>          <list>          <list>        
#>  1     1 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  2     2 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  3     3 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  4     4 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  5     5 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  6     6 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  7     7 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  8     8 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  9     9 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#> 10    10 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#> # ... with 21 more rows, and 2 more variables: Breakfast <list>, Dinner <list>

或者，如果您希望它們作為數字列，請執行以下操作：

pivot_table_1[] <- lapply(pivot_table_1, function(x) {
    unlist(ifelse(lengths(x) == 1, x, list(median(unlist(x)))))})

pivot_table_1
#> # A tibble: 31 x 7
#>      Day `Start Sleeping` `Stop Sleeping` `Start Working` `Stop Working`
#>    <int>            <dbl>           <dbl>           <dbl>          <dbl>
#>  1     1                0             440             490           1005
#>  2     2               20             440             490           1005
#>  3     3               35             440             490           1005
#>  4     4               40             440             490           1005
#>  5     5               50             440               0            965
#>  6     6                0             440               0            965
#>  7     7               40             440             490            965
#>  8     8                0             440             490            965
#>  9     9                0             440             490            965
#> 10    10               40             440             490            965
#> # ... with 21 more rows, and 2 more variables: Breakfast <dbl>, Dinner <dbl>

^{由reprex package (v2.0.1) 創建於 2022-05-22}

Answer 2

tidyr的replace_na()可用於替換列表中的NULL s。 （ NULL是NA的列表列等效項）

library(tidyverse)

replace_na(df, map(keep(df, is.list), ~ list(median(unlist(.x)))))

# # A tibble: 31 × 7
#      Day `Start Sleeping` `Stop Sleeping` `Start Working` `Stop Working` Breakfast Dinner   
#    <int> <list>           <list>          <list>          <list>         <list>    <list>   
#  1     1 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>      <dbl [1]> <dbl [1]>
#  2     2 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>      <dbl [1]> <dbl [1]>
#  3     3 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>      <dbl [1]> <dbl [1]>
#  4     4 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>      <dbl [1]> <dbl [1]>
#  5     5 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>      <dbl [1]> <dbl [1]>
#  6     6 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>      <dbl [1]> <dbl [1]>
#  7     7 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>      <dbl [1]> <dbl [1]>
#  8     8 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>      <dbl [1]> <dbl [1]>
#  9     9 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>      <dbl [1]> <dbl [1]>
# 10    10 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>      <dbl [1]> <dbl [1]>
# # … with 21 more rows

---

如果您希望這些列表列被展平，請嘗試unnest() ：

replace_na(df, map(keep(df, is.list), ~ list(median(unlist(.x))))) %>%
  unnest(where(is.list))

# # A tibble: 31 × 7
#      Day `Start Sleeping` `Stop Sleeping` `Start Working` `Stop Working` Breakfast Dinner
#    <int>            <dbl>           <dbl>           <dbl>          <dbl>     <dbl>  <dbl>
#  1     1                0             440             490           1005       690   1390
#  2     2               20             440             490           1005       645   1360
#  3     3               35             440             490           1005       615   1285
#  4     4               40             440             490           1005       540   1270
#  5     5               50             440               0            965       540   1390
#  6     6                0             440               0            965       540   1140
#  7     7               40             440             490            965       540   1140
#  8     8                0             440             490            965       540   1140
#  9     9                0             440             490            965       540   1130
# 10    10               40             440             490            965       540   1135

Answer 3

另一種可能的解決方案：

library(tidyverse)

df %>% 
  mutate(across(-Day, ~ ifelse(lengths(.x) == 0, median(unlist(.x)), .x))) 

#> # A tibble: 31 × 7
#>      Day `Start Sleeping` `Stop Sleeping` `Start Working` `Stop Working`
#>    <int> <list>           <list>          <list>          <list>        
#>  1     1 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  2     2 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  3     3 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  4     4 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  5     5 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  6     6 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  7     7 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  8     8 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#>  9     9 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#> 10    10 <dbl [1]>        <dbl [1]>       <dbl [1]>       <dbl [1]>     
#> # … with 21 more rows, and 2 more variables: Breakfast <list>, Dinner <list>

---

取消嵌套：

library(tidyverse)

df %>% 
  mutate(across(-Day, ~ ifelse(lengths(.x) == 0, median(unlist(.x)), .x))) %>% 
  unnest(everything())

#> # A tibble: 31 × 7
#>      Day `Start Sleeping` `Stop Sleeping` `Start Working` `Stop Working`
#>    <int>            <dbl>           <dbl>           <dbl>          <dbl>
#>  1     1                0             440             490           1005
#>  2     2               20             440             490           1005
#>  3     3               35             440             490           1005
#>  4     4               40             440             490           1005
#>  5     5               50             440               0            965
#>  6     6                0             440               0            965
#>  7     7               40             440             490            965
#>  8     8                0             440             490            965
#>  9     9                0             440             490            965
#> 10    10               40             440             490            965
#> # … with 21 more rows, and 2 more variables: Breakfast <dbl>, Dinner <dbl>