r - 如何将 单元格重新编码为 tibble 列表列中的嵌套 NA ()？

Question

在带有列表列的小标题中，我如何替换 <NULL>嵌套条目 NA (这将采用 <lgl [1]> 的嵌套形式)？

library(tibble)

tbl_with_null <-
  tibble(letter =  letters[1:10],
       value_1 = list(1, 2, 4, data.frame(a = 1, 2, 3), NULL, 6, 7, c(8, 11, 25), NULL, 10),
       value_2 = list("A", "B", "C", "D", NULL, NULL, NULL, list("H", "B", list(data.frame(id = 1:3))), "I", "J"))

> tbl_with_null
 
## # A tibble: 10 x 3
##    letter value_1          value_2   
##    <chr>  <list>           <list>    
##  1 a      <dbl [1]>        <chr [1]> 
##  2 b      <dbl [1]>        <chr [1]> 
##  3 c      <dbl [1]>        <chr [1]> 
##  4 d      <df[,3] [1 x 3]> <chr [1]> 
##  5 e      <NULL>           <NULL>    
##  6 f      <dbl [1]>        <NULL>    
##  7 g      <dbl [1]>        <NULL>    
##  8 h      <dbl [3]>        <list [3]>
##  9 i      <NULL>           <chr [1]> 
## 10 j      <dbl [1]>        <chr [1]> 

有没有办法对整个 tbl_with_null采取行动替换 <NULL>与 NA要得到:

## # A tibble: 10 x 3
##    letter value_1                 value_2   
##    <chr>  <list>                  <list>    
##  1 a      <dbl [1]>               <chr [1]> 
##  2 b      <dbl [1]>               <chr [1]> 
##  3 c      <dbl [1]>               <chr [1]> 
##  4 d      <df[,3] [1 x 3]>        <chr [1]> 
##  5 e      <lgl [1]> <- NA         <lgl [1]>  # <- NA
##  6 f      <dbl [1]>               <lgl [1]>  # <- NA
##  7 g      <dbl [1]>               <lgl [1]>  # <- NA
##  8 h      <dbl [3]>               <list [3]>
##  9 i      <lgl [1]> <- NA         <chr [1]> 
## 10 j      <dbl [1]>               <chr [1]> 

更新

我在 this solution的基础上取得了一些进展:

tbl_with_null %>%
  mutate(across(c(value_1, value_2), ~replace(., !lengths(.), list(NA))))

## # A tibble: 10 x 3
##    letter value_1          value_2   
##    <chr>  <list>           <list>    
##  1 a      <dbl [1]>        <chr [1]> 
##  2 b      <dbl [1]>        <chr [1]> 
##  3 c      <dbl [1]>        <chr [1]> 
##  4 d      <df[,3] [1 x 3]> <chr [1]> 
##  5 e      <lgl [1]>        <lgl [1]> 
##  6 f      <dbl [1]>        <lgl [1]> 
##  7 g      <dbl [1]>        <lgl [1]> 
##  8 h      <dbl [3]>        <list [3]>
##  9 i      <lgl [1]>        <chr [1]> 
## 10 j      <dbl [1]>        <chr [1]> 

然而 ，这是不够的，因为我正在寻找一种解决方案， 盲目替换 NULL与 NA整个数据帧。如果我们使用 mutate(across(everything(), ~replace(., !lengths(.), list(NA))))我们知道 letters列也变成了列表列，这是无意的。

## # A tibble: 10 x 3
##    letter    value_1          value_2   
##    <list>    <list>           <list>    
##  1 <chr [1]> <dbl [1]>        <chr [1]> 
##  2 <chr [1]> <dbl [1]>        <chr [1]> 
##  3 <chr [1]> <dbl [1]>        <chr [1]> 
##  4 <chr [1]> <df[,3] [1 x 3]> <chr [1]> 
##  5 <chr [1]> <lgl [1]>        <lgl [1]> 
##  6 <chr [1]> <dbl [1]>        <lgl [1]> 
##  7 <chr [1]> <dbl [1]>        <lgl [1]> 
##  8 <chr [1]> <dbl [3]>        <list [3]>
##  9 <chr [1]> <lgl [1]>        <chr [1]> 
## 10 <chr [1]> <dbl [1]>        <chr [1]> 

更新 2

我以为我已经完成了

mutate(across(everything(), ~simplify(replace(., !lengths(.), list(NA)))))

但不幸的是，这在某些情况下会失败，例如以下数据:

tbl_with_no_null <-
  tbl_with_null %>%
  slice(8) %>%
  select(letter, value_1)

## # A tibble: 1 x 2
##   letter value_1  
##   <chr>  <list>   
## 1 h      <dbl [3]>

虽然我期待

tbl_with_no_null %>%
  mutate(across(everything(), ~simplify(replace(., !lengths(.), list(NA)))))

将返回相同的 tbl_with_no_null (因为没有 <NULL> 来替换):

## # A tibble: 1 x 2
##   letter value_1  
##   <chr>  <list>   
## 1 h      <dbl [3]>

但是我得到了错误:

Error: Problem with `mutate()` input `..1`.
x Input `..1` can't be recycled to size 1.
i Input `..1` is `(function (.cols = everything(), .fns = NULL, ..., .names = NULL) ...`.
i Input `..1` must be size 1, not 3.

底线
我正在寻找一种方法来替换 <NULL>与 NA在列表列中，当然，如果没有 <NULL>替换，然后按原样返回输入。

Answer 1

base::rapply不通过 NULL 递归，但你可以使用 rrapply这允许这样做，并且非常有效:

library(rrapply)
rrapply::rrapply(tbl_with_null, function(x) NA, how = "replace", condition = is.null)

# A tibble: 10 x 3
   letter value_1          value_2   
   <chr>  <list>           <list>    
 1 a      <dbl [1]>        <chr [1]> 
 2 b      <dbl [1]>        <chr [1]> 
 3 c      <dbl [1]>        <chr [1]> 
 4 d      <df[,3] [1 x 3]> <chr [1]> 
 5 e      <lgl [1]>        <lgl [1]> 
 6 f      <dbl [1]>        <lgl [1]> 
 7 g      <dbl [1]>        <lgl [1]> 
 8 h      <dbl [3]>        <list [3]>
 9 i      <lgl [1]>        <chr [1]> 
10 j      <dbl [1]>        <chr [1]> 

或者按照@JorisC 的建议。在评论中，使用 class在大型列表上似乎快 25% 的参数:

rrapply(tbl_with_null, classes = "NULL", how = "replace", f = function(x) NA)

只是为了好玩:

eval(parse(text=gsub("NULL","NA",capture.output(dput(tbl_with_null)))))

# A tibble: 10 x 3
   letter value_1          value_2   
   <chr>  <list>           <list>    
 1 a      <dbl [1]>        <chr [1]> 
 2 b      <dbl [1]>        <chr [1]> 
 3 c      <dbl [1]>        <chr [1]> 
 4 d      <df[,3] [1 x 3]> <chr [1]> 
 5 e      <lgl [1]>        <lgl [1]> 
 6 f      <dbl [1]>        <lgl [1]> 
 7 g      <dbl [1]>        <lgl [1]> 
 8 h      <dbl [3]>        <list [3]>
 9 i      <lgl [1]>        <chr [1]> 
10 j      <dbl [1]>        <chr [1]> 

fortunes::fortune(106)

# If the answer is parse() you should usually rethink the question.
#   -- Thomas Lumley
#      R-help (February 2005)

速度比较出人意料，早有预料 parse成为最慢的解决方案:

microbenchmark::microbenchmark(
  rrapply = rrapply::rrapply(tbl_with_null, function(x) NA, how = "replace", condition = is.null),
  parse = eval(parse(text=gsub("NULL","NA",capture.output(dput(tbl_with_null))))),
  dplyr = mutate(tbl_with_null,across(where(is.list), .fns = map_if, .p = is.null, .f = function(x) NA)))
Unit: microseconds
    expr      min       lq       mean    median        uq      max neval cld
 rrapply   25.401   31.801   60.92102   51.2510   58.3010 1053.502   100 a  
   parse  225.001  269.701  327.31600  329.1005  362.4505  687.800   100  b 
   dplyr 2942.501 3207.301 3604.63105 3500.0005 3766.1510 6541.402   100   c