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当条件检查不存在的行时失败

tidyverse r

deschen · 技术社区 · 2 年前

考虑以下数据:

df <- data.frame(group  = c(1, 2, 2, 2),
                 start  = c(2, 7, 7, 7),
                 stop   = c(8, 7, 8, 9),
                 unstop = c(10, 7, 9, 10))

现在,我想以“如果第一行这样做,如果第二行那样做”的形式为每个组设置一个或多或少简单的case_。然而,我得到了一个错误。我认为这是因为组1只有一行,所以无法检查条件:

df |>
  group_by(group) |> 
  mutate(n_rows = n(),
         split_weeks = case_when(n_rows == 1 ~ str_c(start:stop, collapse = ","),
                                 n_rows  > 1 & row_number() == 1 ~ str_c(c(start:stop, unstop:lead(stop)), collapse = ","),
                                 TRUE ~ "fail"))

Error in `mutate()`:
! Problem while computing `split_weeks = case_when(...)`.
â¹ The error occurred in group 1: group = 1.
Caused by error in `unstop:lead(stop)`:
! NA/NaN argument
Run `rlang::last_error()` to see where the error occurred.

知道这是怎么回事吗?

我想这和 lead 函数,因为如果我删除该部分,我“只会收到警告,但至少会得到一个结果。”。

预期产出:

# A tibble: 4 Ã 6
# Groups:   group [2]
  group start  stop unstop n_rows split_weeks  
  <dbl> <dbl> <dbl>  <dbl>  <int> <chr>        
1     1     2     8     10      1 2,3,4,5,6,7,8
2     2     7     7      7      3 7,8          
3     2     7     8      9      3 fail         
4     2     7     9     10      3 fail

2 回复 | 直到 2 年前

Jon Spring 2 年前

这里有一个替代方案,可以产生所需的输出(至少在这种情况下)@安迪·巴克斯特(Andy Baxter)很好地解释了原版失败的原因;即使case_使用第一个case的结果,第二个case抛出一个错误,因此操作失败。你可以通过使用 lead(stop, default = 0) 或者 coalesce(lead(stop), SOMETHING) ,当没有“下一个”值时,这两者都会产生一个可计算的结果(如果没有意义/不需要)。

df |>
  group_by(group) |> 
  mutate(n_rows = n()) %>%
  mutate(split_weeks = case_when(
    n_rows == 1 ~ str_c(start:stop, collapse = ","),
    n_rows  > 1 & row_number() == 1 ~ str_c(unstop:(lead(stop, default = 0)), collapse = ","),
    # n_rows  > 1 & row_number() == 1 ~ str_c(unstop:(coalesce(lead(stop), unstop)), collapse = ","), # Alternative
    TRUE ~ "fail"))

后果

# A tibble: 4 Ã 6
# Groups:   group [2]
  group start  stop unstop n_rows split_weeks  
  <dbl> <dbl> <dbl>  <dbl>  <int> <chr>        
1     1     2     8     10      1 2,3,4,5,6,7,8
2     2     7     7      7      3 7,8          
3     2     7     8      9      3 fail         
4     2     7     9     10      3 fail

Andy Baxter 2 年前

我认为当你要求R找到一个 lead a)在表行末尾之外,或b)在组之外。您可以将默认值0传递给它,该值永远不会被使用,并且会抑制错误,但由于函数试图连接每个 start:stop 和 unstop:lead(stop) 每组所有行的值:

library(tidyverse)

df <- data.frame(group  = c(1, 2, 2, 2),
                 start  = c(2, 7, 7, 7),
                 stop   = c(8, 7, 8, 9),
                 unstop = c(10, 7, 9, 10))


df |>
  group_by(group) |>
  mutate(
    n_rows = n(),
    split_weeks = case_when(
      n_rows == 1 ~ str_c(start:stop, collapse = ","),
      n_rows  > 1 &
        row_number() == 1 ~ str_c(c(start:stop, unstop:lead(stop, default = 0)), collapse = ","),
      TRUE ~ "fail"
    )
  )
#> Warning in start:stop: numerical expression has 3 elements: only the first used

#> Warning in start:stop: numerical expression has 3 elements: only the first used

#> Warning in start:stop: numerical expression has 3 elements: only the first used

#> Warning in start:stop: numerical expression has 3 elements: only the first used
#> Warning in unstop:lead(stop, 0): numerical expression has 3 elements: only the
#> first used

#> Warning in unstop:lead(stop, 0): numerical expression has 3 elements: only the
#> first used
#> # A tibble: 4 Ã 6
#> # Groups:   group [2]
#>   group start  stop unstop n_rows split_weeks  
#>   <dbl> <dbl> <dbl>  <dbl>  <int> <chr>        
#> 1     1     2     8     10      1 2,3,4,5,6,7,8
#> 2     2     7     7      7      3 7,7          
#> 3     2     7     8      9      3 fail         
#> 4     2     7     9     10      3 fail

整理的一种方法是:

查找组外的lead值(默认设置为避免最后一行出现错误)
查找组中的行数和行号
再解组!
按行计算,使R只关注该行中的值
执行串联

结果就是这样,虽然不知道为什么 7,7,8 被放入该单元格(有意义,因为它连接了7到7和7到8):

df |> 
  mutate(lead_stop = lead(stop, default = 0)) |>
  group_by(group) |>
  mutate(
    n_rows = n(),
    rownum = row_number()) |>
  ungroup() |>
  rowwise() |>
  mutate(
    split_weeks = case_when(
      rownum > 1 ~ "fail",
      n_rows == 1 ~ str_c(start:stop, collapse = ","),
      n_rows  > 1 & rownum == 1 ~ str_c(c(start:stop, unstop:lead_stop), collapse = ","),
      TRUE ~ "fail"
    )
  )
#> # A tibble: 4 Ã 8
#> # Rowwise: 
#>   group start  stop unstop lead_stop n_rows rownum split_weeks  
#>   <dbl> <dbl> <dbl>  <dbl>     <dbl>  <int>  <int> <chr>        
#> 1     1     2     8     10         7      1      1 2,3,4,5,6,7,8
#> 2     2     7     7      7         8      3      1 7,7,8        
#> 3     2     7     8      9         9      3      2 fail         
#> 4     2     7     9     10         0      3      3 fail

^{于2022年5月7日由

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