将dplyr转换为data.table

Question

所以我正在尝试翻译一些dplyr代码。我试图从将dplyr转换为data.table的程序包中获得帮助，但仍然无法正常工作。错误是row_number来自dplyr..

我需要dplyr代码中的所有步骤（即使在这里没有用mtcars）

library(dplyr)
library(dtplyr) # from https://github.com/tidyverse/dtplyr
library(data.table)

mtcars %>% 
  distinct(mpg, .keep_all = TRUE) %>% 
  group_by(am) %>% 
  arrange(mpg, .by_group = TRUE) %>% 
  mutate(row_num = LETTERS[row_number()]) %>% 
  ungroup() 

# using dtplyr
dt <- lazy_dt(mtcars)

dt %>% 
  distinct(mpg, .keep_all = TRUE) %>% 
  group_by(am) %>% 
  arrange(mpg, .by_group = TRUE) %>% 
  mutate(row_num = LETTERS[row_number()]) %>% 
  ungroup() %>% 
  show_query()
#> unique(`_DT1`, by = "mpg")[order(am, mpg)][, `:=`(row_num = c("A", 
#> "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", 
#> "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z")[row_number()]), 
#>     keyby = .(am)]

# I then use the query from dtplyr 
DT <- as.data.table(mtcars)
unique(DT, by = "mpg")[order(am, mpg)][, `:=`(row_num = c("A", 
                                                              "B", "C", "D", "E", "F", "G", 
                                                              "H", "I", "J", "K", "L", "M", 
                                                              "N", "O", "P", "Q", "R", "S", 
                                                              "T", "U", "V", "W", "X", "Y", 
                                                              "Z")[row_number()]), keyby = .(am)]

#> row_number() should only be called in a data context

^{由reprex软件包（v0.3.0）创建于2019-07-14}

Answer 1

我可以推荐rowid函数吗？它在“幕后”进行了分组步骤，您可能会发现它看起来更干净：

unique(DT, by='mpg')[order(am, mpg), row_num := LETTERS[rowid(am)]]

如果您喜欢链接，还可以将所有内容都包含在内[]：

DT[ , .SD[1L], by = mpg
   ][order(am, mpg), row_num := LETTERS[rowid(am)]]

Answer 2

我正在尝试对翻译进行一些调整，以便 dtplyr 会自动生成更像您想要的内容：

library(dtplyr)
library(dplyr, warn.conflicts = FALSE)

dt <- lazy_dt(mtcars)

dt %>% 
  distinct(mpg, .keep_all = TRUE) %>% 
  group_by(am) %>% 
  arrange(mpg, .by_group = TRUE) %>% 
  mutate(row_num = LETTERS[row_number()]) %>% 
  ungroup() %>% 
  show_query()
#> unique(`_DT1`, by = "mpg")[order(am, mpg)][, `:=`(row_num = ..LETTERS[seq_len(.N)]), 
#>    keyby = .(am)]

或者像@MichaelChirico 建议的那样避免分组：

dt %>% 
  distinct(mpg, .keep_all = TRUE) %>% 
  arrange(am, mpg) %>% 
  mutate(row_num = LETTERS[row_number(am)]) %>% 
  ungroup() %>% 
  show_query()
#> unique(`_DT1`, by = "mpg")[order(am, mpg)][, `:=`(row_num =  ..LETTERS[frank(am, 
#>    ties.method = "first", na.last = "keep")])]

（使用..前面的LETTERS是 data.table 功能，它清楚地表明您指的是数据框之外的变量；这里可能没有必要，但我认为安全总比抱歉好。）

Answer 3

我们可以用seq_len(.N)

unique(DT, by = "mpg")[order(am, mpg)][, 
     `:=`(row_num = LETTERS[seq_len(.N)]), by = .(am)][]