6

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 的包中获得帮助，但它仍然不起作用。错误row_number来自dplyr..

我需要dplyr代码中的所有步骤（即使它们在这里没有意义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 包（v0.3.0）于 2019 年 7 月 14 日创建

4

4 回答 4

8

我可以推荐rowid函数吗？它在“幕后”执行分组步骤，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)]]

如果你喜欢链接，你也可以把所有东西都放在里面[]：

DT[ , .SD[1L], by = mpg
   ][order(am, mpg), row_num := LETTERS[rowid(am)]]
于 2019-07-14T00:19:57.790 回答
6

我正在尝试对翻译进行一些调整，以便 dtplyr 自动生成更像你想要的东西：

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LETTERSdata.table 功能可以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。）

于 2019-07-15T15:43:18.973 回答
5

我们可以用seq_len(.N)

unique(DT, by = "mpg")[order(am, mpg)][, 
     `:=`(row_num = LETTERS[seq_len(.N)]), by = .(am)][]
于 2019-07-13T23:45:52.477 回答
1

由于data.table语法受到严重批评，下面是两个版本的 akrun 答案，恕我直言，语法更清晰。

当它被多次data.table管道传输时，我发现更难遵循代码，特别是当有一个调用时（在 dplyr 中）。[:=mutate

library(data.table)
dt = as.data.table(mtcars)

dt = unique(dt, by = "mpg")
dt = dt[order(am, mpg)]
dt[, row_num:=LETTERS[seq_len(.N)], by=am]
dt[1:3]

    mpg cyl disp  hp drat   wt  qsec vs am gear carb row_num
1: 10.4   8  472 205 2.93 5.25 17.98  0  0    3    4       A
2: 13.3   8  350 245 3.73 3.84 15.41  0  0    3    4       B
3: 14.3   8  360 245 3.21 3.57 15.84  0  0    3    4       C

另一种选择是使用%>%管道。

library(magrittr)

dt = as.data.table(mtcars)
dt = unique(dt, by = "mpg") %>%
  .[order(am, mpg)] %>%
  .[, row_num:=LETTERS[seq_len(.N)], by=am]
dt[1:3]

#     mpg cyl disp  hp drat   wt  qsec vs am gear carb row_num
# 1: 10.4   8  472 205 2.93 5.25 17.98  0  0    3    4       A
# 2: 13.3   8  350 245 3.73 3.84 15.41  0  0    3    4       B
# 3: 14.3   8  360 245 3.21 3.57 15.84  0  0    3    4       C
于 2019-07-15T19:40:57.130 回答