r - data.table 相当于 tidyr::complete with group_by

Question

我有以下数据框:

library(tidyverse)
df <- data_frame(
  id = c(1, 1, 2, 2), 
  date1 = as.Date(c("2013-01-01", "2013-02-01", "2015-04-01", "2015-05-01")), 
  date2 = as.Date(c("2012-12-09", "2012-12-09", "2015-03-10", "2015-03-10"))
)

# A tibble: 4 x 3
     id      date1      date2
  <dbl>     <date>     <date>
1     1 2013-01-01 2012-12-09
2     1 2013-02-01 2012-12-09
3     2 2015-04-01 2015-03-10
4     2 2015-05-01 2015-03-10

我想完成这个数据框，这样对于每个 id ，都会有另一个 date1 值。这另一个 date1 值计算为下个月。还有一个 date2 值对于所有 id 都是相同的。使用 tidyr::complete 这个 Action 可以这样完成:

df %>% 
  group_by(id) %>% 
  complete(date1 = seq.Date(from = min(date1), length.out = 3, by = "month"), date2 = date2[1])

# A tibble: 6 x 3
# Groups:   id [2]
     id      date1      date2
  <dbl>     <date>     <date>
1     1 2013-01-01 2012-12-09
2     1 2013-02-01 2012-12-09
3     1 2013-03-01 2012-12-09
4     2 2015-04-01 2015-03-10
5     2 2015-05-01 2015-03-10
6     2 2015-06-01 2015-03-10

由于我的原始数据中有大约 150K 组，因此 tidyr 解决方案需要一个多小时才能完成。我假设使用 data.table 可以提高速度。在 data.table 中可以做同样的事情吗？

data.table equivalent of tidyr::complete() 中也有类似的问题，但没有 group_by 子句。

Answer 1

基于一些初始基准测试，data.table 方法似乎更快

library(data.table)
setDT(df)[, .(date1 = seq(min(date1), length.out = 3, by = 'month'), date2 = date2[1]), id]

基准

 df <- data_frame(
  id = rep(1:3000, each = 2), 
  date1 = rep(as.Date(c("2013-01-01", "2013-02-01", "2015-04-01", "2015-05-01")),
  length.out = 6000), 
  date2 = rep(as.Date(c("2012-12-09", "2012-12-09", "2015-03-10", "2015-03-10")), 
   length.out = 6000))

system.time({
df %>% 
  group_by(id) %>% 
  complete(date1 = seq.Date(from = min(date1), 
          length.out = 3, by = "month"), date2 = date2[1])
})
#user  system elapsed 
#64.05   21.27   86.05 

system.time({
setDT(df)[, .(date1 = seq(min(date1), length.out = 3, by = 'month'), date2 = date2[1]), id]
})
#user  system elapsed 
#  0.14    0.00    0.14