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

我希望完成此数据框，以便每个id都有另一个date1值。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，可以按以下方式执行此操作：

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，tidyr解决方案需要花费超过一小时才能完成。我假设使用data.table可以获得速度。可以在data.table中完成同样的事情吗？

在data.table equivalent of tidyr::complete()中提出了类似问题，但没有group_by条款。

Answer 1

根据一些初步基准测试，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

Answer 2

如果你需要速度，尽量保持精简：

library(data.table)
library(lubridate)

> dt[, .SD
     ][, .(date1=max(date1)), .(id, date2)
     ][, date1Inc := date1 + months(1)
     ][, rbind(dt, .SD[, .(id, date1=date1Inc, date2)])
     ][order(id, date1)
     ]

   id      date1      date2
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
>   
>

data.table相当于tidyr :: complete with group_by

2 个答案:

基准