如何根据多个条件延迟日期并在发生后重置延迟

Question

我有一个车站维修的数据框。

工作流程是这样的：机械师去一个站点，他们按下一个按钮，记录一个名为 release 的 action。在他们修好电台后，他们再次按下按钮，现在的动作是 return.

您可以在下面看到 row 1 和 row 2 是一个已完成的任务，需要 Jane Jetson 10 秒完成。

                   dt        name foo_id foo_role bikeId station_name station_id  action
1 2019-12-12 13:05:47 Jane Jetson 106337 Mechanic  12345   FooStation    1234.89 Release
2 2019-12-12 13:05:57 Jane Jetson 106337 Mechanic  12345   FooStation    1234.89  Return
3 2019-12-12 13:06:16    John Doe 106338 Mechanic  12345   FooStation    1234.89 Release
4 2019-12-12 13:06:19    John Doe 106338 Mechanic  12345   FooStation    1234.89  Return
5 2019-12-12 13:07:16    John Doe 106338 Mechanic  12345   FooStation    1234.89 Release
6 2019-12-12 14:07:16    John Doe 106338 Mechanic  56789 Some Station    4567.12 Release

我想要发生的事情：

我想知道每个mechanic使用actionRelease然后跟随Return修复站需要多长时间。
如果 Release 没有 Return，我想用 Sys.time() 从 dt 中减去它。你会看到 row 5 和 row 6

我这样做了：（我不是 100% 确定我需要之前的操作，但我包括以防万一。）

library(dplyr)
library(tidyr)
foo = arrange(foo, foo_id, name, foo_role, bikeId, station_id) %>% 
  group_by(foo_id,name, foo_role, bikeId, station_name,station_id) %>%
  mutate(prev_dt = lag(dt, order_by = foo_id), 
         prev_action = lag(action, order_by=foo_id, default = 'NaN'))
foo$timediffsecs = as.numeric(difftime(foo$dt,foo$prev_dt,units='secs'))

> foo
# A tibble: 6 x 11
# Groups:   foo_id, name, foo_role, bikeId, station_name, station_id [3]
  dt                  name        foo_id foo_role bikeId station_name station_id action  prev_dt             prev_action timediffsecs
  <dttm>              <fct>        <int> <fct>     <int> <fct>             <dbl> <chr>   <dttm>              <chr>              <dbl>
1 2019-12-12 13:05:47 Jane Jetson 106337 Mechanic  12345 FooStation        1235. Release NA                  NaN                   NA
2 2019-12-12 13:05:57 Jane Jetson 106337 Mechanic  12345 FooStation        1235. Return  2019-12-12 13:05:47 Release               10
3 2019-12-12 13:06:16 John Doe    106338 Mechanic  12345 FooStation        1235. Release NA                  NaN                   NA
4 2019-12-12 13:06:19 John Doe    106338 Mechanic  12345 FooStation        1235. Return  2019-12-12 13:06:16 Release                3
5 2019-12-12 13:07:16 John Doe    106338 Mechanic  12345 FooStation        1235. Release 2019-12-12 13:06:19 Return                57
6 2019-12-12 14:07:16 John Doe    106338 Mechanic  56789 Some Station      4567. Release NA                  NaN                   NA

问题:

row 5 是一个新的循环，因为 action Release 和 Return 之前发生过，但 timediffsecs 记录了 57 秒。 row 5 Prev_dt 和 prev_action 应该是 NA 和 timediffsecs = Sys.time() - dt.
row 6 应该有 timediffsecs = Sys.time() - dt

我认为可行的方法：

我将 prev_action NA 更改为 NaN 这样我就可以做一些 if else 语句，但我不太确定如何为此构造一个。我想将 prev_dt 中的 NA 更改为默认 dt，但这样做有问题。我想尝试这个的原因是我可以使用条件语句，但如果不需要，则无需更改 NA。

tl;dr: 我希望 timediffsecs 记录正确的秒数。 row 5 和 row 6 有问题。 row 5 应该是 Sys.time() - dt。 row 6我要returnSys.time() - dt

数据：

structure(list(dt = structure(c(1576173947, 1576173957, 1576173976, 
1576173979, 1576174036, 1576177636), class = c("POSIXct", "POSIXt"
), tzone = ""), name = structure(c(1L, 1L, 2L, 2L, 2L, 2L), .Label = c("Jane Jetson", 
"John Doe"), class = "factor"), foo_id = c(106337L, 106337L, 
106338L, 106338L, 106338L, 106338L), foo_role = structure(c(1L, 
1L, 1L, 1L, 1L, 1L), .Label = "Mechanic", class = "factor"), 
    bikeId = c(12345L, 12345L, 12345L, 12345L, 12345L, 56789L
    ), station_name = structure(c(1L, 1L, 1L, 1L, 1L, 2L), .Label = c("FooStation", 
    "Some Station"), class = "factor"), station_id = c(1234.89, 
    1234.89, 1234.89, 1234.89, 1234.89, 4567.12), action = c("Release", 
    "Return", "Release", "Return", "Release", "Release")), row.names = c(NA, 
-6L), class = "data.frame")

Answer 1

使用 dplyr 的一种方法可能是为每个 name（或 foo_id）创建组，并且每次 "Release" 发生。在这个组中，如果存在 'Return'，我们计算 'Return' 和 'Release' 的时间差，或者计算 'Release' 和当前时间之间的差值。

library(dplyr)

df %>%
  group_by(name, group = cumsum(action == "Release")) %>%
  mutate(timediffsecs = if (any(action == 'Return')) 
            dt[action == 'Return'] - dt[action == 'Release'] else Sys.time() - dt, 
         #If we want to replace Release values with NA
         timediffsecs  = replace(timediffsecs, n() > 1 & action == 'Release', NA))


#   dt                  name        foo_id foo_role bikeId station_name station_id action  group timediffsecs
#  <dttm>              <fct>        <int> <fct>     <int> <fct>             <dbl> <chr>   <int> <drtn>      
#1 2019-12-13 02:05:47 Jane Jetson 106337 Mechanic  12345 FooStation        1235. Release     1     NA secs 
#2 2019-12-13 02:05:57 Jane Jetson 106337 Mechanic  12345 FooStation        1235. Return      1     10 secs 
#3 2019-12-13 02:06:16 John Doe    106338 Mechanic  12345 FooStation        1235. Release     2     NA secs 
#4 2019-12-13 02:06:19 John Doe    106338 Mechanic  12345 FooStation        1235. Return      2      3 secs 
#5 2019-12-13 02:07:16 John Doe    106338 Mechanic  12345 FooStation        1235. Release     3 472603 secs 
#6 2019-12-13 03:07:16 John Doe    106338 Mechanic  56789 Some Station      4567. Release     4 469003 secs

虽然这适用于给定的示例，但可能需要根据数据进行一些调整。

如何根据多个条件延迟日期并在发生后重置延迟

How to lag dates based on multiple conditions and reseting the lag after it occurs

loops

if-statement

r

lag