计数自上次条件满足以来的时间，不满足时重置为 0

【问题标题】：Count time-since last condition met and reset to 0 when not计数自上次条件满足以来的时间，不满足时重置为 0
【发布时间】：2021-12-24 16:49:29
【问题描述】：

我想计算从满足条件开始的时间，但是每次不满足条件时，时间都应该重新回到0。使用dplyr 实现这一目标会很棒，但我愿意接受任何建议。

使用代码更容易查看：

library(dplyr)

d <- structure(list(date = structure(c(17105, 17182, 17275, 17359, 
                                          17437, 17472, 17500, 17539, 
                                          17624, 17658, 17693, 17742, 
                                          17828, 17877, 18004, 18053, 
                                          18087, 18130, 18186, 18214, 
                                          18298, 18415, 18527, 18583, 
                                          18610), 
                                        class = "Date"), 
                    condition = c(FALSE, FALSE, FALSE, FALSE, TRUE, TRUE, TRUE, 
                                  TRUE, TRUE, FALSE, FALSE, FALSE, TRUE, FALSE, 
                                  TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, FALSE, 
                                  TRUE, FALSE, TRUE, TRUE)), 
               class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,-25L)) 
                                                                                                                                                                            
# I can easily calculate the time since the last observation when the 
# condition has been met:
 
dd <- d %>% 
  mutate(time_condition = case_when(
    condition == FALSE ~ date - date, #So it is 0
    condition == TRUE & lag(condition) == FALSE ~ date - date, # again, we want 0
    condition == TRUE & lag(condition) == TRUE ~ date - lag(date)),
    time_condition = as.numeric(time_condition)) 

# This is how it looks now
dd %>% 
  print(n = 25)
#> # A tibble: 25 × 3
#>    date       condition time_condition
#>    <date>     <lgl>              <dbl>
#>  1 2016-10-31 FALSE                  0
#>  2 2017-01-16 FALSE                  0
#>  3 2017-04-19 FALSE                  0
#>  4 2017-07-12 FALSE                  0
#>  5 2017-09-28 TRUE                   0
#>  6 2017-11-02 TRUE                  35
#>  7 2017-11-30 TRUE                  28
#>  8 2018-01-08 TRUE                  39
#>  9 2018-04-03 TRUE                  85
#> 10 2018-05-07 FALSE                  0
#> 11 2018-06-11 FALSE                  0
#> 12 2018-07-30 FALSE                  0
#> 13 2018-10-24 TRUE                   0
#> 14 2018-12-12 FALSE                  0
#> 15 2019-04-18 TRUE                   0
#> 16 2019-06-06 TRUE                  49
#> 17 2019-07-10 TRUE                  34
#> 18 2019-08-22 TRUE                  43
#> 19 2019-10-17 TRUE                  56
#> 20 2019-11-14 TRUE                  28
#> 21 2020-02-06 FALSE                  0
#> 22 2020-06-02 TRUE                   0
#> 23 2020-09-22 FALSE                  0
#> 24 2020-11-17 TRUE                   0
#> 25 2020-12-14 TRUE                  27

我想要的是一种cumsum()，当条件不再满足时重置为 0。数据应如下所示：

should_be <- c(0, 0, 0, 0, 0, 35, 35 + 28, 35 + 28 + 39, 35 + 28 + 39 + 85, 
               0, 0, 0, 0, 0, 0, 49, 49 + 34, 49 + 34 + 43, 49 + 34 + 43 + 56, 
               49 + 34 + 43 + 56 + 28, 0, 0, 0, 0, 27)

dd %>% 
  mutate(time_condition_wanted = should_be) %>% 
  print(n = 25)
#> # A tibble: 25 × 4
#>    date       condition time_condition time_condition_wanted
#>    <date>     <lgl>              <dbl>                 <dbl>
#>  1 2016-10-31 FALSE                  0                     0
#>  2 2017-01-16 FALSE                  0                     0
#>  3 2017-04-19 FALSE                  0                     0
#>  4 2017-07-12 FALSE                  0                     0
#>  5 2017-09-28 TRUE                   0                     0
#>  6 2017-11-02 TRUE                  35                    35
#>  7 2017-11-30 TRUE                  28                    63
#>  8 2018-01-08 TRUE                  39                   102
#>  9 2018-04-03 TRUE                  85                   187
#> 10 2018-05-07 FALSE                  0                     0
#> 11 2018-06-11 FALSE                  0                     0
#> 12 2018-07-30 FALSE                  0                     0
#> 13 2018-10-24 TRUE                   0                     0
#> 14 2018-12-12 FALSE                  0                     0
#> 15 2019-04-18 TRUE                   0                     0
#> 16 2019-06-06 TRUE                  49                    49
#> 17 2019-07-10 TRUE                  34                    83
#> 18 2019-08-22 TRUE                  43                   126
#> 19 2019-10-17 TRUE                  56                   182
#> 20 2019-11-14 TRUE                  28                   210
#> 21 2020-02-06 FALSE                  0                     0
#> 22 2020-06-02 TRUE                   0                     0
#> 23 2020-09-22 FALSE                  0                     0
#> 24 2020-11-17 TRUE                   0                     0
#> 25 2020-12-14 TRUE                  27                    27

由reprex package (v2.0.1) 于 2021 年 11 月 12 日创建

【问题讨论】：

    标签： r date dplyr duration cumsum


    【解决方案1】：

    我刚刚为这个问题想出了（有点复杂）的解决方案，并将其发布在这里以供后代使用。如果有更简单的解决方案，那就太好了。

    library(dplyr)

d <- structure(list(date = structure(c(17105, 17182, 17275, 17359, 
                                          17437, 17472, 17500, 17539, 
                                          17624, 17658, 17693, 17742, 
                                          17828, 17877, 18004, 18053, 
                                          18087, 18130, 18186, 18214, 
                                          18298, 18415, 18527, 18583, 
                                          18610), 
                                        class = "Date"), 
                    condition = c(FALSE, FALSE, FALSE, FALSE, 
                                  TRUE, TRUE, TRUE, TRUE, 
                                  TRUE, FALSE, FALSE, FALSE, 
                                  TRUE, FALSE, TRUE, TRUE, TRUE, 
                                  TRUE, TRUE, TRUE, FALSE, 
                                  TRUE, FALSE, TRUE, TRUE)), 
               class = c("tbl_df", "tbl", "data.frame"), 
               row.names = c(NA,-25L)) 
                                                                                                                                                                            

dd <- d %>% 
  mutate(time_condition = case_when(
    condition == FALSE ~ date - date, #So it is 0
    condition == TRUE & lag(condition) == FALSE ~ date - date, # again, we want 0
    condition == TRUE & lag(condition) == TRUE ~ date - lag(date)),
    time_condition = as.numeric(time_condition)) 


should_be <- c(0, 0, 0, 0, 0, 35, 35 + 28, 35 + 28 + 39, 35 + 28 + 39 + 85, 
               0, 0, 0, 0, 0, 0, 49, 49 + 34, 49 + 34 + 43, 49 + 34 + 43 + 56, 
               49 + 34 + 43 + 56 + 28, 0, 0, 0, 0, 27)

    我可以创建一个change 变量，每次状态更改时都会增加。之后，cumsum() 可以在 group_by 那个 change 变量的数据集上使用。

    
dd %>% 
  mutate(time_condition_wanted = should_be) %>% 
  # create that increases at each status change
  mutate(change = if_else(condition != lag(condition), 1, 0),
         change = if_else(is.na(change), 0, change),
         change = cumsum(change)) %>% 
  group_by(change) %>% 
  mutate(y = cumsum(time_condition)) %>% 
  print(n = 25)
#> # A tibble: 25 × 6
#> # Groups:   change [10]
#>    date       condition time_condition time_condition_wanted change     y
#>    <date>     <lgl>              <dbl>                 <dbl>  <dbl> <dbl>
#>  1 2016-10-31 FALSE                  0                     0      0     0
#>  2 2017-01-16 FALSE                  0                     0      0     0
#>  3 2017-04-19 FALSE                  0                     0      0     0
#>  4 2017-07-12 FALSE                  0                     0      0     0
#>  5 2017-09-28 TRUE                   0                     0      1     0
#>  6 2017-11-02 TRUE                  35                    35      1    35
#>  7 2017-11-30 TRUE                  28                    63      1    63
#>  8 2018-01-08 TRUE                  39                   102      1   102
#>  9 2018-04-03 TRUE                  85                   187      1   187
#> 10 2018-05-07 FALSE                  0                     0      2     0
#> 11 2018-06-11 FALSE                  0                     0      2     0
#> 12 2018-07-30 FALSE                  0                     0      2     0
#> 13 2018-10-24 TRUE                   0                     0      3     0
#> 14 2018-12-12 FALSE                  0                     0      4     0
#> 15 2019-04-18 TRUE                   0                     0      5     0
#> 16 2019-06-06 TRUE                  49                    49      5    49
#> 17 2019-07-10 TRUE                  34                    83      5    83
#> 18 2019-08-22 TRUE                  43                   126      5   126
#> 19 2019-10-17 TRUE                  56                   182      5   182
#> 20 2019-11-14 TRUE                  28                   210      5   210
#> 21 2020-02-06 FALSE                  0                     0      6     0
#> 22 2020-06-02 TRUE                   0                     0      7     0
#> 23 2020-09-22 FALSE                  0                     0      8     0
#> 24 2020-11-17 TRUE                   0                     0      9     0
#> 25 2020-12-14 TRUE                  27                    27      9    27

    由reprex package (v2.0.1) 于 2021 年 11 月 12 日创建

    【讨论】：

      【解决方案2】：

      这是另一个版本-

      library(dplyr)

dd %>%
  group_by(grp = cumsum(time_condition == 0)) %>%
  mutate(result = cumsum(time_condition)) %>%
  ungroup %>%
  select(-grp)

      这会返回 -

      #         date condition time_condition result
#1  2016-10-31     FALSE              0      0
#2  2017-01-16     FALSE              0      0
#3  2017-04-19     FALSE              0      0
#4  2017-07-12     FALSE              0      0
#5  2017-09-28      TRUE              0      0
#6  2017-11-02      TRUE             35     35
#7  2017-11-30      TRUE             28     63
#8  2018-01-08      TRUE             39    102
#9  2018-04-03      TRUE             85    187
#10 2018-05-07     FALSE              0      0
#11 2018-06-11     FALSE              0      0
#12 2018-07-30     FALSE              0      0
#13 2018-10-24      TRUE              0      0
#14 2018-12-12     FALSE              0      0
#15 2019-04-18      TRUE              0      0
#16 2019-06-06      TRUE             49     49
#17 2019-07-10      TRUE             34     83
#18 2019-08-22      TRUE             43    126
#19 2019-10-17      TRUE             56    182
#20 2019-11-14      TRUE             28    210
#21 2020-02-06     FALSE              0      0
#22 2020-06-02      TRUE              0      0
#23 2020-09-22     FALSE              0      0
#24 2020-11-17      TRUE              0      0
#25 2020-12-14      TRUE             27     27

      【讨论】：

      • 非常优雅，谢谢！
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