R中的滞后累积量

我有如下数据:

df <- tibble(
      date= seq.Date(as.Date("2021-01-01"), as.Date("2022-02-01"), by = "month"),
      val1 = c(100, 100, 105, 125, 125, 125, 125, 132, 132, 132, 135, 150, 150, 150),
      val2 = c(100, 100, 100, 125, 125, 125, 125, 125, 125, 125, 125, 150, 150, 150),
      diff = val1-val2)

       date        val1  val2  diff
       <date>     <dbl> <dbl> <dbl>
     1 2021-01-01   100   100     0
     2 2021-02-01   100   100     0
     3 2021-03-01   105   100     5
     4 2021-04-01   125   125     0
     5 2021-05-01   125   125     0
     6 2021-06-01   125   125     0
     7 2021-07-01   125   125     0
     8 2021-08-01   132   125     7
     9 2021-09-01   132   125     7
    10 2021-10-01   132   125     7
    11 2021-11-01   135   125    10
    12 2021-12-01   150   150     0
    13 2022-01-01   150   150     0
    14 2022-02-01   150   150     0

我正在try 生成以下输出:

output <- tibble(
  date= seq.Date(as.Date("2021-01-01"), as.Date("2022-02-01"), by = "month"),
  val1 = c(100, 100, 105, 125, 125, 125, 125, 132, 132, 132, 135, 150, 150, 150),
  val2 = c(100, 100, 100, 125, 125, 125, 125, 125, 125, 125, 125, 150, 150, 150),
  diff = val1-val2,
  diff_calc = c(0, 0, 0, 5, 5, 5, 5, 5, 5, 5, 5, 15, 15, 15))

 date        val1  val2  diff diff_calc
 <date>     <dbl> <dbl> <dbl>     <dbl>
 1 2021-01-01   100   100     0         0
 2 2021-02-01   100   100     0         0
 3 2021-03-01   105   100     5         0
 4 2021-04-01   125   125     0         5
 5 2021-05-01   125   125     0         5
 6 2021-06-01   125   125     0         5
 7 2021-07-01   125   125     0         5
 8 2021-08-01   132   125     7         5
 9 2021-09-01   132   125     7         5
10 2021-10-01   132   125     7         5
11 2021-11-01   135   125    10         5
12 2021-12-01   150   150     0        15
13 2022-01-01   150   150     0        15
14 2022-02-01   150   150     0        15

其中diff_calc是diff中之前唯一值的累积和，除非多个唯一diff值连续出现，否则它应该是最大值加上使用相同逻辑的任何之前diff值的累积和.

这是我之前询问的this question的派生，但我意识到我没有提供最好的例子或描述来满足我在这里需要的内容，所以作为新问题发布.谢谢！