我会结合使用 tidyverse 中可用的技术:

I would use a combination of techniques available in tidyverse:

首先创建一个分组变量(new_id)，将原来的id和new_id结合起来，基于一个分组相加.然后我们可以根据Amount > 5000之和的条件filter.我们可以把这个和filter然后join 或 semi_join 以根据条件进行过滤.

First create a grouping variable (new_id) and use the original id and new_id in combination to add together based on a grouping. Then we can filter by the criteria of the sum of the Amount > 5000. We can take this and filter then join or semi_join to filter based on the criteria.

ids 是一个数据集，它根据 id 和 new_id 和 filter 找到总 Amounts for when Dollars >5000代码>.这将为您提供满足您条件的 id 和 new_id

ids is a dataset that finds the total Amount based on id and new_id and filters for when Dollars > 5000. This gives you the id and new_id that meets your criteria

df <- data.frame(id=c("9","9","9","5","5","4","4","4","4","4","20","20"),
                 Date=c("11/29/2018","11/29/2018","11/29/2018","2/13/2019","2/13/2019",
                        "6/15/2018","6/20/2018","8/17/2018","8/20/2018","8/23/2018","12/25/2018","12/25/2018"), 
                 Buyer= c("John","John","John","Maria","Maria","Sandy","Sandy","Sandy","Sandy","Sandy","Paul","Paul"), 
                 Amount= c(959,1158,596,922,922,1849,4193,4256,65,100,313,99), stringsAsFactors = F) %>% 
  group_by(Buyer,id) %>% mutate(diffs = c(NA, diff(as.Date(Date, format = "%m/%d/%Y")))) 


library(tidyverse)

df1 <- df %>% mutate(Date      = as.Date(Date , format = "%m/%d/%Y"), 
                     tf1       = (id != lag(id, default = 0)),
                     tf2       = (is.na(diffs) | diffs > 5))

df1$new_id <- cumsum(df1$tf1 + df1$tf2 > 0)

>df1
       id    Date       Buyer Amount diffs days_post  tf1   tf2   new_id
       <chr> <date>     <chr>  <dbl> <dbl> <date>     <lgl> <lgl>  <int>
     1 9     2018-11-29 John     959    NA 2018-12-04 TRUE  TRUE       1
     2 9     2018-11-29 John    1158     0 2018-12-04 FALSE FALSE      1
     3 9     2018-11-29 John     596     0 2018-12-04 FALSE FALSE      1
     4 5     2019-02-13 Maria    922    NA 2019-02-18 TRUE  TRUE       2
     5 5     2019-02-13 Maria    922     0 2019-02-18 FALSE FALSE      2
     6 4     2018-06-15 Sandy   1849    NA 2018-06-20 TRUE  TRUE       3
     7 4     2018-06-20 Sandy   4193     5 2018-06-25 FALSE FALSE      3
     8 4     2018-08-17 Sandy   4256    58 2018-08-22 FALSE TRUE       4
     9 4     2018-08-20 Sandy     65     3 2018-08-25 FALSE FALSE      4
    10 4     2018-08-23 Sandy    100     3 2018-08-28 FALSE FALSE      4
    11 20    2018-12-25 Paul     313    NA 2018-12-30 TRUE  TRUE       5
    12 20    2018-12-25 Paul      99     0 2018-12-30 FALSE FALSE      5

ids <- df1 %>% 
       group_by(id, new_id) %>% 
       summarise(dollar = sum(Amount)) %>% 
       ungroup() %>% filter(dollar > 5000)
  id   new_id  dollar
 <chr>  <int>   <dbl>
1 4         3    6042
df1 %>% semi_join(ids)