簡體   English   中英

按r中的連續值分組

[英]group by consecutive values in r

我有一個來自支持票務系統的數據集,它記錄了代理商在分類和響應客戶請求時所做的每次點擊。 系統會為每次點擊分配一個新的hist_id,但代理會點擊幾個字段,觸發表格中的多行,他們認為是單個“交互”。

我的目標是通過對每個組中的第一個和最后一個modify_time值執行diff來計算每個交互的句柄時間。

我目前陷入困境,因為代理人將全天與案件進行多次互動。

這是一個示例數據幀:

hist_id <- c(1234, 2345, 3456, 4567, 5678, 6789, 7890)
case_id <- c(1, 1, 1, 1, 1, 1, 1)
agent_name <- c("John", "John", "John", "Paul", "Paul", "John", "John")
modify_time <- as.POSIXct(c(1510095120, 1510095180, 1510095240, 1510098600, 1510098720, 1510135200, 1510135320), origin = "1970-01-01")
df <- data.frame(hist_id, case_id, agent_name, modify_time)

使用group_id和agent_name上的group by按預期分組符合條件的所有行:

df %>% group_by(case_id, agent_name) %>% mutate(first = first(modify_time), last = last(modify_time), diff = min(difftime(last, first)))

這給了我這個:

    # A tibble: 7 x 7
# Groups:   case_id, agent_name [2]
  hist_id case_id agent_name         modify_time               first                last       diff
    <dbl>   <dbl>     <fctr>              <dttm>              <dttm>              <dttm>     <time>
1    1234       1       John 2017-11-07 16:52:00 2017-11-07 16:52:00 2017-11-08 04:02:00 40200 secs
2    2345       1       John 2017-11-07 16:53:00 2017-11-07 16:52:00 2017-11-08 04:02:00 40200 secs
3    3456       1       John 2017-11-07 16:54:00 2017-11-07 16:52:00 2017-11-08 04:02:00 40200 secs
4    4567       1       Paul 2017-11-07 17:50:00 2017-11-07 17:50:00 2017-11-07 17:52:00   120 secs
5    5678       1       Paul 2017-11-07 17:52:00 2017-11-07 17:50:00 2017-11-07 17:52:00   120 secs
6    6789       1       John 2017-11-08 04:00:00 2017-11-07 16:52:00 2017-11-08 04:02:00 40200 secs
7    7890       1       John 2017-11-08 04:02:00 2017-11-07 16:52:00 2017-11-08 04:02:00 40200 secs

John返回真正的第一個和最后一個modify_times。 但是,我需要對case_id和agent_name的連續匹配進行分組,以便考慮Paul的交互。 所以這里記錄了三個互動:一個來自John,一個來自Paul,另一個來自John。

期望的輸出將是這樣的:

    # A tibble: 7 x 7
# Groups:   case_id, agent_name [2]
  hist_id case_id agent_name         modify_time               first                last       diff
    <dbl>   <dbl>     <fctr>              <dttm>              <dttm>              <dttm>     <time>
1    1234       1       John 2017-11-07 16:52:00 2017-11-07 16:52:00 2017-11-07 16:54:00 120 secs
2    2345       1       John 2017-11-07 16:53:00 2017-11-07 16:52:00 2017-11-07 16:54:00 120 secs
3    3456       1       John 2017-11-07 16:54:00 2017-11-07 16:52:00 2017-11-07 16:54:00 120 secs
4    4567       1       Paul 2017-11-07 17:50:00 2017-11-07 17:50:00 2017-11-07 17:52:00 120 secs
5    5678       1       Paul 2017-11-07 17:52:00 2017-11-07 17:50:00 2017-11-07 17:52:00 120 secs
6    6789       1       John 2017-11-08 04:00:00 2017-11-08 04:00:00 2017-11-08 04:02:00 120 secs
7    7890       1       John 2017-11-08 04:02:00 2017-11-08 04:00:00 2017-11-08 04:02:00 120 secs

這是一個tidyverse方法,它按processing cluster identity以及case_idagent_name對組進行分區:

按順序排列所有單擊,每次hist_id序列遇到到新agent_name的轉換時, hist_id生成一個新的id標志。 cumsum這些標志為每個代理生成一個唯一的prcl_id ,每個集群處理塊。 使用所有三個id,您可以在所需的分區中運行您選擇的突變。

df %>% 
    arrange(hist_id) %>%  # to ensure there are no wrinkles
    mutate(ag_chg_flg = ifelse(lag(agent_name) != agent_name, 1, 0) %>%
               coalesce(0) # to reassign the first click in a case_id to 0 (from NA)
           ) %>% 
    group_by(case_id, agent_name) %>%  
    mutate(prcl_id = cumsum(ag_chg_flg) + 1) %>%  # generate the proc_clst_id (starting at 1) 
    group_by(case_id, agent_name, prcl_id) %>%  # group by the complete composite id
    mutate(first = first(modify_time),
           last = last(modify_time),
           diff = min(difftime(last, first))
           )

哪個讓你:

 # A tibble: 7 x 9 # Groups: case_id, agent_name, prcl_id [3] hist_id case_id agent_name modify_time ag_chg_flg prcl_id first last diff <dbl> <dbl> <fctr> <dttm> <dbl> <dbl> <dttm> <dttm> <time> 1 1234 1 John 2017-11-07 14:52:00 0 1 2017-11-07 14:52:00 2017-11-07 14:54:00 2 mins 2 2345 1 John 2017-11-07 14:53:00 0 1 2017-11-07 14:52:00 2017-11-07 14:54:00 2 mins 3 3456 1 John 2017-11-07 14:54:00 0 1 2017-11-07 14:52:00 2017-11-07 14:54:00 2 mins 4 4567 1 Paul 2017-11-07 15:50:00 1 2 2017-11-07 15:50:00 2017-11-07 15:52:00 2 mins 5 5678 1 Paul 2017-11-07 15:52:00 0 2 2017-11-07 15:50:00 2017-11-07 15:52:00 2 mins 6 6789 1 John 2017-11-08 02:00:00 1 2 2017-11-08 02:00:00 2017-11-08 02:02:00 2 mins 7 7890 1 John 2017-11-08 02:02:00 0 2 2017-11-08 02:00:00 2017-11-08 02:02:00 2 mins 

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM