怎么才能高效地将长话单拆分成按小时分解的多条短话单? - V2EX
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nemo95
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怎么才能高效地将长话单拆分成按小时分解的多条短话单?

  •  
  •   nemo95 2021-08-12 10:51:07 +08:00 1686 次点击
    这是一个创建于 1523 天前的主题,其中的信息可能已经有所发展或是发生改变。

    原始话单是这样的:

    USERID STARTTIME ENDTIME SERVICETYPE CHANNELCODE PROGRAMNAME
    xxxxxxxxxxxxxxxxx1 2021-05-24 19:52:28 2021-05-24 23:56:27 1 精灵宝可梦

    老板让分析每个用户各个时段都在看啥,所以我想吧话单拆分成这样:

    USERID STARTTIME ENDTIME SERVICETYPE CHANNELCODE PROGRAMNAME PERIODTIME
    xxxxxxxxxxxxxxxxx1 2021-05-24 19:52:28 2021-05-24 20:00:00 1 精灵宝可梦 2021-05-24 19:00:00
    xxxxxxxxxxxxxxxxx1 2021-05-24 20:00:00 2021-05-24 21:00:00 1 精灵宝可梦 2021-05-24 20:00:00
    xxxxxxxxxxxxxxxxx1 2021-05-24 21:00:00 2021-05-24 22:00:00 1 精灵宝可梦 2021-05-24 21:00:00
    xxxxxxxxxxxxxxxxx1 2021-05-24 22:00:00 2021-05-24 23:00:00 1 精灵宝可梦 2021-05-24 22:00:00
    xxxxxxxxxxxxxxxxx1 2021-05-24 23:00:00 2021-05-24 23:56:27 1 精灵宝可梦 2021-05-24 23:00:00

    目前的方法是根据起止时间生成时间序列,然后 for 循环生成新的行,再拼接成一个新的 dataframe

    # split_data ... for i in range(0, len_date-1): df_y['Period'] = date_rng[i] df_y['EndTime'] = date_rng[i+1] df_y['StartHour'] = date_rng[i] df_y['EndHour'] = date_rng[i+1] df_x = df_x.append(df_y) ... 

    主函数还要写个 for 循环遍历整个话单

    for i in range(len(df_tmp)): df_x = split_data(df_tmp.iloc[i]) df_t = df_t.append(df_x) df_t 

    这样能获得想要的结果,但话单太多了,跑起来没完没了……

    有没有更好的方法能提高下效率?

    2 条回复    2021-08-12 16:19:12 +08:00
    swulling
        1
    swulling  
       2021-08-12 10:54:17 +08:00
    把话单拆分,多进程跑,不够就多加机器
    princelai
        2
    princelai  
       2021-08-12 16:19:12 +08:00   1
    有办法,https://paste.ubuntu.com/p/Df6JCbsPn6/

    代码:


    结果:
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