按不同级别的数据汇总分组之间的差异答案

【问题标题】：Aggregate difference between group by data at different levels按不同级别的数据汇总分组之间的差异
【发布时间】：2020-07-08 02:01:51
【问题描述】：

这是我的数据框

data = [[1,'A','a','2020-01-01'],
    [1,'A','b','2020-01-02'],
    [1,'B','a','2020-01-03'],
    [2,'A','a','2020-01-04'],
    [2,'A','b','2020-01-05'],
    [2,'A','b','2020-01-06']]

df_1 = pd.DataFrame(data = data,columns = ['id','main','sub_steps','date'])
df_1['date'] = pd.to_datetime(df_1['date'])

我想按id 列分组并计算Main 或sub_steps 发生变化时的时间差。

想要的结果

   id   main sub_steps       date sub_steps date_main_diff date_subStep_diff
0   1    A           a 2020-01-01    [a, b]         0 days            0 days
1   1    A           b 2020-01-02    [a, b]         1 days            0 days
2   1    B           a 2020-01-03       [a]         0 days            0 days
3   2    A           a 2020-01-04 [a, b, b]         0 days            0 days
4   2    A           b 2020-01-05 [a, b, b]         1 days            0 days
5   2    A           b 2020-01-06 [a, b, b]         2 days            1 days

我只能想办法

(df_1.merge(df_1.groupby(['id','Main'])
            .agg({'sub_steps':list,
                'date': df_1.date - df_1.date.shift(1) })
            ,on=['id', 'Main']))

给出错误TypeError: 'NaTType' object is not callable

日期差异列的唯一问题在于我得到了我想要的。

【问题讨论】：

标签： python python-3.x pandas pandas-groupby

【解决方案1】：

我们只能用transform和diff一一做专栏

df['sub_steps1']=df.groupby(['id','main'])['sub_steps'].transform(lambda x : [x.tolist()]*len(x))
 df['date_main_diff']=df.groupby(['id','main']).date.diff().fillna(pd.Timedelta('0 days'))
df['date_main_diff']=df.groupby(['id','main']).date_main_diff.apply(lambda x : x.cumsum())
df['date_subStep_diff']=df.groupby(['id','main','sub_steps']).date.diff().fillna(pd.Timedelta('0 days'))
df['date_subStep_diff']=df.groupby(['id','main','sub_steps']).date_subStep_diff.apply(lambda x : x.cumsum())
df
       id main sub_steps       date sub_steps1 date_main_diff date_subStep_diff
    0   1    A         a 2020-01-01     [a, b]         0 days            0 days
    1   1    A         b 2020-01-02     [a, b]         1 days            0 days
    2   1    B         a 2020-01-03        [a]         0 days            0 days
    3   2    A         a 2020-01-04  [a, b, b]         0 days            0 days
    4   2    A         b 2020-01-05  [a, b, b]         1 days            0 days
    5   2    A         b 2020-01-06  [a, b, b]         2 days            1 days

【讨论】：