從 df 中提取值

Question

問題陳述：

每輛車有多個充電和放電實例，獲取特定日期每輛車的最小充電量、最大充電量、最小放電量和最大放電量。

df1

Date             Time    vehicle_no  soc     SOC Diff
0   2022-10-01  02:27:56    DL21GD0100  80.0    0
1   2022-10-01  02:28:26    DL21GD0100  80.0    Discharging
2   2022-10-01  02:28:56    DL21GD0100  80.0    Discharging
3   2022-10-01  02:29:26    DL21GD0100  80.0    Discharging
4   2022-10-01  02:29:56    DL21GD0100  69.0    Discharging
5    2022-10-01 02:29:56    DL21GD0100  70.0    Charging
6    2022-10-01 02:29:56    DL21GD0100  71.0    Charging
7    2022-10-01 02:29:56    DL21GD0100  72.0    Charging
8   2022-10-01 03:16:00    DL21GD0100   63.0    Discharging
9  2022-10-01  03:16:30    DL21GD0100    23.0    Discharging
10  2022-10-01  04:17:00    DL21GD0100    54.0    Charging
11  2022-10-01  09:17:30    WB25M9298   24.0    Charging
12  2022-10-01  09:18:00    WB25M9298   25.0    Charging

Answer 1

閱讀 3 個不同選項的完整答案

映射嚴格充電/放電

您可以使用groupby.diff來獲得每組的差異，然后numpy.sign和map ：

df['status'] = np.sign(df.groupby('vehicle_no')['soc'].diff()
                       ).map({1: 'Charging', -1: 'Discharging'})

或者使用numpy.select ：

s = df.groupby('vehicle_no')['soc'].diff()

df['status'] = np.select([s>0, s<0], ['Charging', 'Discharging'], np.nan)

Output：

          Date      Time  vehicle_no   soc       status
0   2022-10-01  02:27:56  DL21GD0100  80.0          NaN
2   2022-10-01  02:28:56  DL21GD0100  80.0          NaN
3   2022-10-01  02:29:26  DL21GD0100  80.0          NaN
4   2022-10-01  02:29:56  DL21GD0100  69.0  Discharging
5   2022-10-01  02:29:56  DL21GD0100  70.0     Charging
6   2022-10-01  02:29:56  DL21GD0100  71.0     Charging
7   2022-10-01  02:29:56  DL21GD0100  72.0     Charging
8   2022-10-01  09:16:00   WB25M9298  23.0          NaN
9   2022-10-01  09:16:30   WB25M9298  23.0          NaN
10  2022-10-01  09:17:00   WB25M9298  24.0     Charging
11  2022-10-01  09:17:30   WB25M9298  24.0          NaN
12  2022-10-01  09:18:00   WB25M9298  25.0     Charging

映射充電/放電穩定作為放電

如果您想將等值視為放電：

df['status'] = np.where(df.groupby('vehicle_no')['soc'].diff().gt(0), 'Charging', 'Discharging')

Output：

          Date      Time  vehicle_no   soc       status
0   2022-10-01  02:27:56  DL21GD0100  80.0  Discharging
2   2022-10-01  02:28:56  DL21GD0100  80.0  Discharging
3   2022-10-01  02:29:26  DL21GD0100  80.0  Discharging
4   2022-10-01  02:29:56  DL21GD0100  69.0  Discharging
5   2022-10-01  02:29:56  DL21GD0100  70.0     Charging
6   2022-10-01  02:29:56  DL21GD0100  71.0     Charging
7   2022-10-01  02:29:56  DL21GD0100  72.0     Charging
8   2022-10-01  09:16:00   WB25M9298  23.0  Discharging
9   2022-10-01  09:16:30   WB25M9298  23.0  Discharging
10  2022-10-01  09:17:00   WB25M9298  24.0     Charging
11  2022-10-01  09:17:30   WB25M9298  24.0  Discharging
12  2022-10-01  09:18:00   WB25M9298  25.0     Charging

映射充電/放電穩定如前 state：

d = {1: 'Charging', -1: 'Discharging'}
df['status'] = (df.groupby('vehicle_no')['soc']
                .transform(lambda s: np.sign(s.diff()).map(d).ffill())
                .fillna('Discharging')
               )

Output：

          Date      Time  vehicle_no   soc       status
0   2022-10-01  02:27:56  DL21GD0100  80.0  Discharging
2   2022-10-01  02:28:56  DL21GD0100  80.0  Discharging
3   2022-10-01  02:29:26  DL21GD0100  80.0  Discharging
4   2022-10-01  02:29:56  DL21GD0100  69.0  Discharging
5   2022-10-01  02:29:56  DL21GD0100  70.0     Charging
6   2022-10-01  02:29:56  DL21GD0100  71.0     Charging
7   2022-10-01  02:29:56  DL21GD0100  72.0     Charging
8   2022-10-01  09:16:00   WB25M9298  23.0  Discharging
9   2022-10-01  09:16:30   WB25M9298  23.0  Discharging
10  2022-10-01  09:17:00   WB25M9298  24.0     Charging
11  2022-10-01  09:17:30   WB25M9298  24.0     Charging
12  2022-10-01  09:18:00   WB25M9298  25.0     Charging

Answer 2

嘗試這個-

df1.groupby(['vehicle_no','status']).agg({'soc':[min,max]})