从df中提取值答案

【问题标题】：Extracting values from a df从df中提取值
【发布时间】：2022-11-15 12:28:39
【问题描述】：

问题陈述每辆车有多个充电和放电实例，获取特定日期每辆车的最小充电量、最大充电量、最小放电量和最大放电量

df1

日期时间 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 放电 2 2022-10-01 02:28:56 DL21GD0100 80.0 放电 3 2022-10-01 02:29:26 DL21GD0100 80.0 放电 4 2022-10-01 02:29:56 DL21GD0100 69.0 放电 5 2022-10-01 02:29:56 DL21GD0100 70.0 充电 6 2022-10-01 02:29:56 DL21GD0100 71.0 充电 7 2022-10-01 02:29:56 DL21GD0100 72.0 充电 8 2022-10-01 03:16:00 DL21GD0100 63.0 放电 9 2022-10-01 03:16:30 DL21GD0100 23.0 放电 10 2022-10-01 04:17:00 DL21GD0100 54.0 充电 11 2022-10-01 09:17:30 WB25M9298 24.0 充电 12 2022-10-01 09:18:00 WB25M9298 25.0 充电

【问题讨论】：

标签： python pandas

【解决方案1】：

阅读 3 个不同选项的完整答案

严格映射充电/放电

您可以使用groupby.diff 获取每组的差异，然后使用numpy.sign 和map：

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

或者numpy.select：

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

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

输出：

          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

映射充电/放电稳定作为放电

如果您想将等值视为放电：

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

输出：

          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

映射充电/放电与之前的状态一样稳定：

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

输出：

          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

【讨论】：