column basis中的列值post数据帧分组

Question

我正在试着找出每个国际海事组织号码在启航和到达之间的时间差。

IMO       Name          State      Datetime
8300327 SILVER FJORD    Arrival 13/08/2021 04:51
8300327 SILVER FJORD    Sailing 13/08/2021 22:59
8300327 SILVER FJORD    Arrival 20/08/2021 10:52
8300327 SILVER FJORD    Sailing 20/08/2021 20:24
9340738 FRAMFJORD       Arrival 19/08/2021 11:05
9340738 FRAMFJORD       Sailing 20/08/2021 17:32

对于上面的数据帧，输出应该是

IMO     Name            State     Datetime           Time_int
8300327 SILVER FJORD    Arrival 13/08/2021 04:51    
8300327 SILVER FJORD    Sailing 13/08/2021 22:59    18:08:00
8300327 SILVER FJORD    Arrival 20/08/2021 10:52    
8300327 SILVER FJORD    Sailing 20/08/2021 20:24    09:32:00
9340738 FRAMFJORD       Arrival 19/08/2021 11:05    
9340738 FRAMFJORD       Sailing 20/08/2021 17:32    06:27:00

我已经写了下面的计算代码

def dwell_calc(df):
    if (df['State'] == "Sailing"):
        val = df['Datetime'].diff().dt.seconds.div(3600).fillna(0).reset_index()

        return val


# data.sort_values(['IMO', 'Datetime'], inplace=True)


cond2=(data['State']=='Sailing')
data.loc[cond2, 'time_int'] = dwell_calc(data)

print(data['time_int'])

我收到错误消息：

 if (df['State'] == "Sailing"):
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

请帮助解决使用python查找时间间隔的问题

Answer 1

下面是最终的正确代码

代码：

df['time_diff'] = df.sort_values(["IMO","Datetime"]).groupby(["IMO"],as_index="False")['Datetime'].diff().dt.seconds.div(3600)
cond1=df['State']=="Arrival"
df.loc[cond1,"time_diff"]=0

输出：

IMO      State  Datetime            time_diff
8300327 Arrival 2021-08-13 04:51:00 0
8300327 Sailing 2021-08-13 22:59:00 18.1333333
8300327 Arrival 2021-08-20 10:52:00 0
8300327 Sailing 2021-08-21 02:24:00 15.5333333
8516263 Arrival 2021-08-22 20:10:00 0
8516263 Sailing 2021-08-23 17:25:00 21.25
8802882 Arrival 2021-08-18 07:25:00 0
8802882 Sailing 2021-08-18 22:01:00 14.6

Answer 2

在sort_values()之后尝试diff

df.sort_values(["IMO", "Datetime"])
df.loc[df["State"]=="Sailing", "Time_int"] = df["Datetime"].diff()

>>> df
       IMO          Name    State            Datetime        Time_int
0  8300327  SILVER FJORD  Arrival 2021-08-13 04:51:00             NaT
1  8300327  SILVER FJORD  Sailing 2021-08-13 22:59:00 0 days 18:08:00
2  8300327  SILVER FJORD  Arrival 2021-08-20 10:52:00             NaT
3  8300327  SILVER FJORD  Sailing 2021-08-20 20:24:00 0 days 09:32:00
4  9340738     FRAMFJORD  Arrival 2021-08-19 11:05:00             NaT
5  9340738     FRAMFJORD  Sailing 2021-08-20 17:32:00 1 days 06:27:00

Answer 3

如果您确保数据已排序，则可以计算奇数行和偶数行之间的差异：

df['Time_int'] = df.loc[1::2, 'Datetime'] - df.loc[::2, 'Datetime'].values

输出：

>>> df
       IMO          Name    State            Datetime        Time_int
0  8300327  SILVER FJORD  Arrival 2021-08-13 04:51:00             NaT
1  8300327  SILVER FJORD  Sailing 2021-08-13 22:59:00 0 days 18:08:00
2  8300327  SILVER FJORD  Arrival 2021-08-20 10:52:00             NaT
3  8300327  SILVER FJORD  Sailing 2021-08-21 02:24:00 0 days 15:32:00
4  9340738     FRAMFJORD  Arrival 2021-08-19 11:05:00             NaT
5  9340738     FRAMFJORD  Sailing 2021-08-20 17:32:00 1 days 06:27:00

如果您的数据没有排序，您必须解释如何对前4行进行分组。