Question

我有以下数据类型：

id=["Train A","Train A","Train A","Train B","Train B","Train B"]
arrival_time = ["0"," 2016-05-19 13:50:00","2016-05-19 21:25:00","0","2016-05-24 18:30:00","2016-05-26 12:15:00"]
departure_time = ["2016-05-19 08:25:00","2016-05-19 16:00:00","2016-05-20 07:45:00","2016-05-24 12:50:00","2016-05-25 23:00:00","2016-05-26 19:45:00"]

获取以下数据：

id              arrival_time                departure_time
Train A                 0                  2016-05-19 08:25:00
Train A          2016-05-19 13:50:00       2016-05-19 16:00:00
Train A          2016-05-19 21:25:00       2016-05-20 07:45:00
Train B                    0               2016-05-24 12:50:00
Train B          2016-05-24 18:30:00       2016-05-25 23:00:00
Train B          2016-05-26 12:15:00       2016-05-26 19:45:00

出发时间和到达时间的数据类型为datetime64 [ns]。

如何找到第一排出发时间和第二排到达时间之间的时差？我厌倦了以下代码，但它没有用。例如，找到[2016-05-19 08:25:00]和[2016-05-19 13:50:00]之间的时差。

df['Duration'] = df.departure_time.iloc[i+1] - df.arrival_time.iloc[i]

Answer 1

我认为您需要先转换dates字符串to_datetime，同时0必须转换为NaN：

df = pd.DataFrame({'id': id, 'arrival_time':arrival_time, 'departure_time':departure_time})

df['arrival_time'] = pd.to_datetime(df['arrival_time'].replace('0', np.nan))
#another solution for replace not dates to NaT
#df['arrival_time'] = pd.to_datetime(df['arrival_time'], errors='coerce')
df['departure_time'] = pd.to_datetime(df['departure_time'])
print (df)
         arrival_time      departure_time       id
0                 NaT 2016-05-19 08:25:00  Train A
1 2016-05-19 13:50:00 2016-05-19 16:00:00  Train A
2 2016-05-19 21:25:00 2016-05-20 07:45:00  Train A
3                 NaT 2016-05-24 12:50:00  Train B
4 2016-05-24 18:30:00 2016-05-25 23:00:00  Train B
5 2016-05-26 12:15:00 2016-05-26 19:45:00  Train B

然后departure_time列id每个组groupby arrival_time和<{1}}列。

df['Duration'] = df.groupby('id')['departure_time'].shift() - df['arrival_time']
print (df)
         arrival_time      departure_time       id          Duration
0                 NaT 2016-05-19 08:25:00  Train A               NaT
1 2016-05-19 13:50:00 2016-05-19 16:00:00  Train A -1 days +18:35:00
2 2016-05-19 21:25:00 2016-05-20 07:45:00  Train A -1 days +18:35:00
3                 NaT 2016-05-24 12:50:00  Train B               NaT
4 2016-05-24 18:30:00 2016-05-25 23:00:00  Train B -1 days +18:20:00
5 2016-05-26 12:15:00 2016-05-26 19:45:00  Train B -1 days +10:45:00

或者可能需要交换列以获得正时间值：

df['Duration'] = df['arrival_time'] - df.groupby('id')['departure_time'].shift()
print (df)
         arrival_time      departure_time       id  Duration
0                 NaT 2016-05-19 08:25:00  Train A       NaT
1 2016-05-19 13:50:00 2016-05-19 16:00:00  Train A  05:25:00
2 2016-05-19 21:25:00 2016-05-20 07:45:00  Train A  05:25:00
3                 NaT 2016-05-24 12:50:00  Train B       NaT
4 2016-05-24 18:30:00 2016-05-25 23:00:00  Train B  05:40:00
5 2016-05-26 12:15:00 2016-05-26 19:45:00  Train B  13:15:00

最后可以通过shift将timedelta转换为seconds：

df['Duration'] = (df['arrival_time'] - df.groupby('id')['departure_time'].shift()).dt.total_seconds()
print (df)
         arrival_time      departure_time       id  Duration
0                 NaT 2016-05-19 08:25:00  Train A       NaN
1 2016-05-19 13:50:00 2016-05-19 16:00:00  Train A   19500.0
2 2016-05-19 21:25:00 2016-05-20 07:45:00  Train A   19500.0
3                 NaT 2016-05-24 12:50:00  Train B       NaN
4 2016-05-24 18:30:00 2016-05-25 23:00:00  Train B   20400.0
5 2016-05-26 12:15:00 2016-05-26 19:45:00  Train B   47700.0

如何找到大熊猫两个日期时间的差异？

1 个答案: