python - 如何从日期时间中删除秒数？

Question

我有以下日期，我尝试了以下代码，

df['start_date_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"]
df['start_date_time'] = pd.to_datetime([df['start_date_time']).replace(second = 0)

我收到以下错误:

TypeError: replace() got an unexpected keyword argument 'second'

Answer 1

如果在输出中需要日期时间的解决方案:

df = pd.DataFrame({'start_date_time': ["2016-05-19 08:25:23","2016-05-19 16:00:45"]})
df['start_date_time'] = pd.to_datetime(df['start_date_time'])
print (df)
       start_date_time
0  2016-05-19 08:25:23
1  2016-05-19 16:00:45

使用 Series.dt.floor 按分钟 T或 Min :

df['start_date_time'] = df['start_date_time'].dt.floor('T')

df['start_date_time'] = df['start_date_time'].dt.floor('Min')

---

您可以使用转换为 numpy values先截断seconds通过转换到<M8[m] ，但此解决方案删除了​​可能的时区:

df['start_date_time'] = df['start_date_time'].values.astype('<M8[m]')
print (df)
      start_date_time
0 2016-05-19 08:25:00
1 2016-05-19 16:00:00

另一个解决方案是创建 timedelta来自 second 的系列并减去:

print (pd.to_timedelta(df['start_date_time'].dt.second, unit='s'))
0   00:00:23
1   00:00:45
Name: start_date_time, dtype: timedelta64[ns]

df['start_date_time'] = df['start_date_time'] - 
                        pd.to_timedelta(df['start_date_time'].dt.second, unit='s')
print (df)
      start_date_time
0 2016-05-19 08:25:00
1 2016-05-19 16:00:00

时间:

df = pd.DataFrame({'start_date_time': ["2016-05-19 08:25:23","2016-05-19 16:00:45"]})
df['start_date_time'] = pd.to_datetime(df['start_date_time'])

#20000 rows
df = pd.concat([df]*10000).reset_index(drop=True)


In [28]: %timeit df['start_date_time'] = df['start_date_time'] - pd.to_timedelta(df['start_date_time'].dt.second, unit='s')
4.05 ms ± 130 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [29]: %timeit df['start_date_time1'] = df['start_date_time'].values.astype('<M8[m]')
1.73 ms ± 117 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [30]: %timeit df['start_date_time'] = df['start_date_time'].dt.floor('T')
1.07 ms ± 116 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [31]: %timeit df['start_date_time2'] = df['start_date_time'].apply(lambda t: t.replace(second=0))
183 ms ± 19.7 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

---

如果输出中需要strings repr of datetimes 的解决方案

使用 Series.dt.strftime :

print(df['start_date_time'].dt.strftime('%Y-%m-%d %H:%M'))
0    2016-05-19 08:25
1    2016-05-19 16:00
Name: start_date_time, dtype: object

如有必要，设置 :00到秒:

print(df['start_date_time'].dt.strftime('%Y-%m-%d %H:%M:00'))
0    2016-05-19 08:25:00
1    2016-05-19 16:00:00
Name: start_date_time, dtype: object