如何从日期时间中删除秒? [英] How to remove seconds from datetime?
本文介绍了如何从日期时间中删除秒?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有以下日期,并且尝试了以下代码,
I have the following date and I tried the following code,
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'
推荐答案
在输出中需要 datetimes 的解决方案:
Solutions if need datetimes in output:
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
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')
您可以先使用转换为numpy values
,然后通过强制转换为<M8[m]
截断seconds
,但是此解决方案会删除可能的时区:
You can use convert to numpy values
first and then truncate seconds
by cast to <M8[m]
, but this solution remove possible timezones:
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
另一种解决方案是从timedelta系列> second
并减去:
Another solution is create timedelta
Series from second
and substract:
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)
如果需要输出中日期时间的字符串代表的解决方案
Solutions if need strings repr of datetimes in output
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,将:00
设置为秒:
And if necessary set :00
to seconds:
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
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