将时间四舍五入到最近的一小时

Question

试图将观察到的时间四舍五入到最近的一小时

Example Data: 
observed_time = ['2020-02-20T17:54:00Z', '2020-02-20T18:54:00Z']

slice_begin_time=['2020-02-20T17:50:00Z', '2020-02-20T18:50:00Z', '2020-02-20T19:50:00Z', '2020-02-20T20:50:00Z', '2020-02-20T21:50:00Z']
slice_end_time=['2020-02-20T18:05:00Z', '2020-02-20T19:05:00Z', '2020-02-20T20:05:00Z', '2020-02-20T21:05:00Z', '2020-02-20T22:05:00Z']

### LIBS
from datetime import datetime, timedelta
import pandas as pd

s = pd.Series(pd.to_datetime(observed_time))
for i in range(len(slice_begin_time)):
    s[s.between(pd.Timestamp(slice_begin_time[i]),pd.Timestamp(slice_end_time[i]))] == s.round(freq = 'H')
print(s)

我得到了错误：round（）得到了一个意外的关键字参数'freq'

我确实想把它连成一个系列

Answer 1

希望这有助于：

observed_time = ['2020-02-20T17:54:00Z', '2020-02-20T18:54:00Z']

slice_begin_time=['2020-02-20T17:50:00Z', '2020-02-20T18:50:00Z', '2020-02-20T19:50:00Z', '2020-02-20T20:50:00Z', '2020-02-20T21:50:00Z']
slice_end_time=['2020-02-20T18:05:00Z', '2020-02-20T19:05:00Z', '2020-02-20T20:05:00Z', '2020-02-20T21:05:00Z', '2020-02-20T22:05:00Z']

def match(time):
    for i in range(len(slice_begin_time)):
        if pd.Timestamp(slice_begin_time[i]) <= time <= pd.Timestamp(slice_end_time[i]):
            return time.round('1H')
            break
    return time  # returns the input time in case nothing matches

s = pd.Series(pd.to_datetime(observed_time))
# apply map function match to the series elements    
s = s.map(match)
print(s)

输出：

0   2020-02-20 18:00:00+00:00
1   2020-02-20 19:00:00+00:00
dtype: datetime64[ns, UTC]

如果您只是想将时间序列转换为最近的小时，您可以使用：

s = s.dt.round('1H')