使用 Pandas 的两个时间戳之间的每小时时间序列（以分钟为单位）

Question

我有一系列带有开始时间和结束时间的时间戳。我想生成两个时间戳之间每小时的分钟数：

import pandas as pd

start_time = pd.to_datetime('2013-03-26 21:49:08',infer_datetime_format=True)
end_time = pd.to_datetime('2013-03-27 05:21:00, infer_datetime_format=True)

pd.date_range(start_time, end_time, freq='h')

给出：

DatetimeIndex(['2013-03-26 21:49:08', '2013-03-26 22:49:08',
               '2013-03-26 23:49:08', '2013-03-27 00:49:08',
               '2013-03-27 01:49:08', '2013-03-27 02:49:08',
               '2013-03-27 03:49:08', '2013-03-27 04:49:08'],
              dtype='datetime64[ns]', freq='H')

示例结果：我想计算开始和结束时间之间以小时为界的分钟数，如下所示：

 2013-03-26 21:00:00'  - 10m 52secs
 2013-03-26 22:00:00'  - 60 m 
 2013-03-26 23:00:00'  - 60 m

 2013-03-27 05:00:00'  - 21 m 

我看过 pandas resample，但不完全确定如何实现这一点。任何方向表示赞赏。

Answer 1

构造两个Series，分别对应每小时的开始和结束时间。使用clip_lower 和clip_upper 将它们限制在您想要的时间跨度内，然后减去：

# hourly range, floored to the nearest hour
rng = pd.date_range(start_time.floor('h'), end_time.floor('h'), freq='h')

# get the left and right endpoints for each hour
# clipped to be inclusive of [start_time, end_time]
left = pd.Series(rng, index=rng).clip_lower(start_time)
right = pd.Series(rng + 1, index=rng).clip_upper(end_time)

# construct a series of the lengths
s = right - left

结果输出：

2013-03-26 21:00:00   00:10:52
2013-03-26 22:00:00   01:00:00
2013-03-26 23:00:00   01:00:00
2013-03-27 00:00:00   01:00:00
2013-03-27 01:00:00   01:00:00
2013-03-27 02:00:00   01:00:00
2013-03-27 03:00:00   01:00:00
2013-03-27 04:00:00   01:00:00
2013-03-27 05:00:00   00:21:00
Freq: H, dtype: timedelta64[ns]

Answer 2

这似乎是一个可行的解决方案：

import pandas as pd
import datetime as dt

def bounded_min(t, range_time):
    """ For a given timestamp t and considered time interval range_time,
    return the desired bounded value in minutes and seconds"""

    # min() takes care of the end of the time interval, 
    # max() takes care of the beginning of the interval
    s = (min(t + dt.timedelta(hours=1), range_time.max()) - 
         max(t, range_time.min())).total_seconds() 
    if s%60:
        return "%dm %dsecs" % (s/60, s%60)
    else:
        return "%dm" % (s/60)

start_time = pd.to_datetime('2013-03-26 21:49:08',infer_datetime_format=True)
end_time = pd.to_datetime('2013-03-27 05:21:00', infer_datetime_format=True)

range_time = pd.date_range(start_time, end_time, freq='h')
# Include the end of the time range using the union() trick, as described at:
# https://stackoverflow.com/questions/37890391/how-to-include-end-date-in-pandas-date-range-method
range_time = range_time.union([end_time])

# This is essentially timestamps for beginnings of hours 
index_time = pd.Series(range_time).apply(lambda x: dt.datetime(year=x.year, 
                              month=x.month, 
                              day=x.day,
                              hour=x.hour, 
                              minute=0, 
                              second=0))

bounded_mins = index_time.apply(lambda x: bounded_min(x, range_time))

# Put timestamps and values together
bounded_df = pd.DataFrame(bounded_mins, columns=["Bounded Mins"]).set_index(index_time)

print bounded_df

一定要喜欢强大的 lambda 表达式:)。不过也许有更简单的方法。

输出：

                      Bounded Mins
2013-03-26 21:00:00   10m 52secs
2013-03-26 22:00:00          60m
2013-03-26 23:00:00          60m
2013-03-27 00:00:00          60m
2013-03-27 01:00:00          60m
2013-03-27 02:00:00          60m
2013-03-27 03:00:00          60m
2013-03-27 04:00:00          60m
2013-03-27 05:00:00          21m

Answer 3

在某种 for 循环中使用 datetime.timedelta() 似乎正是您要寻找的。​​p>

https://docs.python.org/2/library/datetime.html#datetime.timedelta