计算列表产量误差的平均时间

Question

我试图使用this线程的代码从列表中计算出平均时间。所有其他代码建议都不适用于我，因为它们考虑的是持续时间而不是时间。

import datetime
import math
import numpy

def datetime_to_radians(x):
    # radians are calculated using a 24-hour circle, not 12-hour, starting at north and moving clockwise
    time_of_day = x.time()
    seconds_from_midnight = 3600 * time_of_day.hour + 60 * time_of_day.minute + time_of_day.second
    radians = float(seconds_from_midnight) / float(12 * 60 * 60) * 2.0 * math.pi
    return radians

def average_angle(angles):
    # angles measured in radians
    x_sum = numpy.sum([math.sin(x) for x in angles])
    y_sum = numpy.sum([math.cos(x) for x in angles])
    x_mean = x_sum / float(len(angles))
    y_mean = y_sum / float(len(angles))
    return numpy.arctan2(x_mean, y_mean)

def radians_to_time_of_day(x):
    # radians are measured clockwise from north and represent time in a 24-hour circle
    seconds_from_midnight = int(float(x) / (2.0 * math.pi) * 12.0 * 60.0 * 60.0)
    hour = seconds_from_midnight / 3600
    minute = (seconds_from_midnight % 3600) / 60
    second = seconds_from_midnight % 60
    return datetime.time(hour, minute, second)

def average_times_of_day(x):
    # input datetime.datetime array and output datetime.time value
    angles = [datetime_to_radians(y) for y in x]
    avg_angle = average_angle(angles)
    return radians_to_time_of_day(avg_angle)

average_times_of_day([datetime.datetime(2017, 6, 9, 0, 10), datetime.datetime(2017, 6, 9, 0, 20)])
# datetime.time(0, 15)

average_times_of_day([datetime.datetime(2017, 6, 9, 23, 50), datetime.datetime(2017, 6, 9, 0, 10)])
# datetime.time(0, 0)

我得到了以下错误：

TypeError: integer argument expected, got float

有人能帮忙吗？

Answer 1

下面是另一种可以处理日期时间列表的解决方案--仅考虑小时和分钟的对象：

from cmath import phase
from cmath import rect
import datetime
from math import degrees
from math import radians

dfList = ([datetime.datetime(2017, 9, 15, 8, 8),
           datetime.datetime(2017, 9, 14, 8, 5),
           datetime.datetime(2017, 9, 13, 6, 56),
           datetime.datetime(2017, 12, 9, 6, 14),
           datetime.datetime(2017, 11, 9, 6, 42)])

dfList = [item.strftime("%H:%M") for item in dfList]


def mean_angle(deg):
    return degrees(phase(sum(rect(1, radians(d)) for d in deg) / len(deg)))


def mean_time(times):
    t = (time.split(':') for time in times)
    seconds = ((int(m) * 60 + int(h) * 3600)
               for h, m in t)
    day = 24 * 60 * 60
    to_angles = [s * 360. / day for s in seconds]
    mean_as_angle = mean_angle(to_angles)
    mean_seconds = mean_as_angle * day / 360.
    if mean_seconds < 0:
        mean_seconds += day
    h, m = divmod(mean_seconds, 3600)
    m, s = divmod(m, 60)
    return '%02i:%02i' % (h, m)


print(mean_time(dfList))

Answer 2

您的代码在Python2.7中运行得很好，但Python3.6却不行。

错误抛到第26行，return datetime.time(hour, minute, second)。

将hour、minute和second的值转换为int()后，工作正常。

def radians_to_time_of_day(x):
    # radians are measured clockwise from north and represent time in a 24-hour circle
    seconds_from_midnight = int(float(x) / (2.0 * math.pi) * 12.0 * 60.0 * 60.0)
    hour = int(seconds_from_midnight / 3600)
    minute = int((seconds_from_midnight % 3600) / 60)
    second = int(seconds_from_midnight % 60)
    return datetime.time(hour, minute, second)

Answer 3

在Python3中，即使两个操作数都是整数，/也执行真(浮动)除法。

使用//代替

def radians_to_time_of_day(x):
    # radians are measured clockwise from north and represent time in a 24-hour circle
    seconds_from_midnight = int(float(x) / (2.0 * math.pi) * 12.0 * 60.0 * 60.0)
    hour = seconds_from_midnight // 3600
    minute = (seconds_from_midnight % 3600) // 60
    second = seconds_from_midnight % 60
    return datetime.time(hour, minute, second)