平均日期数组计算

我想得到以下日期的平均值。我考虑过将所有数据转换为秒，然后对它们进行平均。但可能有更好的方法来做到这一点。

date = ['2016-02-23 09:36:26', '2016-02-24 10:00:32', '2016-02-24 11:28:22', '2016-02-24 11:27:20', '2016-02-24 11:24:15', '2016-02-24 11:20:25', '2016-02-24 11:17:43', '2016-02-24 11:12:03', '2016-02-24 11:09:11', '2016-02-24 11:08:44', '2016-02-24 11:05:28', '2016-02-24 11:03:23', '2016-02-24 10:58:08', '2016-02-24 10:53:59', '2016-02-24 10:49:34', '2016-02-24 10:43:33', '2016-02-24 10:35:27', '2016-02-24 10:31:50', '2016-02-24 10:31:17', '2016-02-24 10:30:05', '2016-02-24 10:29:21']

讨厌的解决方案：

import datetime
import time
import numpy as np

date = ['2016-02-23 09:36:26', '2016-02-24 10:00:32', '2016-02-24 11:28:22', '2016-02-24 11:27:20', '2016-02-24 11:24:15', '2016-02-24 11:20:25', '2016-02-24 11:17:43', '2016-02-24 11:12:03', '2016-02-24 11:09:11', '2016-02-24 11:08:44', '2016-02-24 11:05:28', '2016-02-24 11:03:23', '2016-02-24 10:58:08', '2016-02-24 10:53:59', '2016-02-24 10:49:34', '2016-02-24 10:43:33', '2016-02-24 10:35:27', '2016-02-24 10:31:50', '2016-02-24 10:31:17', '2016-02-24 10:30:05', '2016-02-24 10:29:21']
sec = [time.mktime(datetime.datetime.strptime(d, "%Y-%m-%d %H:%M:%S").timetuple()) for d in date]
mean = datetime.datetime.fromtimestamp(np.mean(sec))
print(mean)

在 NumPy 中所有datetime64[s]s 内部由 8 字节整数表示。 整数表示自纪元以来的秒数。

所以你可以转换date列表到 NumPy 数组datetime64[s]数据类型， 将其视为 dtypei8（8字节整数），取平均值，然后将int转换回adatetime64[s].

import numpy as np

date = ['2016-02-23 09:36:26', '2016-02-24 10:00:32', '2016-02-24 11:28:22', '2016-02-24 11:27:20', '2016-02-24 11:24:15', '2016-02-24 11:20:25', '2016-02-24 11:17:43', '2016-02-24 11:12:03', '2016-02-24 11:09:11', '2016-02-24 11:08:44', '2016-02-24 11:05:28', '2016-02-24 11:03:23', '2016-02-24 10:58:08', '2016-02-24 10:53:59', '2016-02-24 10:49:34', '2016-02-24 10:43:33', '2016-02-24 10:35:27', '2016-02-24 10:31:50', '2016-02-24 10:31:17', '2016-02-24 10:30:05', '2016-02-24 10:29:21']

mean = (np.array(date, dtype='datetime64[s]')
        .view('i8')
        .mean()
        .astype('datetime64[s]'))

print(mean)

prints

2016-02-24T09:43:40-0500