如何将numpy datetime64[ns]转换为python datetime?

我想知道你是否需要所有这些转换工作。使用正确的时间单位A datetime64 可以生成 datetime 直接对象。

我不确定你的 when 变量,但假设它来自 pandas 就像一个 DatetimeIndex :

In [56]: time = pandas.date_range('6/28/2013', periods=5, freq='5D')
In [57]: time
Out[57]: 
DatetimeIndex(['2013-06-28', '2013-07-03', '2013-07-08', '2013-07-13',
               '2013-07-18'],
              dtype='datetime64[ns]', freq='5D')

等效numpy数组

In [58]: time.values
Out[58]: 
array(['2013-06-28T00:00:00.000000000', '2013-07-03T00:00:00.000000000',
       '2013-07-08T00:00:00.000000000', '2013-07-13T00:00:00.000000000',
       '2013-07-18T00:00:00.000000000'], dtype='datetime64[ns]')
In [59]: time.values.tolist()
Out[59]: 
[1372377600000000000,
 1372809600000000000,
 1373241600000000000,
 1373673600000000000,
 1374105600000000000]

用 [ns] 结果是一个大整数,某种类型的“时间戳”。但如果我将时间单位转换为秒,甚至是微秒(us):

In [60]: time.values.astype('datetime64[s]')
Out[60]: 
array(['2013-06-28T00:00:00', '2013-07-03T00:00:00',
       '2013-07-08T00:00:00', '2013-07-13T00:00:00',
       '2013-07-18T00:00:00'], dtype='datetime64[s]')
In [61]: time.values.astype('datetime64[s]').tolist()
Out[61]: 
[datetime.datetime(2013, 6, 28, 0, 0),
 datetime.datetime(2013, 7, 3, 0, 0),
 datetime.datetime(2013, 7, 8, 0, 0),
 datetime.datetime(2013, 7, 13, 0, 0),
 datetime.datetime(2013, 7, 18, 0, 0)]

结果是 日期时间 物体。

def myfunc(lat, lon, when):
    ts = (when - np.datetime64('1970-01-01T00:00:00Z','s')) / np.timedelta64(1, 's')
    date = datetime.utcfromtimestamp(ts)
    print("Numpy date= ", when, " Python date= ", date)
    return float(90) - next_func(lat, lon, date)

尝试此代码

要将numpy datetime64[ns]转换为python datetime,只需尝试以下代码段

from datetime import datetime
datetime.utcfromtimestamp('your_time_stamp')