python - 来自 unix utc 秒的 numpy datetime64

Question

注意:我认为 datetime64 做的是对的。所以我会保留这篇文章以防它有用。

从 numpy 1.7.0 开始，传入 np.datetime64 的秒数被解释为本地时区。有没有一种干净快速的方法可以将 unix utc 秒导入 np.datetime64？我有 50M 的数组，似乎应该有一种方法可以告诉 np.datetime64 我的秒值是 UTC，不是吗？

datetime.datetime.utcfromtimestamp(1338624706)
datetime.datetime(2012, 6, 2, 8, 11, 46)  # this is the time I'm looking for

np.datetime64(1338624706, 's')
numpy.datetime64('2012-06-02T01:11:46-0700')  # Darn you ISO!  Off by 7 hours

dt64 = np.datetime64(1338624706, 's')
dt64.astype(datetime.datetime)
datetime.datetime(2012, 6, 2, 8, 11, 46)  # Wait, did it do the right thing?

# This seems like the best option at the moment,
# but requires building datetime.datetime objects:
dt64 = np.datetime64(datetime.datetime.utcfromtimestamp(1338624706))
numpy.datetime64('2012-06-02T01:11:46.000000-0700') # Show this
dt64.astype(datetime.datetime)
datetime.datetime(2012, 6, 2, 8, 11, 46)  # Looks like it worked

我真的不想求助于字符串操作。如果能够将 unix utc 整数数组或 float 直接转换为正确的 dt64，我会很高兴。

https://stackoverflow.com/a/13704307/417578暗示 numpy 1.8.0 可能会做我想做的事，但是有什么东西可以在 1.7.0 中工作吗？

Answer 1

这是 pandas 的另一种方式(正确处理不同版本的 numpy datetime64 中的怪癖，所以这在 numpy 1.6.2 中有效)——我想你可能需要当前的主人(0.11-dev)

# obviously replace this by your utc seconds
# need to convert to the default in pandas of datetime64[ns]
z = pd.Series([(1338624706 + i)*1e9 for i in range(50)],dtype='datetime64[ns]')

In [35]: z.head()
Out[35]: 
0   2012-06-02 08:11:46
1   2012-06-02 08:11:47
2   2012-06-02 08:11:48
3   2012-06-02 08:11:49
4   2012-06-02 08:11:50
Dtype: datetime64[ns]

# turn it into a DatetimeIndex and localize
lidx = pd.DatetimeIndex(z).tz_localize('UTC')

<class 'pandas.tseries.index.DatetimeIndex'>
[2012-06-02 08:11:46, ..., 2012-06-02 08:12:35]
Length: 50, Freq: None, Timezone: UTC

# now you have a nice object to say convert timezones
In [44]: lidx.tz_convert('US/Eastern')
Out[44]: 
<class 'pandas.tseries.index.DatetimeIndex'>
[2012-06-02 04:11:46, ..., 2012-06-02 04:12:35]
Length: 50, Freq: None, Timezone: US/Eastern