Question

注意：我认为datetime64正在做正确的事情。所以我会留下帖子，以防它有用。

从numpy 1.7.0开始，传入np.datetime64的秒被解释为在本地时区。是否有一种干净，快速的方法将unix utc秒导入np.datetime64？我有50M这些数组的数组，似乎应该有一种方法告诉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

我真的不想诉诸字符串操作。我很高兴能够将unix utc int数组转换成直接浮点数到正确的dt64。

https://stackoverflow.com/a/13704307/417578暗示numpy 1.8.0可能会做我想要的，但是有什么东西可以在1.7.0中运行吗？

Answer 1

大熊猫的另一种方式（正确地处理不同版本的numpy datetime64中的怪癖，所以这适用于numpy 1.6.2） - 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

Answer 2

也许我误解了这个问题，但不是时区只是一个显示问题？

utc_time = datetime.datetime.utcnow()
print utc_time
dt64 =  np.datetime64(utc_time)
print dt64
print dt64.astype(datetime.datetime)


2013-02-24 17:30:53.586297
2013-02-24T11:30:53.586297-0600
2013-02-24 17:30:53.586297

时间没有以任何方式“改变”：

some_time = datetime.datetime.utcfromtimestamp(1338624706)
dt64 = np.datetime64(1338624706,'s')
print dt64.astype(int64)
1338624706

这是numpy 1.7。

来自unix utc秒的numpy datetime64

2 个答案: