有点复杂...
But
d = np.array(1, dtype='timedelta64[D]')
x = x.astype('datetime64[D]')
deltas = np.diff(x, axis=1) / d
np.concatenate([
i + np.arange(j + 1) for i, j in zip(x[:, 0], deltas[:, 0].astype(int))
]).astype('datetime64[ns]')
array(['2017-10-02T00:00:00.000000000', '2017-10-03T00:00:00.000000000',
'2017-10-04T00:00:00.000000000', '2017-10-05T00:00:00.000000000',
'2017-10-06T00:00:00.000000000', '2017-10-07T00:00:00.000000000',
'2017-10-08T00:00:00.000000000', '2017-10-09T00:00:00.000000000',
'2017-10-10T00:00:00.000000000', '2017-10-11T00:00:00.000000000',
'2017-10-12T00:00:00.000000000'], dtype='datetime64[ns]')
怎么运行的
-
d
代表一天
-
x
变成没有时间戳的日期
-
diff
让我知道相差的天数...但是在timedelta
space
- 我除以我的
d
这也是在timedelta
空间和维度消失了……留给我float
我投射到int
- 当我添加对的第一列时
x[:, 0]
对于一个整数数组,我得到一个添加 1 个单位的广播,无论维度是什么x
,即datetime64[D]
。所以我要添加一天。
源自/受 @hpaulj 启发
如果他们发布答案就会删除
d = np.array(1, dtype='timedelta64[D]')
np.concatenate([np.arange(row[0], row[1] + 1, d) for row in x])
array(['2017-10-02T00:00:00.000000000', '2017-10-03T00:00:00.000000000',
'2017-10-04T00:00:00.000000000', '2017-10-05T00:00:00.000000000',
'2017-10-06T00:00:00.000000000', '2017-10-07T00:00:00.000000000',
'2017-10-08T00:00:00.000000000', '2017-10-09T00:00:00.000000000',
'2017-10-10T00:00:00.000000000', '2017-10-11T00:00:00.000000000',
'2017-10-12T00:00:00.000000000'], dtype='datetime64[ns]')