timedelta64[ns]: max() and min() return numpy.int64

[michaelaye]

I guess that max() and min() are not implementing yet fully:

In [28]: diff.head()
Out[28]:
0               NaT
1   00:00:00.128001
2   00:00:00.127999
3   00:00:00.128001
4   00:00:00.127999
Dtype: timedelta64[ns]

In [29]: diff.max()
Out[29]: 1934745001000

In [30]: diff.max().dtype
Out[30]: dtype('int64')

In [31]: diff[0].dtype
Out[31]: dtype('timedelta64[ns]')

In [32]: diff.min().dtype
Out[32]: dtype('int64')

[jreback]

timedeltas IS an int under the hood
but I'll take a look at some point

[jreback]

#2990 should close this, merging soon

[jreback]

this was merged into master....pls check out and let me know (also slight update on the docs, will be avail after 5pm est today I think...on the dev link)

[michaelaye]

Works good, thanks a lot!

[michaelaye]

oh, I noticed once thing odd, unrelated to this:

>>> diff.head()
0               NaT
1   00:00:00.127999
2   00:00:00.128000
3   00:00:00.128001
4   00:00:00.128000
dtype: timedelta64[ns]
>>> diff[0]
106751 days, 23:47:16.854775

but as usual, that's only a display issue, I guess?

[jreback]

see below

the 'NaT' is really a np.datetime64, with really large negative value, its not a 'type'
like the 'NaT' in a datetime64[ns] series (its the same integer value tought)

datetime64[ns] series hold Timestamp types
there isn't an equivalent timedelta64 type (well there is, but its the numpy type np.datetime64)
and it doesn't know any about NaT

but I think numpy 1.7 fixes this (common usage is 1.6.2)

That's why I always wrap a series around the timedelta64s (even as a return value from max, which
really should be a scalar)

In [9]: td = pd.Series(pd.date_range('20130102',periods=6))

In [10]: td = td-td.shift()

In [11]: td
Out[11]: 
0                NaT
1   1 days, 00:00:00
2   1 days, 00:00:00
3   1 days, 00:00:00
4   1 days, 00:00:00
5   1 days, 00:00:00
dtype: timedelta64[ns]

In [12]: td[0]
Out[12]: 106751 days, 23:47:16.854775

In [13]: type(td[0])
Out[13]: numpy.timedelta64