Difference between data type 'datetime64[ns]' and '<M8[ns]'?

Question:

I have created a TimeSeries in pandas:

In [346]: from datetime import datetime

In [347]: dates = [datetime(2011, 1, 2), datetime(2011, 1, 5), datetime(2011, 1, 7),

 .....: datetime(2011, 1, 8), datetime(2011, 1, 10), datetime(2011, 1, 12)]

In [348]: ts = Series(np.random.randn(6), index=dates)

In [349]: ts

Out[349]: 

2011-01-02 0.690002

2011-01-05 1.001543

2011-01-07 -0.503087

2011-01-08 -0.622274

2011-01-10 -0.921169

2011-01-12 -0.726213

I’m following on the example from ‘Python for Data Analysis’ book.

In the following paragraph, the author checks the index type:

In [353]: ts.index.dtype

Out[353]: dtype('datetime64[ns]')

When I do exactly the same operation in the console I get:

ts.index.dtype
dtype('<M8[ns]')

What is the difference between two types 'datetime64[ns]' and '<M8[ns]' ?

And why do I get a different type?

Asked By: LLaP

||

Answers:

datetime64[ns] is a general dtype, while <M8[ns] is a specific dtype. General dtypes map to specific dtypes, but may be different from one installation of NumPy to the next.

On a machine whose byte order is little endian, there is no difference between
np.dtype('datetime64[ns]') and np.dtype('<M8[ns]'):

In [6]: np.dtype('datetime64[ns]') == np.dtype('<M8[ns]')
Out[6]: True

However, on a big endian machine, np.dtype('datetime64[ns]') would equal np.dtype('>M8[ns]').

So datetime64[ns] maps to either <M8[ns] or >M8[ns] depending on the endian-ness of the machine.

There are many other similar examples of general dtypes mapping to specific dtypes:
int64 maps to <i8 or >i8, and int maps to either int32 or int64
depending on the bit architecture of the OS and how NumPy was compiled.


Apparently, the repr of the datetime64 dtype has change since the time the book was written to show the endian-ness of the dtype.

Answered By: unutbu

A bit of background will help understand the nuances of the output.

Numpy has an elaborate hierarchy of data types. The type information is stored as attributes in a data type object, which is an instance of numpy.dtype class. It describes how the bytes in the fixed-size block of memory corresponding to an array item should be interpreted (order of bytes, number of bytes, etc.).

Create an instance of the dtype

In [1]: import numpy as np

In [2]: dt = np.datetime64('1980', 'ns')

In [3]: dt
Out[3]: numpy.datetime64('1980-01-01T00:00:00.000000000')

In [4]: dt.dtype
Out[4]: dtype('<M8[ns]')

Examine the attributes

In [5]: dt.dtype.char
Out[5]: 'M'

In [6]: dt.dtype.name
Out[6]: 'datetime64[ns]'

In [7]: dt.dtype.str
Out[7]: '<M8[ns]'

In [8]: dt.dtype.type
Out[8]: numpy.datetime64

repr and str are string representations of an object, and each can have a different output for the same underlying data type.

In [9]: repr(dt.dtype)
Out[9]: "dtype('<M8[ns]')"

In [10]: str(dt.dtype)
Out[10]: 'datetime64[ns]'

An application (shell, console, debugger etc.) may invoke either one of them, so the output may look different for the same type.

As confusing as this is, there are still more nuances in terms of bit width, type aliases etc. See Data types in Python, Numpy and Pandas for the gory details.

Answered By: ap-osd
Categories: questions Tags: , , ,
Answers are sorted by their score. The answer accepted by the question owner as the best is marked with
at the top-right corner.