如何区分np.ndarray或None？

Question

There is the code:

import numpy as np

def check(x):
    if x == None:
        print('x is None')

check(np.array([1,2]))

x can be None or np.ndarray, so I want to check whether x is None or np.ndarray, but if I pass np.ndarray into check , it will raise a error, because np.ndarray==None should use np.any() or np.all() . So what should I do?

Answer 1

Try this one:

import numpy as np

def check(x):
    if type(x) is np.ndarray:
        print('x is numpy.ndarray')
    else:
        raise ValueError("x is None")

check(np.array([1, 2]))

Answer 2

Checking for ndarray type

Preferred over type(x) : use isinstance .

From PEP 8 :

Object type comparisons should always use isinstance() instead of comparing types directly.

In your example: use if isinstance(x, np.ndarray):

Checking if x is None

Option 1: Use elif x is None: . This explicitly checks that x is None.

Option 2: Use elif not x: . This takes advantage of the "Falsiness" of None , but it also would evaluate to True if x is other "Falsey" values such as np.nan , 0, or an empty data structure.

Answer 3

The reason it raises the value error is that with a recent enough version of Numpy the __eq__ override of arrays does an elementwise object comparison even when comparing to None.

With Numpy 1.12.1:

In [2]: if np.array([1,2,3]) == None:
   ...:     print('Equals None')
   ...: else:
   ...:     print('Does not equal None')
   ...:     
/home/user/Work/SO/bin/ipython:1: FutureWarning: comparison to `None` will result in an elementwise object comparison in the future.
  #!/home/user/Work/SO/bin/python3
Does not equal None

and with Numpy 1.13.1:

In [1]: if np.array([1,2,3]) == None:
   ...:     print('Equals None')
   ...: else:
   ...:     print('Does not equal None')
   ...:     
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-1-06133f6732e1> in <module>()
----> 1 if np.array([1,2,3]) == None:
      2     print('Equals None')
      3 else:
      4     print('Does not equal None')
      5 

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

The error is quite self explanatory. Numpy has a different view on truthiness of an array compared to how plain Python sequences behave.

In order to check if an object is the singleton value None , you should use identity comparison as also explained here :

def check(x):
    if x is None:
        print('x is None')

    else:
        print('x is not None')