ValueError when checking if variable is None or numpy.array

Question

I'd like to check if variable is None or numpy.array. I've implemented check_a function to do this.

def check_a(a):
    if not a:
        print "please initialize a"

a = None
check_a(a)
a = np.array([1,2])
check_a(a)

But, this code raises ValueError. What is the straight forward way?

ValueError                                Traceback (most recent call last)
<ipython-input-41-0201c81c185e> in <module>()
      6 check_a(a)
      7 a = np.array([1,2])
----> 8 check_a(a)

<ipython-input-41-0201c81c185e> in check_a(a)
      1 def check_a(a):
----> 2     if not a:
      3         print "please initialize a"
      4 
      5 a = None

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

This ValueError is one of the most common numpy questions. It means not a produces a boolean array, with (in this case) 2 values. This boolean array cannot be used as an if condition! The is None alternative is good to know, but you should also understand this error. — hpaulj
– hpaulj, Commented Apr 22, 2016 at 5:01
@hpaulj: Not quite - you can't overload not, so the error actually happens when not tries to treat the array as a single boolean and finds out it can't. If it had been ~a, that would have used NumPy's overload and failed when if tries to use the negated array as a single boolean. — user2357112
– user2357112, Commented Apr 19, 2019 at 22:41

user2357112 · Accepted Answer · 2019-04-19 22:50:56Z

289

Using not a to test whether a is None assumes that the other possible values of a have a truth value of True. However, most NumPy arrays don't have a truth value at all, and not cannot be applied to them.

If you want to test whether an object is None, the most general, reliable way is to literally use an is check against None:

if a is None:
    ...
else:
    ...

This doesn't depend on objects having a truth value, so it works with NumPy arrays.

Note that the test has to be is, not ==. is is an object identity test. == is whatever the arguments say it is, and NumPy arrays say it's a broadcasted elementwise equality comparison, producing a boolean array:

>>> a = numpy.arange(5)
>>> a == None
array([False, False, False, False, False])
>>> if a == None:
...     pass
...
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: The truth value of an array with more than one element is ambiguous.
 Use a.any() or a.all()

On the other side of things, if you want to test whether an object is a NumPy array, you can test its type:

# Careful - the type is np.ndarray, not np.array. np.array is a factory function.
if type(a) is np.ndarray:
    ...
else:
    ...

You can also use isinstance, which will also return True for subclasses of that type (if that is what you want). Considering how terrible and incompatible np.matrix is, you may not actually want this:

# Again, ndarray, not array, because array is a factory function.
if isinstance(a, np.ndarray):
    ...
else:
    ...

edited Apr 19, 2019 at 22:50

user2357112

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answered Apr 22, 2016 at 2:54

Jerfov2

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3 Comments

Monica Heddneck Over a year ago

which do you recommend is the "best" solution?

Maitreya Over a year ago

Thanks! The isinstance(a, np.ndarray) way looks cleanest to me. The company I recently joined does the is not None check and it makes my pythonic alarms blare every time I read it.

ZohanPolis Over a year ago

This is a great, clear and detailed answer. I kept checking for if a == None and wondered why it wasn't working. Thanks :)

SKO · Accepted Answer · 2021-01-27 00:54:34Z

-1

To stick to == without consideration of the other type, the following is also possible.
type(a) == type(None)

edited Jan 27, 2021 at 0:54

answered Jan 22, 2021 at 4:39

SKO

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