ValueError when checking if variable is None or numpy.array

梦想与她 提交于 2019-11-28 18:04:50
Jerfov2

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:
    ...    

You can see if object has shape or not

def check_array(x):
    try:
        x.shape
        return True
    except:
        return False

If you are trying to do something very similar a is not None, the same issue comes up: Numpy complains that one must use a.any or a.all. A workaround is to do:

if not (a is None):
    pass

Not too pretty, but it does the job.

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