[python] How can I check whether a numpy array is empty or not?

How can I check whether a numpy array is empty or not?

I used the following code, but this fails if the array contains a zero.

if not self.Definition.all():

Is this the solution?

if self.Definition == array( [] ):

http://www.scipy.org/Tentative_NumPy_Tutorial#head-6a1bc005bd80e1b19f812e1e64e0d25d50f99fe2

NumPy's main object is the homogeneous multidimensional array. In Numpy dimensions are called axes. The number of axes is rank. Numpy's array class is called ndarray. It is also known by the alias array. The more important attributes of an ndarray object are:

ndarray.ndim
the number of axes (dimensions) of the array. In the Python world, the number of dimensions is referred to as rank.

ndarray.shape
the dimensions of the array. This is a tuple of integers indicating the size of the array in each dimension. For a matrix with n rows and m columns, shape will be (n,m). The length of the shape tuple is therefore the rank, or number of dimensions, ndim.

ndarray.size
the total number of elements of the array. This is equal to the product of the elements of shape.

Why would we want to check if an array is empty? Arrays don't grow or shrink in the same that lists do. Starting with a 'empty' array, and growing with np.append is a frequent novice error.

Using a list in if alist: hinges on its boolean value:

In [102]: bool([])                                                                       
Out[102]: False
In [103]: bool([1])                                                                      
Out[103]: True

But trying to do the same with an array produces (in version 1.18):

In [104]: bool(np.array([]))                                                             
/usr/local/bin/ipython3:1: DeprecationWarning: The truth value 
   of an empty array is ambiguous. Returning False, but in 
   future this will result in an error. Use `array.size > 0` to 
   check that an array is not empty.
  #!/usr/bin/python3
Out[104]: False

In [105]: bool(np.array([1]))                                                            
Out[105]: True

and bool(np.array([1,2]) produces the infamous ambiguity error.

edit

The accepted answer suggests size:

In [11]: x = np.array([])
In [12]: x.size
Out[12]: 0

But I (and most others) check the shape more than the size:

In [13]: x.shape
Out[13]: (0,)

Another thing in its favor is that it 'maps' on to an empty list:

In [14]: x.tolist()
Out[14]: []

But there are other other arrays with 0 size, that aren't 'empty' in that last sense:

In [15]: x = np.array([[]])
In [16]: x.size
Out[16]: 0
In [17]: x.shape
Out[17]: (1, 0)
In [18]: x.tolist()
Out[18]: [[]]
In [19]: bool(x.tolist())
Out[19]: True

np.array([[],[]]) is also size 0, but shape (2,0) and len 2.

While the concept of an empty list is well defined, an empty array is not well defined. One empty list is equal to another. The same can't be said for a size 0 array.

The answer really depends on

• what do you mean by 'empty'?
• what are you really test for?

One caveat, though. Note that np.array(None).size returns 1! This is because a.size is equivalent to np.prod(a.shape), np.array(None).shape is (), and an empty product is 1.

>>> import numpy as np
>>> np.array(None).size
1
>>> np.array(None).shape
()
>>> np.prod(())
1.0

Therefore, I use the following to test if a numpy array has elements:

>>> def elements(array):
    ...     return array.ndim and array.size

>>> elements(np.array(None))
0
>>> elements(np.array([]))
0
>>> elements(np.zeros((2,3,4)))
24