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.
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