检查 numpy 数组的长度和宽度是否具有相同数量的元素

Question

Given a numpy array, what is the best way to check if it has the same number of elements across its length and width? So for example, in the following arrays,

arr1 = np.array([[0,1],[2,3],[4,5]])
arr2 = np.array([1,2,None,4])
arr3 = np.array([1,2,3,[4,5]])

arr1 , arr2 are OK but arr3 is not OK.

This SO post partially answers my question but it doesn't quite work because it will say arr2 is not OK because it has None .

Answer 1

here is a piece of code cleverly using try-except to check the required conditions. (This should cover all cases of numerical data). .astype(float) is able to handle None and it converts it to nan. list is not handled by it, so we can use that to differentiate.

def check_arr(array):
    try:
        array = array.astype(float)
        return "OK"
    except:
        return "not OK"
arr1 = np.array([[0,1],[2,3],[4,5]])
arr2 = np.array([1,2,None,4])
arr3 = np.array([1,2,3,[4,5]])

print(f"arr1 is {check_arr(arr1)}")
print(f"arr2 is {check_arr(arr2)}")
print(f"arr3 is {check_arr(arr3)}")

Output:

arr1 is OK
arr2 is OK
arr3 is not OK

A better way would be to do this:

def check_arr(array):
    return "not OK" if array.dtype==object else "OK"

but this won't work in your case as a valid case in your example has the NoneType value, which makes the array datatype object . I would suggest you use np.nan and float arrays instead of None in general for such cases.