arranging a array without loops list comp or recursion

Question

Definition:Arranged array an arranged array is an array of dim 2 , shape is square matrix (NXN) and for every cell in the matrix : A[I,J] > A[I,J+1] AND A[I,J] > A[I+1,J]

I have an assignment to write a func that: gets a numpy array and returns True if - the given array is an arranged array False - otherwise

note: We CANNOT use loops, list comps OR recursion. the point of the task is to use numpy things. assumptions: we can assume that the array isn't empty and has no NA's, also all of the cells are numerics

My code isn't very numpy oriented.. :

def is_square_ordered_matrix(A):
    # Checking if the dimension is 2
    if A.ndim != 2:
        return False
    # Checking if it is a squared matrix
    if A.shape[0] != A.shape[1]:
        return False
    # Saving the original shape to reshape later
    originalDim = A.shape
    # Making it a dim of 1 to use it as a list
    arrayAsList = list((A.reshape((1,originalDim[0]**2)))[0])
    # Keeping original order before sorting
    originalArray = arrayAsList[:]
    # Using the values of the list as keys to see if there are doubles
    valuesDictionary = dict.fromkeys(arrayAsList, 1)
    # If len is different, means there are doubles and i should return False
    if len(arrayAsList) != len(valuesDictionary):
        return False
    # If sorted list is equal to original list it means the original is already ordered and i should return True
    arrayAsList.sort(reverse=True)
    if originalArray == arrayAsList:
        return True
    else:
        return False

True example:

is_square_ordered_matrix(np.arange(8,-1,-1).reshape((3,3)))

False example:

is_square_ordered_matrix(np.arange(9).reshape((3,3)))
is_square_ordered_matrix(np.arange(5,-1,-1).reshape((3,2)))

Answer 1

Simple comparison:

>>> def is_square_ordered_matrix(a):
...     return a.shape[0] == a.shape[1] and np.all(a[:-1] > a[1:]) and np.all(a[:, :-1] > a[:, 1:])
...
>>> is_square_ordered_matrix(np.arange(8,-1,-1).reshape((3,3)))
True
>>> is_square_ordered_matrix(np.arange(9).reshape((3,3)))
False
>>> is_square_ordered_matrix(np.arange(5,-1,-1).reshape((3,2)))
False

First, compare a[:-1] with a[1:] , which will compare the elements of each row with the elements of the next row, and then use np.all to judge:

>>> a = np.arange(8,-1,-1).reshape((3,3))
>>> a[:-1]   # First and second lines
array([[8, 7, 6],
       [5, 4, 3]])
>>> a[1:]    # Second and third lines
array([[5, 4, 3],
       [2, 1, 0]])
>>> a[:-1] > a[1:]
array([[ True,  True,  True],
       [ True,  True,  True]])
>>> np.all(a[:-1] > a[1:])
True

Then compare a[:,:-1] with a[:, 1:] , which will compare the columns:

>>> a[:, :-1]   # First and second columns
array([[8, 7],
       [5, 4],
       [2, 1]])
>>> a[:, 1:]    # Second and third columns
array([[7, 6],
       [4, 3],
       [1, 0]])
>>> a[:, :-1] > a[:, 1:]
array([[ True,  True],
       [ True,  True],
       [ True,  True]])

The result of row comparison and column comparison is the result you want.