Build 2d pyramidal array - Python / NumPy

Approach #1

We can use broadcasting -

def pyramid(n):
    r = np.arange(n)
    d = np.minimum(r,r[::-1])
    return np.minimum.outer(d,d)

Approach #2

We can also use concatenation to create d, like so -

d = np.r_[np.arange(n//2),np.arange(n//2-(n%2==0),-1,-1)]

Thus, giving us an alternative one-liner -

np.minimum.outer(*(2*[np.r_[np.arange(n//2),np.arange(n//2-(n%2==0),-1,-1)]]))

Sample runs -

In [83]: pyramid(5)
Out[83]: 
array([[0, 0, 0, 0, 0],
       [0, 1, 1, 1, 0],
       [0, 1, 2, 1, 0],
       [0, 1, 1, 1, 0],
       [0, 0, 0, 0, 0]])

In [84]: pyramid(6)
Out[84]: 
array([[0, 0, 0, 0, 0, 0],
       [0, 1, 1, 1, 1, 0],
       [0, 1, 2, 2, 1, 0],
       [0, 1, 2, 2, 1, 0],
       [0, 1, 1, 1, 1, 0],
       [0, 0, 0, 0, 0, 0]])

In [85]: pyramid(8)
Out[85]: 
array([[0, 0, 0, 0, 0, 0, 0, 0],
       [0, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 2, 2, 2, 2, 1, 0],
       [0, 1, 2, 3, 3, 2, 1, 0],
       [0, 1, 2, 3, 3, 2, 1, 0],
       [0, 1, 2, 2, 2, 2, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 0],
       [0, 0, 0, 0, 0, 0, 0, 0]])

Use numpy.pad:

import numpy as np

def pyramid(n):
    if n % 2:
        arr = np.zeros((1,1))
        N = int((n-1)/2)
    else:
        arr = np.zeros((2,2))
        N = int(n/2)-1

    for i in range(N):
        arr += 1
        arr = np.pad(arr, 1, mode='constant')
    return arr

Output:

pyramid(6)
array([[0., 0., 0., 0., 0., 0.],
       [0., 1., 1., 1., 1., 0.],
       [0., 1., 2., 2., 1., 0.],
       [0., 1., 2., 2., 1., 0.],
       [0., 1., 1., 1., 1., 0.],
       [0., 0., 0., 0., 0., 0.]])

pyramid(5)
array([[0., 0., 0., 0., 0.],
       [0., 1., 1., 1., 0.],
       [0., 1., 2., 1., 0.],
       [0., 1., 1., 1., 0.],
       [0., 0., 0., 0., 0.]])

numpy.pad(arr, 1, 'constant') returns arr wrapped with 1 layer of zeros.