creating numpy array with coordinates of vertices of N-dimensional cube

Question

What could be an efficient way of creating numpy array containing the coordinates of vertices of the N-dimensional cube centered at origin of coordinates, given the dimension N.

For example, for N=1 it should return np.array([[1],[-1]])

For N=2 it should return np.array([[1,1],[1,-1],[-1,1],[-1,-1]])

For N=3: np.array([[1,1,1],[1,1,-1],[1,-1,1],[1,-1,-1],[-1,1,1],[-1,1,-1],[-1,-1,1],[-1,-1,-1]])

Chris Mueller · Accepted Answer · 2018-09-07 20:40:31Z

5

You can use product from itertools

from itertools import product

def vertices(N): 
    return list(product((1, -1), repeat=N))

print(vertices(1))
# [(1,), (-1,)]

print(vertices(2))
# [(1, 1), (1, -1), (-1, 1), (-1, -1)]

print(vertices(3))
# [(1, 1, 1), (1, 1, -1), (1, -1, 1), (1, -1, -1), (-1, 1, 1), (-1, 1, -1), (-1, -1, 1), (-1, -1, -1)]

answered Sep 7, 2018 at 20:40

Chris Mueller

6,7305 gold badges31 silver badges35 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Warren Weckesser · Accepted Answer · 2018-09-07 21:04:24Z

2

Here's another method: 2*((np.arange(2**N)[:,None] & (1 << np.arange(N))) > 0) - 1

In [25]: N = 1

In [26]: 2*((np.arange(2**N)[:,None] & (1 << np.arange(N))) > 0) - 1
Out[26]: 
array([[-1],
       [ 1]])

In [27]: N = 2

In [28]: 2*((np.arange(2**N)[:,None] & (1 << np.arange(N))) > 0) - 1
Out[28]: 
array([[-1, -1],
       [ 1, -1],
       [-1,  1],
       [ 1,  1]])

In [29]: N = 3

In [30]: 2*((np.arange(2**N)[:,None] & (1 << np.arange(N))) > 0) - 1
Out[30]: 
array([[-1, -1, -1],
       [ 1, -1, -1],
       [-1,  1, -1],
       [ 1,  1, -1],
       [-1, -1,  1],
       [ 1, -1,  1],
       [-1,  1,  1],
       [ 1,  1,  1]])

What it does:

Create an array of integers from 0 to 2**N-1.
Convert the integers to binary (e.g. 3 becomes [1, 1, 0]).
Multiply the binary array by 2 and subtract 1 to convert from 0 and 1 to -1 and 1.

Numpy broadcasting is used to vectorize the operations.

edited Sep 7, 2018 at 21:04

answered Sep 7, 2018 at 20:57

Warren Weckesser

116k20 gold badges207 silver badges224 bronze badges

1 Comment

Warren Weckesser Over a year ago

@kevinkayaks I added a brief explanation to the answer.

kevinkayaks · Accepted Answer · 2018-09-07 20:43:03Z

2

Here is a pure numpy implementation involving meshgrid and stack. It's a little contrived-- would welcome input to make it better.

pts = np.stack(([-1,1],)*N,0)
vertices = (np.array(np.meshgrid(*pts)).T).reshape(2**N,N)
vertices

N=3 returns

array([[-1, -1, -1],
       [-1,  1, -1],
       [ 1, -1, -1],
       [ 1,  1, -1],
       [-1, -1,  1],
       [-1,  1,  1],
       [ 1, -1,  1],
       [ 1,  1,  1]])

N=1 returns

array([[-1],
       [ 1]])

answered Sep 7, 2018 at 20:43

kevinkayaks

2,7561 gold badge17 silver badges30 bronze badges

Collectives™ on Stack Overflow

creating numpy array with coordinates of vertices of N-dimensional cube

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related