Here's a pure-numpy implementation. It's about 5× faster than using itertools.
Python 3:
import numpy as np
def cartesian(arrays, out=None):
"""
Generate a cartesian product of input arrays.
Parameters
----------
arrays : list of array-like
1-D arrays to form the cartesian product of.
out : ndarray
Array to place the cartesian product in.
Returns
-------
out : ndarray
2-D array of shape (M, len(arrays)) containing cartesian products
formed of input arrays.
Examples
--------
>>> cartesian(([1, 2, 3], [4, 5], [6, 7]))
array([[1, 4, 6],
[1, 4, 7],
[1, 5, 6],
[1, 5, 7],
[2, 4, 6],
[2, 4, 7],
[2, 5, 6],
[2, 5, 7],
[3, 4, 6],
[3, 4, 7],
[3, 5, 6],
[3, 5, 7]])
"""
arrays = [np.asarray(x) for x in arrays]
dtype = arrays[0].dtype
n = np.prod([x.size for x in arrays])
if out is None:
out = np.zeros([n, len(arrays)], dtype=dtype)
#m = n / arrays[0].size
m = int(n / arrays[0].size)
out[:,0] = np.repeat(arrays[0], m)
if arrays[1:]:
cartesian(arrays[1:], out=out[0:m, 1:])
for j in range(1, arrays[0].size):
#for j in xrange(1, arrays[0].size):
out[j*m:(j+1)*m, 1:] = out[0:m, 1:]
return out
Python 2:
import numpy as np
def cartesian(arrays, out=None):
arrays = [np.asarray(x) for x in arrays]
dtype = arrays[0].dtype
n = np.prod([x.size for x in arrays])
if out is None:
out = np.zeros([n, len(arrays)], dtype=dtype)
m = n / arrays[0].size
out[:,0] = np.repeat(arrays[0], m)
if arrays[1:]:
cartesian(arrays[1:], out=out[0:m, 1:])
for j in xrange(1, arrays[0].size):
out[j*m:(j+1)*m, 1:] = out[0:m, 1:]
return out