In newer version of numpy (>1.8.x), numpy.meshgrid() provides a much faster implementation: @pv’s solution In [113]: %timeit cartesian(([1, 2, 3], [4, 5], [6, 7])) 10000 loops, best of 3: 135 µs per loop In [114]: cartesian(([1, 2, 3], [4, 5], [6, 7])) Out[114]: array([[1, 4, 6], [1, 4, 7], [1, 5, 6], [1, 5, 7], … Read more
A canonical cartesian_product (almost) There are many approaches to this problem with different properties. Some are faster than others, and some are more general-purpose. After a lot of testing and tweaking, I’ve found that the following function, which calculates an n-dimensional cartesian_product, is faster than most others for many inputs. For a pair of approaches … Read more
you need itertools.product: >>> import itertools >>> a = [[1,2,3],[4,5,6],[7,8,9,10]] >>> list(itertools.product(*a)) [(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 4, 10), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 5, 10), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 6, 10), (2, 4, 7), (2, 4, 8), (2, … Read more
itertools.product Available from Python 2.6. import itertools somelists = [ [1, 2, 3], [‘a’, ‘b’], [4, 5] ] for element in itertools.product(*somelists): print(element) Which is the same as, for element in itertools.product([1, 2, 3], [‘a’, ‘b’], [4, 5]): print(element)