["python","arrays","numpy","multidimensional-array","cartesian-product"]
Using NumPy to build an array of all combinations of two arrays

I'm trying to run over the parameters space of a six-parameter function to study its numerical behavior before trying to do anything complex with it, so I'm searching for an efficient way to do this. My function takes float values given in a 6-dim NumPy array as input. What I tried to do initially was this: First, I created a function that takes two arrays and generate an array with all combinations of values from the two arrays: from numpy import * def comb(a, b): c = [] for i in a: for j in b: c.append(r_[i,j]) return c Then, I used reduce() to apply that to m copies of the same array: def combs(a, m): return reduce(comb, [a]*m) Finally, I evaluate my function like this: values = combs(np.arange(0, 1, 0.1), 6) for val in values: print F(val) This works, but it's way too slow. I know the space of parameters is huge, but this shouldn't be so slow. I have only sampled 106 (a million) points in this example and it took more than 15 seconds just to create the array values. Is there a more efficient way of doing this with NumPy? I can modify the way the function F takes its arguments if it's necessary.
In newer versions of NumPy (>1.8.x), numpy.meshgrid() provides a much faster implementation: For 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], [2, 4, 6], [2, 4, 7], [2, 5, 6], [2, 5, 7], [3, 4, 6], [3, 4, 7], [3, 5, 6], [3, 5, 7]]) numpy.meshgrid() used to be two-dimensional only, but now it is capable of being multidimensional. In this case, three-dimensional: In [115]: %timeit np.array(np.meshgrid([1, 2, 3], [4, 5], [6, 7])).T.reshape(-1,3) 10000 loops, best of 3: 74.1 µs per loop In [116]: np.array(np.meshgrid([1, 2, 3], [4, 5], [6, 7])).T.reshape(-1,3) Out[116]: array([[1, 4, 6], [1, 5, 6], [2, 4, 6], [2, 5, 6], [3, 4, 6], [3, 5, 6], [1, 4, 7], [1, 5, 7], [2, 4, 7], [2, 5, 7], [3, 4, 7], [3, 5, 7]]) Note that the order of the final result is slightly different.