Generating a numpy array with all combinations of numbers that sum to less than a given number

Question

There are several elegant examples of using numpy in Python to generate arrays of all combinations. For example the answer here: Using numpy to build an array of all combinations of two arrays .

Now suppose there is an extra constraint, namely, the sum of all numbers cannot add up to more than a given constant K . Using a generator and itertools.product , for an example with K=3 where we want the combinations of three variables with ranges 0-1,0-3, and 0-2 we can do it a follows:

from itertools import product
K = 3
maxRange = np.array([1,3,2])
states = np.array([i for i in product(*(range(i+1) for i in maxRange)) if sum(i)<=K])

which returns

array([[0, 0, 0],
       [0, 0, 1],
       [0, 0, 2],
       [0, 1, 0],
       [0, 1, 1],
       [0, 1, 2],
       [0, 2, 0],
       [0, 2, 1],
       [0, 3, 0],
       [1, 0, 0],
       [1, 0, 1],
       [1, 0, 2],
       [1, 1, 0],
       [1, 1, 1],
       [1, 2, 0]])

In principle, the approach from https://stackoverflow.com/a/25655090/1479342 can be used to generate all possible combinations without the constraint and then selecting the subset of combinations that sum up to less than K . However, that approach generates much more combinations than necessary, especially if K is relatively small compared to sum(maxRange) .

There must be a way to do this faster and with lower memory usage. How can this be achieved using a vectorized approach (for example using np.indices )?

Answer 1

I don't know what's a numpy approach, but here's a reasonably clean solution. Let A be an array of integers and let k be a number that you are given as input.

Start with an empty array B ; keep the sum of the array B in a variable s (initially set to zero). Apply the following procedure:

• if the sum s of the array B is less than k , then (i) add it to the collection, (ii) and for each element from the original array A , add that element to B and update s , (iii) delete it from A and (iv) recursively apply the procedure; (iv) when the call returns, add the element back to A and update s ; else do nothing.

This bottom-up approach prunes invalid branches early on and only visits the necessary subsets (ie almost only the subsets that sum to less than k ).

Answer 2

Edited

1. For completeness, I'm adding here the OP's code:

    def partition0(max_range, S): K = len(max_range) return np.array([i for i in itertools.product(*(range(i+1) for i in max_range)) if sum(i)<=S]) 

2. The first approach is pure np.indices . It's fast for small input but consumes a lot of memory (OP already pointed out it's not what he meant).

    def partition1(max_range, S): max_range = np.asarray(max_range, dtype = int) a = np.indices(max_range + 1) b = a.sum(axis = 0) <= S return (a[:,b].T) 

3. Recurrent approach seems to be much better than those above:

    def partition2(max_range, max_sum): max_range = np.asarray(max_range, dtype = int).ravel() if(max_range.size == 1): return np.arange(min(max_range[0],max_sum) + 1, dtype = int).reshape(-1,1) P = partition2(max_range[1:], max_sum) # S[i] is the largest summand we can place in front of P[i] S = np.minimum(max_sum - P.sum(axis = 1), max_range[0]) offset, sz = 0, S.size out = np.empty(shape = (sz + S.sum(), P.shape[1]+1), dtype = int) out[:sz,0] = 0 out[:sz,1:] = P for i in range(1, max_range[0]+1): ind, = np.nonzero(S) offset, sz = offset + sz, ind.size out[offset:offset+sz, 0] = i out[offset:offset+sz, 1:] = P[ind] S[ind] -= 1 return out 

4. After a short thought, I was able to take it a bit further. If we know in advance the number of possible partitions, we can allocate enough memory at once. (It's somewhat similar to cartesian in an already linked thread .)

   First, we need a function which counts partitions.

    def number_of_partitions(max_range, max_sum): ''' Returns an array arr of the same shape as max_range, where arr[j] = number of admissible partitions for j summands bounded by max_range[j:] and with sum <= max_sum ''' M = max_sum + 1 N = len(max_range) arr = np.zeros(shape=(M,N), dtype = int) arr[:,-1] = np.where(np.arange(M) <= min(max_range[-1], max_sum), 1, 0) for i in range(N-2,-1,-1): for j in range(max_range[i]+1): arr[j:,i] += arr[:Mj,i+1] return arr.sum(axis = 0) 

   The main function:

    def partition3(max_range, max_sum, out = None, n_part = None): if out is None: max_range = np.asarray(max_range, dtype = int).ravel() n_part = number_of_partitions(max_range, max_sum) out = np.zeros(shape = (n_part[0], max_range.size), dtype = int) if(max_range.size == 1): out[:] = np.arange(min(max_range[0],max_sum) + 1, dtype = int).reshape(-1,1) return out P = partition3(max_range[1:], max_sum, out=out[:n_part[1],1:], n_part = n_part[1:]) # P is now a useful reference S = np.minimum(max_sum - P.sum(axis = 1), max_range[0]) offset, sz = 0, S.size out[:sz,0] = 0 for i in range(1, max_range[0]+1): ind, = np.nonzero(S) offset, sz = offset + sz, ind.size out[offset:offset+sz, 0] = i out[offset:offset+sz, 1:] = P[ind] S[ind] -= 1 return out 

5. Some tests:

    max_range = [3, 4, 6, 3, 4, 6, 3, 4, 6] for f in [partition0, partition1, partition2, partition3]: print(f.__name__ + ':') for max_sum in [5, 15, 25]: print('Sum %2d: ' % max_sum, end = '') %timeit f(max_range, max_sum) print() partition0: Sum 5: 1 loops, best of 3: 859 ms per loop Sum 15: 1 loops, best of 3: 1.39 s per loop Sum 25: 1 loops, best of 3: 3.18 s per loop partition1: Sum 5: 10 loops, best of 3: 176 ms per loop Sum 15: 1 loops, best of 3: 224 ms per loop Sum 25: 1 loops, best of 3: 403 ms per loop partition2: Sum 5: 1000 loops, best of 3: 809 µs per loop Sum 15: 10 loops, best of 3: 62.5 ms per loop Sum 25: 1 loops, best of 3: 262 ms per loop partition3: Sum 5: 1000 loops, best of 3: 853 µs per loop Sum 15: 10 loops, best of 3: 59.1 ms per loop Sum 25: 1 loops, best of 3: 249 ms per loop 

   And something larger:

    %timeit partition0([3,6] * 5, 20) 1 loops, best of 3: 11.9 s per loop %timeit partition1([3,6] * 5, 20) The slowest run took 12.68 times longer than the fastest. This could mean that an intermediate result is being cached 1 loops, best of 3: 2.33 s per loop # MemoryError in another test %timeit partition2([3,6] * 5, 20) 1 loops, best of 3: 877 ms per loop %timeit partition3([3,6] * 5, 20) 1 loops, best of 3: 739 ms per loop