有几个在 Python 中使用 numpy 生成所有组合数组的优雅示例。例如这里的答案:Using numpy to build an array of all combinations of two arrays .
现在假设有一个额外的约束,即所有数字的总和不能超过给定常数K
。使用生成器和 itertools.product
,以 K=3
为例,我们想要三个变量的组合,范围为 0-1、0-3 和 0- 2 我们可以这样做:
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])
返回
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]])
原则上,来自 https://stackoverflow.com/a/25655090/1479342 的方法可用于在没有约束的情况下生成所有可能的组合,然后选择总和小于 K
的组合子集。但是,该方法生成的组合比必要的多得多,尤其是当 K
与 sum(maxRange)
相比相对较小时。
必须有一种方法可以更快地完成此操作并减少内存使用量。如何使用矢量化方法(例如使用 np.indices
)实现这一点?
最佳答案
已编辑
为了完整起见,我在此处添加了 OP 的代码:
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])
第一种方法是纯粹的
np.indices
。它对于小输入来说速度很快,但会消耗大量内存(OP 已经指出这不是他的意思)。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)
递归方法似乎比上面的方法好得多:
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
经过短暂的思考,我能够更进一步。如果我们事先知道可能的分区数,我们可以一次性分配足够的内存。 (它有点类似于 an already linked thread 中的
笛卡尔
。)首先,我们需要一个计算分区的函数。
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[:M-j,i+1] return arr.sum(axis = 0)
主要功能:
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
一些测试:
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
还有更大的东西:
%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
关于python - 生成一个 numpy 数组,其中包含总和小于给定数字的所有数字组合,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36435754/