Question

我试图获得n维概率向量的“网格”---每个条目在0和1之间的向量，并且所有条目加起来为1.我希望每个可能的向量都在哪个坐标中可以采用0到1之间的均匀间隔值的 v 中的任何一个。

为了说明这一点，接下来是一个非常低效的实现，对于n = 3和v = 3：

from itertools import product
grid_redundant = product([0, .5, 1], repeat=3)
grid = [point for point in grid_redundant if sum(point)==1]

现在grid包含[(0, 0, 1), (0, 0.5, 0.5), (0, 1, 0), (0.5, 0, 0.5), (0.5, 0.5, 0), (1, 0, 0)]。

对于更高维度和更细粒度的网格而言，这种“实现”非常糟糕。是否有一种很好的方法可以使用numpy？

我或许可以在动机上添加一点：如果从随机分布中抽样给我足够的极值点，我会非常高兴，但事实并非如此。见this question。我所追求的“网格”不是随机的，而是系统地扫描单纯形（概率向量的空间）。

Answer 1

这是一个递归解决方案。虽然它应该比发布的片段更快，但它不使用NumPy并且不是超级高效的：

import math
from itertools import permutations

def probability_grid(values, n):
    values = set(values)
    # Check if we can extend the probability distribution with zeros
    with_zero = 0. in values
    values.discard(0.)
    if not values:
        raise StopIteration
    values = list(values)
    for p in _probability_grid_rec(values, n, [], 0.):
        if with_zero:
            # Add necessary zeros
            p += (0.,) * (n - len(p))
        if len(p) == n:
            yield from set(permutations(p))  # faster: more_itertools.distinct_permutations(p)

def _probability_grid_rec(values, n, current, current_sum, eps=1e-10):
    if not values or n <= 0:
        if abs(current_sum - 1.) <= eps:
            yield tuple(current)
    else:
        value, *values = values
        inv = 1. / value
        # Skip this value
        yield from _probability_grid_rec(
            values, n, current, current_sum, eps)
        # Add copies of this value
        precision = round(-math.log10(eps))
        adds = int(round((1. - current_sum) / value, precision))
        for i in range(adds):
            current.append(value)
            current_sum += value
            n -= 1
            yield from _probability_grid_rec(
                values, n, current, current_sum, eps)
        # Remove copies of this value
        if adds > 0:
            del current[-adds:]

print(list(probability_grid([0, 0.5, 1.], 3)))

输出：

[(1.0, 0.0, 0.0), (0.0, 1.0, 0.0), (0.0, 0.0, 1.0), (0.5, 0.5, 0.0), (0.0, 0.5, 0.5), (0.5, 0.0, 0.5)]

与发布的方法进行快速比较：

from itertools import product

def probability_grid_basic(values, n):
    grid_redundant = product(values, repeat=n)
    return [point for point in grid_redundant if sum(point)==1]

values = [0, 0.25, 1./3., .5, 1]
n = 6
%timeit list(probability_grid(values, n))
1.61 ms ± 20.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit probability_grid_basic(values, n)
6.27 ms ± 186 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 2

完全通用性，对于高维向量，甚至在接受的答案中使用巧妙的解决方案，这是相当难以管理的。在我自己的情况下，计算所有值的相关子集是值得的。例如，以下函数计算仅具有dimension非零等概率条目的所有n - 维概率向量：

import itertools as it
import numpy as np

def equip_n(dimension, n):
"""
Calculate all possible <dimension>-dimensional probability vectors with n nonzero,
equiprobable entries
"""
combinations  = np.array([comb for comb in it.combinations(range(dimension), n)])
vectors = np.zeros((combinations.shape[0], dimension))
for line, comb in zip(vectors, combinations):
    line[comb] = 1/n
return vectors 

print(equip_n(6, 3))

返回

[[ 0.3333  0.3333  0.3333  0.      0.      0.    ]
 [ 0.3333  0.3333  0.      0.3333  0.      0.    ] 
 [ 0.3333  0.3333  0.      0.      0.3333  0.    ]
 [ 0.3333  0.3333  0.      0.      0.      0.3333]
 [ 0.3333  0.      0.3333  0.3333  0.      0.    ]
 [ 0.3333  0.      0.3333  0.      0.3333  0.    ]
 [ 0.3333  0.      0.3333  0.      0.      0.3333]
 [ 0.3333  0.      0.      0.3333  0.3333  0.    ]
 [ 0.3333  0.      0.      0.3333  0.      0.3333]
 [ 0.3333  0.      0.      0.      0.3333  0.3333]
 [ 0.      0.3333  0.3333  0.3333  0.      0.    ]
 [ 0.      0.3333  0.3333  0.      0.3333  0.    ]
 [ 0.      0.3333  0.3333  0.      0.      0.3333]
 [ 0.      0.3333  0.      0.3333  0.3333  0.    ]
 [ 0.      0.3333  0.      0.3333  0.      0.3333]
 [ 0.      0.3333  0.      0.      0.3333  0.3333]
 [ 0.      0.      0.3333  0.3333  0.3333  0.    ]
 [ 0.      0.      0.3333  0.3333  0.      0.3333]
 [ 0.      0.      0.3333  0.      0.3333  0.3333]
 [ 0.      0.      0.      0.3333  0.3333  0.3333]]

这非常快。 %timeit equip_n(6, 3)返回

15.1 µs ± 74.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

网格概率向量

2 个答案: