【问题标题】:Vectorizing `numpy.random.choice` for given 2D array of probabilities along an axis为沿轴的给定二维概率数组向量化“numpy.random.choice”
【发布时间】:2018-05-23 03:37:13
【问题描述】:

Numpy 具有random.choice 函数,它允许您从分类分布中进行抽样。您将如何在轴上重复此操作?为了说明我的意思,这是我当前的代码:

categorical_distributions = np.array([
    [.1, .3, .6],
    [.2, .4, .4],
])
_, n = categorical_distributions.shape
np.array([np.random.choice(n, p=row)
          for row in categorical_distributions])

理想情况下,我想消除 for 循环。

【问题讨论】:

标签: python numpy random vectorization


【解决方案1】:

这是获取每行随机索引的一种矢量化方法,a 作为概率数组2D -

(a.cumsum(1) > np.random.rand(a.shape[0])[:,None]).argmax(1)

概括以覆盖2D 数组的行和列 -

def random_choice_prob_index(a, axis=1):
    r = np.expand_dims(np.random.rand(a.shape[1-axis]), axis=axis)
    return (a.cumsum(axis=axis) > r).argmax(axis=axis)

让我们通过运行超过一百万次来验证给定的样本 -

In [589]: a = np.array([
     ...:     [.1, .3, .6],
     ...:     [.2, .4, .4],
     ...: ])

In [590]: choices = [random_choice_prob_index(a)[0] for i in range(1000000)]

# This should be close to first row of given sample
In [591]: np.bincount(choices)/float(len(choices))
Out[591]: array([ 0.099781,  0.299436,  0.600783])

运行时测试

原来的循环方式-

def loopy_app(categorical_distributions):
    m, n = categorical_distributions.shape
    out = np.empty(m, dtype=int)
    for i,row in enumerate(categorical_distributions):
        out[i] = np.random.choice(n, p=row)
    return out

更大阵列的计时 -

In [593]: a = np.array([
     ...:     [.1, .3, .6],
     ...:     [.2, .4, .4],
     ...: ])

In [594]: a_big = np.repeat(a,100000,axis=0)

In [595]: %timeit loopy_app(a_big)
1 loop, best of 3: 2.54 s per loop

In [596]: %timeit random_choice_prob_index(a_big)
100 loops, best of 3: 6.44 ms per loop

【讨论】:

  • 优秀的答案。如果没有替换,你将如何实现选择?
  • 感谢您的回答。如果它与任何人相关,此方法基于逆变换采样的思想,请参见例如stephens999.github.io/fiveMinuteStats/…
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