向量化numpy中的for循环

Question

我正在尝试使用numpy在python中计算矩阵z（定义如下）。

这是我当前的解决方案（使用1 for循环）

z = np.zeros((n, k))
for i in range(n):
    v = pi * (1 / math.factorial(x[i])) * np.exp(-1 * lamb) * (lamb ** x[i])
    numerator = np.sum(v)
    c = v / numerator
    z[i, :] = c
return z

是否可以完全矢量化此计算？ 我需要进行数千次迭代的这种计算，并且numpy中的矩阵运算比for循环大得多。

Answer 1

这是E的向量化版本。 它用NumPy广播和基于数组的算法代替了for-loop和标量算法：

def alt_E(x):
    x = x[:, None]
    z = pi * (np.exp(-lamb) * (lamb**x)) / special.factorial(x)
    denom = z.sum(axis=1)[:, None]
    z /= denom
    return z

---

我运行em.py来了解x ， lamb ， pi ， n和k的典型大小。 在这种大小的数据上， alt_E比E快120倍：

In [32]: %timeit E(x)
100 loops, best of 3: 11.5 ms per loop

In [33]: %timeit alt_E(x)
10000 loops, best of 3: 94.7 µs per loop

In [34]: 11500/94.7
Out[34]: 121.43611404435057

---

这是我用于基准测试的设置：

import math
import numpy as np
import scipy.special as special

def alt_E(x):
    x = x[:, None]
    z = pi * (np.exp(-lamb) * (lamb**x)) / special.factorial(x)
    denom = z.sum(axis=1)[:, None]
    z /= denom
    return z


def E(x):
    z = np.zeros((n, k))
    for i in range(n):
        v = pi * (1 / math.factorial(x[i])) * \
            np.exp(-1 * lamb) * (lamb ** x[i])
        numerator = np.sum(v)
        c = v / numerator
        z[i, :] = c
    return z

n = 576
k = 2

x = np.array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5])

lamb = np.array([ 0.84835141, 1.04025989])
pi = np.array([ 0.5806958, 0.4193042])

assert np.allclose(alt_E(x), E(x))

---

顺便说一下，也可以使用scipy.stats.poisson计算E ：

import scipy.stats as stats
pois = stats.poisson(mu=lamb)

def alt_E2(x):
    z = pi * pois.pmf(x[:,None])
    denom = z.sum(axis=1)[:, None]
    z /= denom
    return z

但这并没有更快，至少对于这种长度的数组而言：

In [33]: %timeit alt_E(x)
10000 loops, best of 3: 94.7 µs per loop

In [102]: %timeit alt_E2(x)
1000 loops, best of 3: 278 µs per loop

对于更大的x ， alt_E2更快：

In [104]: x = np.random.random(10000)

In [106]: %timeit alt_E(x)
100 loops, best of 3: 2.18 ms per loop

In [105]: %timeit alt_E2(x)
1000 loops, best of 3: 643 µs per loop