Question

下面定义的函数func()埋在Markov Chain蒙特卡洛过程的内部，这意味着它被称为数百万次。我需要尽可能地提高其性能，但是我还没有找到一种方法来实现它。

掩码（msk）的定义和加权平均值大约占用相同的时间部分。可能没有比numpy更快的方法来获得加权平均值的方法，但是掩模定义至少可以改善吗？

def func(weights, x1, x2, x3, x4):
    for x in (x2, x3, x4):
        # Mask of the distance between the column '-6' of x1 versus arrays
        # x2,x3,x4
        msk = abs(x1[-6] - x[-6]) > 0.01
        # If the distance in this array is larger than the maximum allowed,
        # mask with the values taken from 'x1'.
        x[:, msk] = x1[:, msk]

    # Weighted average for all the arrays.
    avrg_data = np.average(np.array([x1, x2, x3, x4]), weights=weights, axis=0)

    return avrg_data


# Random data with proper shape
x1, x2, x3, x4 = np.random.uniform(1., 10., (4, 10, 1000))
weights = np.random.uniform(0.01, .5, 4)

# Call many times and time it
s = t.time()
for _ in range(10000):
    func(weights, x1, x2, x3, x4)
print(t.time() - s)

Answer 1

我玩弄了您的代码，碰巧尝试不使用np.average直接进行加权平均会更快，看起来像这样。在我的平台上，速度大约快40％。

import time as t
import numpy as np

def func(weights, x1, x2, x3, x4):
    for x in (x2, x3, x4):
        # Mask of the distance between the column '-6' of x1 versus arrays
        # x2,x3,x4
        msk = abs(x1[-6] - x[-6]) > 0.01

        # If the distance in this array is larger than the maximum allowed,
        # mask with the values taken from 'x1'.
        x[:, msk] = x1[:, msk]

    # Weighted average for all the arrays.
    avrg_data = np.average(np.array([x1, x2, x3, x4]), weights=weights, axis=0)

    return avrg_data

def faster_func(weights, x1, x2, x3, x4):
    for x in (x2, x3, x4):
        # Mask of the distance between the column '-6' of x1 versus arrays
        # x2,x3,x4
        msk = abs(x1[-6] - x[-6]) > 0.01

        # If the distance in this array is larger than the maximum allowed,
        # mask with the values taken from 'x1'.
        x[:, msk] = x1[:, msk]

    # Scale weights so they add up to 1, then add based on them
    weights = weights / np.mean(weights) / 4    
    avrg_data = x1*weights[0] + x2*weights[1] + x3*weights[2] + x4*weights[3]

    return avrg_data

# Random data with proper shape
x1, x2, x3, x4 = np.random.uniform(1., 10., (4, 10, 1000))
weights = np.random.uniform(0.01, .5, 4)

# Call many times and time it
for method in (func, faster_func):
    s = t.time()
    for _ in range(10000):
        method(weights, x1, x2, x3, x4)
    print(method, t.time() - s)

# Test that the results are still the same
result1 = func(weights, x1, x2, x3, x4)
result2 = faster_func(weights, x1, x2, x3, x4)
biggest_difference = np.max(abs(result1-result2))
print("Biggest difference between two methods was %.8f" % biggest_difference)

改善掩蔽性能，然后进行加权平均

1 个答案: