python - 脚本时间改进

Question

我正在尝试提高部分脚本的时间效率，但我没有更多想法。我在 Matlab 和 Python 中运行了以下脚本，但 Matlab 的实现比 Python 的实现快四倍。知道如何改进吗？

Python:

import time
import numpy as np

def ComputeGradient(X, y, theta, alpha):
    m = len(y)
    factor = alpha / m
    h = np.dot(X, theta)
    theta = [theta[i] - factor * sum((h-y) * X[:,i]) for i in [0,1]]
    #Also tried this but with worse performances
    #diff = np.tile((h-y)[:, np.newaxis],2)
    #theta = theta - factor * sum(diff * X)
    return theta

if __name__ == '__main__':
    data = np.loadtxt("data_LinReg.txt", delimiter=',')    
    theta = [0, 0]
    alpha = 0.01
    X = data[:,0]
    y = data[:,1]
    X = np.column_stack((np.ones(len(y)), X))
    start_time = time.time()
    for i in range(0, 1500, 1):
        theta = ComputeGradient(X, y, theta, alpha)
    stop_time = time.time()
    print("--- %s seconds ---" % (stop_time - start_time))

--> 0.048s

Matlab:

data = load('data_LinReg.txt');
X = data(:, 1); y = data(:, 2);
m = length(y);
X = [ones(m, 1), data(:,1)]; % Add a column of ones to x
theta = zeros(2, 1);
iterations = 1500;
alpha = 0.01;
tic
for i = 1:1500
   theta = gradientDescent(X, y, theta, alpha);
end
toc

function theta = gradientDescent(X, y, theta, alpha)
   m = length(y); % number of training examples
   h = X * theta;
   t1 = theta(1) - alpha * sum(X(:,1).*(h-y)) / m;
   t2 = theta(2) - alpha * sum(X(:,2).*(h-y)) / m;
   theta = [t1; t2];
end

--> 0.01s

[编辑]:解决方案

一种可能的途径是使用 numpy 向量化而不是 python 根函数。在建议的代码中，将 sum 替换为 np.sum 提高了时间效率，使其更接近 Matlab(0.019s 而不是 0.048s)

此外，我分别测试了向量上的函数:np.dot、np.sum、*(乘积)，所有这些函数似乎都比等效的 Matlab 更快(在某些情况下确实更快)。我想知道为什么它在 Python 中仍然比较慢....

Answer 1

此解决方案提供了一个优化的 MATLAB 实现 -

• gradient-descent 实现的函数内联。
• 预先计算在循环中重复使用的某些值。

代码-

data = load('data_LinReg.txt');

iterations = 1500;
alpha = 0.01;
m = size(data,1);

M = alpha/m; %// scaling factor

%// Pre-compute certain values that are repeatedly used inside the loop
sum_a = M*sum(data(:,1));
sum_p = M*sum(data(:,2));
sum_ap = M*sum(data(:,1).*data(:,2));
sum_sqa = M*sum(data(:,1).^2);
one_minus_alpha = 1 - alpha;
one_minus_sum_sqa = 1 - sum_sqa;

%// Start processing
t1n0 = 0;
t2n0 = 0;
for i = 1:iterations
    temp = t1n0*one_minus_alpha - t2n0*sum_a + sum_p;
    t2n0 = t2n0*one_minus_sum_sqa - t1n0*sum_a + sum_ap;
    t1n0 = temp;
end
theta = [t1n0;t2n0];

快速测试表明，这比问题中发布的 MATLAB 代码有明显的加速。

现在，我对 python 不太熟悉，但我认为这个 MATLAB 代码可以很容易地移植到 python。