MATLAB矩阵乘法性能比NumPy快5倍

Question

我在MATLAB和Python中设置了两个相同的测试,用于矩阵乘法和广播.对于Python,我使用了NumPy,对于MATLAB,我使用了使用BLAS 的mtimesx库.

MATLAB

close all; clear;

N = 1000 + 100; % a few initial runs to be trimmed off at the end

a = 100;
b = 30;
c = 40;
d = 50;
A = rand(b, c, a);
B = rand(c, d, a);
C = zeros(b, d, a);

times = zeros(1, N);
for ii = 1:N
    tic
    C = mtimesx(A,B);
    times(ii) = toc;
end

times = times(101:end) * 1e3;

plot(times);
grid on;
title(median(times));

蟒蛇

import timeit
import numpy as np
import matplotlib.pyplot as plt


N = 1000 + 100  # a few initial runs to be trimmed off at the end

a = 100
b = 30
c = 40
d = 50
A = np.arange(a * b * c).reshape([a, b, c])
B = np.arange(a * c * d).reshape([a, c, d])
C = np.empty(a * b * d).reshape([a, b, d])

times = np.empty(N)

for i in range(N):
    start = timeit.default_timer()
    C = A @ B
    times[i] = timeit.default_timer() - start

times = times[101:] * 1e3

plt.plot(times, linewidth=0.5)
plt.grid()
plt.title(np.median(times))
plt.show()

• 我的Python是默认安装的pip,使用OpenBLAS.
• 我正在运行英特尔NUC i3.

MATLAB代码在1ms运行,而Python在5.8ms运行,我无法弄清楚为什么,因为它们似乎都在使用BLAS.

---

编辑

来自Anaconda:

In [7]: np.__config__.show()
mkl_info:
    libraries = ['mkl_rt']
    library_dirs = [...]
    define_macros = [('SCIPY_MKL_H', None), ('HAVE_CBLAS', None)]
    include_dirs = [...]
blas_mkl_info:
    libraries = ['mkl_rt']
    library_dirs = [...]
    define_macros = [('SCIPY_MKL_H', None), ('HAVE_CBLAS', None)]
    include_dirs = [...]
blas_opt_info:
    libraries = ['mkl_rt']
    library_dirs = [...]
    define_macros = [('SCIPY_MKL_H', None), ('HAVE_CBLAS', None)]
    include_dirs = [...]
lapack_mkl_info:
    libraries = ['mkl_rt']
    library_dirs = [...]
    define_macros = [('SCIPY_MKL_H', None), ('HAVE_CBLAS', None)]
    include_dirs = [...]
lapack_opt_info:
    libraries = ['mkl_rt']
    library_dirs = [...]
    define_macros = [('SCIPY_MKL_H', None), ('HAVE_CBLAS', None)]
    include_dirs = [...]

从numpy使用pip

In [2]: np.__config__.show()
blas_mkl_info:
NOT AVAILABLE
blis_info:
NOT AVAILABLE
openblas_info:
    library_dirs = [...]
    libraries = ['openblas']
    language = f77
    define_macros = [('HAVE_CBLAS', None)]
blas_opt_info:
    library_dirs = [...]
    libraries = ['openblas']
    language = f77
    define_macros = [('HAVE_CBLAS', None)]
lapack_mkl_info:
NOT AVAILABLE
openblas_lapack_info:
    library_dirs = [...]
    libraries = ['openblas']
    language = f77
    define_macros = [('HAVE_CBLAS', None)]
lapack_opt_info:
    library_dirs = [...]
    libraries = ['openblas']
    language = f77
    define_macros = [('HAVE_CBLAS', None)]

---

编辑2 我试图取代C = A @ B以np.matmul(A, B, out=C)并获得2个糟糕的时间,例如大约11毫秒.这真的很奇怪.

Answer 1

您的MATLAB代码使用浮点数组,但NumPy代码使用整数数组.这使得时间有显着差异.对于MATLAB和NumPy之间的"苹果到苹果"比较,Python/NumPy代码也必须使用浮点数组.

然而,这不是唯一重要的问题.在NumPy github网站上的问题7569(以及问题8957中)中讨论的NumPy存在缺陷."堆叠"数组的矩阵乘法不使用快速BLAS例程来执行乘法.这意味着具有两个以上维度的数组的乘法可能比预期慢得多.

2-d数组的乘法确实使用快速例程,因此您可以通过在循环中将各个2-d数组相乘来解决此问题.出人意料的是,尽管一个Python循环的开销,它比快@,matmul或者einsum在许多情况下,给整个堆叠阵列.

这是NumPy问题中显示的函数的变体,它在Python循环中执行矩阵乘法:

def xmul(A, B):
    """
    Multiply stacked matrices A (with shape (s, m, n)) by stacked
    matrices B (with shape (s, n, p)) to produce an array with
    shape (s, m, p).

    Mathematically equivalent to A @ B, but faster in many cases.

    The arguments are not validated.  The code assumes that A and B
    are numpy arrays with the same data type and with shapes described
    above.
    """
    out = np.empty((a.shape[0], a.shape[1], b.shape[2]), dtype=a.dtype)
    for j in range(a.shape[0]):
        np.matmul(a[j], b[j], out=out[j])
    return out

我的NumPy安装也使用MKL(它是Anaconda发行版的一部分).下面是一个定时比较A @ B和xmul(A, B)使用的浮点值的数组:

In [204]: A = np.random.rand(100, 30, 40)

In [205]: B = np.random.rand(100, 40, 50)

In [206]: %timeit A @ B
4.76 ms ± 6.37 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [207]: %timeit xmul(A, B)
582 µs ± 35.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

即使xmul使用Python循环,它也需要大约1/8的时间A @ B.