Question

我已经在MATLAB和Python中设置了两个相同的关于广播矩阵乘法的测试。对于Python，我使用NumPy；对于MATLAB，我使用mtimesx库，该库使用BLAS。

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));

Python

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pip使用OpenBLAS安装的默认Python。
我正在Intel NUC i3上运行。

MATLAB代码的运行时间为1毫秒，而Python的运行时间为5.8毫秒，我不知道为什么，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 我尝试将C = A @ B替换为np.matmul(A, B, out=C)，并得到了2倍的更差的时间，例如大约11ms。这真的很奇怪。

Answer 1

您的MATLAB代码使用浮点数组，但是NumPy代码使用整数数组。这在时间上有很大的不同。对于MATLAB与NumPy之间的“一对一比较”，Python / NumPy代码还必须使用浮点数组。

但这不是唯一的重要问题。在NumPy github站点的issue 7569（然后在issue 8957）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。这意味着二维以上数组的乘法可能比预期的慢得多。

二维数组的乘法确实使用快速例程，因此您可以通过将单个二维数组乘以一个循环来解决此问题。令人惊讶的是，尽管有Python循环的开销，但在许多情况下，它比应用于完整堆栈数组的@，matmul或einsum快。

这是NumPy问题中显示的函数的变体，该函数在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）。以下是使用浮点值数组的A @ B和xmul(A, B)的时序比较：

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，也大约需要A @ B的1/8。

Answer 2

我认为这是内存排序的问题。 Matlab的zeros(a, b, c)就像numpy的zeros((a, b, c), order='F')，这不是默认值。

当然，正如您已经确定的，@在与mtimesx不同的轴上进行操作。为了使比较公平，应确保数组按matlab顺序排列，然后转置以处理语义差异

# note: `order` in reshape actually changes the resulting array data,
# not just its memory layout
A = np.arange(a * b * c).reshape([b, c, a], order='F').transpose((2, 0, 1))
B = np.arange(a * c * d).reshape([c, d, a], order='F').transpose((2, 0, 1))

Answer 3

您能否再次使用最近发布的NumPy 1.16？我们将matmul重构为在内部二维中使用BLAS，这可以加快代码的速度。

MATLAB矩阵乘法性能比NumPy快5倍

3 个答案: