Numpy/Scipy中的多线程整数矩阵乘法

Question

做一些类似的事情

import numpy as np
a = np.random.rand(10**4, 10**4)
b = np.dot(a, a)

使用多个内核，运行良好。

a = np.random.randint(2, size=(n, n)).astype(np.int8)

array: np.random.randint(2, size=shape).astype(dtype)

dtype    shape          %time (average)

float32 (2000, 2000)    62.5 ms
float32 (3000, 3000)    219 ms
float32 (4000, 4000)    328 ms
float32 (10000, 10000)  4.09 s

int8    (2000, 2000)    13 seconds
int8    (3000, 3000)    3min 26s
int8    (4000, 4000)    12min 20s
int8    (10000, 10000)  It didn't finish in 6 hours

float16 (2000, 2000)    2min 25s
float16 (3000, 3000)    Not tested
float16 (4000, 4000)    Not tested
float16 (10000, 10000)  Not tested

import scipy.linalg.blas as blas
a = np.random.randint(2, size=(n, n)).astype(np.int8)
b = blas.sgemm(alpha=1.0, a=a, b=a)

所以，如果我要做整数矩阵乘法，我得做下面的一个：

1. 使用numpy慢得让人痛苦的np.dot并庆幸我可以保留8位整数。
2. 使用Scipy的SGEMM并使用4倍内存。
3. 使用numpy的np.float16并且只使用2倍内存，但要注意的是，np.dot在float16数组上的速度要比在float32数组上慢得多，比int8慢得多。
4. 为多线程整数矩阵乘法找到一个优化的库（其实Mathematica就是这么做的，但我更喜欢Python的解决方案），理想情况下支持1位数组，虽然8位数组也没问题……（我的目标实际上是在有限域z/2z上做矩阵的乘法，我知道我可以用Sage做这件事，这是非常Pythonic的，但是，再一次，有严格的Python吗？）

我能按方案4做吗？这样的图书馆存在吗？

Answer 1

注意，当这个答案变老时，numpy可能会获得优化的整数支持。请验证此答案是否在您的安装程序中运行得更快。

• 选项5-滚动自定义解决方案：将矩阵产品划分为几个子产品，并并行执行这些子产品。这可以用标准的Python模块相对容易地实现。使用numpy.dot来计算子产品，它释放全局解释器锁。因此，可以使用相对轻量级的线程，并且可以从主线程访问阵列以提高内存效率。

实施：

import numpy as np
from numpy.testing import assert_array_equal
import threading
from time import time


def blockshaped(arr, nrows, ncols):
    """
    Return an array of shape (nrows, ncols, n, m) where
    n * nrows, m * ncols = arr.shape.
    This should be a view of the original array.
    """
    h, w = arr.shape
    n, m = h // nrows, w // ncols
    return arr.reshape(nrows, n, ncols, m).swapaxes(1, 2)


def do_dot(a, b, out):
    #np.dot(a, b, out)  # does not work. maybe because out is not C-contiguous?
    out[:] = np.dot(a, b)  # less efficient because the output is stored in a temporary array?


def pardot(a, b, nblocks, mblocks, dot_func=do_dot):
    """
    Return the matrix product a * b.
    The product is split into nblocks * mblocks partitions that are performed
    in parallel threads.
    """
    n_jobs = nblocks * mblocks
    print('running {} jobs in parallel'.format(n_jobs))

    out = np.empty((a.shape[0], b.shape[1]), dtype=a.dtype)

    out_blocks = blockshaped(out, nblocks, mblocks)
    a_blocks = blockshaped(a, nblocks, 1)
    b_blocks = blockshaped(b, 1, mblocks)

    threads = []
    for i in range(nblocks):
        for j in range(mblocks):
            th = threading.Thread(target=dot_func, 
                                  args=(a_blocks[i, 0, :, :], 
                                        b_blocks[0, j, :, :], 
                                        out_blocks[i, j, :, :]))
            th.start()
            threads.append(th)

    for th in threads:
        th.join()

    return out


if __name__ == '__main__':
    a = np.ones((4, 3), dtype=int)
    b = np.arange(18, dtype=int).reshape(3, 6)
    assert_array_equal(pardot(a, b, 2, 2), np.dot(a, b))

    a = np.random.randn(1500, 1500).astype(int)

    start = time()
    pardot(a, a, 2, 4)
    time_par = time() - start
    print('pardot: {:.2f} seconds taken'.format(time_par))

    start = time()
    np.dot(a, a)
    time_dot = time() - start
    print('np.dot: {:.2f} seconds taken'.format(time_dot))
    

通过这个实现，我获得了大约x4的加速，这是我的机器中的内核的物理数量：

running 8 jobs in parallel
pardot: 5.45 seconds taken
np.dot: 22.30 seconds taken