Question

这必须在这里处理类似的问题：Calling BLAS / LAPACK directly using the SciPy interface and Cython但是不同，因为我在这里使用SciPy示例中的实际代码_test_dgemm：https://github.com/scipy/scipy/blob/master/scipy/linalg/cython_blas.pyx这是非常快的（输入矩阵输入时比numpy.dot快5倍，否则快20倍。如果传递Mx1 1xN向量，则不会产生任何结果。它产生与numpy.dot相同的值，并传递矩阵。我已经最小化了代码，因为为了清晰起见，没有发布任何答案。这是dgemm.pyx.：

import numpy as np
cimport numpy as np
from scipy.linalg.cython_blas cimport dgemm
from cython cimport boundscheck

@boundscheck(False)
cpdef int fast_dgemm(double[:,::1] a, double[:,::1] b, double[:,::1] c, double alpha=1.0, double beta=0.0) nogil except -1:

    cdef:
        char *transa = 'n'
        char *transb = 'n'
        int m, n, k, lda, ldb, ldc
        double *a0=&a[0,0]
        double *b0=&b[0,0]
        double *c0=&c[0,0]

    ldb = (&a[1,0]) - a0 if a.shape[0] > 1 else 1
    lda = (&b[1,0]) - b0 if b.shape[0] > 1 else 1

    k = b.shape[0]
    if k != a.shape[1]:
        with gil:
            raise ValueError("Shape mismatch in input arrays.")
    m = b.shape[1]
    n = a.shape[0]
    if n != c.shape[0] or m != c.shape[1]:
        with gil:
            raise ValueError("Output array does not have the correct shape.")
    ldc = (&c[1,0]) - c0 if c.shape[0] > 1 else 1
    dgemm(transa, transb, &m, &n, &k, &alpha, b0, &lda, a0,
               &ldb, &beta, c0, &ldc)
    return 0

以下是一个示例测试脚本：

import numpy as np;

a=np.random.randn(1000);
b=np.random.randn(1000);
a.resize(len(a),1);
a=np.array(a, order='c');
b.resize(1,len(b)); 
b=np.array(b, order='c');
c = np.empty((a.shape[0],b.shape[1]), float, order='c'); 

from dgemm import _test_dgemm; 
_test_dgemm(a,b,c);

如果您想在Windows上使用Python 3.5 x64进行播放，请setup.py通过命令提示符键入python setup.py build_ext --inplace --compiler=msvc

来构建它

from Cython.Distutils import build_ext
import numpy as np
import os

try:
    from setuptools import setup
    from setuptools import Extension
except ImportError:
    from distutils.core import setup
    from distutils.extension import Extension

module = 'dgemm'

ext_modules = [Extension(module, sources=[module + '.pyx'],
              include_dirs=['C://Program Files (x86)//Windows Kits//10//Include//10.0.10240.0//ucrt','C://Program Files (x86)//Microsoft Visual Studio 14.0//VC//include','C://Program Files (x86)//Windows Kits//8.1//Include//shared'],
              library_dirs=['C://Program Files (x86)//Windows Kits//8.1//bin//x64', 'C://Windows//System32', 'C://Program Files (x86)//Microsoft Visual Studio 14.0//VC//lib//amd64', 'C://Program Files (x86)//Windows Kits//8.1//Lib//winv6.3//um//x64', 'C://Program Files (x86)//Windows Kits//10//Lib//10.0.10240.0//ucrt//x64'],
              extra_compile_args=['/Ot', '/favor:INTEL64', '/EHsc', '/GA'],
              language='c++')]

setup(
    name = module,
    ext_modules = ext_modules,
    cmdclass = {'build_ext': build_ext},
    include_dirs = [np.get_include(), os.path.join(np.get_include(), 'numpy')]
    )

非常感谢任何帮助！

Answer 1

如果我看对了，你会尝试将fortran-routines用于带有c-memory-layout的数组。

即使你明显知道，我想首先详细说明行主要顺序（c-memory-layout）和列主要顺序（fortran-memory-layout），以便推断出我的答案。

因此，如果我们有一个SCREEN 12 CLS PRINT "" PRINT "" PRINT "" PRINT "" PRINT "" PRINT " POKELITE - By Mark " PRINT "" PRINT "" INPUT "Join or Host a game? ", hostorjoin$ hostorjoin$ = UCASE$(hostorjoin$) IF hostorjoin$ = "JOIN" THEN GOTO JOIN IF hostorjoin$ = "HOST" THEN GOTO HOST ' If neither "HOST" nor "JOIN" is specified, what happens? HOST: server& = _OPENHOST("TCP/IP:300") PRINT "Waiting for connection..." PRINT "! Remember: If playing locally, give the other player your IPv4 Address !" DO connection& = _OPENCONNECTION(server&) LOOP UNTIL connection& <> 0 PRINT "" PRINT "2nd Player Joined!" SLEEP 2 GOTO GAME JOIN: INPUT "Enter Server IPv4 Address (Example: 192.168.1.25): ", joinip$ connection& = _OPENCLIENT("TCP/IP:300:" + joinip$) IF connection& = 0 THEN PRINT "Connection failed!": SLEEP 2: CLS: GOTO JOIN GOTO GAME GAME: CLS INPUT "Enter your name: ", playerName$ IF playerName$ = "" THEN GOTO GAME PRINT "Waiting for other player..." ' Send name to opponent and wait for opponent's name. PUT connection&, , playerName$ DO GET connection&, , opponentName$ LOOP UNTIL opponentName$ <> "" PRINT "You: "; playerName$ PRINT "Opponent:"; opponentName$矩阵（即2行3列）2x3，并将其存储在一些连续的内存中，我们得到：

这意味着如果我们得到一个连续的记忆，它代表一个行主要顺序的矩阵，并将其解释为列主要顺序的矩阵，我们将得到一个完全不同的矩阵！

但是，我们要看一下我们可以轻松看到的转置矩阵row-major-order(A) = A11, A12, A13, A21, A22, A23 col-major-order(A) = A11, A21, A12, A22, A13, A33：

A^t

这意味着，如果我们想以行 - 主顺序得到矩阵row-major-order(A) = col-major-order(A^t) col-major-order(A) = row-major-order(A^t)作为结果，那么blas-routine应该按列主要顺序写入转置矩阵C（之后）这一切我们无法改变）进入这个记忆。但是，C和C^t=(AB)^t=B^t*A^t和B^t是按主列顺序重新解释的原始矩阵。

现在，让A^t成为A - 矩阵和n x k B - 矩阵，dgemm例程的调用应如下：< / p>

k x m

正如您所看到的，您在代码中切换了一些dgemm(transa, transb, &m, &n, &k, &alpha, b0, &m, a0, &k, &beta, c0, &m)和n。

使用Cython的Scipy cython_blas接口不能处理向量Mx1 1xN

1 个答案: