python - 如何在python中对包含eigvalsh的复杂代码进行矢量化

Question

我有以下代码（对不起，它不是太小，我已经尝试从原始代码中减少它）。

eval_s()基本上，我在运行以下方法/函数时遇到性能问题：

1) 找到 4x4 厄密矩阵的 4 个特征值eigvalsh()

2) 将特征值的倒数相加成一个变量result

3) 我对由 参数化的许多矩阵重复第 1 步和第 2 步x,y,z，将累积和存储在 中result。

我在步骤 3 中重复计算（查找特征值和求和）的次数取决于ksep我的代码中的一个变量，我需要这个数字来增加我的实际代码（即，ksep必须减少）。但是其中的计算eval_s()有一个 for 循环x,y,z，我猜这真的会减慢速度。[试着ksep=0.5明白我的意思。]

有没有办法对我的示例代码中指示的方法（或者一般来说，涉及查找参数化矩阵的特征值的函数）进行矢量化？

代码：

import numpy as np
import sympy as sp
import itertools as it
from sympy.abc import x, y, z


class Solver:
    def __init__(self, vmat):
        self._vfunc = sp.lambdify((x, y, z),
                                  expr=vmat,
                                  modules='numpy')
        self._q_count, self._qs = None, []  # these depend on ksep!

    ################################################################
    # How to vectorize this?
    def eval_s(self, stiff):
        assert len(self._qs) == self._q_count, "Run 'populate_qs' first!"
        result = 0
        for k in self._qs:
            evs = np.linalg.eigvalsh(self._vfunc(*k))
            result += np.sum(np.divide(1., (stiff + evs)))
        return result.real - 4 * self._q_count
    ################################################################

    def populate_qs(self, ksep: float = 1.7):
        self._qs = [(kx, ky, kz) for kx, ky, kz
                    in it.product(np.arange(-3*np.pi, 3.01*np.pi, ksep),
                                  np.arange(-3*np.pi, 3.01*np.pi, ksep),
                                  np.arange(-3*np.pi, 3.01*np.pi, ksep))]
        self._q_count = len(self._qs)


def test():
    vmat = sp.Matrix([[1, sp.cos(x/4+y/4), sp.cos(x/4+z/4), sp.cos(y/4+z/4)],
                      [sp.cos(x/4+y/4), 1, sp.cos(y/4-z/4), sp.cos(x/4 - z/4)],
                      [sp.cos(x/4+z/4), sp.cos(y/4-z/4), 1, sp.cos(x/4-y/4)],
                      [sp.cos(y/4+z/4), sp.cos(x/4-z/4), sp.cos(x/4-y/4), 1]]) * 2
    solver = Solver(vmat)
    solver.populate_qs(ksep=1.7)  # <---- Performance starts to worsen (in eval_s) when ksep is reduced!
    print(solver.eval_s(0.65))


if __name__ == "__main__":
    import timeit
    print(timeit.timeit("test()", setup="from __main__ import test", number=100))

ps 代码的 sympy 部分可能看起来很奇怪，但它在我的原始代码中是有目的的。

标签： pythonpython-3.xnumpyvectorization

Answer 1

您可以，方法如下：

def eval_s_vectorized(self, stiff):
    assert len(self._qs) == self._q_count, "Run 'populate_qs' first!"
    mats = np.stack([self._vfunc(*k) for k in self._qs], axis=0)
    evs = np.linalg.eigvalsh(mats)
    result = np.sum(np.divide(1., (stiff + evs)))
    return result.real - 4 * self._q_count

这仍然使 Sympy 表达式的评估未向量化。这部分向量化有点棘手，主要是因为1输入矩阵中的 s。您可以通过修改来制作代码的完全矢量化版本，Solver以便将标量常量替换为数组常量vmat：

import itertools as it
import numpy as np
import sympy as sp
from sympy.abc import x, y, z
from sympy.core.numbers import Number
from sympy.utilities.lambdify import implemented_function

xones = implemented_function('xones', lambda x: np.ones(len(x)))
lfuncs = {'xones': xones}

def vectorizemat(mat):
    ret = mat.copy()
    # get the first element of the set of symbols that mat uses
    for x in mat.free_symbols: break
    for i,j in it.product(*(range(s) for s in mat.shape)):
        if isinstance(mat[i,j], Number):
            ret[i,j] = xones(x) * mat[i,j]
    return ret

class Solver:
    def __init__(self, vmat):
        self._vfunc = sp.lambdify((x, y, z),
                                  expr=vectorizemat(vmat),
                                  modules=[lfuncs, 'numpy'])
        self._q_count, self._qs = None, []  # these depend on ksep!

    def eval_s_vectorized_completely(self, stiff):
        assert len(self._qs) == self._q_count, "Run 'populate_qs' first!"
        evs = np.linalg.eigvalsh(self._vfunc(*self._qs.T).T)
        result = np.sum(np.divide(1., (stiff + evs)))
        return result.real - 4 * self._q_count

    def populate_qs(self, ksep: float = 1.7):
        self._qs = np.array([(kx, ky, kz) for kx, ky, kz
                    in it.product(np.arange(-3*np.pi, 3.01*np.pi, ksep),
                                  np.arange(-3*np.pi, 3.01*np.pi, ksep),
                                  np.arange(-3*np.pi, 3.01*np.pi, ksep))])
        self._q_count = len(self._qs)

测试/计时

对于小型ksep，矢量化版本比原始版本快约 2 倍，完全矢量化版本快约 20 倍：

# old version for ksep=.3
import timeit
print(timeit.timeit("test()", setup="from __main__ import test", number=10))
-85240.46154500882
-85240.46154500882
-85240.46154500882
-85240.46154500882
-85240.46154500882
-85240.46154500882
-85240.46154500882
-85240.46154500882
-85240.46154500882
-85240.46154500882
118.42847006605007

# vectorized version for ksep=.3
import timeit
print(timeit.timeit("test()", setup="from __main__ import test", number=10))
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
64.95763925800566

# completely vectorized version for ksep=.3
import timeit
print(timeit.timeit("test()", setup="from __main__ import test", number=10))
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
5.648927717003971

矢量化版本的结果中的舍入误差与原始版本略有不同。这可能是由于result计算总和的方式不同。