我有以下代码（抱歉，它不是太小，我已经尝试从原来减少它） .

基本上，我在运行 eval_s() 方法/函数时遇到性能问题，其中I：

1）用 eigvalsh() 找到4x4埃尔米特矩阵的4个特征值

2）将特征值的倒数加到变量 result 中

3）并且我为 x,y,z 参数化的许多矩阵重复步骤1和2，将累积和存储在 result 中 .

我在第3步中重复计算（找到特征值和求和）的次数取决于我的代码中的变量 ksep ，我需要在实际代码中增加这个数字（即 ksep 必须减少） . 但是 eval_s() 中的计算在 x,y,z 上有一个for循环，我猜测它真的会让事情变慢 . [试试 ksep=0.5 看看我的意思 . ]

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

Code:

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))

附：代码的同情部分可能看起来很奇怪，但它在我的原始代码中起作用 .

2 回答

你可以，这是如何：

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表达式的评估无法实现 . 这部分向量化有点棘手，主要是因为输入矩阵中的 1 . 您可以通过修改 Solver 来创建代码的完全矢量化版本，以便它用 vmat 中的数组常量替换标量常量：

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 ksep ，矢量化版本比原始版本快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 . 这可能是由于计算 result 中的和的差异所致 .

回复于 2024-05-13T19:09:05+08:00

@tel完成了大部分工作 . 以下是如何在20倍速度上获得另外2倍的加速 .

手动执行线性代数 . 当我尝试时，我感到震惊的是，小矩阵上的numpy是多么浪费：

>>> from timeit import timeit

# using eigvalsh
>>> print(timeit("test(False, 0.1)", setup="from __main__ import test", number=3))
-2301206.495955009
-2301206.495955009
-2301206.495955009
55.794611917983275
>>> print(timeit("test(False, 0.3)", setup="from __main__ import test", number=5))
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
-85240.46154498367
3.400342195003759

# by hand
>>> print(timeit("test(True, 0.1)", setup="from __main__ import test", number=3))
-2301206.495955076
-2301206.495955076
-2301206.495955076
26.67294767702697
>>> print(timeit("test(True, 0.3)", setup="from __main__ import test", number=5))
-85240.46154498379
-85240.46154498379
-85240.46154498379
-85240.46154498379
-85240.46154498379
1.5047460949863307

请注意，加速的一部分可能被共享代码掩盖，仅在线性代数上似乎更多，尽管我没有太敏锐地检查 .

一个警告：我正在使用Schur补码对矩阵的2by2分割来计算逆的对角元素 . 如果Schur补码不存在，即如果左上或右下子矩阵不可逆，则这将失败 .

这是修改后的代码：

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()
    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):
        vmat = vectorizemat(vmat)
        self._vfunc = sp.lambdify((x, y, z),
                                  expr=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!"
        mats = self._vfunc(*self._qs.T).T
        evs = np.linalg.eigvalsh(mats)
        result = np.sum(np.divide(1., (stiff + evs)))
        return result.real - 4 * self._q_count

    def eval_s_pp(self, stiff):
        assert len(self._qs) == self._q_count, "Run 'populate_qs' first!"
        mats = self._vfunc(*self._qs.T).T
        np.einsum('...ii->...i', mats)[...] += stiff
        (A, B), (C, D) = mats.reshape(-1, 2, 2, 2, 2).transpose(1, 3, 0, 2, 4)
        res = 0
        for AA, BB, CC, DD in ((A, B, C, D), (D, C, B, A)):
            (a, b), (c, d) = DD.transpose(1, 2, 0)
            rdet = 1 / (a*d - b*b)[:, None]
            iD = DD[..., ::-1, ::-1].copy()
            iD.reshape(-1, 4)[..., 1:3] *= -rdet
            np.einsum('...ii->...i', iD)[...] *= rdet
            (Aa, Ab), (Ac, Ad) = AA.transpose(1, 2, 0)
            (Ba, Bb), (Bc, Bd) = BB.transpose(1, 2, 0)
            (Da, Db), (Dc, Dd) = iD.transpose(1, 2, 0)
            a = Aa - Ba*Ba*Da - 2*Bb*Ba*Db - Bb*Bb*Dd
            d = Ad - Bd*Bd*Dd - 2*Bc*Bd*Db - Bc*Bc*Da
            b = Ab - Ba*Bc*Da - Ba*Bd*Db - Bb*Bd*Dd - Bb*Bc*Dc
            res += ((a + d) / (a*d - b*b)).sum()
        return res - 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)


def test(manual=False, ksep=0.3):
    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=ksep)  # <---- Performance starts to worsen (in eval_s) when ksep is reduced!
    if manual:
        print(solver.eval_s_pp(0.65))
    else:
        print(solver.eval_s_vectorized_completely(0.65))

回复于 2024-05-13T19:09:05+08:00

如何在python中矢量化包含eigvalsh的复杂代码

2 回答

测试/计时

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