python - 如何在 numpy 中优化此函数的计算？

Question

我想在 numpy 中实现以下问题，这是我的代码。

我已经用一个 for 循环尝试了以下 numpy 代码来解决这个问题。我想知道是否有更有效的方法来进行此计算？我真的很感激!

k, d = X.shape
m = Y.shape[0]

c1 = 2.0*sigma**2
c2 = 0.5*np.log(np.pi*c1)
c3 = np.log(1.0/k)

L_B = np.zeros((m,))
for i in xrange(m):
    if i % 100 == 0:
        print i
    L_B[i] = np.log(np.sum(np.exp(np.sum(-np.divide(
                np.power(X-Y[i,:],2), c1)-c2,1)+c3)))

print np.mean(L_B)

我想到了 np.expand_dims(X, 2).repeat(Y.shape[0], 2)-Y 通过创建 3D 张量，因此可以通过以下计算来完成广播，但是当 m 很大时，这会浪费大量内存。

我也相信 np.einsum() 只使用 for 循环，所以可能效率不高，如果我错了请纠正我。

有什么想法吗？

Answer 1

优化阶段#1

我的第一级优化是在引入新轴时使用循环代码直接转换为基于广播的代码，因此内存效率不高，如下所列-

p1 = (-((X[:,None] - Y)**2)/c1)-c2
p11 = p1.sum(2)
p2 = np.exp(p11+c3)
out = np.log(p2.sum(0)).mean()

优化阶段#2

考虑到我们打算分离出对常量的操作，引入了一些优化，我最终得到了以下 -

c10 = -c1
c20 = X.shape[1]*c2

subs = (X[:,None] - Y)**2
p00 = subs.sum(2)
p10 = p00/c10
p11 = p10-c20
p2 = np.exp(p11+c3)
out = np.log(p2.sum(0)).mean()

优化阶段#3

进一步研究并查看可以优化操作的地方，我最终使用了 Scipy's cdist取代平方和sum-reduction的繁重工作。这应该是非常有效的内存，并为我们提供了最终的实现，如下所示 -

from scipy.spatial.distance import cdist

# Setup constants
c10 = -c1
c20 = X.shape[1]*c2
c30 = c20-c3
c40 = np.exp(c30)
c50 = np.log(c40)

# Get stagewise operations corresponding to loopy ones
p1 = cdist(X, Y, 'sqeuclidean')
p2 = np.exp(p1/c10).sum(0)
out = np.log(p2).mean() - c50

---

运行时测试

方法-

def loopy_app(X, Y, sigma):
    k, d = X.shape
    m = Y.shape[0]

    c1 = 2.0*sigma**2
    c2 = 0.5*np.log(np.pi*c1)
    c3 = np.log(1.0/k)

    L_B = np.zeros((m,))
    for i in xrange(m):
        L_B[i] = np.log(np.sum(np.exp(np.sum(-np.divide(
                    np.power(X-Y[i,:],2), c1)-c2,1)+c3)))

    return np.mean(L_B)

def vectorized_app(X, Y, sigma):
    # Setup constants
    k, d = D_A.shape
    c1 = 2.0*sigma**2
    c2 = 0.5*np.log(np.pi*c1)
    c3 = np.log(1.0/k)

    c10 = -c1
    c20 = X.shape[1]*c2
    c30 = c20-c3
    c40 = np.exp(c30)
    c50 = np.log(c40)

    # Get stagewise operations corresponding to loopy ones
    p1 = cdist(X, Y, 'sqeuclidean')
    p2 = np.exp(p1/c10).sum(0)
    out = np.log(p2).mean() - c50
    return out

时间和验证-

In [294]: # Setup inputs with m(=D_B.shape[0]) being a large number
     ...: X = np.random.randint(0,9,(100,10))
     ...: Y = np.random.randint(0,9,(10000,10))
     ...: sigma = 2.34
     ...: 

In [295]: np.allclose(loopy_app(X, Y, sigma),vectorized_app(X, Y, sigma))
Out[295]: True

In [296]: %timeit loopy_app(X, Y, sigma)
1 loops, best of 3: 225 ms per loop

In [297]: %timeit vectorized_app(X, Y, sigma)
10 loops, best of 3: 23.6 ms per loop

In [298]: # Setup inputs with m(=Y.shape[0]) being a much large number
     ...: X = np.random.randint(0,9,(100,10))
     ...: Y = np.random.randint(0,9,(100000,10))
     ...: sigma = 2.34
     ...: 

In [299]: np.allclose(loopy_app(X, Y, sigma),vectorized_app(X, Y, sigma))
Out[299]: True

In [300]: %timeit loopy_app(X, Y, sigma)
1 loops, best of 3: 2.27 s per loop

In [301]: %timeit vectorized_app(X, Y, sigma)
1 loops, best of 3: 243 ms per loop

大约 10x 那里加速了!