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？非常感谢！

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通过创建一个3D张量，所以下面的计算可以通过广播来完成，但是当m很大时会浪费很多内存。

我也认为 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后直接将循环代码转换为基于 broadcasting 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，进行了一些优化，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

更进一步，看到可以优化操作的地方，我最终使用 Scipy's cdist 来代替平方和 sum-reduction 的重量级工作。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

接近 -

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

在10x附近加速了！

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