使用numexpr的欧几里德范数

Question

我需要用numexpr重写这个代码，它是计算矩阵数据行x cols和向量1x cols的欧几里德范数矩阵。

d = ((data-vec)**2).sum(axis=1)

怎么做呢？也许还有另一种更快的方法？

我使用hdf5时遇到的问题，以及从中读取的数据矩阵。例如，这段代码会产生错误:对象没有对齐。

#naive numpy solution, can be parallel?
def test_bruteforce_knn():
    h5f = tables.open_file(fileName)

    t0= time.time()
    d = np.empty((rows*batches,))
    for i in range(batches):
        d[i*rows:(i+1)*rows] = ((h5f.root.carray[i*rows:(i+1)*rows]-vec)**2).sum(axis=1)
    print (time.time()-t0)
    ndx = d.argsort()
    print ndx[:k]

    h5f.close()

#using some tricks (don't work error: objects are not aligned )
def test_bruteforce_knn():
    h5f = tables.open_file(fileName)

    t0= time.time()
    d = np.empty((rows*batches,))
    for i in range(batches):
        d[i*rows:(i+1)*rows] = (np.einsum('ij,ij->i', h5f.root.carray[i*rows:(i+1)*rows],
        h5f.root.carray[i*rows:(i+1)*rows]) 
        + np.dot(vec, vec)
        -2 * np.dot(h5f.root.carray[i*rows:(i+1)*rows], vec))
    print (time.time()-t0)
    ndx = d.argsort()
    print ndx[:k]

    h5f.close()

使用numexpr:似乎numexpr不理解h5f.root.carrayi*rows：(i+1)*行，它必须重新分配吗？

import numexpr as ne

def test_bruteforce_knn():
    h5f = tables.open_file(fileName)

    t0= time.time()
    d = np.empty((rows*batches,))
    for i in range(batches):
        ne.evaluate("sum((h5f.root.carray[i*rows:(i+1)*rows] - vec) ** 2, axis=1)")
    print (time.time()-t0)
    ndx = d.argsort()
    print ndx[:k]

    h5f.close()

Answer 1

有一种可能的快速方法(对于非常大的数组)是只使用NumPy的，这是在scikit中使用的--学习：

def squared_row_norms(X):
    # From http://stackoverflow.com/q/19094441/166749
    return np.einsum('ij,ij->i', X, X)

def squared_euclidean_distances(data, vec):
    data2 = squared_row_norms(data)
    vec2 = squared_row_norms(vec)
    d = np.dot(data, vec.T).ravel()
    d *= -2
    d += data2
    d += vec2
    return d

这是基于这样一个事实：(x -y)2=x 2+y 2-2 2xy，即使对矢量也是如此。

测试：

>>> data = np.random.randn(10, 40)
>>> vec = np.random.randn(1, 40)
>>> ((data - vec) ** 2).sum(axis=1)
array([  96.75712686,   69.45894306,  100.71998244,   80.97797154,
         84.8832107 ,   82.28910021,   67.48309433,   81.94813371,
         64.68162331,   77.43265692])
>>> squared_euclidean_distances(data, vec)
array([  96.75712686,   69.45894306,  100.71998244,   80.97797154,
         84.8832107 ,   82.28910021,   67.48309433,   81.94813371,
         64.68162331,   77.43265692])
>>> from sklearn.metrics.pairwise import euclidean_distances
>>> euclidean_distances(data, vec, squared=True).ravel()
array([  96.75712686,   69.45894306,  100.71998244,   80.97797154,
         84.8832107 ,   82.28910021,   67.48309433,   81.94813371,
         64.68162331,   77.43265692])

配置文件：

>>> data = np.random.randn(1000, 40)
>>> vec = np.random.randn(1, 40)
>>> %timeit ((data - vec)**2).sum(axis=1)
10000 loops, best of 3: 114 us per loop
>>> %timeit squared_euclidean_distances(data, vec)
10000 loops, best of 3: 52.5 us per loop

使用numexpr也是可能的，但它似乎对1000个点没有任何加速作用(在10000点时，它也不会更好)：

>>> %timeit ne.evaluate("sum((data - vec) ** 2, axis=1)")
10000 loops, best of 3: 142 us per loop