Question

我有以下函数返回计算最近邻居的数组：

def p_batch(U,X,Y):
    return [nearest(u,X,Y) for u in U]

我想用numpy替换for循环。我一直在研究numpy.vectorize（），因为这似乎是正确的方法，但我无法让它工作。这是我到目前为止所尝试的：

def n_batch(U,X,Y):
    vbatch = np.vectorize(nearest)
    return vbatch(U,X,Y)

任何人都可以在我出错的地方给我一个暗示吗？

修改

最近的实施：

def nearest(u,X,Y):
    return Y[np.argmin(np.sqrt(np.sum(np.square(np.subtract(u,X)),axis=1)))]

U，X，Y的函数（M = 20，N = 100，d = 50）：

U = numpy.random.mtrand.RandomState(123).uniform(0,1,[M,d])
X = numpy.random.mtrand.RandomState(456).uniform(0,1,[N,d])
Y = numpy.random.mtrand.RandomState(789).randint(0,2,[N])

Answer 1

方法＃1

您可以使用Scipy's cdist生成所有这些欧几里德距离，然后只需使用argmin并将其编入Y -

from scipy.spatial.distance import cdist

out = Y[cdist(U,X).argmin(1)]

示例运行 -

In [76]: M,N,d = 5,6,3
    ...: U = np.random.mtrand.RandomState(123).uniform(0,1,[M,d])
    ...: X = np.random.mtrand.RandomState(456).uniform(0,1,[N,d])
    ...: Y = np.random.mtrand.RandomState(789).randint(0,2,[N])
    ...: 

# Using a loop comprehension to verify values
In [77]: [nearest(U[i], X,Y) for i in range(len(U))]
Out[77]: [1, 0, 0, 1, 1]

In [78]: Y[cdist(U,X).argmin(1)]
Out[78]: array([1, 0, 0, 1, 1])

方法＃2

sklearn.metrics.pairwise_distances_argmin_min直接向我们提供argmin索引的另一种方式 -

from sklearn.metrics import pairwise

Y[pairwise.pairwise_distances_argmin(U,X)]

使用M=20,N=100,d=50 -

进行运行时测试

In [90]: M,N,d = 20,100,50
    ...: U = np.random.mtrand.RandomState(123).uniform(0,1,[M,d])
    ...: X = np.random.mtrand.RandomState(456).uniform(0,1,[N,d])
    ...: Y = np.random.mtrand.RandomState(789).randint(0,2,[N])
    ...:

在cdist和pairwise_distances_argmin -

之间进行测试

In [91]: %timeit cdist(U,X).argmin(1)
10000 loops, best of 3: 55.2 µs per loop

In [92]: %timeit pairwise.pairwise_distances_argmin(U,X)
10000 loops, best of 3: 90.6 µs per loop

针对循环版本的计时 -

In [93]: %timeit [nearest(U[i], X,Y) for i in range(len(U))]
1000 loops, best of 3: 298 µs per loop

In [94]: %timeit Y[cdist(U,X).argmin(1)]
10000 loops, best of 3: 55.6 µs per loop

In [95]: %timeit Y[pairwise.pairwise_distances_argmin(U,X)]
10000 loops, best of 3: 91.1 µs per loop

In [96]: 298.0/55.6   # Speedup with cdist over loopy one
Out[96]: 5.359712230215827

矢量化最近邻居计算

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