python - Numpy 有效的对抗所有人

Question

我有一个 3xN 的 3d 坐标数组，我想有效地计算所有条目的距离矩阵。 是否有任何有效的循环策略而不是可以应用的嵌套循环？

当前的伪代码实现:

for i,coord in enumerate(coords):
    for j,coords2 in enumerate(coords):
        if i != j:
             dist[i,j] = numpy.norm(coord - coord2)

Answer 1

要准确重现您的结果:

>>> import scipy.spatial as sp
>>> import numpy as np
>>> a=np.random.rand(5,3) #Note this is the transpose of your array.
>>> a
array([[ 0.83921304,  0.72659404,  0.50434178],  #0
       [ 0.99883826,  0.91739731,  0.9435401 ],  #1
       [ 0.94327962,  0.57665875,  0.85853404],  #2
       [ 0.30053567,  0.44458829,  0.35677649],  #3
       [ 0.01345765,  0.49247883,  0.11496977]]) #4
>>> sp.distance.cdist(a,a)
array([[ 0.        ,  0.50475862,  0.39845025,  0.62568048,  0.94249268],
       [ 0.50475862,  0.        ,  0.35554966,  1.02735895,  1.35575051],
       [ 0.39845025,  0.35554966,  0.        ,  0.82602847,  1.1935422 ],
       [ 0.62568048,  1.02735895,  0.82602847,  0.        ,  0.3783884 ],
       [ 0.94249268,  1.35575051,  1.1935422 ,  0.3783884 ,  0.        ]])

为了在不重复计算的情况下更有效地做到这一点，并且只计算唯一的对:

>>> sp.distance.pdist(a)
array([ 0.50475862,  0.39845025,  0.62568048,  0.94249268,  0.35554966,
        1.02735895,  1.35575051,  0.82602847,  1.1935422 ,  0.3783884 ])
#pairs: [(0, 1), (0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4), (2, 3),
#         (2, 4), (3, 4)]

注意两个数组之间的关系。 cdist 数组可以通过以下方式复制:

>>> out=np.zeros((a.shape[0],a.shape[0]))
>>> dists=sp.distance.pdist(a)
>>> out[np.triu_indices(a.shape[0],1)]=dists
>>> out+=out.T

>>> out
array([[ 0.        ,  0.50475862,  0.39845025,  0.62568048,  0.94249268],
       [ 0.50475862,  0.        ,  0.35554966,  1.02735895,  1.35575051],
       [ 0.39845025,  0.35554966,  0.        ,  0.82602847,  1.1935422 ],
       [ 0.62568048,  1.02735895,  0.82602847,  0.        ,  0.3783884 ],
       [ 0.94249268,  1.35575051,  1.1935422 ,  0.3783884 ,  0.        ]])

---

一些有点令人惊讶的时机-

设置:

def pdist_toarray(a):
    out=np.zeros((a.shape[0],a.shape[0]))
    dists=sp.distance.pdist(a)

    out[np.triu_indices(a.shape[0],1)]=dists
    return out+out.T

def looping(a):
    out=np.zeros((a.shape[0],a.shape[0]))
    for i in xrange(a.shape[0]):
        for j in xrange(a.shape[0]):
            out[i,j]=np.linalg.norm(a[i]-a[j])
    return out

时间:

arr=np.random.rand(1000,3)

%timeit sp.distance.pdist(arr)
100 loops, best of 3: 4.26 ms per loop

%timeit sp.distance.cdist(arr,arr)
100 loops, best of 3: 9.31 ms per loop

%timeit pdist_toarray(arr)
10 loops, best of 3: 66.2 ms per loop

%timeit looping(arr)
1 loops, best of 3: 16.7 s per loop

因此，如果您想要返回方阵，则应该使用 cdist，如果您只想让对使用 pdist。与 cdist 相比，具有 1000 个元素的数组的循环速度慢了约 4000 倍，而具有 10 个元素的数组的循环速度慢了约 70 倍。