python - 脾气暴躁的人

我有一个3dN的3d坐标数组，我想有效地计算所有条目的距离矩阵。
有没有一种有效的循环策略，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)

最佳答案

要完全重现您的结果，请执行以下操作:

>>> 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.        ]])

为了更有效地执行此操作而不重复计算，而仅计算唯一对:

>>> 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重现:

>>> 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，如果您想返回平方数组，并且只希望两对使用cdist，则应该使用pdist。与cdist相比，对于具有1000个元素的数组，循环的速度要慢约4000倍，对于具有10个元素的数组，循环的速度要慢约70倍。

关于python - 脾气暴躁的人，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/18537878/