scipy 成对距离与 X.X+Y.Y - X.Y^t 之间的差异

Question

假设我们有数据

d1 = np.random.uniform(low=0, high=2, size=(3,2))
d2 = np.random.uniform(low=3, high=5, size=(3,2))
X = np.vstack((d1,d2))

X
array([[ 1.4930674 ,  1.64890721],
       [ 0.40456265,  0.62262546],
       [ 0.86893397,  1.3590808 ],
       [ 4.04177045,  4.40938126],
       [ 3.01396153,  4.60005842],
       [ 3.2144552 ,  4.65539323]])

我想比较两种生成成对距离的方法：

假设X和Y相同：

(X-Y)^2 = X.X + Y.Y - 2*X.Y^t

这是第一种方法，因为它在 scikit-learn 中用于计算成对距离，后来用于核矩阵。

import numpy as np
def cal_pdist1(X):
       Y = X
       XX = np.einsum('ij,ij->i', X, X)[np.newaxis, :]
       YY = XX.T
       distances = -2*np.dot(X, Y.T)
       distances += XX
       distances += YY
       return(distances)

cal_pdist1(X)
array([[  0.        ,   2.2380968 ,   0.47354188,  14.11610424,
         11.02241244,  12.00213414],
       [  2.2380968 ,   0.        ,   0.75800718,  27.56880003,
         22.62893544,  24.15871196],
       [  0.47354188,   0.75800718,   0.        ,  19.37122424,
         15.1050792 ,  16.36714548],
       [ 14.11610424,  27.56880003,  19.37122424,   0.        ,
          1.09274896,   0.74497242],
       [ 11.02241244,  22.62893544,  15.1050792 ,   1.09274896,
          0.        ,   0.04325965],
       [ 12.00213414,  24.15871196,  16.36714548,   0.74497242,
          0.04325965,   0.        ]])

现在，如果我使用 scipy 成对距离函数，我会得到

import scipy, scipy.spatial
pd_sparse = scipy.spatial.distance.pdist(X, metric='seuclidean')
scipy.spatial.distance.squareform(pd_sparse)
array([[ 0.        ,  0.92916653,  0.45646989,  2.29444795,  1.89740167,
         2.00059442],
       [ 0.92916653,  0.        ,  0.50798432,  3.22211357,  2.78788236,
         2.90062103],
       [ 0.45646989,  0.50798432,  0.        ,  2.72720831,  2.28001564,
         2.39338343],
       [ 2.29444795,  3.22211357,  2.72720831,  0.        ,  0.71411943,
         0.58399694],
       [ 1.89740167,  2.78788236,  2.28001564,  0.71411943,  0.        ,
         0.14102567],
       [ 2.00059442,  2.90062103,  2.39338343,  0.58399694,  0.14102567,
         0.        ]])

结果完全不同！他们不应该是一样的吗？

Answer 1

pdist(..., metric='seuclidean') 计算 standardized Euclidean 距离，而不是 squared Euclidean 距离（cal_pdist returns).

来自 the docs:

Y = pdist(X, 'seuclidean', V=None)

Computes the standardized Euclidean distance. The standardized Euclidean distance between two n-vectors u and v is
   __________________
  √∑(ui−vi)^2 / V[xi]
V is the variance vector; V[i] is the variance computed over all the i’th components of the points. If not passed, it is automatically computed.

尝试传递 metric='sqeuclidean'，您将看到两个函数 return 相同的结果在舍入误差范围内。

scipy 成对距离与 X.X+Y.Y - X.Y^t 之间的差异

Difference between scipy pairwise distance and X.X+Y.Y - X.Y^t

python

distance

scipy

numpy