Question

我有两个numpy数组如下。

X = np.array([-0.34095692,-0.34044722,-0.27155318,-0.21320583,-0.44657865,-0.19587836, -0.29414279, -0.3948753 ,-0.21655774 , -0.34857087])
Y = np.array([0.16305762,0.38554548, 0.10412536, -0.57981103, 0.17927523, -0.22612216, -0.34569697, 0.30463137,0.01301744,-0.42661108])

这些是10个用户的x和y协调。我需要找到每个用户之间的相似性。例如：

x1 = -0.34095692
y1 = 0.16305762
x2 = -0.34044722
y2 = 0.38554548

Euclidean distance = (|x1-y1|^2 + |x2-y2|^2)^1/2

所以最终我想得到一个如下矩阵：帮助我实现这个目标。

enter image description here

Answer 1

完成工作的简短代码段：

A = (X-Y)**2
p, q = np.meshgrid(np.arange(10), np.arange(10))
np.sqrt(A[p]-A[q])

修改：说明

A只是一个预先计算的向量，包含所有平方差异。
魔法在np.meshgrid：此函数的目的是生成两个不同数组中的所有值对。这不是最好的解决方案，因为您将得到整个矩阵，但对于您拥有的样本数量来说，这并不是什么大问题。生成的值将对应于A。
指数化部分A[p]也是一种魔力。自己尝试一下，了解它的行为。
这里的矩阵充满了nan，但这就是你要求的。真正的欧几里德距离为+，而不是-。

 array([[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]])

array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
   [2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
   [3, 3, 3, 3, 3, 3, 3, 3, 3, 3],
   [4, 4, 4, 4, 4, 4, 4, 4, 4, 4],
   [5, 5, 5, 5, 5, 5, 5, 5, 5, 5],
   [6, 6, 6, 6, 6, 6, 6, 6, 6, 6],
   [7, 7, 7, 7, 7, 7, 7, 7, 7, 7],
   [8, 8, 8, 8, 8, 8, 8, 8, 8, 8],
   [9, 9, 9, 9, 9, 9, 9, 9, 9, 9]])

Answer 2

使用zip(X, Y)获取坐标对，如果您想获得点之间的欧几里德距离，则应为(|x1-x2|^2+|y1-y2|^2)^0.5，而不是(|x1-y1|^2 - |x2-y2|^2)^1/2：

In [125]: coords=zip(X, Y)

In [126]: from scipy import spatial
     ...: dists=spatial.distance.cdist(coords, coords)

In [127]: dists
Out[127]: 
array([[ 0.        ,  0.22248844,  0.09104884,  0.75377329,  0.10685954,
         0.41534165,  0.5109039 ,  0.15149362,  0.19490308,  0.58971785],
       [ 0.22248844,  0.        ,  0.28973034,  0.9737061 ,  0.23197262,
         0.62852005,  0.73270705,  0.09751671,  0.39258852,  0.81219719],
       [ 0.09104884,  0.28973034,  0.        ,  0.68642072,  0.19047682,
         0.33880688,  0.45038919,  0.23539542,  0.1064197 ,  0.53629553],
       [ 0.75377329,  0.9737061 ,  0.68642072,  0.        ,  0.79415038,
         0.35411306,  0.24770988,  0.90290761,  0.59283795,  0.20443561],
       [ 0.10685954,  0.23197262,  0.19047682,  0.79415038,  0.        ,
         0.47665258,  0.54665574,  0.13560014,  0.28381556,  0.61376196],
       [ 0.41534165,  0.62852005,  0.33880688,  0.35411306,  0.47665258,
         0.        ,  0.15477091,  0.56683251,  0.24003205,  0.25201351],
       [ 0.5109039 ,  0.73270705,  0.45038919,  0.24770988,  0.54665574,
         0.15477091,  0.        ,  0.65808357,  0.36700881,  0.09751671],
       [ 0.15149362,  0.09751671,  0.23539542,  0.90290761,  0.13560014,
         0.56683251,  0.65808357,  0.        ,  0.34181257,  0.73270705],
       [ 0.19490308,  0.39258852,  0.1064197 ,  0.59283795,  0.28381556,
         0.24003205,  0.36700881,  0.34181257,  0.        ,  0.45902146],
       [ 0.58971785,  0.81219719,  0.53629553,  0.20443561,  0.61376196,
         0.25201351,  0.09751671,  0.73270705,  0.45902146,  0.        ]])

要获取此数组的上三角形，请使用numpy.triu：

In [128]: np.triu(dists)
Out[128]: 
array([[ 0.        ,  0.22248844,  0.09104884,  0.75377329,  0.10685954,
         0.41534165,  0.5109039 ,  0.15149362,  0.19490308,  0.58971785],
       [ 0.        ,  0.        ,  0.28973034,  0.9737061 ,  0.23197262,
         0.62852005,  0.73270705,  0.09751671,  0.39258852,  0.81219719],
       [ 0.        ,  0.        ,  0.        ,  0.68642072,  0.19047682,
         0.33880688,  0.45038919,  0.23539542,  0.1064197 ,  0.53629553],
       [ 0.        ,  0.        ,  0.        ,  0.        ,  0.79415038,
         0.35411306,  0.24770988,  0.90290761,  0.59283795,  0.20443561],
       [ 0.        ,  0.        ,  0.        ,  0.        ,  0.        ,
         0.47665258,  0.54665574,  0.13560014,  0.28381556,  0.61376196],
       [ 0.        ,  0.        ,  0.        ,  0.        ,  0.        ,
         0.        ,  0.15477091,  0.56683251,  0.24003205,  0.25201351],
       [ 0.        ,  0.        ,  0.        ,  0.        ,  0.        ,
         0.        ,  0.        ,  0.65808357,  0.36700881,  0.09751671],
       [ 0.        ,  0.        ,  0.        ,  0.        ,  0.        ,
         0.        ,  0.        ,  0.        ,  0.34181257,  0.73270705],
       [ 0.        ,  0.        ,  0.        ,  0.        ,  0.        ,
         0.        ,  0.        ,  0.        ,  0.        ,  0.45902146],
       [ 0.        ,  0.        ,  0.        ,  0.        ,  0.        ,
         0.        ,  0.        ,  0.        ,  0.        ,  0.        ]])

计算两个numpy数组中的欧几里德距离

2 个答案: