在Python(NumPy)中高效计算相似度矩阵

问题描述

让X为Bxn numpy矩阵，即

import numpy as np
B = 10
n = 2
X = np.random.random((B, n))

现在，我对计算形状为BxB的所谓内核(甚至相似性)矩阵K感兴趣，其第{i,j}个元素的给出如下:

Now, I'm interested in computing the so-called kernel (or even similarity) matrix K, which is of shape BxB, and its {i,j}-th element is given as follows:

K(i，j)= fun(x_i，x_j)

K(i,j) = fun(x_i, x_j)

其中，x_t表示矩阵X的第t行，而fun是x_i，x_j的某些函数.例如，该函数可以是所谓的RBF函数，即

where x_t denotes the t-th row of matrix X and fun is some function of x_i, x_j. For instance, this function could be the so-called RBF function, i.e.,

K(i，j)= exp(-| x_i-x_j | ^ 2).

K(i,j) = exp(-|x_i - x_j|^2).

为此，一个简单的方法如下:

For doing so, a naive way would be the following:

K = np.zeros((B, B))
for i in range(X.shape[0]):
    x_i = X[i, :]
    for j in range(X.shape[0]):
        x_j = X[j, :]
        K[i, j] = np.exp(-np.linalg.norm(x_i - x_j, 2) ** 2)

为了效率，我想要以向量化的方式执行上述操作.你能帮忙吗?

What I want is to do the above operation in a vectorized way, for the sake of efficiency. Could you help?

推荐答案

我不确定您是否可以仅使用numpy做到这一点.我会使用方法 cdist 来自scipy库，如下所示:

I'm not sure that you can due this using only numpy. I would use the method cdist from the scipy library, something like this:

import numpy as np
from scipy.spatial.distance import cdist
B=5
X=np.random.rand(B*B).reshape((B,B))
dist = cdist(X, X, metric='euclidean')
K = np.exp(dist)

dist
array([[ 0.        ,  1.2659804 ,  0.98231231,  0.80089176,  1.19326493],
       [ 1.2659804 ,  0.        ,  0.72658078,  0.80618767,  0.3776364 ],
       [ 0.98231231,  0.72658078,  0.        ,  0.70205336,  0.81352455],
       [ 0.80089176,  0.80618767,  0.70205336,  0.        ,  0.60025858],
       [ 1.19326493,  0.3776364 ,  0.81352455,  0.60025858,  0.        ]])
K
array([[ 1.        ,  3.5465681 ,  2.67062441,  2.22752646,  3.29783084],
       [ 3.5465681 ,  1.        ,  2.06799756,  2.23935453,  1.45883242],
       [ 2.67062441,  2.06799756,  1.        ,  2.01789192,  2.25584482],
       [ 2.22752646,  2.23935453,  2.01789192,  1.        ,  1.82259002],
       [ 3.29783084,  1.45883242,  2.25584482,  1.82259002,  1.        ]])

希望这可以为您提供帮助.做得好

Hoping this can help you. Good work

编辑您也可以仅将numpy数组用于theano实现:

EDITYou can also use only numpy array, for a theano implementaion:

dist = (X ** 2).sum(1).reshape((X.shape[0], 1)) + (X ** 2).sum(1).reshape((1, X.shape[0])) - 2 * X.dot(X.T)

应该可以！

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