常見的距離算法和相似度計(jì)算方法

>>> pdist = nn.PairwiseDistance(p=2)>>> input1 = torch.randn(100, 128)>>> input2 = torch.randn(100, 128)>>> output = pdist(input1, input2)

C代碼包： emd.h, emd.c, emd.i
OpenCV：實(shí)現(xiàn)了EMD api,     pip install --upgrade setuptools    pip install numpy Matplotlib    pip install opencv-python
import numpy as npimport cv
#p、q是兩個矩陣，第一列表示權(quán)值，后面三列表示直方圖或數(shù)量p=np.asarray([[0.4,100,40,22],        [0.3,211,20,2],        [0.2,32,190,150],        [0.1,2,100,100]],np.float32)q=np.array([[0.5,0,0,0],        [0.3,50,100,80],        [0.2,255,255,255]],np.float32)pp=cv.fromarray(p)qq=cv.fromarray(q)emd=cv.CalcEMD2(pp,qq,cv.CV_DIST_L2)

import numpy as npdef mashi_distance(x,y):    print x    print y    #馬氏距離要求樣本數(shù)要大于維數(shù)，否則無法求協(xié)方差矩陣    #此處進(jìn)行轉(zhuǎn)置，表示10個樣本，每個樣本2維    X=np.vstack([x,y])
    print X    XT=X.T
    print XT
    #方法一：根據(jù)公式求解    S=np.cov(X)   #兩個維度之間協(xié)方差矩陣    SI = np.linalg.inv(S) #協(xié)方差矩陣的逆矩陣    #馬氏距離計(jì)算兩個樣本之間的距離，此處共有4個樣本，兩兩組合，共有6個距離。    n=XT.shape[0]    d1=[]    for i in range(0,n):        for j in range(i+1,n):            delta=XT[i]-XT[j]            d=np.sqrt(np.dot(np.dot(delta,SI),delta.T))            print d            d1.append(d)

>>> input1 = torch.randn(100, 128)>>> input2 = torch.randn(100, 128)>>> cos = nn.CosineSimilarity(dim=1, eps=1e-6)>>> output = cos(input1, input2)

import numpy as npx=np.random.random(8)y=np.random.random(8)
#方法一：根據(jù)公式求解x_=x-np.mean(x)y_=y-np.mean(y)d1=np.dot(x_,y_)/(np.linalg.norm(x_)*np.linalg.norm(y_))
#方法二：根據(jù)numpy庫求解X=np.vstack([x,y])d2=np.corrcoef(X)[0][1]

>>> torch.nn.functional.kl_div(input, target, size_average=None, reduce=None, reduction='mean')>>> F.kl_div(q.log(),p,reduction='sum')#函數(shù)中的 p q 位置相反(也就是想要計(jì)算D(p||q)，要寫成kl_div（q.log（），p）的形式)，而且q要先取 log#reduction 是選擇對各部分結(jié)果做什么操作，

import numpy as npfrom scipy.spatial.distance import pdistx=np.random.random(8)>0.5y=np.random.random(8)>0.5
x=np.asarray(x,np.int32)y=np.asarray(y,np.int32)
#方法一：根據(jù)公式求解up=np.double(np.bitwise_and((x != y),np.bitwise_or(x != 0, y != 0)).sum())down=np.double(np.bitwise_or(x != 0, y != 0).sum())d1=(up/down)

#方法二：根據(jù)scipy庫求解X=np.vstack([x,y])d2=pdist(X,'jaccard')

import numpy as npdef tanimoto_coefficient(p_vec, q_vec):    """    This method implements the cosine tanimoto coefficient metric    :param p_vec: vector one    :param q_vec: vector two    :return: the tanimoto coefficient between vector one and two    """    pq = np.dot(p_vec, q_vec)    p_square = np.linalg.norm(p_vec)    q_square = np.linalg.norm(q_vec)    return pq / (p_square + q_square - pq)

#標(biāo)準(zhǔn)化互信息from sklearn import metricsif __name__ == '__main__':    A = [1, 1, 1, 2, 3, 3]    B = [1, 2, 3, 1, 2, 3]    result_NMI=metrics.normalized_mutual_info_score(A, B)    print("result_NMI:",result_NMI)