# -*- coding:utf-8 -*- import numpy as np import face_mysql class matrix: def __init__(self): pass # 两个矩阵的欧式距离 def EuclideanDistances(self, A, B): BT = B.transpose() # vecProd = A * BT vecProd = np.dot(A, BT) # print(vecProd) SqA = A ** 2 # print(SqA) sumSqA = np.matrix(np.sum(SqA, axis=1)) sumSqAEx = np.tile(sumSqA.transpose(), (1, vecProd.shape[1])) # print(sumSqAEx) SqB = B ** 2 sumSqB = np.sum(SqB, axis=1) sumSqBEx = np.tile(sumSqB, (vecProd.shape[0], 1)) SqED = sumSqBEx + sumSqAEx - 2 * vecProd SqED[SqED < 0] = 0.0 ED = np.sqrt(SqED) return ED.transpose() # def get_socres(self, A, ugroup): # 设置每次处理的最大数据库记录数 #如果数据库中记录太多时可分批进行处理 maxlen = 128 fmysql = face_mysql.face_mysql() results = np.array(fmysql.findall_facejson(ugroup)) #如果没有数据库找到时 直接返回空的list if results.shape[0] == 0: return [],[],[] pic_scores_all = [] #获取数据库中的入库时的图片名称 pic_names在数据库中存的是数组列索引4这个位置 pic_names = results[:, 4] #获取入库时图片对象的uid pic_uid在数据库中存的是数组列索引2这个位置 pic_uid = results[:, 2] for i in range(0, len(results), maxlen): pic_vectors = results[i:i + maxlen, 3] # 效率待优化,现在是每行处理 pic_vectors = [[float(j) for j in i.split(',')] for i in pic_vectors] pic_socores = self.EuclideanDistances(A, np.array(pic_vectors)) pic_socores_list = np.array(pic_socores).tolist() pic_scores_all.extend(pic_socores_list) pic_scores_all = np.array(pic_scores_all).transpose() # 获取距离最近的值 # np.argsort() 返回排序后的索引 pic_min_scores = np.amin(pic_scores_all, axis=1) pic_min_names = [] pic_min_uid = [] for i in range(0, len(pic_min_scores)): # 获取最小值的index index = np.where(pic_scores_all[i] == pic_min_scores[i]) # print(int(index[0])) # 有多个符合条件的只取第一个 pic_min_names.append(pic_names[index[0][0]]) pic_min_uid.append(pic_uid[index[0][0]]) # print(pic_min_names) return pic_min_scores.tolist(), pic_min_names, pic_min_uid