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- # -*- 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
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