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KNN Python实现

KNN Python实现
'''
k近邻(kNN)算法的工作机制比较简单,根据某种距离测度找出距离给定待测样本距离最小的k个训练样本,根据k个训练样本进行预测。
分类问题:k个点中出现频率最高的类别作为待测样本的类别
回归问题:通常以k个训练样本的平均值作为待测样本的预测值
kNN模型三要素:距离测度、k值的选择、分类或回归决策方式
'''
import numpy as np
class KNNClassfier(object):

    def __init__(self, k=5, distance='euc'):
        self.k = k
        self.distance = distance
        self.x = None
        self.y = None
    def fit(self,X, Y):
        '''
        X : array-like [n_samples,shape]
        Y : array-like [n_samples,1]
        '''        
        self.x = X
        self.y = Y
    def predict(self,X_test):
        '''
        X_test : array-like [n_samples,shape]
        Y_test : array-like [n_samples,1]
        output : array-like [n_samples,1]
        '''  
        output = np.zeros((X_test.shape[0],1))
        for i in range(X_test.shape[0]):
            dis = [] 
            for j in range(self.x.shape[0]):
                if self.distance == 'euc': # 欧式距离
                    dis.append(np.linalg.norm(X_test[i]-self.x[j,:]))
            labels = []
            index=sorted(range(len(dis)), key=dis.__getitem__)
            for j in range(self.k):
                labels.append(self.y[index[j]])
            counts = []
            for label in labels:
                counts.append(labels.count(label))
            output[i] = labels[np.argmax(counts)]
        return output
    def score(self,x,y):
        pred = self.predict(x)
        err = 0.0
        for i in range(x.shape[0]):
            if pred[i]!=y[i]:
                err = err+1
        return 1-float(err/x.shape[0])


if __name__ == '__main__':
    from sklearn import datasets
    iris = datasets.load_iris()
    x = iris.data
    y = iris.target
    # x = np.array([[0.5,0.4],[0.1,0.2],[0.7,0.8],[0.2,0.1],[0.4,0.6],[0.9,0.9],[1,1]]).reshape(-1,2)
    # y = np.array([0,1,0,1,0,1,1]).reshape(-1,1)
    clf = KNNClassfier(k=3)
    clf.fit(x,y)
    print('myknn score:',clf.score(x,y))
    from sklearn.neighbors import KNeighborsClassifier
    clf_sklearn = KNeighborsClassifier(n_neighbors=3)
    clf_sklearn.fit(x,y)
    print('sklearn score:',clf_sklearn.score(x,y))

手写数字识别

from sklearn import datasets
from KNN import KNNClassfier
import matplotlib.pyplot as plt 
import numpy as  np
import time

digits = datasets.load_digits()
x = digits.data
y = digits.target

myknn_start_time = time.time()
clf = KNNClassfier(k=5)
clf.fit(x,y)
print('myknn score:',clf.score(x,y))
myknn_end_time = time.time()

from sklearn.neighbors import KNeighborsClassifier
sklearnknn_start_time = time.time()
clf_sklearn = KNeighborsClassifier(n_neighbors=5)
clf_sklearn.fit(x,y)
print('sklearn score:',clf_sklearn.score(x,y))
sklearnknn_end_time = time.time()

print('myknn uses time:',myknn_end_time-myknn_start_time)
print('sklearn uses time:',sklearnknn_end_time-sklearnknn_start_time)

可以看出处理较大数据集时,本人编写的kNN时间开销非常大,原因在于每次查找k个近邻点时都将扫描整个数据集,计算量很大,因此
k近邻(kNN)的实现还需要考虑如何最快的查找出k个近邻点,为了减少距离计算次数,可通过构造kd树,减少对大部分点的搜索、计算,kd树的构造可参考《统计学习方法》-李航

 

posted on 2019-01-09 21:13  贝壳w  阅读(2760)  评论(0编辑  收藏  举报

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