K-NN回归算法

from sklearn.datasets import load_iris
import numpy  as np
import matplotlib.pyplot as plt
       
iris = load_iris()
iris_data = iris.data
iris_target = iris.target
print(iris.feature_names)

X = iris_data[:,0:2]
y = iris_data[:,3]
#['sepal length (cm)', 'sepal width (cm)', 'petal length (cm)', 'petal width (cm)']

#We'll try to predict the petal length based on the sepal length and width.
#We'll also fit a regular linear regression to see how well the k-NN regression does in comparison

#线性回归
from sklearn.linear_model import LinearRegression
lr = LinearRegression()
lr.fit(X, y)
print ("The MSE is: {:.2}".format(np.power(y - lr.predict(X),2).mean()))



#K-NN 回归
from sklearn.neighbors import KNeighborsRegressor
knnr = KNeighborsRegressor(n_neighbors=10)
knnr.fit(X, y)
print ("The MSE is: {:.2}".format(np.power(y - knnr.predict(X),2).mean()))

#仅仅显示预测函数如何使用而已
print(knnr.predict(np.array([3.0,5.0]).reshape(1,-1)))

#Let's look at what the k-NN regression does when we tell it to use the closest 10 points for regression:
f, ax = plt.subplots(nrows=2, figsize=(7, 10))
ax[0].set_title("Predictions")
ax[0].scatter(X[:, 0], X[:, 1], s=lr.predict(X)*80, label='LRPredictions', color='c', edgecolors='black')
ax[1].scatter(X[:, 0], X[:, 1], s=knnr.predict(X)*80, label='k-NNPredictions', color='m', edgecolors='black')
ax[0].legend()
ax[1].legend()
f.show()

#针对某一个类别(KNN的效果优于线性)
setosa_idx = np.where(iris.target_names=='setosa')
setosa_mask = (iris.target == setosa_idx[0])
print(y[setosa_mask][:20])
print(knnr.predict(X)[setosa_mask][:20])
print(lr.predict(X)[setosa_mask][:20])

#针对某一个具体的点
#The k-NN regression is very simply calculated taking the average of the k closest point to the point being tested.
#Let's manually predict a single point:
example_point = X[0]
'''
原始真值
>>> X[0]
array([ 5.1,  3.5])
>>> y[0]
0.20000000000000001
'''

from sklearn.metrics import pairwise
distances_to_example = pairwise.pairwise_distances(X)[0]  #X[0]和其它150个元素（包括自己）的距离 
ten_closest_points = X[np.argsort(distances_to_example)][:10] #排序后，寻找10个距离最小的索引
ten_closest_y = y[np.argsort(distances_to_example)][:10]#所这些最下的10个已知数找出来
print(ten_closest_y.mean())

#We can see that this is very close to what was expected.

['sepal length (cm)', 'sepal width (cm)', 'petal length (cm)', 'petal width (cm)']
The MSE is: 0.15
The MSE is: 0.069
[ 0.2]
[ 0.2  0.2  0.2  0.2  0.2  0.4  0.3  0.2  0.2  0.1  0.2  0.2  0.1  0.1  0.2
  0.4  0.4  0.3  0.3  0.3]
[ 0.28  0.17  0.21  0.2   0.31  0.27  0.21  0.31  0.19  0.17  0.29  0.28
  0.17  0.19  0.26  0.27  0.27  0.28  0.27  0.31]
[ 0.44636645  0.53893889  0.29846368  0.27338255  0.32612885  0.47403161
  0.13064785  0.42128532  0.22322028  0.49136065  0.56918808  0.27596658
  0.46627952  0.10298268  0.71709085  0.45411854  0.47403161  0.44636645
  0.73958795  0.30363175]
0.28