包的导入以及图像上中文字体设置、图像清晰度设置

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

plt.rcParams['font.sans-serif'] = ['STfangsong']
plt.rcParams['axes.unicode_minus'] = False

%config InlineBackend.figure_format = 'svg'

常用函数

array1 = np.arange(1, 10).reshape(3, 3)
array1

array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

array2 = np.array([[1, 1, 1], [2, 2, 2], [3, 3, 3]])
array2

array([[1, 1, 1],
       [2, 2, 2],
       [3, 3, 3]])

# 水平方向拼接
np.hstack((array1, array2))

array([[1, 2, 3, 1, 1, 1],
       [4, 5, 6, 2, 2, 2],
       [7, 8, 9, 3, 3, 3]])

# 垂直方向拼接
np.vstack((array1, array2))

array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9],
       [1, 1, 1],
       [2, 2, 2],
       [3, 3, 3]])

# 沿着指定的轴拼接
np.concatenate((array1, array2))

array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9],
       [1, 1, 1],
       [2, 2, 2],
       [3, 3, 3]])

np.concatenate((array1, array2), axis=1)

array([[1, 2, 3, 1, 1, 1],
       [4, 5, 6, 2, 2, 2],
       [7, 8, 9, 3, 3, 3]])

# 垂直方向拆分
np.vsplit(array2, 3)

[array([[1, 1, 1]]), array([[2, 2, 2]]), array([[3, 3, 3]])]

# 水平方向拆分
np.hsplit(array2, 3)

[array([[1],
        [2],
        [3]]),
 array([[1],
        [2],
        [3]]),
 array([[1],
        [2],
        [3]])]

# 在末尾追加元素
np.append(array1, 10)

array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10])

# 在指定位置插入元素
np.insert(array1, 0, 0)

array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

array1[array1 % 3 == 0]

array([3, 6, 9])

# 根据条件筛选数据
np.extract(array1 % 3 == 0, array1)

array([3, 6, 9])

# 根据条件和公式获取数据
x = np.arange(10)
condlist = [x < 3, x > 5]
choicelist = [x, x ** 2]
np.select(condlist, choicelist, default=np.nan)

array([ 0.,  1.,  2., nan, nan, nan, 36., 49., 64., 81.])

# 根据条件和公式获取数据
np.where(x < 5, x, 10 * x)

array([ 0,  1,  2,  3,  4, 50, 60, 70, 80, 90])

def fib(n):
    a, b = 0, 1
    for _ in range(n):
        a, b = b, a + b
        yield a

        
gen = fib(20)
# 通过迭代器（生成器）创建数组对象
array3 = np.fromiter(gen, dtype=np.int64, count=10)
array3

array([ 1,  1,  2,  3,  5,  8, 13, 21, 34, 55], dtype=int64)

# 调整数组的大小
np.resize(array1, (4, 4))

array([[1, 2, 3, 4],
       [5, 6, 7, 8],
       [9, 1, 2, 3],
       [4, 5, 6, 7]])

向量

点积运算

A ⋅ B = a 1 b 1 + a 2 b 2 = ∣ A ∣ ∣ B ∣ c o s θ A \cdot B = a_1b_1 + a_2b_2 = \lvert A \rvert \lvert B \rvert cos \thetaA⋅B=a1​b1​+a2​b2​=∣A∣∣B∣cosθ

A ⋅ B = ∑ i = 1 n a i b i = ∣ A ∣ ∣ B ∣ c o s θ A \cdot B = \sum_{i=1}^{n} a_ib_i = \lvert A \rvert \lvert B \rvert cos \thetaA⋅B=i=1∑n​ai​bi​=∣A∣∣B∣cosθ

v1 = np.array([3, 5])
v2 = np.array([1, 3])
# inner_prod = np.dot(v1, v2)
inner_prod = np.inner(v1, v2)
print('向量点积:', inner_prod)

向量点积: 18

说明：在欧几里得几何中，两个笛卡尔坐标向量的点积也称为内积（inner product），但是内积的含义要高于点积，点积相当于是内积在欧几里得空间 $ \mathbb{R}^n $ 的特例，而内积可以推广到赋范向量空间。

v1_norm = np.linalg.norm(v1)
v2_norm = np.linalg.norm(v2)
print('v1的模:', np.round(v1_norm, 6))
print('v2的模:', np.round(v2_norm, 6))

v1的模: 5.830952
v2的模: 3.162278

cos_theta = inner_prod / (v1_norm * v2_norm)
print('向量夹角余弦值:', cos_theta)
print('夹角:', np.arccos(cos_theta) * 180 / np.pi)

向量夹角余弦值: 0.9761870601839526
夹角: 12.52880770915155

行列式

d e t ( A ) = ∑ n ! ± a 1 α a 2 β a 3 γ ⋯ a n ω det(A) = \sum_{n!} \pm a_{1\alpha}a_{2\beta}a_{3\gamma} \cdots a_{n\omega}det(A)=n!∑​±a1α​a2β​a3γ​⋯anω​

array4 = np.stack((v1, v2))
array4

array([[3, 5],
       [1, 3]])

d e t ∣ 3 5 1 3 ∣ = 4 det \begin{vmatrix} 3 & 5 \ 1 & 3 \end{vmatrix} = 4det∣∣∣∣​31​53​∣∣∣∣​=4

# 计算行列式的值
np.round(np.linalg.det(array4), 2)

4.0

d e t ∣ 1 2 3 4 5 6 7 8 9 ∣ = 0 det \begin{vmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{vmatrix} = 0det∣∣∣∣∣∣​147​258​369​∣∣∣∣∣∣​=0

np.linalg.det(array1)

0.0

矩阵

array1 = np.arange(1, 10).reshape((3, 3))
array1

array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

对上面的array1做一组线性变换，就知道为什么它的秩是2了。

∣ 1 2 3 4 5 6 7 8 9 ∣ → ∣ 1 2 3 0 − 3 − 6 0 − 6 − 12 ∣ \begin{vmatrix} 1 & 2 & 3\ 4 & 5 & 6\ 7 & 8 & 9 \end{vmatrix} \quad \to \quad \begin{vmatrix} 1 & 2 & 3\ 0 & -3 & -6\ 0 & -6 & -12 \end{vmatrix}∣∣∣∣∣∣​147​258​369​∣∣∣∣∣∣​→∣∣∣∣∣∣​100​2−3−6​3−6−12​∣∣∣∣∣∣​

# 求逆矩阵
# LinAlgError ---> Singluar matrix ---> 奇异矩阵不能求逆
# np.linalg.inv(array1)

array2 = np.array([[1, 2], [3, 4]])
array2

array([[1, 2],
       [3, 4]])

array3 = np.linalg.inv(array2)
array3

array([[-2. ,  1. ],
       [ 1.5, -0.5]])

A ⋅ A − 1 = I A \cdot A^{-1} = IA⋅A−1=I

np.round(array2 @ array3, 2)

array([[1., 0.],
       [0., 1.]])

# 求矩阵的秩
np.linalg.matrix_rank(array1)

2

array1[2, 2] = 8
array1

array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 8]])

np.linalg.matrix_rank(array1)

3

解线性方程：

{ 3 x + y = 9 x + 2 y = 8 \begin{cases} 3x + y = 9 \ x + 2y = 8 \end{cases}{3x+y=9x+2y=8​

A = np.array([[3, 1], [1, 2]])
b = np.array([9, 8]).reshape(-1, 1)

np.linalg.solve(A, b)

array([[2.],
       [3.]])

A x = b A − 1 A x = A − 1 b I x = A − 1 b Ax = b\ A^{-1}Ax = A^{-1}b\ Ix = A^{-1}bAx=bA−1Ax=A−1bIx=A−1b

A_1 = np.linalg.inv(A)
A_1

array([[ 0.4, -0.2],
       [-0.2,  0.6]])

A_1 @ b

array([[2.],
       [3.]])

最小二乘解

!pip install scikit-learn

Looking in indexes: https://pypi.doubanio.com/simple
Requirement already satisfied: scikit-learn in d:\programs\python\python38\lib\site-packages (0.24.2)
Requirement already satisfied: threadpoolctl>=2.0.0 in d:\programs\python\python38\lib\site-packages (from scikit-learn) (2.2.0)
Requirement already satisfied: numpy>=1.13.3 in d:\programs\python\python38\lib\site-packages (from scikit-learn) (1.21.2)
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Requirement already satisfied: scipy>=0.19.1 in d:\programs\python\python38\lib\site-packages (from scikit-learn) (1.7.1)

from sklearn.datasets import load_boston

# 获取波士顿房价数据
dataset = load_boston()
print(dataset.DESCR)

.. _boston_dataset:

Boston house prices dataset
---------------------------

**Data Set Characteristics:**  

    :Number of Instances: 506 

    :Number of Attributes: 13 numeric/categorical predictive. Median Value (attribute 14) is usually the target.

    :Attribute Information (in order):
        - CRIM     per capita crime rate by town
        - ZN       proportion of residential land zoned for lots over 25,000 sq.ft.
        - INDUS    proportion of non-retail business acres per town
        - CHAS     Charles River dummy variable (= 1 if tract bounds river; 0 otherwise)
        - NOX      nitric oxides concentration (parts per 10 million)
        - RM       average number of rooms per dwelling
        - AGE      proportion of owner-occupied units built prior to 1940
        - DIS      weighted distances to five Boston employment centres
        - RAD      index of accessibility to radial highways
        - TAX      full-value property-tax rate per $10,000
        - PTRATIO  pupil-teacher ratio by town
        - B        1000(Bk - 0.63)^2 where Bk is the proportion of black people by town
        - LSTAT    % lower status of the population
        - MEDV     Median value of owner-occupied homes in $1000's

    :Missing Attribute Values: None

    :Creator: Harrison, D. and Rubinfeld, D.L.

This is a copy of UCI ML housing dataset.
https://archive.ics.uci.edu/ml/machine-learning-databases/housing/

​
This dataset was taken from the StatLib library which is maintained at Carnegie Mellon University.

The Boston house-price data of Harrison, D. and Rubinfeld, D.L. 'Hedonic
prices and the demand for clean air', J. Environ. Economics & Management,
vol.5, 81-102, 1978.   Used in Belsley, Kuh & Welsch, 'Regression diagnostics
...', Wiley, 1980.   N.B. Various transformations are used in the table on
pages 244-261 of the latter.

The Boston house-price data has been used in many machine learning papers that address regression
problems.   
     
.. topic:: References

   - Belsley, Kuh & Welsch, 'Regression diagnostics: Identifying Influential Data and Sources of Collinearity', Wiley, 1980. 244-261.
   - Quinlan,R. (1993). Combining Instance-Based and Model-Based Learning. In Proceedings on the Tenth International Conference of Machine Learning, 236-243, University of Massachusetts, Amherst. Morgan Kaufmann.

​

dataset.data.shape

(506, 13)

dataset.feature_names

array(['CRIM', 'ZN', 'INDUS', 'CHAS', 'NOX', 'RM', 'AGE', 'DIS', 'RAD',
       'TAX', 'PTRATIO', 'B', 'LSTAT'], dtype='<U7')

# 用波士顿房价数据创建DataFrame对象
df = pd.DataFrame(data=dataset.data, columns=dataset.feature_names)
df

506 rows × 13 columns

# 添加房价列
df['PRICE'] = dataset.target
df

506 rows × 14 columns

# 计算协方差
df.cov()

# 计算皮尔逊相关系数
np.round(df.corr(), 2)

rooms = df['RM'].values
prices = df['PRICE'].values
history_data = {room: price for room, price in zip(rooms, prices)}
history_data

{6.575: 24.0,
 6.421: 21.6,
 7.185: 34.9,
 6.998: 33.4,
 7.147: 36.2,
 6.43: 28.7,
 6.012: 22.9,
 6.172: 27.1,
 5.631: 16.5,
 6.004: 20.3,
 6.377: 15.0,
 6.009: 21.7,
 5.889: 21.7,
 5.949: 20.4,
 6.096: 13.5,
 5.834: 19.9,
 5.935: 8.4,
 5.99: 17.5,
 5.456: 20.2,
 5.727: 18.2,
 5.57: 13.6,
 5.965: 19.6,
 6.142: 15.2,
 5.813: 16.6,
 5.924: 15.6,
 5.599: 13.9,
 6.047: 14.8,
 6.495: 26.4,
 6.674: 21.0,
 5.713: 20.1,
 6.072: 14.5,
 5.95: 13.2,
 5.701: 13.1,
 5.933: 18.9,
 5.841: 20.0,
 5.85: 21.0,
 5.966: 16.0,
 6.595: 30.8,
 7.024: 34.9,
 6.77: 26.6,
 6.169: 25.3,
 6.211: 25.0,
 6.069: 21.2,
 5.682: 19.3,
 5.786: 20.0,
 6.03: 11.9,
 5.399: 14.4,
 5.602: 19.4,
 5.963: 19.7,
 6.115: 20.5,
 6.511: 25.0,
 5.998: 23.4,
 5.888: 23.3,
 7.249: 35.4,
 6.383: 24.7,
 6.816: 31.6,
 6.145: 23.3,
 5.927: 19.6,
 5.741: 18.7,
 6.456: 22.2,
 6.762: 25.0,
 7.104: 33.0,
 6.29: 23.5,
 5.787: 19.4,
 5.878: 22.0,
 5.594: 17.4,
 5.885: 20.9,
 6.417: 13.0,
 5.961: 20.5,
 6.065: 22.8,
 6.245: 23.4,
 6.273: 24.1,
 6.286: 21.4,
 6.279: 20.0,
 6.14: 20.8,
 6.232: 21.2,
 5.874: 20.3,
 6.727: 27.5,
 6.619: 23.9,
 6.302: 24.8,
 6.167: 19.9,
 6.389: 23.9,
 6.63: 27.9,
 6.015: 22.5,
 6.121: 22.2,
 7.007: 23.6,
 7.079: 28.7,
 6.405: 12.5,
 6.442: 22.9,
 6.249: 20.6,
 6.625: 28.4,
 6.163: 21.4,
 8.069: 38.7,
 7.82: 45.4,
 7.416: 33.2,
 6.781: 26.5,
 6.137: 19.3,
 5.851: 19.5,
 5.836: 19.5,
 6.127: 22.7,
 6.474: 19.8,
 6.229: 21.4,
 6.195: 21.7,
 6.715: 22.8,
 5.913: 18.8,
 6.092: 18.7,
 6.254: 18.5,
 5.928: 18.3,
 6.176: 21.2,
 6.021: 19.2,
 5.872: 20.4,
 5.731: 19.3,
 5.87: 22.0,
 5.856: 21.1,
 5.879: 18.8,
 5.986: 21.4,
 5.613: 15.7,
 5.693: 16.2,
 6.431: 24.6,
 5.637: 14.3,
 6.458: 19.2,
 6.326: 24.4,
 6.372: 23.0,
 5.822: 18.4,
 5.757: 15.0,
 6.335: 18.1,
 5.942: 17.4,
 6.454: 17.1,
 5.857: 13.3,
 6.151: 17.8,
 6.174: 14.0,
 5.019: 14.4,
 5.403: 13.4,
 5.468: 15.6,
 4.903: 11.8,
 6.13: 13.8,
 5.628: 15.6,
 4.926: 14.6,
 5.186: 17.8,
 5.597: 15.4,
 6.122: 22.1,
 5.404: 19.3,
 5.012: 15.3,
 5.709: 19.4,
 6.129: 17.0,
 6.152: 8.7,
 5.272: 13.1,
 6.943: 41.3,
 6.066: 24.3,
 6.51: 23.3,
 6.25: 27.0,
 7.489: 50.0,
 7.802: 50.0,
 8.375: 50.0,
 5.854: 10.8,
 6.101: 25.0,
 7.929: 50.0,
 5.877: 23.8,
 6.319: 23.8,
 6.402: 22.3,
 5.875: 50.0,
 5.88: 19.1,
 5.572: 23.1,
 6.416: 23.6,
 5.859: 22.6,
 6.546: 29.4,
 6.02: 23.2,
 6.315: 22.3,
 6.86: 29.9,
 6.98: 29.8,
 7.765: 39.8,
 6.144: 19.8,
 7.155: 37.9,
 6.563: 32.5,
 5.604: 26.4,
 6.153: 29.6,
 7.831: 50.0,
 6.782: 7.5,
 6.556: 29.8,
 6.951: 26.7,
 6.739: 30.5,
 7.178: 36.4,
 6.8: 31.1,
 6.604: 29.1,
 7.875: 50.0,
 7.287: 33.3,
 7.107: 30.3,
 7.274: 34.6,
 6.975: 34.9,
 7.135: 32.9,
 6.162: 13.3,
 7.61: 42.3,
 7.853: 48.5,
 8.034: 50.0,
 5.891: 22.6,
 5.783: 22.5,
 6.064: 24.4,
 5.344: 20.0,
 5.96: 21.7,
 5.807: 22.4,
 6.375: 28.1,
 5.412: 23.7,
 6.182: 25.0,
 6.642: 28.7,
 5.951: 21.5,
 6.373: 23.0,
 6.164: 21.7,
 6.879: 27.5,
 6.618: 30.1,
 8.266: 44.8,
 8.725: 50.0,
 8.04: 37.6,
 7.163: 31.6,
 7.686: 46.7,
 6.552: 31.5,
 5.981: 24.3,
 7.412: 31.7,
 8.337: 41.7,
 8.247: 48.3,
 6.726: 29.0,
 6.086: 24.0,
 6.631: 25.1,
 7.358: 31.5,
 6.481: 23.7,
 6.606: 23.3,
 6.897: 22.0,
 6.095: 20.1,
 6.358: 22.2,
 6.393: 23.7,
 5.593: 17.6,
 5.605: 18.5,
 6.108: 21.9,
 6.226: 20.5,
 6.433: 24.5,
 6.718: 26.2,
 6.487: 24.4,
 6.438: 24.8,
 6.957: 29.6,
 8.259: 42.8,
 5.876: 20.9,
 7.454: 44.0,
 8.704: 50.0,
 7.333: 36.0,
 6.842: 30.1,
 7.203: 33.8,
 7.52: 43.1,
 8.398: 48.8,
 7.327: 31.0,
 7.206: 36.5,
 5.56: 22.8,
 7.014: 30.7,
 8.297: 50.0,
 7.47: 43.5,
 5.92: 20.7,
 6.24: 25.2,
 6.538: 24.4,
 7.691: 35.2,
 6.758: 32.4,
 6.854: 32.0,
 7.267: 33.2,
 6.826: 33.1,
 6.482: 29.1,
 6.812: 35.1,
 6.968: 10.4,
 7.645: 46.0,
 7.923: 50.0,
 7.088: 32.2,
 6.453: 22.0,
 6.23: 20.1,
 6.209: 21.4,
 6.565: 24.8,
 6.861: 28.5,
 7.148: 37.3,
 6.678: 28.6,
 6.549: 27.1,
 5.79: 20.3,
 6.345: 22.5,
 7.041: 29.0,
 6.871: 24.8,
 6.59: 22.0,
 6.982: 33.1,
 7.236: 36.1,
 6.616: 28.4,
 7.42: 33.4,
 6.849: 28.2,
 6.635: 24.5,
 5.972: 20.3,
 4.973: 16.1,
 6.023: 19.4,
 6.266: 21.6,
 6.567: 23.8,
 5.705: 16.2,
 5.914: 17.8,
 5.782: 19.8,
 6.382: 23.1,
 6.113: 21.0,
 6.426: 23.8,
 6.376: 17.7,
 6.041: 20.4,
 5.708: 18.5,
 6.415: 25.0,
 6.312: 21.2,
 6.083: 22.2,
 5.868: 19.3,
 6.333: 22.6,
 5.706: 17.1,
 6.031: 19.4,
 6.316: 22.2,
 6.31: 20.7,
 6.037: 21.1,
 5.869: 19.5,
 5.895: 18.5,
 6.059: 20.6,
 5.985: 19.0,
 5.968: 18.7,
 7.241: 32.7,
 6.54: 16.5,
 6.696: 23.9,
 6.874: 31.2,
 6.014: 17.5,
 5.898: 17.2,
 6.516: 23.1,
 6.939: 26.6,
 6.49: 22.9,
 6.579: 24.1,
 5.884: 18.6,
 6.728: 14.9,
 5.663: 18.2,
 5.936: 13.5,
 6.212: 17.8,
 6.395: 21.7,
 6.112: 22.6,
 6.398: 25.0,
 6.251: 12.6,
 5.362: 20.8,
 5.803: 16.8,
 8.78: 21.9,
 3.561: 27.5,
 4.963: 21.9,
 3.863: 23.1,
 4.97: 50.0,
 6.683: 50.0,
 7.016: 50.0,
 6.216: 50.0,
 4.906: 13.8,
 4.138: 11.9,
 7.313: 15.0,
 6.649: 13.9,
 6.794: 22.0,
 6.38: 9.5,
 6.223: 10.2,
 6.545: 10.9,
 5.536: 11.3,
 5.52: 12.3,
 4.368: 8.8,
 5.277: 7.2,
 4.652: 10.5,
 5.0: 7.4,
 4.88: 10.2,
 5.39: 19.7,
 6.051: 23.2,
 5.036: 9.7,
 6.193: 11.0,
 5.887: 12.7,
 6.471: 13.1,
 5.747: 8.5,
 5.453: 5.0,
 5.852: 6.3,
 5.987: 5.6,
 6.343: 7.2,
 6.404: 12.1,
 5.349: 8.3,
 5.531: 8.5,
 5.683: 5.0,
 5.608: 27.9,
 5.617: 17.2,
 6.852: 27.5,
 6.657: 17.2,
 4.628: 17.9,
 5.155: 16.3,
 4.519: 7.0,
 6.434: 7.2,
 5.304: 12.0,
 5.957: 8.8,
 6.824: 8.4,
 6.411: 16.7,
 6.006: 14.2,
 5.648: 20.8,
 6.103: 13.4,
 5.565: 11.7,
 5.896: 8.3,
 5.837: 10.2,
 6.202: 10.9,
 6.348: 14.5,
 6.833: 14.1,
 6.425: 16.1,
 6.436: 14.3,
 6.208: 11.7,
 6.629: 13.4,
 6.461: 9.6,
 5.627: 12.8,
 5.818: 10.5,
 6.406: 17.1,
 6.219: 18.4,
 6.485: 15.4,
 6.459: 11.8,
 6.341: 14.9,
 6.185: 14.6,
 6.749: 13.4,
 6.655: 15.2,
 6.297: 16.1,
 7.393: 17.8,
 6.525: 14.1,
 5.976: 12.7,
 6.301: 14.9,
 6.081: 20.0,
 6.701: 16.4,
 6.317: 19.5,
 6.513: 20.2,
 5.759: 19.9,
 5.952: 19.0,
 6.003: 19.1,
 5.926: 24.5,
 6.437: 23.2,
 5.427: 13.8,
 6.484: 16.7,
 6.242: 23.0,
 6.75: 23.7,
 7.061: 25.0,
 5.762: 21.8,
 5.871: 20.6,
 6.114: 19.1,
 5.905: 20.6,
 5.454: 15.2,
 5.414: 7.0,
 5.093: 8.1,
 5.983: 20.1,
 5.707: 21.8,
 5.67: 23.1,
 5.794: 18.3,
 6.019: 21.2,
 5.569: 17.5,
 6.027: 16.8,
 6.593: 22.4,
 6.12: 20.6,
 6.976: 23.9}

通过计算皮尔逊相关系数，发现房间数和房价存在正相关，接下来我们通过学习历史数据，最终实现用房间数预测房价的目标。

import heapq

nums = [35, 98, 76, 12, 55, 68, 47, 92]
print(heapq.nlargest(3, nums))
print(heapq.nsmallest(3, nums))

[98, 92, 76]
[12, 35, 47]

import heapq

# kNN算法
def predict_price_by_knn(history_data, room, k=5):
    # keys = sorted(history_data, key=lambda x: (x - room) ** 2)[:k]
    keys = heapq.nsmallest(k, history_data, key=lambda x: (x - room) ** 2)
    return np.mean([history_data[key] for key in keys])

# 预测房价
np.round(predict_price_by_knn(history_data, 6.25), 2)

20.42

np.round(predict_price_by_knn(history_data, 5.125), 2)

13.26

# 通过散点图研究变量的关系
plt.scatter(rooms, prices)
plt.show()

​

​

通过上面的图，我们发现房间数和房价呈现出线性关系，接下来我们尝试用一个线性函数来实现对房价的预测。

损失函数

回归方程：x xx 代表房间数，y yy 就是要预测的房价。
y = a x + b y = ax + by=ax+b

现在我们的问题是找到一组a和b，让预测达到最佳的效果（误差最小就是最佳）。

均方误差：让均方误差最小的 a aa 和 b bb 就是最佳拟合。
M S E = 1 N ∑ ( y i ^ − y i ) 2 MSE = \frac{1} {N} \sum (\hat{y_i} - y_i)^2MSE=N1​∑(yi​^​−yi​)2

def get_loss(x, y, a, b):
    """损失函数"""
    y_hat = a * x + b
    return np.mean((y_hat - y) ** 2)

# 通过蒙特卡罗模拟找到实现最佳拟合的a和b的值
import random

best_a, best_b = None, None
min_loss = np.inf

for _ in range(1000):
    # 随机产生a和b的值
    a = random.random() * 200 - 100
    b = random.random() * 200 - 100
    # 计算损失（MSE）
    curr_loss = get_loss(rooms, prices, a, b)
    # 让损失更小的a和b就是更好的拟合
    if curr_loss < min_loss:
        min_loss = curr_loss
        best_a, best_b = a, b
print(best_a, best_b)
print(min_loss)

12.414266461017732 -56.48722240398021
50.00741553150247

梯度下降

损失函数是凹函数，找到使函数最小的a和b的值，可以用下面的方法：

a ′ = a + ( − 1 ) × ∂ l o s s ( a , b ) ∂ a × Δ a^\prime = a + (-1) \times \frac {\partial loss(a, b)} {\partial a} \times \Deltaa′=a+(−1)×∂a∂loss(a,b)​×Δ
b ′ = b + ( − 1 ) × ∂ l o s s ( a , b ) ∂ b × Δ b^\prime = b + (-1) \times \frac {\partial loss(a, b)} {\partial b} \times \Deltab′=b+(−1)×∂b∂loss(a,b)​×Δ

对于求MSE的损失函数来说，可以用下面的公式计算偏导数：

f ( a , b ) = 1 N ∑ i = 1 N ( y i − ( a x i + b ) ) 2 f(a, b) = \frac {1} {N} \sum_{i=1}^{N}(y_i - (ax_i + b))^2f(a,b)=N1​i=1∑N​(yi​−(axi​+b))2
∂ f ( a , b ) ∂ a = 2 N ∑ i = 1 N ( − x i y i + x i 2 a + x i b ) \frac {\partial {f(a, b)}} {\partial {a}} = \frac {2} {N} \sum_{i=1}^{N}(-x_iy_i + x_i^2a + x_ib)∂a∂f(a,b)​=N2​i=1∑N​(−xi​yi​+xi2​a+xi​b)
∂ f ( a , b ) ∂ b = 2 N ∑ i = 1 N ( − y i + x i a + b ) \frac {\partial {f(a, b)}} {\partial {b}} = \frac {2} {N} \sum_{i=1}^{N}(-y_i + x_ia + b)∂b∂f(a,b)​=N2​i=1∑N​(−yi​+xi​a+b)

# 求a的偏导数
def partial_a(x, y, a, b):
    return 2 * np.mean((y - a * x - b) * (-x))

# 求b的偏导数
def partial_b(x, y, a, b):
    return 2 * np.mean(-y + a * x + b)

# 通过梯度下降的方式向拐点逼近
# 这种方式能够更快的找到最佳拟合的a和b
# a和b的初始值可以随意设定，delta的值要足够小
a, b = 35, -35
delta = 0.01

for _ in range(100):
    a = a - partial_a(rooms, prices, a, b) * delta
    b = b - partial_b(rooms, prices, a, b) * delta
print(a, b)
print(get_loss(rooms, prices, a, b))

9.276809660789766 -35.781905844032686
43.61576735104159

# 通过线性回归方程预测房价
def predict_price_by_regression(a, b, x):
    return a * x + b

# 预测房价
print(np.round(predict_price_by_regression(best_a, best_b, 6.25), 2))
print(np.round(predict_price_by_regression(a, b, 6.25), 2))

21.1
22.2

print(np.round(predict_price_by_regression(best_a, best_b, 5.12), 2))
print(np.round(predict_price_by_regression(a, b, 5.12), 2))

7.07
11.72

# 比较两条拟合曲线
y_hat1 = best_a * rooms + best_b
y_hat2 = a * rooms + b
plt.scatter(rooms, prices)
plt.plot(rooms, y_hat1, color='red', linewidth=4)
plt.plot(rooms, y_hat2, color='green', linewidth=4)
plt.show()

​

​

最小二乘解就是用已经得到的历史数据（x和y的值）找到能够最佳拟合这些历史数据的a和b。

y = a x + b y = ax + by=ax+b

对于上面的方程，相当于x是变量a的系数，1是变量b的系数。

lstsq函数参数说明

lstsq函数的第一个参数是$ \begin{bmatrix} x \ 1 \ \end{bmatrix} ^T $，第二个参数就是y，rcond参数暂时不管，直接设置为None。

# lstsq函数的第一个参数
param1 = np.vstack([rooms, np.ones(rooms.size)]).T
param1

array([[6.575, 1.   ],
       [6.421, 1.   ],
       [7.185, 1.   ],
       ...,
       [6.976, 1.   ],
       [6.794, 1.   ],
       [6.03 , 1.   ]])

# lstsq函数的第二个参数
param2 = prices
param2

array([24. , 21.6, 34.7, 33.4, 36.2, 28.7, 22.9, 27.1, 16.5, 18.9, 15. ,
       18.9, 21.7, 20.4, 18.2, 19.9, 23.1, 17.5, 20.2, 18.2, 13.6, 19.6,
       15.2, 14.5, 15.6, 13.9, 16.6, 14.8, 18.4, 21. , 12.7, 14.5, 13.2,
       13.1, 13.5, 18.9, 20. , 21. , 24.7, 30.8, 34.9, 26.6, 25.3, 24.7,
       21.2, 19.3, 20. , 16.6, 14.4, 19.4, 19.7, 20.5, 25. , 23.4, 18.9,
       35.4, 24.7, 31.6, 23.3, 19.6, 18.7, 16. , 22.2, 25. , 33. , 23.5,
       19.4, 22. , 17.4, 20.9, 24.2, 21.7, 22.8, 23.4, 24.1, 21.4, 20. ,
       20.8, 21.2, 20.3, 28. , 23.9, 24.8, 22.9, 23.9, 26.6, 22.5, 22.2,
       23.6, 28.7, 22.6, 22. , 22.9, 25. , 20.6, 28.4, 21.4, 38.7, 43.8,
       33.2, 27.5, 26.5, 18.6, 19.3, 20.1, 19.5, 19.5, 20.4, 19.8, 19.4,
       21.7, 22.8, 18.8, 18.7, 18.5, 18.3, 21.2, 19.2, 20.4, 19.3, 22. ,
       20.3, 20.5, 17.3, 18.8, 21.4, 15.7, 16.2, 18. , 14.3, 19.2, 19.6,
       23. , 18.4, 15.6, 18.1, 17.4, 17.1, 13.3, 17.8, 14. , 14.4, 13.4,
       15.6, 11.8, 13.8, 15.6, 14.6, 17.8, 15.4, 21.5, 19.6, 15.3, 19.4,
       17. , 15.6, 13.1, 41.3, 24.3, 23.3, 27. , 50. , 50. , 50. , 22.7,
       25. , 50. , 23.8, 23.8, 22.3, 17.4, 19.1, 23.1, 23.6, 22.6, 29.4,
       23.2, 24.6, 29.9, 37.2, 39.8, 36.2, 37.9, 32.5, 26.4, 29.6, 50. ,
       32. , 29.8, 34.9, 37. , 30.5, 36.4, 31.1, 29.1, 50. , 33.3, 30.3,
       34.6, 34.9, 32.9, 24.1, 42.3, 48.5, 50. , 22.6, 24.4, 22.5, 24.4,
       20. , 21.7, 19.3, 22.4, 28.1, 23.7, 25. , 23.3, 28.7, 21.5, 23. ,
       26.7, 21.7, 27.5, 30.1, 44.8, 50. , 37.6, 31.6, 46.7, 31.5, 24.3,
       31.7, 41.7, 48.3, 29. , 24. , 25.1, 31.5, 23.7, 23.3, 22. , 20.1,
       22.2, 23.7, 17.6, 18.5, 24.3, 20.5, 24.5, 26.2, 24.4, 24.8, 29.6,
       42.8, 21.9, 20.9, 44. , 50. , 36. , 30.1, 33.8, 43.1, 48.8, 31. ,
       36.5, 22.8, 30.7, 50. , 43.5, 20.7, 21.1, 25.2, 24.4, 35.2, 32.4,
       32. , 33.2, 33.1, 29.1, 35.1, 45.4, 35.4, 46. , 50. , 32.2, 22. ,
       20.1, 23.2, 22.3, 24.8, 28.5, 37.3, 27.9, 23.9, 21.7, 28.6, 27.1,
       20.3, 22.5, 29. , 24.8, 22. , 26.4, 33.1, 36.1, 28.4, 33.4, 28.2,
       22.8, 20.3, 16.1, 22.1, 19.4, 21.6, 23.8, 16.2, 17.8, 19.8, 23.1,
       21. , 23.8, 23.1, 20.4, 18.5, 25. , 24.6, 23. , 22.2, 19.3, 22.6,
       19.8, 17.1, 19.4, 22.2, 20.7, 21.1, 19.5, 18.5, 20.6, 19. , 18.7,
       32.7, 16.5, 23.9, 31.2, 17.5, 17.2, 23.1, 24.5, 26.6, 22.9, 24.1,
       18.6, 30.1, 18.2, 20.6, 17.8, 21.7, 22.7, 22.6, 25. , 19.9, 20.8,
       16.8, 21.9, 27.5, 21.9, 23.1, 50. , 50. , 50. , 50. , 50. , 13.8,
       13.8, 15. , 13.9, 13.3, 13.1, 10.2, 10.4, 10.9, 11.3, 12.3,  8.8,
        7.2, 10.5,  7.4, 10.2, 11.5, 15.1, 23.2,  9.7, 13.8, 12.7, 13.1,
       12.5,  8.5,  5. ,  6.3,  5.6,  7.2, 12.1,  8.3,  8.5,  5. , 11.9,
       27.9, 17.2, 27.5, 15. , 17.2, 17.9, 16.3,  7. ,  7.2,  7.5, 10.4,
        8.8,  8.4, 16.7, 14.2, 20.8, 13.4, 11.7,  8.3, 10.2, 10.9, 11. ,
        9.5, 14.5, 14.1, 16.1, 14.3, 11.7, 13.4,  9.6,  8.7,  8.4, 12.8,
       10.5, 17.1, 18.4, 15.4, 10.8, 11.8, 14.9, 12.6, 14.1, 13. , 13.4,
       15.2, 16.1, 17.8, 14.9, 14.1, 12.7, 13.5, 14.9, 20. , 16.4, 17.7,
       19.5, 20.2, 21.4, 19.9, 19. , 19.1, 19.1, 20.1, 19.9, 19.6, 23.2,
       29.8, 13.8, 13.3, 16.7, 12. , 14.6, 21.4, 23. , 23.7, 25. , 21.8,
       20.6, 21.2, 19.1, 20.6, 15.2,  7. ,  8.1, 13.6, 20.1, 21.8, 24.5,
       23.1, 19.7, 18.3, 21.2, 17.5, 16.8, 22.4, 20.6, 23.9, 22. , 11.9])

lstsq函数返回值说明

lstsq函数返回的是一个四元组，四元组中的第一个元素就是要求解的方程的系数，四元组中的第二个元素是误差平方和。

# rcond参数直接设置为None（暂不解释）
result = np.linalg.lstsq(param1, param2, rcond=None)
result

(array([  9.10210898, -34.67062078]),
 array([22061.87919621]),
 2,
 array([143.99484122,   2.46656609]))

a, b = result[0]
mse = result[1][0] / rooms.size
print(a, b)
print(mse)

9.102108981180313 -34.67062077643857
43.600551771169584

# 比较两条拟合曲线
plt.scatter(rooms, prices)
# 梯度下降法给出的a和b预测出的房价
plt.plot(rooms, y_hat2, color='red', linewidth=4)
# lstsq函数给出的a和b预测出的房价
y_hat3 = a * rooms + b
plt.plot(rooms, y_hat3, color='green', linewidth=4)
plt.show()

​

​