Aula 15: Regressão Linear
| Idade | Altura |
|---|---|
| 18 | 76.1 |
| 19 | 77 |
| 20 | 78.1 |
| 24 | 79.9 |
| 25 | 81.1 |
| 27 | 81.8 |
| 28 | 82.8 |
| 29 | 83.5 |
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
x = np.array([18,19,20,24,25,27,28,29]).reshape(-1, 1)
y = np.array([76.1, 77, 78.1, 79.9, 81.1, 81.8, 82.8, 83.5]).reshape(-1, 1)
model = LinearRegression()
model.fit(x, y)
plt.scatter(x, y)
plt.plot(x,model.predict(x), color="red")
| Idade | Altura |
|---|---|
| 18 | 76.1 |
| 19 | 77 |
| 20 | 78.1 |
| 24 | 79.9 |
| 25 | 81.1 |
| 27 | 81.8 |
| 28 | 82.8 |
| 29 | 83.5 |
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
x = np.array([18,19,20,24,25,27,28,29]).reshape(-1, 1)
y = np.array([76.1, 77, 78.1, 79.9, 81.1, 81.8, 82.8, 83.5]).reshape(-1, 1)
model = LinearRegression()
model.fit(x, y)
plt.scatter(x, y)
plt.plot(x,model.predict(x), color="red")
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
x = np.array([18,19,20,24,25,27,28,29]).reshape(-1, 1)
y = np.array([76.1, 77, 78.1, 79.9, 81.1, 81.8, 82.8, 83.5]).reshape(-1, 1)
model = LinearRegression()
model.fit(x, y)
plt.scatter(x, y)
plt.plot(x,model.predict(x), color="red")
Reta \(r\colon y = \beta_1 x + \beta_0\)
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
x = np.array([18,19,20,24,25,27,28,29]).reshape(-1, 1)
y = np.array([76.1, 77, 78.1, 79.9, 81.1, 81.8, 82.8, 83.5]).reshape(-1, 1)
model = LinearRegression()
model.fit(x, y)
plt.scatter(x, y)
predicted=model.predict(x)
plt.plot(x,predicted, color="red")
plt.ylim=range(76,83)
plt.annotate("$y_2$", (x[2],y[2]), xytext=(x[2],y[2]+.5))
plt.annotate("$\hat{y}_2$", (x[2],predicted[2]), xytext=(x[2],predicted[2]-.5) )
plt.annotate("$e_2$", (x[2],predicted[2]+.2), xytext=(x[2]-.8,predicted[2]+.2), arrowprops=dict(arrowstyle='->', color='blue', linewidth=2))
for i in range(8):
plt.vlines(x[i], y[i], predicted[i])
plt.show()
Reta \(r\colon y = \beta_1 x + \beta_0\)
Resíduos:
\(e_1 = y_1 - \hat{y}_1\) \(e_2 = y_2 - \hat{y}_2\)
\[\vdots\]
\(e_n = y_n - \hat{y}_n\)
\(e_n = y_n - (\beta_1 x_n - \beta_0)\)
import numpy as np
import itertools
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
x = np.array([5,7,8,13,3,17,2,10,4,11,12,9,6]).reshape((-1,1))
y = np.array([99,86,87,88,111,86,103,92,94,78,77,85,86]).reshape((-1,1))
model = LinearRegression()
model.fit(x, y)
plt.scatter(x, y)
predicted=model.predict(x)
plt.plot(x,predicted, color="red")
plt.plot(x,np.repeat(94, 13), color="green")
for i in range(13):
plt.vlines(x[i], y[i], predicted[i], color="red")
for i in range(13):
plt.vlines(x[i], y[i], 94, color="green")
plt.show()
Note
Nosso objetivo é escolher a reta que minimiza os resíduos
\[E = \sum_{i=1}^n (y_i -\beta_1x_0 - \beta_0)^2\]
import numpy as np
import itertools
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
x = np.array([5,7,8,13,3,17,2,10,4,11,12,9,6]).reshape((-1,1))
y = np.array([99,86,87,88,111,86,103,92,94,78,77,85,86]).reshape((-1,1))
model = LinearRegression()
model.fit(x, y)
plt.scatter(x, y)
predicted=model.predict(x)
plt.plot(x,predicted, color="red")
for i in range(13):
plt.vlines(x[i], y[i], predicted[i], color="red")
plt.show()
\[A \underbrace{x}_{\text{coeficientes}} = b + \underbrace{e}_{\text{resíduos}}\]
Ou seja, o erro é dado por \[e = Ax+b\]
Consideremos \(\hat{x}\) como a melhor solução possível, ou seja \[e = A\hat{x} - b\] é o menor erro possível
Perceba que \(A\hat{x}=p\), onde \(p\) é um vetor formado pela combinação linear das colunas de \(A\), ou seja, \(p\in C(A)\) 
\(Ax=p\) será perpendicular ao vetor \(e\). Sendo assim, \(p^\top e = 0\). Como \(e = A\hat{x}-b\) temos
\[\begin{align} (A\hat{x})^\top (A\hat{x}-b) &= 0 \\ x^\top \underbrace{A^\top (A\hat{x} - b)}_{0} &= 0 \end{align}\]
\[ A^\top A \hat{x} - A^\top = 0\] ou
\[ A^\top A \hat{x} = A^\top b\]
\[ A^\top A \hat{x} = A^\top b\]
Sejam
\[A = \begin{bmatrix}1 & x_1\\ \vdots & \vdots\\ 1 & x_n\end{bmatrix} \]
e \[ b = \begin{bmatrix}y_1 \\ \vdots\\ y_n\end{bmatrix}\]
Vamos analisar a matriz \(A^\top A\)
\[\begin{align} A^\top A &= \begin{bmatrix} 1 & 1 & \ldots & 1\\ x_1& x_2& \ldots & x_n \end{bmatrix} \begin{bmatrix} 1 & x_1\\ 1 & x_2 \\ \vdots & \vdots \\ 1 & x_n \end{bmatrix}\\ &= \begin{bmatrix} \sum_{i=1}^n 1 & \sum_{i=1}^n x_i \\ \sum_{i=1}^n x_i & \sum_{i=1}^n x_i^2 \end{bmatrix}\\ &=\begin{bmatrix} n & \sum_{i=1}^n x_i \\ \sum_{i=1}^n x_i & \sum_{i=1}^n x_i^2 \end{bmatrix} \end{align}\]
Agora vamos analisar a matriz \(A^\top b\) \[\begin{align} A^\top b &= \begin{bmatrix} 1 & 1 & \ldots & 1\\ x_1& x_2& \ldots & x_n \end{bmatrix} \begin{bmatrix} y_1 \\ \vdots\\ y_n \end{bmatrix}\\ &= \begin{bmatrix} \sum_{i=1}^n y_i \\ \sum_{i=1}^n x_iy_i \end{bmatrix} \end{align}\]
Como solucionar \(A^\top A \hat{x} = A^\top b\) para \(\hat{x}\)?
Método 1:
\[\begin{align} A^\top A\hat{x} &= A^\top b \\ (A^\top A)^{-1}A^\top A\hat{x} &= (A^\top A)^{-1}A^\top b\\ \hat{x} &= (A^\top A)^{-1}A^\top b \end{align}\]
A inversa de uma matrix \(2\times 2\) pode ser denotada pelo produto do recíproco do determinante da matrix pela adjunta da matrix, ou seja
\[ \begin{bmatrix} a & b\\ c & d \end{bmatrix}^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \]
Sendo assim, \[ (A^\top A)^{-1} = \frac{1}{n\sum x_i^2 - \left(\sum x_i\right)^2} \begin{bmatrix} \sum x_i^2 & -\sum x_i \\ - \sum x_i & n \end{bmatrix} \]
Como \(\hat{x} = (A^\top A)^{-1} A^\top b\), temos
\[\begin{align} \hat{x} &= \frac{1}{n\sum x_i^2 - \left(\sum x_i\right)^2} \begin{bmatrix} \sum x_i^2 & -\sum x_i \\ - \sum x_i & n \end{bmatrix} \begin{bmatrix} \sum y_i \\ \sum x_iy_i \end{bmatrix} \\ &= \frac{1}{n\sum x_i^2 - \left(\sum x_i\right)^2} \begin{bmatrix} \sum x_i^2 \sum y_i - \sum x_i \sum x_i y_i\\ n\sum x_i y_i - \sum x_i \sum y_i \end{bmatrix} \end{align}\]
\[ \beta_1 = \frac{n \sum x_iy_i - \sum x_i \sum y_i}{n\sum x_i^2 - (\sum x_i)^2}\]
\[ \beta_0 = \frac{\sum y_i - \beta_1 \sum x_i}{n}\]
Método 2: \[ A^\top A \hat{x} = A^\top b\]
\[ \begin{bmatrix} n & \sum x_i \\ \sum x_i & \sum x_i^2 \end{bmatrix} \begin{bmatrix} \beta_0\\ \beta_1 \end{bmatrix}= \begin{bmatrix} \sum y_i \\ \sum x_iy_i \end{bmatrix} \]
Fazendo eliminação de Gauss temos
\[ \begin{bmatrix} n & \sum x_i \\ 0 & -\frac{1}{n}(\sum x_i)^2 + \sum x_i^2 \end{bmatrix} \begin{bmatrix} \beta_0\\ \beta_1 \end{bmatrix} = \begin{bmatrix} \sum y_i \\ -\frac{1}{n}\sum x_i \sum y_i + \sum x_iy_i \end{bmatrix} \]
\[ \beta_1 = \frac{n \sum x_iy_i - \sum x_i \sum y_i}{n\sum x_i^2 - (\sum x_i)^2}\]
\[ \beta_0 = \frac{\sum y_i - \beta_1 \sum x_i}{n}\]
As equações acima podem ser reorganizadas da seguinte forma \[\beta_1 = \frac{\sum_{i=1}^{n} ( x_i - \bar{x})(y_i - \bar{y}) }{\sum_{i=1}^{n} ( x_i - \bar{x})^2} \]
\[\beta_0 = \bar{y} - \beta_1 \bar{x} \] ## Avaliando a qualidade do ajuste{.scrollable}
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import StandardScaler
x = np.array([5,7,8,13,3,17,2,10,4,11,12,9,6])
y = np.array([99,86,87,88,111,86,103,92,94,78,77,85,86])
num = x.shape[0]*np.sum(x*y) - np.sum(x)*np.sum(y)
den = x.shape[0]*np.sum(x**2) - np.sum(x)**2
b1 = num/den
b0 = (np.sum(y) - b1*np.sum(x))/x.shape[0]
y_hat = b0 + x*b1
plt.scatter(x, y)
plt.plot(x,y_hat, color="red")
for i in range(13):
plt.vlines(x[i], y[i], predicted[i], color="red")
plt.show()
xy = np.stack((x,y),axis=1)
x_std = StandardScaler().fit_transform(xy)
print('correlacao:', np.corrcoef(x,y))
print('correlacao (manual):', (x_std.T @ x_std)/x_std.shape[0])
correlacao: [[ 1. -0.6973462]
[-0.6973462 1. ]]
correlacao (manual): [[ 1. -0.6973462]
[-0.6973462 1. ]]import numpy as np
import matplotlib.pyplot as plt
from sklearn import datasets
from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import StandardScaler
from sklearn.linear_model import LinearRegression
dataset = datasets.load_diabetes(as_frame=True)
trigliceridios = np.array(dataset['data']['s5']).reshape((-1,1))
diabetes = np.array(dataset['target']).reshape((-1,1))
idade = np.array(dataset['data']['age']).reshape((-1,1))
model = LinearRegression()
model.fit(trigliceridios,diabetes)
model2 = LinearRegression()
model2.fit(idade,diabetes)
r_squared = model.score(trigliceridios, diabetes)
r_squared2 = model2.score(idade, diabetes)
plt.scatter(trigliceridios,diabetes)
plt.plot(trigliceridios,model.predict(trigliceridios),color="red")
plt.show()
print('trigliceridios',r_squared)
print('idade', r_squared2)
plt.scatter(idade,diabetes)
plt.plot(idade,model2.predict(idade),color="red")
trigliceridios 0.3202231084297208
idade 0.03530218264671636
