Created
August 15, 2017 16:43
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Script 15min en clase para obtener la regresión con una linea, usa la lógica de multiplicar matrices A-1*y = thetas
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#Regresion lineal, solo jala con linea, implementar el solicitar un grado de polinomio | |
import matplotlib.pyplot as plt | |
import numpy as np | |
x = [ 500,1300,2000,3800,4000,5500,6800,7400,10000] | |
y = [ 1000,2000,3000,4500,5000,5100,5300,5400,6000 ] | |
m = len(x) | |
Ex = sum(i for i in x) | |
ExCuad = sum(i*i for i in x) | |
Ey = sum(i for i in y) | |
Exy = 0 | |
for i in range(0,len(x)): | |
Exy += x[i]*y[i] | |
A = np.array([[m,Ex], | |
[Ex,ExCuad]]) | |
Y = np.array([Ey,Exy]) | |
Ainv = np.linalg.inv(A) | |
thetas = Ainv.dot(Y) | |
fx = lambda _x: thetas[0] + thetas[1]*_x | |
# Plot the data | |
plt.scatter(x,y, label='Valores Tamano/Casa') | |
plt.plot(x, [ fx(i) for i in x], label='Polinomio') | |
# Add a legend | |
plt.legend() | |
# Show the plot | |
plt.show() | |
#Otra solucion: | |
#Transpuesta del vector de datos X multiplicada por datos X | |
#Sacamos inversa a ese resultado | |
#Multiplicamos por X transpuesta | |
#Multiplicamos Matriz Y | |
#Metodo del gradiente descendente | |
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