西瓜書《機器學習》課後答案——Chapter3_3.5
阿新 • • 發佈:2019-01-03
程式設計實現線性判別分析,並給出西瓜資料集3.0alpha上的結果。
"""
Author: Victoria
Created on: 2017.9.15 11:45
"""
import xlrd
import numpy as np
import matplotlib.pyplot as plt
def LDA(X0, X1):
"""
Get the optimal params of LDA model given training data.
Input:
X0: np.array with shape [N1, d]
X1: np.array with shape [N2, d]
Return:
omega: np.array with shape [1, d]. Optimal params of LDA.
"""
#shape [1, d]
mean0 = np.mean(X0, axis=0, keepdims=True)
mean1 = np.mean(X1, axis=0, keepdims=True)
Sw = (X0-mean0).T.dot(X0-mean0) + (X1-mean1).T.dot(X1-mean1)
omega = np.linalg.inv(Sw).dot((mean0-mean1).T)
return omega
if __name__=="__main__":
#read data from xls
work_book = xlrd.open_workbook("3.0alpha.xlsx" )
sheet = work_book.sheet_by_name("Sheet1")
x1 = sheet.row_values(0)
x2 = sheet.row_values(1)
p_x1 = x1[0:8]
p_x2 = x2[0:8]
n_x1 = x1[8:]
n_x2 = x2[8:]
X0 = np.vstack([n_x1, n_x2]).T
X1 = np.vstack([p_x1, p_x2]).T
print X0
#LDA
omega = LDA(X0, X1)
#plot
plt.plot(p_x1, p_x2, "bo")
plt.plot(n_x1, n_x2, "r+")
lda_left = 0
lda_right = -(omega[0]*0.9) / omega[1]
plt.plot([0, 0.9], [lda_left, lda_right], 'g-')
plt.xlabel('density')
plt.ylabel('sugar rate')
plt.title("LDA")
plt.show()