python实现二维函数高次拟合

在参加“数据挖掘”比赛中遇到了关于函数高次拟合的问题,然后就整理了一下源码,以便后期的学习与改进。在本次“数据挖掘”比赛中感觉收获最大的还是对于神经网络的认识,在接近一周的时间里,研究了进40种神经网络模型,虽然在持续一周的挖掘比赛把自己折磨的惨不忍睹,但是收获颇丰。现在想想也挺欣慰自己在这段时间里接受新知识的能力。关于神经网络方面的理解会在后续博文中补充(刚提交完论文,还没来得及整理),先分享一下高次拟合方面的知识。

# coding=utf-8 import matplotlib.pyplot as plt import numpy as np import scipy as sp import csv from scipy.stats import norm from sklearn.pipeline import Pipeline from sklearn.linear_model import LinearRegression from sklearn.preprocessing import PolynomialFeatures from sklearn import linear_model''''' 数据导入 ''' def loadDataSet(fileName): dataMat = [] labelMat = [] csvfile = file(fileName, 'rb') reader = csv.reader(csvfile) b = 0 for line in reader: if line[50] is '': b += 1 else: dataMat.append(float(line[41])/100*20+30) labelMat.append(float(line[25])*100)csvfile.close() print "absence time number: %d" % b return dataMat,labelMatxArr,yArr = loadDataSet('data.csv') x = np.array(xArr) y = np.array(yArr) # x = np.arange(0, 1, 0.002) # y = norm.rvs(0, size=500, scale=0.1) # y = y + x ** 2def rmse(y_test, y): return sp.sqrt(sp.mean((y_test - y) ** 2))def R2(y_test, y_true): return 1 - ((y_test - y_true) ** 2).sum() / ((y_true - y_true.mean()) ** 2).sum()def R22(y_test, y_true): y_mean = np.array(y_true) y_mean[:] = y_mean.mean() return 1 - rmse(y_test, y_true) / rmse(y_mean, y_true)plt.scatter(x, y, s=5) #分别进行1,2,3,6次拟合 degree = [1, 2,3, 6] y_test = [] y_test = np.array(y_test)for d in degree: #普通 # clf = Pipeline([('poly', PolynomialFeatures(degree=d)), #('linear', LinearRegression(fit_intercept=False))]) # clf.fit(x[:, np.newaxis], y)# 岭回归 clf = Pipeline([('poly', PolynomialFeatures(degree=d)), ('linear', linear_model.Ridge())]) clf.fit(x[:, np.newaxis], y) y_test = clf.predict(x[:, np.newaxis])print('多项式参数%s' %clf.named_steps['linear'].coef_) print('rmse=%.2f, R2=%.2f, R22=%.2f, clf.score=%.2f' % (rmse(y_test, y), R2(y_test, y), R22(y_test, y), clf.score(x[:, np.newaxis], y)))plt.plot(x, y_test, linewidth=2)plt.grid() plt.legend(['1', '2','3', '6'], loc='upper left') plt.show()

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