1. 参考别人写的:
#-*- coding:utf-8 -*-import mathimport matplotlib.pyplot as pltdef linefit(x , y): N = float(len(x)) sx,sy,sxx,syy,sxy=0,0,0,0,0 for i in range(0,int(N)): sx += x[i] sy += y[i] sxx += x[i]*x[i] syy += y[i]*y[i] sxy += x[i]*y[i] a = (sy*sx/N -sxy)/( sx*sx/N -sxx) b = (sy - a*sx)/N r = abs(sy*sx/N-sxy)/math.sqrt((sxx-sx*sx/N)*(syy-sy*sy/N)) return a,b,rif __name__ == '__main__': x=[ 1 ,2 ,3 ,4 ,5 ,6] y=[ 2.5 ,3.51 ,4.45 ,5.52 ,6.47 ,7.51] a,b,r=linefit(x,y) print("X=",x) print("Y=",y) print("拟合结果: y = %10.5f x + %10.5f , r=%10.5f" % (a,b,r) ) plt.plot(x, y, "r:", linewidth=2) plt.grid(True) plt.show()
2. 不用拟合,直接显示一个一元函数
#-*- coding:utf-8 -*-import numpy as npimport matplotlib.pyplot as pltimport mathf = lambda x:5*x+4tx = np.linspace(0,10,50)print txplt.plot(tx, f(tx), "r-", linewidth=2)plt.grid(True)plt.show()