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转自：http://blog.itpub.net/12199764/viewspace-1743145/
项目中有涉及趋势预测的工作，整理一下这3种拟合方法：
1、线性拟合-使用math
import math
def 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,r
if __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) )
#结果为：y = 0.97222 x + 1.59056 , r= 0.98591
1、线性拟合-使用numpy
import numpy as np
X=[ 1 ,2 ,3 ,4 ,5 ,6]
Y=[ 2.5 ,3.51 ,4.45 ,5.52 ,6.47 ,7.51]
z1 = np.polyfit(X, Y, 1) #一次多项式拟合，相当于线性拟合
p1 = np.poly1d(z1)
print z1 #[ 1. 1.49333333]
print p1 # 1 x + 1.493
2、二次多项式拟合
import numpy
def polyfit(x, y, degree):
results = {}
coeffs = numpy.polyfit(x, y, degree)
results[‘polynomial‘] = coeffs.tolist()
# r-squared
p = numpy.poly1d(coeffs)
# fit values, and mean
yhat = p(x) # or [p(z) for z in x]
ybar = numpy.sum(y)/len(y) # or sum(y)/len(y)
ssreg = numpy.sum((yhat-ybar)**2) # or sum([ (yihat - ybar)**2 for yihat in yhat])
sstot = numpy.sum((y - ybar)**2) # or sum([ (yi - ybar)**2 for yi in y])
results[‘determination‘] = ssreg / sstot #准确率
return results
x=[ 1 ,2 ,3 ,4 ,5 ,6]
y=[ 2.5 ,3.51 ,4.45 ,5.52 ,6.47 ,7.2]
z1 = polyfit(x, y, 2)
print z1
3、对数函数拟合-这个是最难的，baidu上都找不到，google了半天才找到的。指数、幂数拟合啥的，都用这个，把func改写一下就行
from scipy import log as log print pcov
import numpy
from scipy import log
from scipy.optimize import curve_fit
def func(x, a, b):
y = a * log(x) + b
return y
def polyfit(x, y, degree):
results = {}
#coeffs = numpy.polyfit(x, y, degree)
popt, pcov = curve_fit(func, x, y)
results[‘polynomial‘] = popt
# r-squared
yhat = func(x ,popt[0] ,popt[1] ) # or [p(z) for z in x]
ybar = numpy.sum(y)/len(y) # or sum(y)/len(y)
ssreg = numpy.sum((yhat-ybar)**2) # or sum([ (yihat - ybar)**2 for yihat in yhat])
sstot = numpy.sum((y - ybar)**2) # or sum([ (yi - ybar)**2 for yi in y])
results[‘determination‘] = ssreg / sstot
return results
x=[ 1 ,2 ,3 ,4 ,5 ,6]
y=[ 2.5 ,3.51 ,4.45 ,5.52 ,6.47 ,7.51]
z1 = polyfit(x, y, 2)
print z1
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原文地址：http://www.cnblogs.com/gslyyq/p/5043847.html
