在Python中拟合分段函数 [英] Fitting piecewise function in Python

问题描述

我正在尝试将分段定义的函数拟合到Python中的数据集.我已经搜索了很长时间，但是无论是否可行，我都没有找到答案.

I'm trying to fit a piecewise defined function to a data set in Python. I've searched for quite a while now, but I haven't found an answer whether it is possible or not.

要了解我要做什么，请看下面的示例(对我而言不起作用).在这里，我尝试将位移的绝对值函数(f(x)= | x-p |)拟合到以p为拟合参数的数据集.

To get an impression of what I am trying to do, look at the following example (which is not working for me). Here I'm trying to fit a shifted absolute value function (f(x) = |x-p|) to a dataset with p as the fit parameter.

import scipy.optimize as so
import numpy as np

def fitfunc(x,p):
   if x>p:
      return x-p
   else:
      return -(x-p)

fitfunc = np.vectorize(fitfunc) #vectorize so you can use func with array

x=np.arange(1,10)
y=fitfunc(x,6)+0.1*np.random.randn(len(x))

popt, pcov = so.curve_fit(fitfunc, x, y) #fitting routine that gives error

在Python中有什么方法可以做到这一点吗?

Is there any way of accomplishing this in Python?

在R中执行此操作的一种方法是:

A way of doing this in R is :

# Fit of a absolute value function f(x)=|x-p|

f.lr <- function(x,p) {
    ifelse(x>p, x-p,-(x-p))
}
x <- seq(0,10)  #
y <- f.lr(x,6) + rnorm (length(x),0,2)
plot(y ~ x)
fit.lr <- nls(y ~ f.lr(x,p), start = list(p = 0), trace = T, control = list(warnOnly = T,minFactor = 1/2048))
summary(fit.lr)
coefficients(fit.lr)
p.fit <- coefficients(fit.lr)["p"]
x_fine <- seq(0,10,length.out=1000)
lines(x_fine,f.lr(x_fine,p.fit),type='l',col='red')
lines(x,f.lr(x,6),type='l',col='blue')

经过更多研究，我找到了一种方法.在这种解决方案中，我不喜欢必须自己定义错误函数这一事实.此外，我不太确定为什么必须采用这种lambda样式.因此，非常欢迎任何建议或更复杂的解决方案.

After even more research I found a way of doing it. In this solution, I don't like the fact that I have to define the error function myself. Further I'm not really sure why it has to be in this lambda-style. Therefore any kind of suggestions or more sophisticated solutions are very welcome.

import scipy.optimize as so
import numpy as np
import matplotlib.pyplot as plt

def fitfunc(p,x): return x - p if x > p else p - x 

def array_fitfunc(p,x):
    y = np.zeros(x.shape)
    for i in range(len(y)):
        y[i]=fitfunc(x[i],p)
    return y

errfunc = lambda p, x, y: array_fitfunc(p, x) - y # Distance to the target function

x=np.arange(1,10)
x_fine=np.arange(1,10,0.1)
y=array_fitfunc(6,x)+1*np.random.randn(len(x)) #data with noise

p1, success = so.leastsq(errfunc, -100, args=(x, y), epsfcn=1.) # -100 is the initial value for p; epsfcn sets the step width

plt.plot(x,y,'o') # fit data
plt.plot(x_fine,array_fitfunc(6,x_fine),'r-') #original function
plt.plot(x_fine,array_fitfunc(p1[0],x_fine),'b-') #fitted version
plt.show()

推荐答案

在这里完成此操作，我将分享我自己的最终解决方案.为了贴近我的原始问题，您只需要自己定义向量化函数即可，而不必使用np.vectorize.

To finish this up here, I'll share my own final solution to the problem. In order to stay close to my original question, you just have to define the vectorized function yourself and not use np.vectorize.

import scipy.optimize as so
import numpy as np

def fitfunc(x,p):
   if x>p:
      return x-p
   else:
      return -(x-p)

fitfunc_vec = np.vectorize(fitfunc) #vectorize so you can use func with array

def fitfunc_vec_self(x,p):
  y = np.zeros(x.shape)
  for i in range(len(y)):
    y[i]=fitfunc(x[i],p)
  return y


x=np.arange(1,10)
y=fitfunc_vec_self(x,6)+0.1*np.random.randn(len(x))

popt, pcov = so.curve_fit(fitfunc_vec_self, x, y) #fitting routine that gives error
print popt
print pcov

输出:

[ 6.03608994]
[[ 0.00124934]]

这篇关于在Python中拟合分段函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！