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，但我还没有找到答案是否可行.

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.

要了解我正在尝试做什么，请查看以下示例(这对我不起作用).在这里，我试图将移位绝对值函数 (f(x) = |x-p|) 拟合到以 p 作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

有什么办法可以在 Python 中实现这一点吗?

Is there any way of accomplishing this in Python?

在 R 中这样做的一种方法是:

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')

经过更多的研究，我找到了一种方法.在这个解决方案中，我不喜欢我必须自己定义错误函数的事实.此外，我不确定为什么它必须采用这种 lambda 风格.因此，非常欢迎任何类型的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

为了到此为止，我将分享我自己对问题的最终解决方案.为了接近我原来的问题，您只需要自己定义矢量化函数，而不是使用 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]]

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