Question

您好我正试图弄清楚如何使用分段线性函数拟合这些值。我已经读过这个问题，但我无法前进（How to apply piecewise linear fit in Python?）。在这个例子中展示了如何为2段案例实现分段函数。但我需要在如图所示的三段情况下进行。

我写了这段代码：

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


x1 = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ,11, 12, 13, 14, 15,16,17,18,19,20,21], dtype=float)
y1 = np.array([5, 7, 9, 11, 13, 15, 28.92, 42.81, 56.7, 70.59, 84.47, 98.36, 112.25, 126.14, 140.03,145,147,149,151,153,155])



x = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ,11, 12, 13, 14, 15], dtype=float)
y = np.array([5, 7, 9, 11, 13, 15, 28.92, 42.81, 56.7, 70.59, 84.47, 98.36, 112.25, 126.14, 140.03])

def piecewise(x,x0,x1,y0,y1,k0,k1,k2):
    return np.piecewise(x , [x <= x0, (x>= x1)] , [lambda x:k0*x + y0-k0*x0, lambda x:k1*(x-(x1+x0))-y1, lambda x:k2*x + y1-k2*x1])

p , e = optimize.curve_fit(piecewise_linear, x1, y1)
xd = np.linspace(0, 15, 100)
plt.figure()
plt.plot(x1, y1, "o")
plt.plot(xd, piecewise_linear(xd, *p))

但这是输出

有什么建议吗？我相信问题出在return np.piecewise(x , [x <= x0, (x>= x1)] , [lambda x:k0*x + y0-k0*x0, lambda x:k1*(x-(x1+x0))-y1, lambda x:k2*x + y1-k2*x1])，尤其是第二个lambda。

编辑1：

如果我尝试不同的数据，那么A.L.提供的解决方案我得不到好的结果。

我得到了这个结果：

与

x=[ 16.01690476,  16.13801587,  14.63628571,  15.32664399,
        15.8145    ,  15.71507143,  15.56107143,  15.553     ,
        15.08734524,  14.97275   ,  15.51958333,  16.61981859,
        16.36589286,  14.78708333,  14.41565476,  13.47763158,
        13.42412281,  12.95551378,  13.66601504,  13.63315789,
        13.21463659,  13.53464286,  14.60130952,  14.7774881 ,
        13.04319048,  12.53385965,  12.65745614,  13.90535714,
        14.82412281,  14.6565    ,  15.09541667,  13.41434524,
        13.66033333,  14.57964286,  13.55416667,  13.43041667,
        13.01137566,  12.76429825,  11.55241667,  11.0634881 ,
        10.92729762,  11.21625   ,  10.72092857,  11.80380952,
        12.55233333,  12.11307143,  11.78892857,  12.45458333,
        11.05539286,  10.69214286,  10.32566667,  11.3439881 ,
         9.69563492,  10.72535714,  10.26180272,   7.77272727,
         6.37704082,   8.49666667,   8.5389881 ,   5.68547619,
         7.00616667,   8.22015873,  10.20315476,  15.35736842,
        12.25158333,  11.09622153,  10.4118254 ,   9.8602381 ,
        10.16727273,  15.10858333,  13.82215539,  12.44719298,
        10.92341667,  11.44565476,  11.43333333,  10.5045    ,
        11.14357143,  10.37625   ,   8.93421769,   9.48444444,
        10.43483333,  10.8659881 ,  10.96166667,  10.12872619,
         9.64663265,   9.29979762,   9.67173469,   8.978322  ,
         9.10419501,   9.45411565,  10.46411565,   7.95739229,
         8.72616667,   7.03892857,   7.32547619,   7.56441667,
         6.61022676,   9.09014739,  10.78141667,  10.85918367,
        11.11665476,  10.141     ,   9.17760771,   8.27968254,
        11.02625   ,  12.34809524,  11.17807018,  11.25416667,
        11.29236905,   9.28357143,   9.77033333,  11.52086168,
         9.8625    ,  12.60281955,  12.42785714,  12.11902256,
        13.1       ,  13.02791667,  13.87779449,  15.09857143,
        13.93935185,  13.69821429,  13.39880952,  12.45692982,
        12.76921053,  13.23708333,  13.71666667,  15.39807143,
        15.27916667,  14.66464286,  13.38694444,  10.97555556,
        10.02191667,  11.99608333,  14.26325   ,  15.40991667,
        15.12908333,  15.76265476,  12.12763158,  15.01641667,
        14.39602381,  12.98532143,  14.98807018,  18.30547619,
        16.7564966 ,  16.82982143,  19.8487013 ,  19.18600907]

和

y=[ 2.36846863,  2.73722628,  2.77177583,  2.63930636,  2.80864749,
        2.57066667,  2.65277287,  2.57162347,  2.76295667,  2.79835391,
        2.60431154,  2.17326401,  2.67740698,  2.47138153,  2.49882574,
        2.60987338,  2.69935565,  2.60755362,  2.77702029,  2.62996942,
        2.45959517,  2.52750434,  2.73833005,  2.52009   ,  2.80933226,
        1.63807085,  2.49230099,  2.55441614,  3.19256506,  2.52609288,
        1.02931596,  2.40266963,  2.3306463 ,  2.69094276,  2.60779985,
        2.48351648,  2.45131766,  2.40526763,  2.03952569,  1.86217009,
        1.79971848,  1.91772218,  1.85895421,  2.32725731,  2.28189713,
        2.11835833,  2.09636517,  2.2230303 ,  1.85863317,  1.77550406,
        1.68862391,  1.79187765,  1.70887476,  1.81911193,  1.74802483,
        1.65776432,  1.58012849,  1.67781494,  1.62451541,  1.60555884,
        1.56172214,  1.60083809,  1.65256994,  2.74794704,  2.27089627,
        1.80364982,  1.51412482,  1.77738757,  1.56979564,  2.46538633,
        2.37679625,  2.40389294,  2.04165763,  1.82086407,  1.90609219,
        1.87480978,  1.8877854 ,  1.76080074,  1.68369028,  1.57419297,
        1.66470126,  1.74522552,  1.72459756,  1.65510503,  1.72131148,
        1.6254417 ,  1.57091907,  1.68755268,  1.70307911,  1.59445121,
        1.74393783,  1.72913779,  1.66883237,  1.59859545,  1.62335831,
        1.73378184,  1.62621588,  1.79532164,  1.78289992,  1.79475101,
        1.7826266 ,  1.68778918,  1.64484127,  1.62332696,  1.75372393,
        1.99038021,  1.87268137,  1.86124502,  1.82435911,  1.62927102,
        1.66443723,  1.86743516,  1.62745098,  2.20200312,  2.09641026,
        2.26649111,  2.63271605,  2.18050721,  2.57138433,  2.51833359,
        2.74684184,  2.57209998,  2.63762019,  2.30027877,  2.28471286,
        2.40323668,  2.37103313,  2.16414489,  1.01027109,  2.64181007,
        2.45467765,  2.05773672,  1.73624917,  2.05233688,  2.70820669,
        2.65594222,  2.67445635,  2.37212985,  2.48221803,  2.77655216,
        2.62839879,  2.26481307,  2.58005799,  2.1188172 ,  2.14017268,
        2.16459571,  1.95083406,  1.46224418]

Answer 1

拟合分段线性函数是一个非线性优化问题，可能具有局部优化。您看到的结果可能是优化算法陷入困境的局部优化之一。

解决此问题的一种方法是使用不同的初始值重复您的优化算法并采用最佳拟合。我使用平均绝对误差（MAE）来比较彼此的不同拟合。

perr = np.sum(np.abs(y1-piecewise(x1, *p)))

我也改变了你的分段功能，因为它对我来说有点混乱。但它仍然是以前的分段功能

进一步认为你忘了将x和xd数组扩展为21的值。（这就是绿线早期结束的原因）。

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


def piecewise(x,x0,x1,y0,y1,k0,k1,k2):
    return np.piecewise(x , [x <= x0, np.logical_and(x0<x, x<= x1),x>x1] , [lambda x:k0*x + y0, lambda x:k1*(x-x0)+y1+k0*x0,
                                                                            lambda x:k2*(x-x1) + y0+y1+k0*x0+k1*(x1-x0)])

x1 = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ,11, 12, 13, 14, 15,16,17,18,19,20,21], dtype=float)
y1 = np.array([5, 7, 9, 11, 13, 15, 28.92, 42.81, 56.7, 70.59, 84.47, 98.36, 112.25, 126.14, 140.03,145,147,149,151,153,155])
x = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ,11, 12, 13, 14, 15,16,17,18,19,20,21], dtype=float)
y = np.array([5, 7, 9, 11, 13, 15, 28.92, 42.81, 56.7, 70.59, 84.47, 98.36, 112.25, 126.14, 140.03,145,147,149,151,153,155])


perr_min = np.inf
p_best = None
for n in range(100):
    k = np.random.rand(7)*20
    p , e = optimize.curve_fit(piecewise, x1, y1,p0=k)
    perr = np.sum(np.abs(y1-piecewise(x1, *p)))
    if(perr < perr_min):
        perr_min = perr
        p_best = p

xd = np.linspace(0, 21, 100)
plt.figure()
plt.plot(x1, y1, "o")
y_out = piecewise(xd, *p_best)
plt.plot(xd, y_out)
plt.show()

这给了我：

，p = [6.34259491 15.00000023 2.97272604 7.05498314 2.00751828 13.88881542 1.99960597]

<强> EDIT1

您编辑了自己的问题，这是编辑过的问题的答案。对不起Iam new at stackoverlfow并不确定我是否应该发布另一个答案

在第二个数据集中，您为数据添加了噪音。在我看来，有两种噪音。一种高斯分布，它将点放置在靠近下面的分段线和异常噪声的位置，这些噪声使点远离原始的底线。

根据p，您使用的优化算法根据p优化以下内容： E =总和（平方（y-分段（x，p））） http://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.curve_fit.html#scipy.optimize.curve_fit

高斯噪声不是很成问题。您使用的优化间接假设这种高斯噪声（通过最小化最小平方误差）并尽可能地拟合线。真正的问题在于异常值。

问题在于离群值远离原始函数。即使优化尝试最佳参数，能量函数E也不会是最小的，因为你的异常值远离原始函数，并且这个距离是偶数平方所以它将函数E的最小值偏离真实参数你的功能。

那么解决方案是什么？摆脱异常值。

自动化的方法是ransac https://en.wikipedia.org/wiki/RANSAC

简介：您选择原始数据的随机子集。你希望子集没有异常值。您将函数与子集相匹配并丢弃与拟合函数相差远的点。如果在此步骤中存活了足够的分数，则可以获取所有幸存点并重复拟合。这个错误＆＃34; inlier＆＃34; set是衡量你的健康质量的标准。然后你重复整个过程并采取最佳的最终拟合。

我相应地调整了我的脚本：

from scipy import optimize
import matplotlib.pyplot as plt
import numpy as np
def piecewise(x,x0,x1,y0,y1,k0,k1,k2):
    return np.piecewise(x , [x <= x0, np.logical_and(x0<x, x<= x1),x>x1] , [lambda x:k0*x + y0, lambda x:k1*(x-x0)+y1+k0*x0,
                                                                            lambda x:k2*(x-x1) + y0+y1+k0*x0+k1*(x1-x0)])

x = np.array(x)
y = np.array(y)

x1 = x
y1 = y



perr_min = np.inf
p_best = None
for n in range(100):
    idx = np.random.choice(np.arange(len(x)), 10, replace=False)
    x_sample = x[idx]
    y_sample = y[idx]
    k = np.random.rand(7)*20
    try:
        p , e = optimize.curve_fit(piecewise, x_sample,y_sample ,p0=k)
        each_error = np.abs(y-piecewise(x, *p))
        x_inliner = x[each_error < 1]
        y_inlier = y[each_error < 1]
        if(x_inliner.shape[0] < 0.8 * x.shape[0]):
            continue

        p_inlier , e_inlier = optimize.curve_fit(piecewise, x_inliner,y_inlier ,p0=p)
        perr = np.sum(np.abs(y-piecewise(x, *p_inlier)))


        if(perr < perr_min):
            perr_min = perr
            p_best = p_inlier
    except RuntimeError:
        pass

xd = np.linspace(0, 21, 100)
plt.figure()
plt.plot(x, y, "o")
y_out = piecewise(xd, *p_best)
plt.plot(xd, y_out)
print p_best
plt.show()

重复100次后，我得到以下结果：

分段回归Python

1 个答案: