How to fit a piecewise (alternating linear and constant segments) function to a parabolic function?

Question

I do have a function, for example , but this can be something else as well, like a quadratic or logarithmic function. I am only interested in the domain of . The parameters of the function (a and k in this case) are known as well.

My goal is to fit a continuous piece-wise function to this, which contains alternating segments of linear functions (i.e. sloped straight segments, each with intercept of 0) and constants (i.e. horizontal segments joining the sloped segments together). The first and last segments are both sloped. And the number of segments should be pre-selected between around 9-29 (that is 5-15 linear steps + 4-14 constant plateaus).

Formally

• The input function:

• The fitted piecewise function:

I am looking for the optimal resulting parameters (c,r,b) (in terms of least squares) if the segment numbers (n) are specified beforehand. The resulting constants (c) and the breakpoints (r) should be whole natural numbers, and the slopes (b) round two decimal point values.

I have tried to do the fitting numerically using the pwlf package using a segmented constant models, and further processed the resulting constant model with some graphical intuition to "slice" the constant steps with the slopes. It works to some extent, but I am sure this is suboptimal from both fitting perspective and computational efficiency. It takes multiple minutes to generate a fitting with 8 slopes on the range of 1-50000. I am sure there must be a better way to do this.

My idea would be to instead using only numerical methods/ML, the fact that we have the algebraic form of the input function could be exploited in some way to at least to use algebraic transforms (integrals) to get to a simpler optimization problem.

import numpy as np
import matplotlib.pyplot as plt
import pwlf

# The input function
def input_func(x,k,a):
    return np.power(x,1/a)*k

x = np.arange(1,5e4)
y = input_func(x, 1.8, 1.3)

plt.plot(x,y);

def pw_fit(func, x_r, no_seg, *fparams):
    
    # working on the specified range
    x = np.arange(1,x_r)
    
    y_input = func(x, *fparams)
    
    my_pwlf = pwlf.PiecewiseLinFit(x, y_input, degree=0)
    res = my_pwlf.fit(no_seg)
    yHat = my_pwlf.predict(x)
    
    # Function values at the breakpoints
    y_isec = func(res, *fparams)
    
    # Slope values at the breakpoints
    slopes = np.round(y_isec / res, decimals=2)
    slopes = slopes[1:]
    # For the first slope value, I use the intersection of the first constant plateau and the input function
    slopes = np.insert(slopes,0,np.round(y_input[np.argwhere(np.diff(np.sign(y_input - yHat))).flatten()[0]] / np.argwhere(np.diff(np.sign(y_input - yHat))).flatten()[0], decimals=2))

    plateaus = np.unique(np.round(yHat))
    
    # If due to rounding slope values (to two decimals), there is no change in a subsequent step, I just remove those segments
    to_del = np.argwhere(np.diff(slopes) == 0).flatten()
    slopes = np.delete(slopes,to_del + 1)
    plateaus = np.delete(plateaus,to_del)
    
    breakpoints = [np.ceil(plateaus[0]/slopes[0])]
    for idx, j in enumerate(slopes[1:-1]):
        breakpoints.append(np.floor(plateaus[idx]/j))
        breakpoints.append(np.ceil(plateaus[idx+1]/j))
    breakpoints.append(np.floor(plateaus[-1]/slopes[-1]))

    return slopes, plateaus, breakpoints

slo, plat, breaks = pw_fit(input_func, 50000, 8, 1.8, 1.3)

# The piecewise function itself
def pw_calc(x, slopes, plateaus, breaks):
    x = x.astype('float')
    
    cond_list = [x < breaks[0]]
    
    for idx, j in enumerate(breaks[:-1]):
        cond_list.append((j <= x) & (x < breaks[idx+1]))
        
    cond_list.append(breaks[-1] <= x)
    
    func_list = [lambda x: x * slopes[0]]
    
    for idx, j in enumerate(slopes[1:]):
        func_list.append(plateaus[idx])
        func_list.append(lambda x, j=j: x * j)
          
    return np.piecewise(x, cond_list, func_list)

y_output = pw_calc(x, slo, plat, breaks)

plt.plot(x,y,y_output);

Answer 1

(Not important, but I think the fitted piecewise function is not continuous as it is. Intervals should be x<=r1; r1<x<=r2; ....)

As Anatolyg has pointed out, it looks to me that in the optimal solution (for the function posted at least, and probably for any where the derivative is different from zero), the horizantal segments will collapse to a point or the minimum segment length (in this case 1).

EDIT---------------------------------------------

The behavior above could only be valid if the slopes could have an intercept. If the intercepts are zero, as posted in the question, one consideration must be taken into account: Is the initial parabolic function defined in zero or nearby? Imagine the function y=0.001 *sqrt(x-1000), then the segments defined as b*x will have a slope close to zero and will be so similar to the constant segments that the best fit will be just the line that without intercept that fits better all the function.

Provided that the function is defined in zero or nearby, you can start by approximating the curve just by linear segments (with intercepts):

1. divide the function domain in N intervals(equal intervals or whose size is a function of the average curvature (or second derivative) of the function along the domain).

2. linear fit/regression in each intervals

3. for each interval, if a point (or bunch of points) in the extreme of any interval is better fitted by the line of the neighbor interval than the line of its interval, this point is assigned to the neighbor interval.

4. Repeat from 2) until no extreme points are moved.

Linear regressions might be optimized not to calculate all the covariance matrixes from scratch on each iteration, but just adding the contributions of the moved points to the previous covariance matrixes.

Then each linear segment (LS_i) is replaced by a combination of a small constant segment at the beginning (Cb_i), a linear segment without intercept (S_i), and another constant segment at the end (Ce_i). This segments are easy to calculate as S_i will contain the middle point of LS_i, and Cb_i and Ce_i will have respectively the begin and end values of the segment LS_i. Then the intervals of each segment has to be calculated as an intersection between lines. With this, the constant end segment will be collinear with the constant begin segment from the next interval so they will merge, resulting in a series of constant and linear segments interleaved.

But this would be a floating point start solution. Next, you will have to apply all the roundings which will mess up quite a lot all the segments as the conditions integer intervals and linear segments without slope can be very confronting. In fact, b,c,r are not totally independent. If c_i and r_i+1 are known, then b_i+1 is already fixed

If nothing is broken so far, the final task will be to minimize the error/cost function (I assume that it will be the integral of the error between the parabolic function and the segments). My guess is that gradients here will be quite a pain, as if you change for example one c_i, all the rest of the b_j and c_j will have to adapt as well due to the integer intervals restriction. However, if you can generalize the derivatives between parameters ( how much do I have to adapt b_i+1 if c_i changes a unit), you can propagate the change of one parameter to all other parameters and have kind of a gradient. Then for each interval, you can estimate what would be the ideal parameter and averaging all intervals calculate the best gradient step. Let me illustrate this:

Assuming first that r parameters are fixed, if I change c₁ by one unit, b₂ changes by 0.1, c₂ changes by -0.2 and b₃ changes by 0.2. This would be the gradient.

Then I estimate, comparing with the parabolic curve, that c₁ should increase 0.5 (to reduce the cost by 10 points), b₂ should increase 0.2 (to reduce the cost by 5 points), c₂ should increase 0.2 (to reduce the cost by 6 points) and b₃ should increase 0.1 (to reduce the cost by 9 points).

Finally, the gradient step would be (0.5/1·10 + 0.2/0.1·5 - 0.2/(-0.2)·6 + 0.1/0.2·9)/(10 + 5 + 6 + 9)~= 0.45. Thus, c₁ would increase 0.45 units, b₂ would increase 0.45·0.1, and so on.

When you add the r parameters to the pot, as integer intervals do not have an proper derivative, calculation is not straightforward. However, you can consider r parameters as floating points, calculate and apply the gradient step and then apply the roundings.

Answer 2

We can integrate the squared error function for linear and constant pieces and let SciPy optimize it. Python 3:

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

xl = 1
xh = 50000
a = 1.3
p = 1 / a
n = 8


def split_b_and_c(bc):
    return bc[::2], bc[1::2]


def solve_for_r(b, c):
    r = np.empty(2 * n)
    r[0] = xl
    r[1:-1:2] = c / b[:-1]
    r[2::2] = c / b[1:]
    r[-1] = xh
    return r


def linear_residual_integral(b, x):
    return (
        (x ** (2 * p + 1)) / (2 * p + 1)
        - 2 * b * x ** (p + 2) / (p + 2)
        + b ** 2 * x ** 3 / 3
    )


def constant_residual_integral(c, x):
    return x ** (2 * p + 1) / (2 * p + 1) - 2 * c * x ** (p + 1) / (p + 1) + c ** 2 * x


def squared_error(bc):
    b, c = split_b_and_c(bc)
    r = solve_for_r(b, c)
    linear = np.sum(
        linear_residual_integral(b, r[1::2]) - linear_residual_integral(b, r[::2])
    )
    constant = np.sum(
        constant_residual_integral(c, r[2::2])
        - constant_residual_integral(c, r[1:-1:2])
    )
    return linear + constant


def evaluate(x, b, c, r):
    i = 0
    while x > r[i + 1]:
        i += 1
    return b[i // 2] * x if i % 2 == 0 else c[i // 2]


def main():
    bc0 = (xl + (xh - xl) * np.arange(1, 4 * n - 2, 2) / (4 * n - 2)) ** (
        p - 1 + np.arange(2 * n - 1) % 2
    )
    bc = scipy.optimize.minimize(
        squared_error, bc0, bounds=[(1e-06, None) for i in range(2 * n - 1)]
    ).x
    b, c = split_b_and_c(bc)
    r = solve_for_r(b, c)
    X = np.linspace(xl, xh, 1000)
    Y = [evaluate(x, b, c, r) for x in X]
    plt.plot(X, X ** p)
    plt.plot(X, Y)
    plt.show()


if __name__ == "__main__":
    main()

Answer 3

I have tried to come up with a new solution myself, based on the idea of @Amo Robb, where I have partitioned the domain, and curve fitted a dual - constant and linear - piece together (with the help of np.maximum). I have used the 1 / f(x)' as the function to designate the breakpoints, but I know this is arbitrary and does not provide a global optimum. Maybe there is some optimal function for these breakpoints. But this solution is OK for me, as it might be appropriate to have a better fit at the first segments, at the expense of the error for the later segments. (The task itself is actually a cost based retail margin calculation {supply price -> added margin}, as the retail POS software can only work with such piecewise margin function).
The answer from @David Eisenstat is correct optimal solution if the parameters are allowed to be floats. Unfortunately the POS software can not use floats. It is OK to round up c-s and r-s afterwards. But the b-s should be rounded to two decimals, as those are inputted as percents, and this constraint would ruin the optimal solution with long floats. I will try to further improve my solution with both Amo's and David's valuable input. Thank You for that!

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

# The input function f(x)
def input_func(x,k,a):
    return np.power(x,1/a) * k

# 1 / f(x)'
def one_per_der(x,k,a):
    return a / (k * np.power(x, 1/a-1))

# 1 / f(x)' inverted
def one_per_der_inv(x,k,a):
    return np.power(a / (x*k), a / (1-a))

def segment_fit(start,end,y,first_val):
    b, _ = curve_fit(lambda x,b: np.maximum(first_val, b*x), np.arange(start,end), y[start-1:end-1])
    b = float(np.round(b, decimals=2))
    bp = np.round(first_val / b)
    last_val = np.round(b * end)
    return b, bp, last_val

def pw_fit(end_range, no_seg, **fparams):    

    y_bps = np.linspace(one_per_der(1, **fparams), one_per_der(end_range,**fparams) , no_seg+1)[1:]
    x_bps = np.round(one_per_der_inv(y_bps, **fparams))

    y = input_func(x, **fparams)
    
    slopes = [np.round(float(curve_fit(lambda x,b: x * b, np.arange(1,x_bps[0]), y[:int(x_bps[0])-1])[0]), decimals = 2)]
    plats = [np.round(x_bps[0] * slopes[0])]

    bps = []

    for i, xbp in enumerate(x_bps[1:]):
        b, bp, last_val = segment_fit(int(x_bps[i]+1), int(xbp), y, plats[i])
        slopes.append(b); bps.append(bp); plats.append(last_val)

    breaks = sorted(list(x_bps) + bps)[:-1]
    
    # If due to rounding slope values (to two decimals), there is no change in a subsequent step, I just remove those segments
    to_del = np.argwhere(np.diff(slopes) == 0).flatten()
    breaks_to_del = np.concatenate((to_del * 2, to_del * 2 + 1))

    slopes = np.delete(slopes,to_del + 1)
    plats = np.delete(plats[:-1],to_del)
    breaks = np.delete(breaks,breaks_to_del)
    
    return slopes, plats, breaks

def pw_calc(x, slopes, plateaus, breaks):
    x = x.astype('float')
    
    cond_list = [x < breaks[0]]
    
    for idx, j in enumerate(breaks[:-1]):
        cond_list.append((j <= x) & (x < breaks[idx+1]))
        
    cond_list.append(breaks[-1] <= x)
    
    func_list = [lambda x: x * slopes[0]]
    
    for idx, j in enumerate(slopes[1:]):
        func_list.append(plateaus[idx])
        func_list.append(lambda x, j=j: x * j)
          
    return np.piecewise(x, cond_list, func_list)

fparams = {'k':1.8, 'a':1.2}
end_range = 5e4
no_steps = 10

x = np.arange(1, end_range)
y = input_func(x, **fparams)

slopes, plats, breaks = pw_fit(end_range, no_steps, **fparams)

y_output = pw_calc(x, slopes, plats, breaks)

plt.plot(x,y_output,y);