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);
(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):
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).
linear fit/regression in each intervals
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.
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 (LSi) is replaced by a combination of a small constant segment at the beginning (Cbi), a linear segment without intercept (Si), and another constant segment at the end (Cei). This segments are easy to calculate as Si will contain the middle point of LSi, and Cbi and Cei will have respectively the begin and end values of the segment LSi. 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 ci and ri+1 are known, then bi+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 ci, all the rest of the bj and cj 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 bi+1 if ci 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.
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()
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);
I was trying to obtain the Mathieu characteristic values for a specific problem. I do not have any problem obtaining them, and I have read the documentation from Scipy regarding these functions. The problem is that I know for a fact that the points I am obtaining are not right. My script to obtain the characteristic values I need is below:
import numpy as np
import matplotlib.pyplot as plt
from scipy.special import mathieu_a, mathieu_b, mathieu_cem, mathieu_sem
M = 1.0
g = 1.0
l = 1.0
h = 0.06
U0 = M * g * l
q = 4 * M * l**2 * U0 / h**2
def energy(n, q):
if n % 2 == 0:
return (h**2 / (8 * M * l**2)) * mathieu_a(n, q) + U0
else:
return (h**2 / (8 * M * l**2)) * mathieu_b(n + 1, q) + U0
n_list = np.arange(0, 80, 1)
e_n = [energy(i, q) for i in n_list]
plt.plot(n_list, e_n, '.')
The resulting plot of these values is this one. There is a zone where it appears to be "noise" or a numerical error, and I know that those jumps must not occur. In reality, around x= 40 to x > 40, the points should behave like a staircase of two consecutive points, similar to what can be seen between 70 < x < 80. And the values that x can take for this case are only positive integers.
I saw that the implementation of the Mathieu function has some problems, see here. But this was six years ago! In the answer to this question they use the NAG Library for Python, but it is not exactly open-source.
Is there a way I can still use these functions from Scipy without having this problem? Or is it related to the precision I am using to obtain the Mathieu characteristic value?
I have a coupled system of differential equations that I've already solved with Euler in Excel. Now I want to make it more precise with an ODE-solver in python.
However, there must be a mistake in my code because the curves look different than in Excel. I don't expect the curves to reach 1 and 0 in the end.
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
# define reactor
def reactor(x,z):
n_a = x[0]
n_b = x[1]
n_c = x[2]
dn_adz = A * (-1) * B * (n_a/(n_a + n_b + n_c)) / (1 + C * (n_c/(n_a + n_b + n_c)))
dn_bdz = A * (1) * B * (n_a/(n_a + n_b + n_c)) / (1 + C * (n_c/(n_a + n_b + n_c)))
dn_cdz = A * (1) * B * (n_a/(n_a + n_b + n_c)) / (1 + C * (n_c/(n_a + n_b + n_c)))
dxdz = [dn_adz,dn_bdz,dn_cdz]
return dxdz
# initial conditions
n_a0 = 0.5775
n_b0 = 0.0
n_c0 = 0.0
x0 = [n_a0, n_b0, n_c0]
# parameters
A = 0.12
B = 3.1e-9
C = 4.02e15
# number of steps
n = 100
# z step interval (m)
z = np.linspace(0,0.0274,n)
# solve ODEs
x = odeint(reactor,x0,z)
# Plot the results
plt.plot(z,x[:,0],'b-')
plt.plot(z,x[:,1],'r--')
plt.plot(z,x[:,2],'k:')
plt.show()
Is is a problem with the initial condition that stays constant and does not change from step to step?
Should it be like in Excel with Euler, where the next step uses the conditions/values of the precious step?
From the structure of the right sides you get constant combinations of the state variables, n_a+n_b=n_a0+n_b0 and n_a+n_c=n_a0+n_c0. This means that the dynamic reduces to the one-dimensional dynamic of n_a.
By the first equation, the derivative of n_a is negative for positive n_a, so that the solution is falling towards n_a=0. By the constants of the dynamics, n_b converges to n_a0+n_b0 and n_c converges to n_a0+n_c0.
It is unclear how you get convergence towards 1 in some components, as that is not supported by the initial conditions. Apart from that, the described odeint result fits this qualitative behavior.
For instance if my numbers were: A=[1,2,3] B=[4,5,6] and C=[5,10,11] then the closest function I can get to would be 2A+B. Any idea about what kind of optimization pattern would be needed for something like this?
Assuming you want to find real values p and q such that the distance between pA + qB and C is minimal.
pA + qB = pi + 2pj + 3pk + 4qi + 5qj + 6qk = i(p + 4q) + j(2p + 5q) + k(3p + 6q)
The distance is given by:
f(p,q) = √((p + 4q - 5)² + (2p + 5q - 10)² + (3p + 6q - 11)²)
Take its partial derivatives and find the local minimum.
This can be set up as a system of linear equations and solved using numpy
import numpy as np
A=[1,2,3]
B=[4,5,6]
C=[5,10,11]
# Set up as a system of linear equations to solve with
# min mean square error, i.e. find x, y such that:
# Solve Ax + By = C
# As linear equations become
# [A; B] [x y] = C
# With M = [A; B] we have (i.e. A, B are the column vectors of M)
M = np.matrix((list(zip(A, B))))
# Find Least Square Error Solution
sol = np.linalg.lstsq(M, C, rcond=None)[0] # Solve for [x y]
print("Solution (Least Min Square Error\n", sol) # [2.11111111 0.88888889]
approx = np.dot(M, sol)
print("approximation\n", approx) # [[ 5.66666667 8.66666667 11.66666667]]
I'm tring to approximate an empirical cumulative distribution function (ECDF I want to approximate) with a smooth function (with less than 5 parameter) such as the generalized logistic function.
However, using scipy.optimize.curve_fit, the fitting operation gives really bad approximations or it doesn't work at all (depending on the initial values). The variable series represents my data stored as pandas.Series.
from scipy.optimize import curve_fit
def fit_ecdf(x):
x = np.sort(x)
def result(v):
return np.searchsorted(x, v, side='right') / x.size
return result
ecdf = fit_ecdf(series)
def genlogistic(x, B, M, Q, v):
return 1 / (1 + Q * np.exp(-B * (x - M))) ** (1 / v)
params = curve_fit(genlogistic, xdata = series, ydata = ecdf(series), p0 = (0.1, 10.0, 0.1, 0.1))[0]
Should I use another type of function for the fit?
Are there any code mistakes?
UPDATE - 1
As asked, I link to a csv containing the data.
UPDATE - 2
After a lot of search and trial and error I find out this function
f(x; a, b, c) = 1 - 1 / (1 + (x / b) ** a) ** c
with a = 4.61320000, b = 2.94570952, c = 0.5886922
which fits a lot better than the other one. The only problem is the little step that the ECDF shows near x=1. How can I modify f to improve the quality of the fit? I was thinking of adding some sort of function that is "relevant" only in those kind of points. Here are the graphical results of the fit where the solid blue line is the ECDF and the dotted line represents the (x, f(x)) points.
I find out how to deal with that little step near x=1. As expressed in the question, adding some sort of function that is significant only in that interval was the game changer.
The "step" ends at about (1.7, 0.04) so I needed a sort of function that flattens for x > 1.7 and has y = 0.04 as asymptote. The natural choice (just to stay on point) was to take a function like f(x) = 1/exp(x).
Thanks to JamesPhillips, I also picked up the proper data for the regression (no double values = no overweighted points).
Python Code
from scipy.optimize import curve_fit
def fit_ecdf(x):
x = np.sort(x)
def result(v):
return np.searchsorted(x, v, side = 'right') / x.size
return result
ecdf = fit_ecdf(series)
unique_series = series.unique().tolist()
def cdf_interpolation(x, a, b, c, d):
f_1 = 0.95 + (0 - 0.95) / (1 + (x / b) ** a) ** c + 0.05
f_2 = (0 - 0.05)/(np.exp(d * x))
return f_1 + f_2
params = curve_fit(cdf_interpolation,
xdata = unique_series ,
ydata = ecdf(unique_series),
p0 = (6.0, 3.0, 0.4, 1.0))[0]
Parameters
a = 6.03256462
b = 2.89418871
c = 0.42997956
d = 1.06864006
Graphical results
I got an OK fit for a 5-parameter logistic equation (see image and code) using unique values, not sure if the low end curve is sufficient for your needs, please check.
import numpy as np
def Sigmoidal_FiveParameterLogistic_model(x_in): # from zunzun.com
# coefficients
a = 9.9220221252324947E-01
b = -3.1572339989462903E+00
c = 2.2303376075685142E+00
d = 2.6271495036080207E-02
f = 3.4399008905318986E+00
return d + (a - d) / np.power(1.0 + np.power(x_in / c, b), f)