问题
I am using scipy.optimize fsolve for finding roots of two equations. fsolve works well for some range of b values (0.1 to 0.6), but fails for values like 0.9 or 0.99.
I have tried moving to least_squares or minimize, but get a tuple error whilst providing initial conditions.
Including edits from below findings be self:
from scipy.optimize import fsolve
import scipy.stats as st
from numpy import *
import numpy as np
def rod(var1, var2, mu, sigma):
return (st.lognorm.ppf(var1, s = sigma, scale = np.exp(mu), loc = sigma))/(st.lognorm.ppf(var2, s = sigma, scale = np.exp(mu), loc = sigma))
def fs_spfs(var1, mu, sigma):
return (st.lognorm.ppf(var1, s = sigma, scale = np.exp(mu), loc = sigma))
a = 44.0
b = 0.5 #fsolve works for 0.5, 0.9, 0.99 but not for 0.95, incidentally works for 0.950001
c = 1.26
def f(x):
y = np.zeros(2)
y[0] = ((fs_spfs((1-b), x[0], x[1]) - a))
y[1] = (((fs_spfs(0.9, x[0], x[1])/fs_spfs(0.1, x[0], x[1])) - c))
print(y)
return y
x0 = np.array([1., 0.01])
solution = fsolve(f, x0)
print( "(x, y) = (" + str(solution[0]) + ", " + str(solution[1]) + ")")
Results with b = 0.5
b = 0.5
(x, y) = (3.7821340072441982, 0.09035467410258388)
fs_spfs((1-b), solution[0], solution[1]) # expected answer = 44.
43.99999999999982
rod(0.9, 0.1, solution[0], solution[1]) # exptected answer = 1.26
1.2599999999999958
Results with b = 0.9
b = 0.9
(x, y) = (3.8979025451494755, 0.09033430819655046)
fs_spfs((1-b), solution[0], solution[1]) # expected answer = 44.
43.999999999989164
rod(0.9, 0.1, solution[0], solution[1]) # exptected answer = 1.26
1.2600000000001814
Works for b = 0.99 as well, but fails for b = 0.95. Incidentally works for b = 0.950001
回答1:
Following initial condition seems to be working for majority of common cases:
x0 = np.array([0.000001, 0.0000001])
Works for values till 0.999, still fails for 0.9999.
来源:https://stackoverflow.com/questions/56239990/scipy-fsolve-fails-for-few-values-in-input-and-how-to-improve-solver-convergenc