[英]Minimize objective function using limfit.minimize in Python
I am having a problem with package lmfit.minimize
minimization procedure. 我遇到了包
lmfit.minimize
最小化过程的问题。 Actually, I could not create a correct objective function for my problem. 实际上,我无法为我的问题创建正确的目标函数。
Problem definition 问题定义
yn = a_11*x1**2 + a_12*x2**2 + ... + a_m*xn**2
,where xn
- unknowns, a_m
- coefficients. yn = a_11*x1**2 + a_12*x2**2 + ... + a_m*xn**2
,其中xn
- 未知数, a_m
- 系数。 n = 1..N, m = 1..M N=5
for x1,..,x5
and M=3
for y1, y2, y3
. y1, y2, y3
x1,..,x5
对于x1,..,x5
, M=3
, N=5
。 I need to find the optimum: x1, x2,...,x5
so that it can satisfy the y
我需要找到
optimum: x1, x2,...,x5
以便它可以满足y
My question: 我的问题:
ValueError: operands could not be broadcast together with shapes (3,) (3,5)
. ValueError: operands could not be broadcast together with shapes (3,) (3,5)
。 My code: 我的代码:
import numpy as np
from lmfit import Parameters, minimize
def func(x,a):
return np.dot(a, x**2)
def residual(pars, a, y):
vals = pars.valuesdict()
x = vals['x']
model = func(x,a)
return y - model
def main():
# simple one: a(M,N) = a(3,5)
a = np.array([ [ 0, 0, 1, 1, 1 ],
[ 1, 0, 1, 0, 1 ],
[ 0, 1, 0, 1, 0 ] ])
# true values of x
x_true = np.array([10, 13, 5, 8, 40])
# data without noise
y = func(x_true,a)
#************************************
# Apriori x0
x0 = np.array([2, 3, 1, 4, 20])
fit_params = Parameters()
fit_params.add('x', value=x0)
out = minimize(residual, fit_params, args=(a, y))
print out
if __name__ == '__main__':
main()
Directly using scipy.optimize.minimize()
the code below solves this problem. 直接使用
scipy.optimize.minimize()
,下面的代码解决了这个问题。 Note that with more points yn
you will tend to get the same result as x_true
, otherwise more than one solution exists. 请注意,更多的积分
yn
你往往会得到相同的结果x_true
,否则不止一个解决方案的存在。 You can minimize the effect of the ill-constrained optimization by adding boundaries (see the bounds
parameter used below). 您可以通过添加边界来最小化不受约束的优化的影响(请参阅下面使用的
bounds
参数)。
import numpy as np
from scipy.optimize import minimize
def residual(x, a, y):
s = ((y - a.dot(x**2))**2).sum()
return s
def main():
M = 3
N = 5
a = np.random.random((M, N))
x_true = np.array([10, 13, 5, 8, 40])
y = a.dot(x_true**2)
x0 = np.array([2, 3, 1, 4, 20])
bounds = [[0, None] for x in x0]
out = minimize(residual, x0=x0, args=(a, y), method='L-BFGS-B', bounds=bounds)
print(out.x)
If M>=N
you could also use scipy.optimize.leastsq
for this task: 如果
M>=N
您还可以使用scipy.optimize.leastsq
执行此任务:
import numpy as np
from scipy.optimize import leastsq
def residual(x, a, y):
return y - a.dot(x**2)
def main():
M = 5
N = 5
a = np.random.random((M, N))
x_true = np.array([10, 13, 5, 8, 40])
y = a.dot(x_true**2)
x0 = np.array([2, 3, 1, 4, 20])
out = leastsq(residual, x0=x0, args=(a, y))
print(out[0])
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