如何解决Scipy.Curve 拟合指数函数
我正在尝试拟合方程的曲线：
y = ( (np.exp(-k2*(t+A))) - ((k1/v)*Co) )/ -k2
where A = (-np.log((k1/v)*Co))/k2
主管给我一个数据集，它看起来像一个粗略的指数，在它的顶部变平成一条水平直线。当我拟合方程时，我只收到曲线拟合的一条直线和相应的警告：
<ipython-input-24-7e57039f2862>:36: RuntimeWarning: overflow encountered in exp
return ( (np.exp(-k2*(t+A))) - ((k1/v)*Co) )/ -k2
我使用的代码如下：
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy.optimize import differential_evolution
xData_forfit = [1.07683e+13,1.16162e+13,1.24611e+13,1.31921e+13,1.40400e+13,2.65830e+13,2.79396e+13,2.86676e+13,2.95155e+13,3.03605e+13,3.12055e+13,3.20534e+13,3.27814e+13,3.36293e+13,3.44772e+13,3.53251e+13,3.61730e+13,3.77459e+13,3.85909e+13,3.94388e+13,4.02838e+13,4.11317e+13,4.19767e+13,4.27076e+13,5.52477e+13,5.64143e+13,5.72622e+13,5.81071e+13,5.89550e+13,5.98000e+13,6.05280e+13,6.13759e+13,6.22209e+13,6.30658e+13,6.39137e+13,6.46418e+13,6.55101e+13,6.63551e+13,6.72030e+13,6.80480e+13,6.88929e+13,6.97408e+13,7.04688e+13,7.13167e+13,7.21617e+13,8.50497e+13,8.58947e+13,8.67426e+13,8.75876e+13,8.83185e+13,9.00114e+13,9.08563e+13,9.17013e+13]
yData_forfit = [1375.409524,1378.095238,1412.552381,1382.904762,1495.2,1352.4,1907.971429,1953.52381,1857.352381,1873.990476,1925.114286,1957.085714,2030.52381,1989.8,2042.733333,2060.095238,2134.361905,2200.742857,2342.72381,2456.047619,2604.542857,2707.971429,2759.87619,2880.52381,3009.590476,3118.771429,3051.52381,3019.771429,3003.561905,3083.0,3082.885714,2799.866667,3012.419048,3013.266667,3106.714286,3090.47619,3216.638095,3108.447619,3199.304762,3154.257143,3112.419048,3284.066667,3185.942857,3157.380952,3158.47619,3464.257143,3434.67619,3291.457143,2851.371429,3251.904762,3056.152381,3455.07619,3386.942857]
def fnct_to_opt(t,k2,k1):
#EXPERIMENTAL CONSTANTS
v = 105
Co = 1500
A = (-np.log((k1/v)*Co))/k2
return ( (np.exp(-k2*(t+A))) - ((k1/v)*Co) )/ -k2
initial_k2k1 = [100,1*10**-3]
constants = curve_fit(fnct_to_opt,xData_forfit,yData_forfit,p0=initial_k2k1)
k2_fit = constants[0][0]
k1_fit = constants[0][1]
fit = []
for i in xData_forfit:
fit.append(fnct_to_opt(i,k2_fit,k1_fit))
plt.plot(xData_forfit,'or',ms='2')
plt.plot(xData_forfit,fit)
这给了我这个情节作为结果：
据我所知，由于 np.exp 项的值太大，代码没有产生有用的输出，但我不知道如何诊断此溢出的来源或如何解决问题。任何帮助将不胜感激，谢谢。
解决方法
溢出恰好发生在错误消息告诉您的位置：在 ('except_check','data_check')
('abc','mask')
的 return 表达式中。我让你在错误点之前打印有问题的值；这会告诉你问题所在。
在错误点，fnct_to_opt 中的值在 e+13 到 e+14 的范围内。 A 无关紧要； t 有点低于 -10000.0
因此，k2 的参数中的值远远超出了函数可以处理的域。只需向您的函数添加一行并观察结果：
np.exp我认为问题可能出在优化函数上，从某种意义上说可能是一个错误。
例如：
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy.optimize import differential_evolution
xData_forfit = [1.07683e+13,1.16162e+13,1.24611e+13,1.31921e+13,1.40400e+13,2.65830e+13,2.79396e+13,2.86676e+13,2.95155e+13,3.03605e+13,3.12055e+13,3.20534e+13,3.27814e+13,3.36293e+13,3.44772e+13,3.53251e+13,3.61730e+13,3.77459e+13,3.85909e+13,3.94388e+13,4.02838e+13,4.11317e+13,4.19767e+13,4.27076e+13,5.52477e+13,5.64143e+13,5.72622e+13,5.81071e+13,5.89550e+13,5.98000e+13,6.05280e+13,6.13759e+13,6.22209e+13,6.30658e+13,6.39137e+13,6.46418e+13,6.55101e+13,6.63551e+13,6.72030e+13,6.80480e+13,6.88929e+13,6.97408e+13,7.04688e+13,7.13167e+13,7.21617e+13,8.50497e+13,8.58947e+13,8.67426e+13,8.75876e+13,8.83185e+13,9.00114e+13,9.08563e+13,9.17013e+13]
yData_forfit = [1375.409524,1378.095238,1412.552381,1382.904762,1495.2,1352.4,1907.971429,1953.52381,1857.352381,1873.990476,1925.114286,1957.085714,2030.52381,1989.8,2042.733333,2060.095238,2134.361905,2200.742857,2342.72381,2456.047619,2604.542857,2707.971429,2759.87619,2880.52381,3009.590476,3118.771429,3051.52381,3019.771429,3003.561905,3083.0,3082.885714,2799.866667,3012.419048,3013.266667,3106.714286,3090.47619,3216.638095,3108.447619,3199.304762,3154.257143,3112.419048,3284.066667,3185.942857,3157.380952,3158.47619,3464.257143,3434.67619,3291.457143,2851.371429,3251.904762,3056.152381,3455.07619,3386.942857]
def fnct_to_opt(t,k2,k1):
#EXPERIMENTAL CONSTANTS
v = 105
Co = 1500
#A = (-np.log((k1/v)*Co))/k2
#return ( (np.exp(-k2*(t+A))) - ((k1/v)*Co) )/ -k2
#A = (np.log((k1/v)*Co))/k2
return k2/np.log(t) + k1
initial_k2k1 = [10,1]
constants = curve_fit(fnct_to_opt,xData_forfit,yData_forfit,p0=initial_k2k1)
k2_fit = constants[0][0]
k1_fit = constants[0][1]
#v_fit = constants[0][2]
#Co_fit = constants[0][3]
fit = []
for i in xData_forfit:
fit.append(fnct_to_opt(i,k2_fit,k1_fit))
plt.plot(xData_forfit,'or',ms='2')
plt.plot(xData_forfit,fit)
所以我放置了一个更简单但背后有更清晰直觉的函数。例如，在原版中，我认为使用这些符号和指数根本不会实现形状。但是在我看来，指数放错了位置，所以我将其更改为日志。添加一个常量和一个比例参数。我建议仔细检查原始功能。推导可能存在问题。我不认为是计算问题。
这更接近预期。
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