问题描述
我正在尝试计算两种形式的指数与某些 x，y 数据的最佳拟合（可以从）
代码如下：
from scipy.optimize import curve_fit
import numpy as np
＃获取x，y数据
数据= np.loadtxt（'data.txt'，unpack = True）
xdata，ydata = data [0]，data [1]
＃定义第一个指数函数
def func （x，a，b，c）：
返回a * np.exp（b * x）+ c
＃获取参数估计值
popt，pcov = curve_fit（func ，xdata，ydata）
print popt
＃定义第二个指数函数（一个以上的参数）
def func2（x，a，b，c，d） ：
返回a * np.exp（b * x + c）+ d
＃获取参数估计
popt2，pcov2 = curve_fit（func2，xdata，ydata）
print popt2
第一个指数给出的值完全相同es as zunzun.com（），价格为 popt ：
[7.67760545e-15 1.52175476e + 00 2.15705939e-02]
，但是第二个给出的值显然对于 popt2 是错误的：
[-1.26136676e + 02 -8.13233297e-01 -6.66772692e + 01 3.63133641e-02]
这是zunzun.com的值（）以获取相同的第二个功能：
a = 6.2426224704624871E-15
b = 1.5217697532005228E + 00
c = 2.0660424037614489E-01
d = 2.1570805929514186E-02
我尝试按照此处推荐的方法制作列表数组，但这无济于事。我在这里做什么错了？
加1
我猜想问题与缺少初始值有关（我在喂我的函数（如此处所述：）
如果我像这样将第一个指数的估计值馈入第二个指数（使新参数 d 最初为零）：
popt2，pcov2 = curve_fit（func2，xdata，ydata，p0 = [popt [0]，popt [1]，popt [2] ，0]）
与zunzun.com相比，我得到的结果很合理，但仍然不正确：
[1.22560853e-14 1.52176160e + 00 -4.67859961e-01 2.15706930e-02]
现在问题变成了：我该如何为第二个函数提供更多的原因
请注意，估算中的 a = 0 在您的第一个模型中。因此，他们只是在估计一个常数。因此，在第一种情况下， b 和 b 和 c
在我上次查看它时，Zunzun还使用差分进化作为全局求解器。 Scipy现在已经成为了看起来不错的全局优化器，这在可能出现局部最小值的情况下值得尝试。
我的便宜方式，因为参数不在您的示例中没有很大的范围：尝试随机起始值
np.random.seed（1）
err_last = 20
最佳=无
对于范围（10）中的i：
开始= np.random.uniform（-10，10，size = 4）
＃获取参数估计
尝试：
popt2，pcov2 = curve_fit（func2，xdata，ydata，p0 = start）
，除了RuntimeError：
Continue
err =（ （ydata-func2（xdata，* popt2））** 2）.sum（）
如果err< err_last：
err_last =错误
打印err
最佳= popt2
za = 6.2426224704624871E-15
zb = 1.5217697532005228E + 00
zc = 2.0660424037614489E-01
zd = 2.1570805929514186E-02
zz = np.array（[za，zb，zc，zd]）
打印' zz'，zz
print'cf'，最佳
print'zz'，（（ydata-func2（xdata，* zz））** 2）.sum（）
打印'cf'，err_last
最后一部分打印（zz是zunzun，cf是curve_fit）
zz [6.24262247e-15 1.52176975e + 00 2.06604240e-01 2.15708059e-02]
cf [1.24791299e -16 1.52176944e + 00 4.11911831e + 00 2.15708019e-02]
zz 9.52135153898
cf 9.52135153904
b 和 c 的参数与Zunzun不同，但残差平方和相同。 / p>
加法
a * np.exp（b * x + c）+ d = np.exp（b * x +（c + np.log（a）））+ d
或
a * np.exp（b * x + c）+ d =（a * np .exp（c））* np.exp（b * x）+ d
第二个功能与第一个功能并没有什么不同功能。 a 和 c 没有单独标识。因此，使用派生信息的优化器也会遇到问题，因为如果我正确看到的话，雅可比行列在某些方向上是奇异的。
I'm trying to compute the best fit of two forms of an exponential to some x, y data (the data file can be downloaded from here)
Here's the code:
from scipy.optimize import curve_fit
import numpy as np
# Get x,y data
data = np.loadtxt('data.txt', unpack=True)
xdata, ydata = data[0], data[1]
# Define first exponential function
def func(x, a, b, c):
return a * np.exp(b * x) + c
# Get parameters estimate
popt, pcov = curve_fit(func, xdata, ydata)
print popt
# Define second exponential function (one more parameter)
def func2(x, a, b, c, d):
return a * np.exp(b * x + c) + d
# Get parameters estimate
popt2, pcov2 = curve_fit(func2, xdata, ydata)
print popt2
The first exponential gives the exact same values as zunzun.com (PDF here) for popt:
[ 7.67760545e-15 1.52175476e+00 2.15705939e-02]
but the second gives values that are clearly wrong for popt2:
[ -1.26136676e+02 -8.13233297e-01 -6.66772692e+01 3.63133641e-02]
This are zunzun.com values (PDF here) for that same second function:
a = 6.2426224704624871E-15
b = 1.5217697532005228E+00
c = 2.0660424037614489E-01
d = 2.1570805929514186E-02
I tried making the lists arrays as reccomended here Strange result with python's (scipy) curve fitting, but that didn't help. What am I doing wrong here?
Add 1
I'm guessing the problem has to do with the lack of initial values I'm feeding my function (as explained here: gaussian fit with scipy.optimize.curve_fit in python with wrong results)
If I feed the estimates from the first exponential to the second one like so (making the new parameter d be initially zero):
popt2, pcov2 = curve_fit(func2, xdata, ydata, p0 = [popt[0], popt[1], popt[2], 0])
I get results that are much reasonable but still wrong compared to zunzun.com:
[ 1.22560853e-14 1.52176160e+00 -4.67859961e-01 2.15706930e-02]
So now the question changes to: how can I feed my second function more reasonable parameters automatically?
Note that a=0 in the estimate by zunzun and in your first model. So they are just estimating a constant. So, b in the first case and b and c in the second case are irrelevant and not identified.
Zunzun also uses differential evolution as a global solver, the last time I looked at it. Scipy now has basinhopping as global optimizer that looks pretty good, that is worth a try in cases where local minima are possible.
My "cheap" way, since the parameters don't have a huge range in your example: try random starting values
np.random.seed(1)
err_last = 20
best = None
for i in range(10):
start = np.random.uniform(-10, 10, size=4)
# Get parameters estimate
try:
popt2, pcov2 = curve_fit(func2, xdata, ydata, p0=start)
except RuntimeError:
continue
err = ((ydata - func2(xdata, *popt2))**2).sum()
if err < err_last:
err_last = err
print err
best = popt2
za = 6.2426224704624871E-15
zb = 1.5217697532005228E+00
zc = 2.0660424037614489E-01
zd = 2.1570805929514186E-02
zz = np.array([za,zb,zc,zd])
print 'zz', zz
print 'cf', best
print 'zz', ((ydata - func2(xdata, *zz))**2).sum()
print 'cf', err_last
The last part prints (zz is zunzun, cf is curve_fit)
zz [ 6.24262247e-15 1.52176975e+00 2.06604240e-01 2.15708059e-02]
cf [ 1.24791299e-16 1.52176944e+00 4.11911831e+00 2.15708019e-02]
zz 9.52135153898
cf 9.52135153904
Different parameters than Zunzun for b and c, but the same residual sum of squares.
Addition
a * np.exp(b * x + c) + d = np.exp(b * x + (c + np.log(a))) + d
or
a * np.exp(b * x + c) + d = (a * np.exp(c)) * np.exp(b * x) + d
The second function isn't really different from the first function. a and c are not separately identified. So optimizers, that use the derivative information, will also have problems because the Jacobian is singular in some directions, if I see this correctly.
这篇关于曲线拟合因指数而失败，但zunzun正确的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！