在Python中拟合指数 [英] Fitting an exponent in Python
问题描述
我正在尝试在python中适应指数衰减。我试过使用scipy.optimize.curve_fit，但它完全失败...
I am trying to fit an exponential decay in python. I've tried using scipy.optimize.curve_fit, but it completely fails...
x
Out[18]:
array([ 1.06001000e+04, 1.18721000e+04, 1.32966000e+04,
1.48926000e+04, 1.66801000e+04, 1.86816000e+04,
2.09236000e+04, 2.34351000e+04, 2.62481000e+04,
2.93981000e+04, 3.29261000e+04, 3.68781000e+04,
4.13041000e+04, 4.62611000e+04, 5.18136000e+04,
5.80321000e+04, 6.49966000e+04, 7.27971000e+04,
8.15341000e+04, 9.13196000e+04, 1.02279100e+05,
1.14554100e+05, 1.28302600e+05, 1.43701100e+05,
1.60947600e+05, 1.80264100e+05, 2.01898600e+05,
2.26129600e+05, 2.53268600e+05, 2.83664600e+05,
3.17709100e+05, 3.55839100e+05, 3.98545100e+05,
4.46377100e+05, 4.99949600e+05, 5.59951100e+05,
6.27154100e+05, 7.02422600e+05, 7.86724100e+05,
8.81143100e+05, 9.86894100e+05, 1.10533660e+06,
1.23799410e+06, 1.38657310e+06, 1.55298310e+06,
1.73936510e+06, 1.94811610e+06, 2.18192010e+06,
2.44378460e+06, 2.73707660e+06, 3.06556810e+06,
3.43348410e+06, 3.84555560e+06, 4.30708210e+06,
4.82399910e+06, 5.40295410e+06, 6.05139210e+06,
6.77765310e+06, 7.59107710e+06, 8.50212410e+06,
9.52251060e+06, 1.06653596e+07, 1.19453686e+07,
1.33789981e+07, 1.49846856e+07, 1.67830806e+07,
1.87973106e+07, 2.10532796e+07, 2.35800001e+07,
2.64099671e+07])
y
Out[19]:
array([ 7.21779435e-06, 6.88096911e-06, 6.44766520e-06,
6.06220818e-06, 5.59156825e-06, 5.27746585e-06,
4.90419458e-06, 4.57028098e-06, 4.19594740e-06,
3.87213247e-06, 3.53253198e-06, 3.21746863e-06,
2.96593379e-06, 2.69902818e-06, 2.45720479e-06,
2.22894945e-06, 2.00554860e-06, 1.78755768e-06,
1.60389345e-06, 1.43594942e-06, 1.27660849e-06,
1.12632772e-06, 9.93404773e-07, 8.78887840e-07,
7.68431386e-07, 6.69981141e-07, 5.88274963e-07,
5.12602683e-07, 4.47113130e-07, 3.91898528e-07,
3.42875999e-07, 3.00697454e-07, 2.63373855e-07,
2.35082385e-07, 2.06185600e-07, 1.81771840e-07,
1.60044617e-07, 1.42299315e-07, 1.26392523e-07,
1.12661361e-07, 1.01275721e-07, 9.01458593e-08,
8.09207343e-08, 7.38619000e-08, 6.76745276e-08,
6.17079129e-08, 5.68279252e-08, 5.34049900e-08,
5.05521909e-08, 4.76524243e-08, 4.36574532e-08,
4.05941897e-08, 3.78241485e-08, 3.51867595e-08,
3.34753821e-08, 3.13213498e-08, 2.96139649e-08,
2.74616096e-08, 2.49946165e-08, 2.23428677e-08,
2.04127328e-08, 1.84783950e-08, 1.65030587e-08,
1.47845483e-08, 1.35851162e-08, 1.15353701e-08,
9.18553778e-09, 7.01208306e-09, 5.04006337e-09,
nan])
def exp_func(x, a, b, c):
...: return a * np.exp(-b * x) + c
...:
curve_fit(exp_func, x, y)
Out[21]: (array([ 1., 1., 1.]), inf)
我知道参数的初始猜测非常很重要，但我猜不出来（我只知道b接近1）。因此，如果有人可以指出猜测参数的方法或不需要猜测的方法，我将不胜感激。
这是数据的对数图：
I know that the initial guess for the parameters is very important, but I can't guess them (I only know that b is close to 1). So, if someone could point out either a way of guessing the parameters or a method that would not require the guessing, I would be very grateful. This is the loglog plot of the data:
编辑：
实际上，我意识到所有问题都是由于数据不是指数式的，而是幂函数引起的，因此使用幂函数而不是指数就可以开箱即用。
So, actually I realized that all the problems arise from data not being exponential, but rather power-function, so with power function instead of an exponent it works reasonably out of the box.
推荐答案
您可能想忽略 nan ：
valid = np.logical_not(np.isnan(x + y))
optimize.curve_fit(exp_func, x[valid], y[valid])
为避免数值不稳定，您应采用某种方式标准化输入数据，例如将 x 乘以 1e6 。当然，您将需要相应地更正结果参数。
To avoid numerical instabilities, you should somehow normalize the input data, e.g. by multiplying x with 1e6. Of course, you will need to correct the resulting parameters accordingly.
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