使用 Scipy 进行曲线拟合会导致意外结果
我有一个可以适合给定函数的数据集。我使用 Desmos 图形计算器进行了计算，并得到了预期的结果。但是当我使用 Matplotlib 实现它时，我得到了一条完全不同的曲线，并且我无法找出代码中的错误。这是我的代码：
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
data = np.loadtxt('data.txt')
def func(x, a, b):
return a/(np.sin((x-b)/2))**4
x = data[:, 0]
y = data[:, 1]
popt, pcov = curve_fit(func, x, y)
plt.plot(x, y, 'ko', label="Observed data points")
plt.plot(x, func(x, *popt), 'r-', label="Fitted Curve")
plt.legend()
plt.show()
这是预期的图（使用 Desmos 绘制）： 
以下是我从 Matplotlib 获得的内容： 
如有任何帮助，我们将不胜感激。这是供参考的data.txt：
| X | Y |
|---|---|
| -0.523598776 | 0.530580093 |
| -0.436332313 | 1.016423844 |
| -0.34906585 | 6.38245854 |
| -0.261799388 | 18.70139225 |
| -0.174532925 | 30.7389007 |
| 0.174532925 | 13.59465343 |
| 0.261799388 | 4.011313119 |
| 0.34906585 | 0.805865977 |
| 0.436332313 | 0.50894953 |
| 0.523598776 | 0.253654518 |
I have a data set which I can fit to a given function. I did it using Desmos graphing calculator and got the expected results. But when I implement it using Matplotlib, I get a totally different curve and I can't figure out the error in my code. Here is my code:
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
data = np.loadtxt('data.txt')
def func(x, a, b):
return a/(np.sin((x-b)/2))**4
x = data[:, 0]
y = data[:, 1]
popt, pcov = curve_fit(func, x, y)
plt.plot(x, y, 'ko', label="Observed data points")
plt.plot(x, func(x, *popt), 'r-', label="Fitted Curve")
plt.legend()
plt.show()
Here is the expected plot (plotted using Desmos):
And here is what I got from Matplotlib:
Any help would be appreciated. This is the data.txt for reference:
| X | Y |
|---|---|
| -0.523598776 | 0.530580093 |
| -0.436332313 | 1.016423844 |
| -0.34906585 | 6.38245854 |
| -0.261799388 | 18.70139225 |
| -0.174532925 | 30.7389007 |
| 0.174532925 | 13.59465343 |
| 0.261799388 | 4.011313119 |
| 0.34906585 | 0.805865977 |
| 0.436332313 | 0.50894953 |
| 0.523598776 | 0.253654518 |
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x==b处有一个奇点（您的函数趋于无穷大）。这使得curve_fit很难找到“交叉”任何数据点的b值。由于您的数据大约以零为中心，因此您会更幸运地提供一个初始值，例如You have a singularity at
x==b(your function tends to infinity). That makes it very difficult forcurve_fitto find a value ofbthat "crosses" any of your data points. Since your data is about zero centered, you'll have better luck providing a an initial value such as