我正在尝试使用 UnivariateSpline 函数为某些数据(原始布拉格峰)生成拟合。这是我的代码:
import matplotlib.pyplot as plt
from scipy import interpolate
import numpy as np
spline = interpolate.UnivariateSpline(xi, yi, s = 0)
measured_depth_i = np.linspace(0.4, 10.4, 1000)
measured_dose_i = spline(measured_depth_i)
plt.plot(measured_depth, measured_dose, 'ro', label='Original Points')
plt.plot(measured_depth_i, measured_depth_i, label='Interpolated Points' )
其中 xi =
[ 0.4 , 0.48, 0.63, 0.79, 0.93, 1.08, 1.22, 1.38,
1.53, 1.68, 1.83, 1.99, 2.14, 2.29, 2.43, 2.59,
2.73, 2.89, 3.04, 3.19, 3.34, 3.48, 3.64, 3.8 ,
3.94, 4.09, 4.24, 4.39, 4.55, 4.7 , 4.85, 5. ,
5.15, 5.3 , 5.45, 5.59, 5.75, 5.9 , 6.05, 6.21,
6.35, 6.5 , 6.66, 6.8 , 6.95, 7.11, 7.24, 7.4 ,
7.55, 7.71, 7.86, 8. , 8.16, 8.31, 8.46, 8.61,
8.76, 8.91, 9.06, 9.22, 9.36, 9.52, 9.67, 9.81,
9.97, 10.12, 10.27, 10.4 ]
和 yi =
[ 2.16457961e+01, 2.16457961e+01, 2.16010733e+01,
2.18470483e+01, 2.20483005e+01, 2.22942755e+01,
2.24731664e+01, 2.26967800e+01, 2.29427549e+01,
2.30545617e+01, 2.32781753e+01, 2.34570662e+01,
2.36583184e+01, 2.37924866e+01, 2.39713775e+01,
2.42620751e+01, 2.44856887e+01, 2.46869410e+01,
2.49552773e+01, 2.51341682e+01, 2.53577818e+01,
2.56932021e+01, 2.58497317e+01, 2.61404293e+01,
2.63864043e+01, 2.65876565e+01, 2.68559928e+01,
2.72361360e+01, 2.74821109e+01, 2.78622540e+01,
2.81529517e+01, 2.84436494e+01, 2.88461538e+01,
2.91815742e+01, 2.95617174e+01, 3.00089445e+01,
3.04114490e+01, 3.09033989e+01, 3.13059034e+01,
3.18872987e+01, 3.24463327e+01, 3.31395349e+01,
3.37656530e+01, 3.44364937e+01, 3.51296959e+01,
3.60017889e+01, 3.68515206e+01, 3.78801431e+01,
3.91323792e+01, 4.04293381e+01, 4.19946333e+01,
4.37164580e+01, 4.57513417e+01, 4.83228980e+01,
5.16100179e+01, 5.58363148e+01, 6.13595707e+01,
6.93202147e+01, 8.35420394e+01, 1.00000000e+02,
8.47271914e+01, 3.60688730e+01, 6.21645796e+00,
3.35420394e-01, 2.23613596e-02, 2.23613596e-02,
2.23613596e-02, 2.23613596e-02]
我的插值曲线似乎仍然是一条直线，并且与我的数据完全不匹配，无论我的 s 值如何变化。谁能解释为什么我不能很好地适应我的数据？
最佳答案
诀窍是在绘图语句中正确获取变量名称:
spline = interpolate.UnivariateSpline(xi, yi, s = 0)
measured_depth_i = np.linspace(0.4, 10.4, 1000)
measured_dose_i = spline(measured_depth_i)
plt.plot(xi, yi, 'ro', label='Original Points')_
plt.plot(measured_depth_i, measured_dose_i, label='Interpolated Points' )_
结果图如下:
关于python - 将 UnivariateSpline 与 SCIPY 结合使用，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25151547/