用于多元样条插值的Python库

问题描述 投票:0回答:1

是否有用于多元插值的python库?现在,我有三个自变量和一个因变量。我的数据如下:

X1=[3,3,3.1,3.1,4.2,5.2,6.3,2.3,7.4,8.4,5.4,3.4,3.4,3.4,...]
X2=[12.1,12.7,18.5,18.3,18.4,18.6,24.2,24.4,24.3,24.5,30.9,30.7,30.3,30.4,6.1,6.2,...]
X3=[0.3,9.2,0.3,9.4,0.1,9.8,0.4,9.3,0.7,9.7,18.3,27.4,0.6,9.44,...]
Y=[-5.890,-5.894,2.888,-3.8706,2.1516,-2.7334,1.4723,-2.1049,0.9167,-1.7281,-2.091,-6.7394,0.8777,-1.7046,...]
and len(X1)=len(X2)=len(X3)=len(Y)=400

我想拟合或内插数据,以便在给定任意x1, x2, x3值的情况下,函数f(x1,x2,x3)将产生估计的y值。像给定的x1=4.11, x2=10.34, and x3=10.78,一样,函数将产生-8.7567(best estimate).,我想函数将是多项式的。那么,样条插值也许是这里的最佳选择?

python scipy interpolation spline
1个回答
0
投票

scipy.optimize工作曲线。在此代码中,估计是线性函数,但可能会更好。

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit

X1=[3,3,3.1,3.1,4.2,5.2,6.3,2.3,7.4,8.4,5.4,3.4,3.4,3.4]
X2=[12.1,12.7,18.5,18.3,18.4,18.6,24.2,24.4,24.3,24.5,30.9,30.7,30.3,30.4]
X3=[0.3,9.2,0.3,9.4,0.1,9.8,0.4,9.3,0.7,9.7,18.3,27.4,0.6,9.44]
Y=[-5.890,-5.894,2.888,-3.8706,2.1516,-2.7334,1.4723,-2.1049,0.9167,-1.7281,-2.091,-6.7394,0.8777,-1.7046]

def fitFunc(x, a, b, c, d):
    return a + b*x[0] + c*x[1] + d*x[2]

fitParams, fitCovariances = curve_fit(fitFunc, [X1, X2, X3], Y)
print(' fit coefficients:\n', fitParams)
# fit coefficients:
#  [-6.11934208  0.21643939  0.26186705 -0.33794415]

然后使用fitParams[0] + fitParams[1] * x1 + fitParams[2] * x2 + fitParams[3] * x3估计为y。

# get single y
def estimate(x1, x2, x3):
    return fitParams[0] + fitParams[1] * x1 + fitParams[2] * x2 + fitParams[3] * x3

将结果与原始y比较。

Y_estimated = [estimate(X1[i], X2[i], X3[i]) for i in range(len(X1))]

fig, ax = plt.subplots()
ax.scatter(Y, Y_estimated)

lims = [
    np.min([ax.get_xlim(), ax.get_ylim()]),  # min of both axes
    np.max([ax.get_xlim(), ax.get_ylim()]),  # max of both axes
]

ax.set_xlabel('Y')
ax.set_ylabel('Y_estimated')
ax.plot(lims, lims, 'k-', alpha=0.75, zorder=0)
ax.set_aspect('equal')

enter image description here

参考scipystackoverflow-multifitstackoverflow-plot xy

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