多元样条插值Python库
有用于多元插值的python库吗?现在我有三个自变量和一个因变量。我的数据如下所示:多元样条插值Python库,python,scipy,interpolation,spline,Python,Scipy,Interpolation,Spline,有用于多元插值的python库吗?现在我有三个自变量和一个因变量。我的数据如下所示: 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
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
我想拟合或插值数据,以便给定任意
x1,x2,x3
值,函数f(x1,x2,x3)
将产生一个估计的y
值。就像给定的x1=4.11,x2=10.34,x3=10.78一样,
函数将产生-8.7567(最佳估计)。
我想象函数将是多项式。也许样条插值是最好的选择 曲线拟合在scipy.optimize WORD中。在这段代码中,估计是线性函数,但可能更好
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进行比较
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')
参考