双线性四边形插值的Python实现

本文介绍了双线性四边形插值的Python实现的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在尝试执行双线性四边形插值.因此，我有四个具有已知值的节点，并且我想通过插值找到一个位于这四个节点之间的值，但是这四个节点没有形成矩形. 4节点草图

I'm trying to perform bilinear quadrilateral interpolation. So I have four nodes with known values and I want to find a value that lies in between those four nodes by interpolation, but the four nodes do not form a rectangle. 4-node sketch

我找到了解决此问题的几种方法，但是没有一种方法已经在Python中实现.在某处已经存在完成的python实现吗?如果不是，您会推荐以下两种解决方案中的哪一种?还是您会推荐另一种方法?

I found several ways to solve this, but none of them is implemented in Python already. Does there exist somewhere an already finished python implementation? If not which of the two solutions below would you recommend? Or would you recommend another approach?

**************不同的解决方案*******************

**************Different solutions*******************

解决方案1:

我在这里找到 www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.Ch16.d/IFEM.Ch16.pdf ，我应该求解以下方程组:方程组，其中Ni为: N定义.

I found here, www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.Ch16.d/IFEM.Ch16.pdf, that I should solve the following set of equations: set of equations with Ni being: N definition.

最后，这导致求解以下形式的一组方程:

Finally this results in solving a set of equations of the form:

a*x+b*y+c*xy=z1 d*x+e*y+f*xy=z2

，其中x和y是未知数.可以使用fsolve从数字上解决.

with x and y being the unknowns. This could be solved numerically using fsolve.

解决方案2:

此内容已在此处完整说明: math.stackexchange/Questions/828392/不规则网格的空间插值

This one is completely explained here: math.stackexchange/questions/828392/spatial-interpolation-for-irregular-grid

但是它非常复杂，我认为将花费更多的时间来编写它.

but it's quite complex and I think it will take me longer to code it.

推荐答案

由于缺少答案，我选择了第一个选项.您可以在下面找到代码.始终欢迎提出改进此代码的建议.

Due to a lack of answers I went for the first option. You can find the code below. Recommendations to improve this code are always welcome.

import numpy as np from scipy.optimize import fsolve def interpolate_quatrilateral(pt1,pt2,pt3,pt4,pt): '''Interpolates a value in a quatrilateral figure defined by 4 points. Each point is a tuple with 3 elements, x-coo,y-coo and value. point1 is the lower left corner, point 2 the lower right corner, point 3 the upper right corner and point 4 the upper left corner. args is a list of coordinates in the following order: x1,x2,x3,x4 and x (x-coo of point to be interpolated) and y1,y2... code based on the theory found here: www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.Ch16.d/IFEM.Ch16.pdf''' coos = (pt1[0],pt2[0],pt3[0],pt4[0],pt[0], pt1[1],pt2[1],pt3[1],pt4[1],pt[1]) #coordinates of the points merged in tuple guess = np.array([0,0]) #The center of the quadrilateral seem like a good place to start [eta, mu] = fsolve(func=find_local_coo_equations, x0=guess, args=coos) densities = (pt1[2], pt2[2], pt3[2], pt4[2]) density = find_density(eta,mu,densities) return density def find_local_coo_equations(guess, *args): '''This function creates the transformed coordinate equations of the quatrilateral.''' eta = guess[0] mu = guess[1] eq=[0,0]#Initialize eq eq[0] = 1 / 4 * (args[0] + args[1] + args[2] + args[3]) - args[4] + \ 1 / 4 * (-args[0] - args[1] + args[2] + args[3]) * mu + \ 1 / 4 * (-args[0] + args[1] + args[2] - args[3]) * eta + \ 1 / 4 * (args[0] - args[1] + args[2] - args[3]) * mu * eta eq[1] = 1 / 4 * (args[5] + args[6] + args[7] + args[8]) - args[9] + \ 1 / 4 * (-args[5] - args[6] + args[7] + args[8]) * mu + \ 1 / 4 * (-args[5] + args[6] + args[7] - args[8]) * eta + \ 1 / 4 * (args[5] - args[6] + args[7] - args[8]) * mu * eta return eq def find_density(eta,mu,densities): '''Finds the final density based on the eta and mu local coordinates calculated earlier and the densities of the 4 points''' N1 = 1/4*(1-eta)*(1-mu) N2 = 1/4*(1+eta)*(1-mu) N3 = 1/4*(1+eta)*(1+mu) N4 = 1/4*(1-eta)*(1+mu) density = densities[0]*N1+densities[1]*N2+densities[2]*N3+densities[3]*N4 return density pt1= (0,0,1) pt2= (1,0,1) pt3= (1,1,2) pt4= (0,1,2) pt= (0.5,0.5) print(interpolate_quatrilateral(pt1,pt2,pt3,pt4,pt))