使用常规网格将变量插值到Python scipy不在常规网格上的位置interpolate.interpn值错误

Question

我有一个netcdf文件变量，它是常规网格数据的时间，高度，lon和lat函数：U [时间，高度，lon，lat]。我想将此变量插值到不在常规网格（它位于网格点之间）的lon_new，lat_new的已定义位置。我希望能够根据单个插值位置获得变量U [0,0，lon_new，lat_new]。

我读了scipy插值函数，并认为scipy.interpolate.interpn是我想要使用的函数。我试图做一个这个功能的简单例子，但不断收到错误。

x_points = [1,2,3,4] #lets call this list of lons on the grid 
y_points = [1,2,3,4] #lets call this list of lats on the grid 

#Get the lon,lat pairs 
point_pairs=[] 
for i in x_points: 
    for j in y_points: 
     points = [i,j] 
     point_pairs.append(points) 
print point_pairs 
print np.shape(point_pairs) 

[[1, 1], [1, 2], [1, 3], [1, 4], [2, 1], [2, 2], [2, 3], [2, 4], [3, 1], [3, 2], [3, 3], [3, 4], [4, 1], [4, 2], [4, 3], [4, 4]] 
(16L, 2L) 

xi = (2.5,2.5) #point in between grid points that I am interested in getting  the interpolated value 
xi=np.array(xi) 
print xi 
print np.shape(xi) 

[ 2.5 2.5] 
(2L,) 

values = np.ones(16) #array of values at every grid point Let's say I loop over every grid point and get the value at each one 
print values 

[ 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1.] 

interpolated_value = interpolate.interpn(point_pairs, values, xi, method='linear') 

ValueError: There are 16 point arrays, but values has 1 dimensions 

Answer 1

您可以使用scipy中的任何适当的多元插值函数。在您的示例下面进行更正会产生正确的结果。

# -*- coding: utf-8 -*- 

import numpy as np 
from scipy import interpolate 

x_points = np.array([1, 2, 3, 4]) 
y_points = np.array([1, 2, 3, 4]) 
values = np.ones((4, 4)) # 2 dimensional array 
xi = np.array([2.5, 2.5]) 

interpolated_value = interpolate.interpn((x_points, y_points), values, xi, method='linear') 
print(interpolated_value)