我想使用python执行blinear插值。我要对高度进行插值的示例gps点是：
B = 54.4786674627
L = 17.0470721369
使用具有已知坐标和高度值的四个相邻点：
n = [(54.5, 17.041667, 31.993), (54.5, 17.083333, 31.911), (54.458333, 17.041667, 31.945), (54.458333, 17.083333, 31.866)]
z01 z11
z
z00 z10
这是我的原始尝试：
import math
z00 = n[0][2]
z01 = n[1][2]
z10 = n[2][2]
z11 = n[3][2]
c = 0.016667 #grid spacing
x0 = 56 #latitude of origin of grid
y0 = 13 #longitude of origin of grid
i = math.floor((L-y0)/c)
j = math.floor((B-x0)/c)
t = (B - x0)/c - j
z0 = (1-t)*z00 + t*z10
z1 = (1-t)*z01 + t*z11
s = (L-y0)/c - i
z = (1-s)*z0 + s*z1
其中z0和z1
z01 z0 z11
z
z00 z1 z10
我得到31.964，但从其他软件得到31.961。我的脚本正确吗？您可以提供另一种方法吗？
def bilinear_interpolation(x, y, points):
'''Interpolate (x,y) from values associated with four points.
The four points are a list of four triplets: (x, y, value).
The four points can be in any order. They should form a rectangle.
>>> bilinear_interpolation(12, 5.5,
... [(10, 4, 100),
... (20, 4, 200),
... (10, 6, 150),
... (20, 6, 300)])
165.0
'''
# See formula at: http://en.wikipedia.org/wiki/Bilinear_interpolation
points = sorted(points) # order points by x, then by y
(x1, y1, q11), (_x1, y2, q12), (x2, _y1, q21), (_x2, _y2, q22) = points
if x1 != _x1 or x2 != _x2 or y1 != _y1 or y2 != _y2:
raise ValueError('points do not form a rectangle')
if not x1 <= x <= x2 or not y1 <= y <= y2:
raise ValueError('(x, y) not within the rectangle')
return (q11 * (x2 - x) * (y2 - y) +
q21 * (x - x1) * (y2 - y) +
q12 * (x2 - x) * (y - y1) +
q22 * (x - x1) * (y - y1)
) / ((x2 - x1) * (y2 - y1) + 0.0)
您可以通过添加以下代码来运行测试代码：
if __name__ == '__main__': import doctest doctest.testmod()
在数据集上运行插值将产生：
>>> n = [(54.5, 17.041667, 31.993), (54.5, 17.083333, 31.911), (54.458333, 17.041667, 31.945), (54.458333, 17.083333, 31.866), ] >>> bilinear_interpolation(54.4786674627, 17.0470721369, n) 31.95798688313631
>>> import numpy as np
>>> from scipy.interpolate import griddata
>>> n = np.array([(54.5, 17.041667, 31.993),
(54.5, 17.083333, 31.911),
(54.458333, 17.041667, 31.945),
(54.458333, 17.083333, 31.866)])
>>> griddata(n[:,0:2], n[:,2], [(54.4786674627, 17.0470721369)], method='linear')
array([ 31.95817681])
from bisect import bisect_left
class BilinearInterpolation(object):
""" Bilinear interpolation. """
def __init__(self, x_index, y_index, values):
self.x_index = x_index
self.y_index = y_index
self.values = values
def __call__(self, x, y):
# local lookups
x_index, y_index, values = self.x_index, self.y_index, self.values
i = bisect_left(x_index, x) - 1
j = bisect_left(y_index, y) - 1
x1, x2 = x_index[i:i + 2]
y1, y2 = y_index[j:j + 2]
z11, z12 = values[j][i:i + 2]
z21, z22 = values[j + 1][i:i + 2]
return (z11 * (x2 - x) * (y2 - y) +
z21 * (x - x1) * (y2 - y) +
z12 * (x2 - x) * (y - y1) +
z22 * (x - x1) * (y - y1)) / ((x2 - x1) * (y2 - y1))
您可以这样使用它：
table = BilinearInterpolation( x_index=(54.458333, 54.5), y_index=(17.041667, 17.083333), values=((31.945, 31.866), (31.993, 31.911)) ) print(table(54.4786674627, 17.0470721369)) # 31.957986883136307
此版本没有错误检查，如果尝试在索引边界（或更高）使用它，将会遇到麻烦。有关代码的完整版本，包括错误检查和可选的外推，请查看here。
floor函数的要点是，通常您希望插入一个其坐标位于两个离散坐标之间的值。但是，您似乎已经有了最接近点的实际实际坐标值，这使数学运算变得简单。z00 = n[0][2]
z01 = n[1][2]
z10 = n[2][2]
z11 = n[3][2]
# Let's assume L is your x-coordinate and B is the Y-coordinate
dx = n[2][0] - n[0][0] # The x-gap between your sample points
dy = n[1][1] - n[0][1] # The Y-gap between your sample points
dx1 = (L - n[0][0]) / dx # How close is your point to the left?
dx2 = 1 - dx1 # How close is your point to the right?
dy1 = (B - n[0][1]) / dy # How close is your point to the bottom?
dy2 = 1 - dy1 # How close is your point to the top?
left = (z00 * dy1) + (z01 * dy2) # First interpolate along the y-axis
right = (z10 * dy1) + (z11 * dy2)
z = (left * dx1) + (right * dx2) # Then along the x-axis
从您的示例进行翻译时，可能会有一些错误的逻辑，但是要点是，可以根据每个点与插值目标点的距离比其邻近点的距离更近，对每个点进行加权。