如何在Python中执行双线性插值
如何在Python中执行双线性插值
提问于 2011-12-29 05:29:39
我想使用python执行线性插值。
我要为其插入高程的示例gps点为：
B = 54.4786674627
L = 17.0470721369使用四个具有已知坐标和高度值的邻接点：
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*z1z0和z1在哪里
z01 z0 z11
z
z00 z1 z10我得到31.964，但从其他软件中我得到31.961。
我的脚本正确吗？
你能提供另一种方法吗？
回答 7
Stack Overflow用户
回答已采纳
发布于 2011-12-29 07:14:56
这里有一个你可以使用的可重用的函数。它包括文档测试和数据验证：
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内容来运行测试代码：
if __name__ == '__main__':
import doctest
doctest.testmod()在数据集上运行插值会产生以下结果：
>>> 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)
31.95798688313631Stack Overflow用户
发布于 2011-12-29 06:59:42
不确定这是否有用，但在使用scipy进行线性插值时，我得到了一个不同的值：
>>> 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])Stack Overflow用户
发布于 2013-12-27 03:03:05
受here的启发，我想出了以下代码片段。该API针对多次重用同一个表进行了优化：
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))你可以这样使用它：
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这个版本没有错误检查，如果您试图在索引边界(或更远的边界)使用它，您将遇到麻烦。有关代码的完整版本，包括错误检查和可选外推，请查看here。
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原文链接：
https://stackoverflow.com/questions/8661537
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