性能测试

Question

我想用python进行blinear插值我要插入高度的示例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)]

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为

z01  z0  z11

     z
z00  z1   z10

我得到31.964但是从其他软件得到31.961 我的剧本是否正确？
你能提供另一种方法吗？

Answer 1

这是您可以使用的可重复使用的功能。它包括doctests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：

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：

>>> 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

Answer 2

不确定这是否有用，但在使用scipy进行线性插值时，我得到了不同的值：

>>> 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])

Answer 3

您还可以参考interp function in matplotlib。

Answer 4

受到here的启发，我提出了以下代码段。 API经过优化，可以在同一个表中重复使用很多次：

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))

你可以像这样使用它：

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，如果您尝试在索引边界（或更远）使用它，则会遇到麻烦。对于完整版本的代码，包括错误检查和可选的外推，请查看here。

Answer 5

我认为执行floor函数的重点在于，您通常希望插入一个坐标位于两个离散坐标之间的值。但是，您似乎已经拥有了最近点的实际实际坐标值，这使得数学变得简单。

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

从你的例子中翻译可能有一些错误的逻辑，但它的要点是你可以根据插值目标点与其他邻居的距离来对每个点进行加权。

Answer 6

基于此公式的numpy实现：

def bilinear_interpolation(x,y,x_,y_,val):

    a = 1 /((x_[1] - x_[0]) * (y_[1] - y_[0]))
    xx = np.array([[x_[1]-x],[x-x_[0]]],dtype='float32')
    f = np.array(val).reshape(2,2)
    yy = np.array([[y_[1]-y],[y-y_[0]]],dtype='float32')
    b = np.matmul(f,yy)

    return a * np.matmul(xx.T, b)

输入：在这里，x_是[x0,x1]的列表，y_是[y0,y1]的列表

bilinear_interpolation(x=54.4786674627,
                       y=17.0470721369,
                       x_=[54.458333,54.5],
                       y_=[17.041667,17.083333],
                       val=[31.993,31.911,31.945,31.866])

输出：

array([[31.95912739]])

Answer 7

我建议以下解决方案：

def bilinear_interpolation(x, y, z01, z11, z00, z10):

    def linear_interpolation(x, z0, z1):
        return z0 * x + z1 * (1 - x)

    return linear_interpolation(y, linear_interpolation(x, z01, z11),
                                   linear_interpolation(x, z00, z10))

Answer 8

这是与定义的here相同的解决方案，但应用于某些功能并与Scipy中的interp2d比较。我们使用numba库使插值函数比Scipy实现更快。

import numpy as np
from scipy.interpolate import interp2d
import matplotlib.pyplot as plt

from numba import jit, prange
@jit(nopython=True, fastmath=True, nogil=True, cache=True, parallel=True)
def bilinear_interpolation(x_in, y_in, f_in, x_out, y_out):
    f_out = np.zeros((y_out.size, x_out.size))
    
    for i in prange(f_out.shape[1]):
        idx = np.searchsorted(x_in, x_out[i])
        
        x1 = x_in[idx-1]
        x2 = x_in[idx]
        x = x_out[i]
        
        for j in prange(f_out.shape[0]):
            idy = np.searchsorted(y_in, y_out[j])
            y1 = y_in[idy-1]
            y2 = y_in[idy]
            y = y_out[j]

            
            f11 = f_in[idy-1, idx-1]
            f21 = f_in[idy-1, idx]
            f12 = f_in[idy, idx-1]
            f22 = f_in[idy, idx]
            

            
            f_out[j, i] = ((f11 * (x2 - x) * (y2 - y) +
                            f21 * (x - x1) * (y2 - y) +
                            f12 * (x2 - x) * (y - y1) +
                            f22 * (x - x1) * (y - y1)) /
                           ((x2 - x1) * (y2 - y1)))
    
    return f_out

我们使用很大的插值数组来评估每种方法的性能。

示例函数是

$z(x, y)=\sin \left(\frac{\pi x}{2}\right) e^{y / 2}$

x = np.linspace(0, 4, 13)
y = np.array([0, 2, 3, 3.5, 3.75, 3.875, 3.9375, 4])
X, Y = np.meshgrid(x, y)
Z = np.sin(np.pi*X/2) * np.exp(Y/2)

x2 = np.linspace(0, 4, 1000)
y2 = np.linspace(0, 4, 1000)
Z2 = bilinear_interpolation(x, y, Z, x2, y2)

fun = interp2d(x, y, Z, kind='linear')
Z3 = fun(x2, y2)


fig, ax = plt.subplots(nrows=1, ncols=3, figsize=(10, 6))
ax[0].pcolormesh(X, Y, Z, shading='auto')
ax[0].set_title("Original function")
X2, Y2 = np.meshgrid(x2, y2)
ax[1].pcolormesh(X2, Y2, Z2, shading='auto')
ax[1].set_title("bilinear interpolation")

ax[2].pcolormesh(X2, Y2, Z3, shading='auto')
ax[2].set_title("Scipy bilinear function")

plt.show()

性能测试

不带numba库的Python

在这种情况下，

bilinear_interpolation函数与numba版本相同，除了我们在for循环中使用python normal prange更改了range并删除了函数装饰器jit

%timeit bilinear_interpolation(x, y, Z, x2, y2)

每个循环给出7.15 s±107 ms（平均±标准偏差的7次运行，每个循环1次）

带有numba numba的Python

%timeit bilinear_interpolation(x, y, Z, x2, y2)

每个循环提供2.65 ms±70.5 µs（平均值±标准偏差，共运行7次，每个循环100个循环）

科学实施

%%timeit
f = interp2d(x, y, Z, kind='linear')
Z2 = f(x2, y2)

每个循环提供6.63 ms±145 µs（平均±标准偏差，共运行7次，每个循环100个循环）

性能测试是在'Intel（R）Core（TM）i7-8700K CPU @ 3.70GHz'上进行的

如何在Python中执行双线性插值

8 个答案:

性能测试

不带numba库的Python

带有numba numba的Python

科学实施