Question

我有三列非结构化数据，并希望对它们进行双变量样条拟合。我对Python中的类不太好，所以我不明白如何做到这一点。为了表明我的问题，我做了一个简单的代码：

#! /usr/bin/env python3

import numpy as np
from scipy import interpolate

#an array of 3 columns:
a=np.zeros((200, 3))
a[:,0]=np.random.uniform(0,1,200)
a[:,1]=np.random.uniform(3,5,200)
a[:,2]=np.random.uniform(10,12,200)

#find the boundries
min_x, max_x = np.amin(a[:,0]), np.amax(a[:,0])
min_y, max_y = np.amin(a[:,1]), np.amax(a[:,1])

#Set the resolution:
x_res=1000
y_res=int( ( (max_y-min_y) / (max_x-min_x) )*x_res )

#Make a grid
grid_x, grid_y = np.mgrid[min_x:max_x:x_res*1j, min_y:max_y:y_res*1j]

sbsp=interpolate.SmoothBivariateSpline(a[:,0], a[:,1], a[:,2])

b=sbsp.ev(4,5)
#c=sbsp.ev(grid_x, grid_y)
print(b)

这给出了一个点的插值，但是如果你注释掉倒数第二行，它就不起作用了。如果有人能指导我如何在网格上进行样条插值，我将非常感激。提前谢谢。

Answer 1

方法ev(x,y)要求x和y为一维数组。在您的代码中，grid_x和grid_y是2D。

您可以尝试以下方法：

c=sbsp.ev(grid_x[0,0], grid_y[0,0])

Answer 2

我一直在努力解决类似问题，并最终解决了它。对我而言，关键是使用凹凸的linspace创建两个一维网格，一个用于x，一个用于y。并使用https://docs.scipy.org/doc/scipy/reference/generated/scipy.interpolate.SmoothBivariateSpline.call.html#scipy.interpolate.SmoothBivariateSpline.call中的.__call__在网格上进行评估。

在下面的脚本中，我评估一个样条线或两个x，y，z数组，每个数组将作为图像输出。

#!/bin/python3.7

"""

Use spline interpolation to on grid of x,y,z value where z is either xdiff or ydiff for use as imagemagick 2D displacement maps

"""

import numpy as np
from scipy.interpolate import SmoothBivariateSpline
from skimage import io


# python lists of x,y dst control points and zx=xsrc-xdiff, zy=ysrc-ydiff offsets to be interpolated over full image of size 129x129
x = [8.5, 20.5, 33.5, 48.5, 64.5, 80.5, 95.5, 109.5, 121.5, 5.5, 17.5, 31.5, 46.5, 64.5, 81.5, 97.5, 111.5, 123.5, 2.5, 14.5, 29.5, 45.5, 64.5, 83.5, 99.5, 113.5, 125.5, 1.5, 12.5, 26.5, 43.5, 64.5, 85.5, 103.5, 116.5, 127.5, 0.5, 11.5, 24.5, 41.5, 64.5, 87.5, 103.5, 117.5, 128.5, 1.5, 12.5, 25.5, 42.5, 64.5, 86.5, 103.5, 116.5, 127.5, 2.5, 14.5, 28.5, 45.5, 64.5, 83.5, 100.5, 114.5, 125.5, 5.5, 17.5, 30.5, 46.5, 64.5, 81.5, 97.5, 111.5, 123.5, 8.5, 19.5, 33.5, 48.5, 64.5, 80.5, 95.5, 109.5, 121.5]
y = [7.5, 5.5, 3.5, 1.5, 1.5, 1.5, 3.5, 5.5, 7.5, 20.5, 16.5, 14.5, 12.5, 11.5, 12.5, 15.5, 16.5, 19.5, 33.5, 31.5, 28.5, 26.5, 24.5, 26.5, 28.5, 31.5, 33.5, 48.5, 47.5, 45.5, 42.5, 40.5, 42.5, 45.5, 46.5, 48.5, 64.5, 64.5, 64.5, 64.5, 64.5, 64.5, 64.5, 64.5, 64.5, 80.5, 81.5, 83.5, 86.5, 87.5, 86.5, 83.5, 81.5, 80.5, 95.5, 97.5, 100.5, 103.5, 104.5, 102.5, 100.5, 97.5, 95.5, 109.5, 111.5, 114.5, 116.5, 117.5, 116.5, 114.5, 111.5, 109.5, 121.5, 123.5, 125.5, 127.5, 127.5, 127.5, 125.5, 123.5, 120.5]
zx = [119.5, 123.5, 126.5, 127.5, 127.5, 127.5, 128.5, 130.5, 134.5, 122.5, 126.5, 128.5, 129.5, 127.5, 126.5, 126.5, 128.5, 132.5, 125.5, 129.5, 130.5, 130.5, 127.5, 124.5, 124.5, 126.5, 130.5, 126.5, 131.5, 133.5, 132.5, 127.5, 122.5, 120.5, 123.5, 128.5, 127.5, 132.5, 135.5, 134.5, 127.5, 120.5, 120.5, 122.5, 127.5, 126.5, 131.5, 134.5, 133.5, 127.5, 121.5, 120.5, 123.5, 128.5, 125.5, 129.5, 131.5, 130.5, 127.5, 124.5, 123.5, 125.5, 130.5, 122.5, 126.5, 129.5, 129.5, 127.5, 126.5, 126.5, 128.5, 132.5, 119.5, 124.5, 126.5, 127.5, 127.5, 127.5, 128.5, 130.5, 134.5]
zy = [120.5, 122.5, 124.5, 126.5, 126.5, 126.5, 124.5, 122.5, 120.5, 123.5, 127.5, 129.5, 131.5, 132.5, 131.5, 128.5, 127.5, 124.5, 126.5, 128.5, 131.5, 133.5, 135.5, 133.5, 131.5, 128.5, 126.5, 127.5, 128.5, 130.5, 133.5, 135.5, 133.5, 130.5, 129.5, 127.5, 127.5, 127.5, 127.5, 127.5, 127.5, 127.5, 127.5, 127.5, 127.5, 127.5, 126.5, 124.5, 121.5, 120.5, 121.5, 124.5, 126.5, 127.5, 128.5, 126.5, 123.5, 120.5, 119.5, 121.5, 123.5, 126.5, 128.5, 130.5, 128.5, 125.5, 123.5, 122.5, 123.5, 125.5, 128.5, 130.5, 134.5, 132.5, 130.5, 128.5, 128.5, 128.5, 130.5, 132.5, 135.5]

# convert python lists to numpy arrays
ax = np.asarray(x)
ay = np.asarray(y)
azx = np.asarray(zx)
azy = np.asarray(zy)


# define bbox of interpolated data
# bbox=[minx, maxx, miny, maxy]
bbox=[0, 129, 0, 129]

# convert bbox to numpy array
abbox = np.asarray(bbox)

# do interpolations
xd = SmoothBivariateSpline(ax, ay, azx, w=None, bbox=abbox, kx=3, ky=3)
yd = SmoothBivariateSpline(ax, ay, azy, w=None, bbox=abbox, kx=3, ky=3)

# define integer grid onto which to interpolate
grid_x=np.linspace(0, 129, num=129)
grid_y=np.linspace(0, 129, num=129)


# evaluate at grid points
xdisplace = xd.__call__(grid_x, grid_y, grid=True)
ydisplace = yd.__call__(grid_x, grid_y, grid=True)

# save output using skimage
io.imsave("xdimgs.png", xdisplace.astype('uint8'))
io.imsave("ydimgs.png", ydisplace.astype('uint8'))

# view output using skimage
io.imshow(xdisplace.astype('uint8')) 
io.show()
io.imshow(ydisplace.astype('uint8')) 
io.show()

也许这会对您的脚本有所帮助。

通过网格评估SmoothBivariateSpline

2 个答案: