编辑:更新的事情,以反映你的澄清上述。你的问题现在更清楚了,谢谢!
基本上,你只是想在任意点插入一个二维数组。
scipy.ndimage.map_coordinates
是你想要的……
我理解您的问题,您有一个二维数组,其中“z”值的范围从一些xmin到xmax,以及每个方向的ymin到ymax。
在这些边界坐标之外的任何内容,都要从数组的边返回值。
地图坐标有几个选项可以处理网格边界以外的点,但它们都不能完全满足您的需要。相反,我们可以在边界之外设置任何东西,使其位于边缘,并像往常一样使用地图坐标。
因此,要使用地图坐标,您需要转动实际坐标:
| <1 2 3 4 5+
-------|----------------------------
<10000 | 3.6 6.5 9.1 11.5 13.8
20000 | 3.9 7.3 10.0 13.1 15.9
20000+ | 4.5 9.2 12.2 14.8 18.2
索引坐标:
| 0 1 2 3 4
-------|----------------------------
0 | 3.6 6.5 9.1 11.5 13.8
1 | 3.9 7.3 10.0 13.1 15.9
2 | 4.5 9.2 12.2 14.8 18.2
请注意,您的边界在每个方向上的行为都不同…在x方向上,事情表现得很平稳,但在y方向上,你要求一个“硬”休息,其中y=20000-->3.9,但y=20000.000001-->4.5。
举个例子:
import numpy as np
from scipy.ndimage import map_coordinates
#-- Setup ---------------------------
z = np.array([ [3.6, 6.5, 9.1, 11.5, 13.8],
[3.9, 7.3, 10.0, 13.1, 15.9],
[4.5, 9.2, 12.2, 14.8, 18.2] ])
ny, nx = z.shape
xmin, xmax = 1, 5
ymin, ymax = 10000, 20000
# Points we want to interpolate at
x1, y1 = 1.3, 25000
x2, y2 = 0.2, 50000
x3, y3 = 2.5, 15000
# To make our lives easier down the road, let's
# turn these into arrays of x & y coords
xi = np.array([x1, x2, x3], dtype=np.float)
yi = np.array([y1, y2, y3], dtype=np.float)
# Now, we'll set points outside the boundaries to lie along an edge
xi[xi > xmax] = xmax
xi[xi < xmin] = xmin
# To deal with the "hard" break, we'll have to treat y differently,
# so we're ust setting the min here...
yi[yi < ymin] = ymin
# We need to convert these to (float) indicies
# (xi should range from 0 to (nx - 1), etc)
xi = (nx - 1) * (xi - xmin) / (xmax - xmin)
# Deal with the "hard" break in the y-direction
yi = (ny - 2) * (yi - ymin) / (ymax - ymin)
yi[yi > 1] = 2.0
# Now we actually interpolate
# map_coordinates does cubic interpolation by default,
# use "order=1" to preform bilinear interpolation instead...
z1, z2, z3 = map_coordinates(z, [yi, xi])
# Display the results
for X, Y, Z in zip((x1, x2, x3), (y1, y2, y3), (z1, z2, z3)):
print X, ',', Y, '-->', Z
这就产生了:
1.3 , 25000 --> 5.1807375
0.2 , 50000 --> 4.5
2.5 , 15000 --> 8.12252371652
希望这有助于…