在下面的代码中，python为arr_b返回正确的插值，但不为arr_a返回。
事件虽然，我一直在看这个问题大约一天，我真的不确定发生了什么。
出于某种原因，对于arr_a，即使我玩弄数据和输入，两个D_interpolate也会一直返回[0]。
如何修复我的代码，以便实际插入arr_a并返回正确的结果？
import numpy as np
from scipy.ndimage import map_coordinates
def twoD_interpolate(arr, xmin, xmax, ymin, ymax, x1, y1):
"""
interpolate in two dimensions with "hard edges"
"""
ny, nx = arr.shape # Note the order of ny and xy
x1 = np.atleast_1d(x1)
y1 = np.atleast_1d(y1)
# Mask upper and lower boundaries using @Jamies suggestion
np.clip(x1, xmin, xmax, out=x1)
np.clip(y1, ymin, ymax, out=y1)
# Change coordinates to match your array.
x1 = (x1 - xmin) * (xmax - xmin) / float(nx - 1)
y1 = (y1 - ymin) * (ymax - ymin) / float(ny - 1)
# order=1 is required to return your examples.
return map_coordinates(arr, np.vstack((y1, x1)), order=1)
# test data
arr_a = np.array([[0.7, 1.7, 2.5, 2.8, 2.9],
[1.9, 2.9, 3.7, 4.0, 4.2],
[1.4, 2.0, 2.5, 2.7, 3.9],
[1.1, 1.3, 1.6, 1.9, 2.0],
[0.6, 0.9, 1.1, 1.3, 1.4],
[0.6, 0.7, 0.9, 1.1, 1.2],
[0.5, 0.7, 0.9, 0.9, 1.1],
[0.5, 0.6, 0.7, 0.7, 0.9],
[0.5, 0.6, 0.6, 0.6, 0.7]])
arr_b = np.array([[6.4, 5.60, 4.8, 4.15, 3.5, 2.85, 2.2],
[5.3, 4.50, 3.7, 3.05, 2.4, 1.75, 1.1],
[4.7, 3.85, 3.0, 2.35, 1.7, 1.05, 0.4],
[4.2, 3.40, 2.6, 1.95, 1.3, 0.65, 0.0]])
# Test the second array
print twoD_interpolate(arr_b, 0, 6, 9, 12, 4, 11)
# Test first area
print twoD_interpolate(
arr_a, 0, 500, 0, 2000, 0, 2000)
print arr_a[0]
print twoD_interpolate(
arr_a_60, 0, 500, 0, 2000, 0, 2000)[0]
print twoD_interpolate(
arr_a, 20, 100, 100, 1600, 902, 50)
print twoD_interpolate(
arr_a, 100, 1600, 20, 100, 902, 50)
print twoD_interpolate(
arr_a, 100, 1600, 20, 100, 50, 902)
## Output
[ 1.7]
[ 0.]
[ 0.7 1.7 2.5 2.8 2.9]
0.0
[ 0.]
[ 0.]
[ 0.]
arr = np.array([[12.8, 20.0, 23.8, 26.2, 27.4, 28.6],
[10.0, 13.6, 15.8, 17.4, 18.2, 18.8],
[5.5, 7.7, 8.7, 9.5, 10.1, 10.3],
[3.3, 4.7, 5.1, 5.5, 5.7, 6.1]])
twoD_interpolate(arr, 0, 1, 1400, 3200, 0.5, 1684)
# above should return 21 but is returning 3.44
答案 0 :(得分：2)
这实际上是我在原问题中的错。
如果我们检查其尝试插入twoD_interpolate(arr, 0, 1, 1400, 3200, 0.5, 1684)的位置，我们会得到arr[ 170400, 0.1]作为要查找的值mode='nearest'到arr[ -1 , 0.1]。注意我切换了x和y以获取数组中显示的位置。
这对应于值arr[-1,0] = 3.3和arr[-1,1] = 4.7的插值，因此插值看起来像3.3 * .9 + 4.7 * .1 = 3.44。
问题大步迈进。如果我们采用从50到250的数组：
>>> a=np.arange(50,300,50)
>>> a
array([ 50, 100, 150, 200, 250])
>>> stride=float(a.max()-a.min())/(a.shape[0]-1)
>>> stride
50.0
>>> (75-a.min()) * stride
1250.0 #Not what we want!
>>> (75-a.min()) / stride
0.5 #There we go
>>> (175-a.min()) / stride
2.5 #Looks good
我们可以使用map_coordinates：
#Input array from the above.
print map_coordinates(arr, np.array([[.5,2.5,1250]]), order=1, mode='nearest')
[ 75 175 250] #First two are correct, last is incorrect.
所以我们真正需要的是(x-xmin) / stride，对于前面的例子，步幅为1，所以没关系。
以下是代码：
def twoD_interpolate(arr, xmin, xmax, ymin, ymax, x1, y1):
"""
interpolate in two dimensions with "hard edges"
"""
arr = np.atleast_2d(arr)
ny, nx = arr.shape # Note the order of ny and xy
x1 = np.atleast_1d(x1)
y1 = np.atleast_1d(y1)
# Change coordinates to match your array.
if nx==1:
x1 = np.zeros_like(x1.shape)
else:
x_stride = (xmax-xmin)/float(nx-1)
x1 = (x1 - xmin) / x_stride
if ny==1:
y1 = np.zeros_like(y1.shape)
else:
y_stride = (ymax-ymin)/float(ny-1)
y1 = (y1 - ymin) / y_stride
# order=1 is required to return your examples and mode=nearest prevents the need of clip.
return map_coordinates(arr, np.vstack((y1, x1)), order=1, mode='nearest')
请注意，mode='nearest'不需要剪辑。
print twoD_interpolate(arr, 0, 1, 1400, 3200, 0.5, 1684)
[ 21.024]
print twoD_interpolate(arr, 0, 1, 1400, 3200, 0, 50000)
[ 3.3]
print twoD_interpolate(arr, 0, 1, 1400, 3200, .5, 50000)
[ 5.3]
检查1D或伪1D的阵列。除非输入数组的形状正确，否则将仅插入x维度：
arr = np.arange(50,300,50)
print twoD_interpolate(arr, 50, 250, 0, 5, 75, 0)
[75]
arr = np.arange(50,300,50)[None,:]
print twoD_interpolate(arr, 50, 250, 0, 5, 75, 0)
[75]
arr = np.arange(50,300,50)
print twoD_interpolate(arr, 0, 5, 50, 250, 0, 75)
[50] #Still interpolates the `x` dimension.
arr = np.arange(50,300,50)[:,None]
print twoD_interpolate(arr, 0, 5, 50, 250, 0, 75)
[75]