为什么Scipy的ndimage.map_coordinates对于某些数组不返回任何值或错误的结果? [英] Why is Scipy's ndimage.map_coordinates returning no values or wrong results for some arrays?

问题描述

在以下代码中，python为arr_b返回正确的插值，但没有为arr_a返回正确的插值.

In the following code, python is returning the correct interpolated value for arr_b but not for arr_a.

不过，事件，我已经研究这个问题大约一天了，我真的不确定发生了什么.

Event though, I've been looking at this problem for about a day now, I really am not sure what's going on.

由于某种原因，对于arr_a，twoD_interpolate始终返回[0]，即使我在数据和输入中玩弄或弄乱.

For some reason, for arr_a, twoD_interpolate keeps returning [0] even if I play around or mess around with the data and input.

如何修复我的代码，以便它实际上是在arr_a上插值并返回正确的结果?

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.]

代码返回的值不正确:

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

推荐答案

这实际上是我对原始问题的错.

This is actually my fault in the original question.

如果我们检查其位置试图插入twoD_interpolate(arr, 0, 1, 1400, 3200, 0.5, 1684)的位置，则会得到arr[ 170400, 0.1]作为要查找的值，该值将被mode='nearest'裁剪为arr[ -1 , 0.1].请注意，我切换了x和y以获取其在数组中出现的位置.

If we examine the position it is trying to interpolate twoD_interpolate(arr, 0, 1, 1400, 3200, 0.5, 1684) we get arr[ 170400, 0.1] as the value to find which will be clipped by mode='nearest' to arr[ -1 , 0.1]. Note I switched the x and y to get the positions as it would appear in an array.

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.

This corresponds to a interpolation from the values arr[-1,0] = 3.3 and arr[-1,1] = 4.7 so the interpolation looks like 3.3 * .9 + 4.7 * .1 = 3.44.

问题大步向前.如果我们采用从50到250的数组:

The issues comes in the stride. If we take an array that goes from 50 to 250:

>>> 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进行查看:

We can view this using 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，对于前面的示例，步幅为1，所以没关系.

So what we really need is (x-xmin) / stride, for previous examples the stride was 1 so it did not matter.

代码应为:

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]

检查一维或伪一维数组.除非输入数组的形状正确，否则将仅插值x维度:

Checking for arrays that are either 1D or pseudo 1D. Will interpolate the x dimension only unless the input array is of the proper shape:

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]

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