使用Python中的最近邻方法，根据给定的缩放因子来扩大2D数组

huangapple
go评论99阅读模式
英文:

Enlarging 2D array with a given rescaling factor using nearest-neighbour in Python

问题

我正在尝试使用Python中的最近邻算法编写一个以给定缩放因子放大2D数组的代码。

例如，我有一个如下所示的数组：

  1. [[1, 2],
  2. [3, 4]]

我想要使用最近邻算法和给定的缩放因子来放大这个数组。

让我一步一步来解释。假设缩放因子是3。放大后的数组应该如下所示：

  1. [[1, 0, 0, 2, 0, 0],
  2. [0, 0, 0, 0, 0, 0],
  3. [0, 0, 0, 0, 0, 0],
  4. [3, 0, 0, 4, 0, 0],
  5. [0, 0, 0, 0, 0, 0],
  6. [0, 0, 0, 0, 0, 0]]

填充空元素后，它应该如下所示：

  1. [[1, 1, 2, 2, 2, 2],
  2. [1, 1, 2, 2, 2, 2],
  3. [3, 3, 4, 4, 4, 4],
  4. [3, 3, 4, 4, 4, 4],
  5. [3, 3, 4, 4, 4, 4],
  6. [3, 3, 4, 4, 4, 4]]

这就是输出应该看起来的样子。 (0,2) 的值是 2 而不是 1，因为它的最近邻是 (0,3) 处的 2，而不是 (0,0) 处的 1。

如何实现这个？很容易创建以下类似的数组：

  1. [[1, 1, 1, 2, 2, 2],
  2. [1, 1, 1, 2, 2, 2], 
  3. [1, 1, 1, 2, 2, 2], 
  4. [3, 3, 3, 4, 4, 4], 
  5. [3, 3, 3, 4, 4, 4], 
  6. [3, 3, 3, 4, 4, 4]]

但这不是我想要的。

英文:

I'm trying to write a code that enlarges 2D array with a given rescaling factor in Python using nearest-neighbour algorithm.

For example, I have an array that looks like below.

  1. [[1, 2],
  2. [3, 4]]

And what I want to do is enlarging this array with NN algorithm and a given rescaling factor.

Let me explain step by step. Let's assume that the rescaling factor is 3. The enlarged array should look like below:

  1. [[1, 0, 0, 2, 0, 0],
  2. [0, 0, 0, 0, 0, 0],
  3. [0, 0, 0, 0, 0, 0],
  4. [3, 0, 0, 4, 0, 0],
  5. [0, 0, 0, 0, 0, 0],
  6. [0, 0, 0, 0, 0, 0]]

And after filling the empty elements, it should look like below.

  1. [[1, 1, 2, 2, 2, 2],
  2. [1, 1, 2, 2, 2, 2],
  3. [3, 3, 4, 4, 4, 4],
  4. [3, 3, 4, 4, 4, 4],
  5. [3, 3, 4, 4, 4, 4],
  6. [3, 3, 4, 4, 4, 4]]

This is what output should look like. (0,2) is 2 instead of 1 because its nearest neighbour is 2 at (0,3) not 1 at (0,0).

How can I achieve this?

It was easy to create an array like below:

  1. [[1, 1, 1, 2, 2, 2],
  2. [1, 1, 1, 2, 2, 2], 
  3. [1, 1, 1, 2, 2, 2], 
  4. [3, 3, 3, 4, 4, 4], 
  5. [3, 3, 3, 4, 4, 4], 
  6. [3, 3, 3, 4, 4, 4]]

But It is not what I wanted.

答案1

得分: 0

首先需要创建填充数组，但我们将使用 np.nan 填充数组，以便进行下一步的插值。因为如果在填充之前已经有元素 0，那么当我们使用 0 计算 mask 时，这将导致错误的 mask。以下是填充函数：

  1. def pad_data(arr, padlen):
  2.     m, n = arr.shape
  3.     out = np.empty((m * padlen, n * padlen)) * np.nan
  4.     for i in range(m):
  5.         for j in range(n):
  6.             out[i * padlen, j * padlen] = arr[i, j]
  7.     return out

然后，我们需要使用 scipy 中的 NearestNDInterpolator 进行最近邻插值。完整的代码如下：

  1. import numpy as np
  2. from scipy.interpolate import NearestNDInterpolator
  4. def pad_data(arr, padlen):
  5.     m, n = arr.shape
  6.     out = np.empty((m * padlen, n * padlen)) * np.nan
  7.     for i in range(m):
  8.         for j in range(n):
  9.             out[i * padlen, j * padlen] = arr[i, j]
  10.     return out
  12. arr = np.array([[1, 2], [3, 4]])
  13. arr_pad = pad_data(arr, 3)
  15. mask = np.where(~np.isnan(arr_pad))
  16. interp = NearestNDInterpolator(np.transpose(mask), arr_pad[mask])
  17. filled_data = interp(*np.indices(arr_pad.shape))
  18. filled_data

这将给出以下结果：

  1. array([[1., 1., 2., 2., 2., 2.],
  2.        [1., 1., 2., 2., 2., 2.],
  3.        [3., 3., 4., 4., 4., 4.],
  4.        [3., 3., 4., 4., 4., 4.],
  5.        [3., 3., 4., 4., 4., 4.],
  6.        [3., 3., 4., 4., 4., 4.]])
英文:

First need to create the padded array, but we will pad the array with np.nan for the interpolation of the next step. Cause if you already have element 0 before padding, then when we calculate the mask with 0s, this will give us a wrong mask. Here is the function for padding :

  1. def pad_data(arr,padlen):
  2.     m,n = arr.shape
  3.     out= np.empty((m*padlen, n*padlen)) * np.nan
  4.     for i in range(m):
  5.         for j in range(n):
  6.             out[i*padlen, j*padlen] = arr[i,j]
  7.     return out

Then we need to use the NearestNDInterpolator in scipy for the nearest interpolation. The full code as below:

  1. import numpy as np
  2. from scipy.interpolate import NearestNDInterpolator
  4. def pad_data(arr,padlen):
  5.     m,n = arr.shape
  6.     out= np.empty((m*padlen, n*padlen)) * np.nan
  7.     for i in range(m):
  8.         for j in range(n):
  9.             out[i*padlen, j*padlen] = arr[i,j]
  10.     return out
  14. arr = np.array([[1, 2],[3, 4]])
  15. arr_pad = pad_data(arr,3)
  17. mask = np.where(~np.isnan(arr_pad))
  18. interp = NearestNDInterpolator(np.transpose(mask), arr_pad[mask])
  19. filled_data = interp(*np.indices(arr_pad.shape))
  20. filled_data

Gives you :

  1. array([[1., 1., 2., 2., 2., 2.],
  2.        [1., 1., 2., 2., 2., 2.],
  3.        [3., 3., 4., 4., 4., 4.],
  4.        [3., 3., 4., 4., 4., 4.],
  5.        [3., 3., 4., 4., 4., 4.],
  6.        [3., 3., 4., 4., 4., 4.]])

huangapple
  • 本文由 发表于 2023年4月4日 17:06:41
  • 转载请务必保留本文链接：https://go.coder-hub.com/75927486.html
  • arrays
  • nearest-neighbor
  • python
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定