数组中唯一值的索引 [英] Indices of unique values in array

问题描述

我有一个二维Numpy数组，包含从0到n的值. 我想得到一个长度为n的列表，以使该列表的第i个元素是值i + 1(不包括0)的所有索引的数组.

I have a 2D Numpy array containing values from 0 to n. I want to get a list of length n, such that the i'th element of that list is an array of all the indices with value i+1 (0 is excluded).

例如，对于输入

array([[1, 0, 1],
   [2, 2, 0]])

我希望得到

[array([[0, 0], [0, 2]]), array([[1,0], [1,1]])]

我发现了以下相关问题: 获取所有索引的列表numpy数组中重复元素的数量 这可能会有所帮助，但是我希望找到一个更直接的解决方案，该解决方案不需要对数组进行展平和排序，并且尽可能高效.

I found this related question: Get a list of all indices of repeated elements in a numpy array which may be helpful, but I hoped to find a more direct solution that doesn't require flattening and sorting the array and that is as efficient as possible.

推荐答案

这是一种矢量化方法，适用于任意维数的数组.此解决方案的想法是扩展

Here's a vectorized approach, which works for arrays of an arbitrary amount of dimensions. The idea of this solution is to extend the functionality of the return_index method in np.unique, and return an array of arrays, each containing the N-dimensional indices of unique values in a numpy array.

对于更紧凑的解决方案，我在以下各个步骤中定义了以下功能以及一些解释:

For a more compact solution, I've defined the following function along with some explanations throughout the different steps:

def ndix_unique(x):
    """
    Returns an N-dimensional array of indices
    of the unique values in x
    ----------
    x: np.array
       Array with arbitrary dimensions
    Returns
    -------
    - 1D-array of sorted unique values
    - Array of arrays. Each array contains the indices where a
      given value in x is found
    """
    x_flat = x.ravel()
    ix_flat = np.argsort(x_flat)
    u, ix_u = np.unique(x_flat[ix_flat], return_index=True)
    ix_ndim = np.unravel_index(ix_flat, x.shape)
    ix_ndim = np.c_[ix_ndim] if x.ndim > 1 else ix_flat
    return u, np.split(ix_ndim, ix_u[1:])

---

检查问题数组-

---

Checking with the array from the question -

a = np.array([[1, 0, 1],[2, 2, 0]])

vals, ixs = ndix_unique(a)

print(vals)
array([0, 1, 2])

print(ixs)
[array([[0, 1],
        [1, 2]]), 
 array([[0, 0],
        [0, 2]]), 
 array([[1, 0],
        [1, 1]])]

让我们尝试其他情况:

a = np.array([[1,1,4],[2,2,1],[3,3,1]])

vals, ixs = ndix_unique(a)

print(vals)
array([1, 2, 3, 4])

print(ixs)
array([array([[0, 0],
              [0, 1],
              [1, 2],
              [2, 2]]),
       array([[1, 0],
              [1, 1]]), 
       array([[2, 0],
              [2, 1]]),
       array([[0, 2]])], dtype=object)

对于 1D 阵列:

a = np.array([1,5,4,3,3])

vals, ixs = ndix_unique(a)

print(vals)
array([1, 3, 4, 5])

print(ixs)
array([array([0]), array([3, 4]), array([2]), array([1])], dtype=object)

最后一个带有 3D ndarray的示例:

Finally another example with a 3D ndarray:

a = np.array([[[1,1,2]],[[2,3,4]]])

vals, ixs = ndix_unique(a)

print(vals)
array([1, 2, 3, 4])

print(ixs)
array([array([[0, 0, 0],
              [0, 0, 1]]),
       array([[0, 0, 2],
              [1, 0, 0]]), 
       array([[1, 0, 1]]),
       array([[1, 0, 2]])], dtype=object)

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