沿numpy数组中值的轴按元素计数 [英] element-wise count along axis of values in numpy array

问题描述

如何沿给定轴获得 numpy 数组中每个元素出现次数的元素计数?逐元素"是指数组的每个值都应该转换为它出现的次数.

How can I get an element-wise count of each element's number of occurrences in a numpy array, along a given axis? By "element-wise," I mean each value of the array should be converted to the number of times it appears.

简单的二维输入:

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

应该输出:

[[3, 3, 3],
 [3, 3, 3],
 [1, 1, 1]]

该解决方案还需要相对于给定轴起作用.例如，如果我的输入数组 a 具有形状 (4, 2, 3, 3)，我认为它是一个 3x3 矩阵的 4x2 矩阵"，运行 solution(a) 应该吐出上述形式的 (4, 2, 3, 3) 解决方案，其中每个 3x3 子矩阵"包含相对于该子矩阵的相应元素的计数，而不是整个 numpy 数组.

The solution also needs to work relative to a given axis. For example, if my input array a has shape (4, 2, 3, 3), which I think of as "a 4x2 matrix of 3x3 matrices," running solution(a) should spit out a (4, 2, 3, 3) solution of the form above, where each 3x3 "submatrix" contains counts of the corresponding elements relative to that submatrix alone, rather than the entire numpy array at large.

更复杂的例子:假设我以 a 上面的例子输入并调用 skimage.util.shape.view_as_windows(a, (2, 2)).这给了我形状 (2, 2, 2, 2):

More complex example: suppose I take the example input above a and call skimage.util.shape.view_as_windows(a, (2, 2)). This gives me array b of shape (2, 2, 2, 2):

[[[[1 1]
   [2 2]]

  [[1 1]
   [2 2]]]


 [[[2 2]
   [3 4]]

  [[2 2]
   [4 5]]]]

然后solution(b)应该输出:

[[[[2 2]
   [2 2]]

  [[2 2]
   [2 2]]]


 [[[2 2]
   [1 1]]

  [[2 2]
   [1 1]]]]

因此，即使值 1 在 a 中出现 3 次，在 b 中出现 4 次，它在每个 2x2 窗口中只出现两次.

So even though the value 1 occurs 3 times in a and 4 times in b, it only occurs twice in each 2x2 window.

推荐答案

入门方法

我们可以使用np.unique 获取出现次数，并从 0 开始标记每个元素，让我们用所需输出的标签索引这些计数，就像这样 -

We can use np.unique to get the counts of occurrences and also tag each element from 0 onwards, letting us index into those counts with the tags for the desired output, like so -

In [43]: a
Out[43]: 
array([[1, 1, 1],
       [2, 2, 2],
       [3, 4, 5]])

In [44]: _,ids,c = np.unique(a, return_counts=True, return_inverse=True)

In [45]: c[ids].reshape(a.shape)
Out[45]: 
array([[3, 3, 3],
       [3, 3, 3],
       [1, 1, 1]])

对于输入数组中的正整数，我们也可以使用 np.bincount -

For positive integers numbers in input array, we can also use np.bincount -

In [73]: c = np.bincount(a.ravel())

In [74]: c[a]
Out[74]: 
array([[3, 3, 3],
       [3, 3, 3],
       [1, 1, 1]])

对于负整数，只需将其中的最小值偏移即可.

For negative integers numbers, simply offset by the minimum in it.

扩展到通用的 n-dims

让我们为此使用 bincount -

In [107]: ar
Out[107]: 
array([[[1, 1, 1],
        [2, 2, 2],
        [3, 4, 5]],

       [[2, 3, 5],
        [4, 3, 4],
        [3, 1, 2]]])

In [104]: ar2D = ar.reshape(-1,ar.shape[-2]*ar.shape[-1])

# bincount2D_vectorized from https://stackoverflow.com/a/46256361/ @Divakar
In [105]: c = bincount2D_vectorized(ar2D)

In [106]: c[np.arange(ar2D.shape[0])[:,None], ar2D].reshape(ar.shape)
Out[106]: 
array([[[3, 3, 3],
        [3, 3, 3],
        [1, 1, 1]],

       [[2, 3, 1],
        [2, 3, 2],
        [3, 1, 2]]])

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