numpy的：一个阵列的增量元素给予递增所需的索引 [英] Numpy: increment elements of an array given the indices required to increment

问题描述

我试图把第二阶张成二进制三阶张量。给定一个二阶张量amxn numpy的数组：A，我需要把各要素值：X，A和用矢量替换为：V，其尺寸等于A的最大值，但增加值1以v的对应于x值的指数（即v [X] = 1）。在矩阵，其中涉及增量给定指数：我一直在关注这个问题，在生产由2维坐标给定指数与增量的数组。我一直在阅读的答案，尝试使用np.ravel_multi_index（）和np.bincount（）做相同，但与3维坐标，但是我一直得到一个ValueError错误：在坐标数组无效条目。这是我一直使用的是什么：

I am trying to turn a second order tensor into a binary third order tensor. Given a second order tensor as a m x n numpy array: A, I need to take each element value: x, in A and replace it with a vector: v, with dimensions equal to the maximum value of A, but with a value of 1 incremented at the index of v corresponding to the value x (i.e. v[x] = 1). I have been following this question: Increment given indices in a matrix, which addresses producing an array with increments at indices given by 2 dimensional coordinates. I have been reading the answers and trying to use np.ravel_multi_index() and np.bincount() to do the same but with 3 dimensional coordinates, however I keep on getting a ValueError: "invalid entry in coordinates array". This is what I have been using:

def expand_to_tensor_3(array):
    (x, y) = array.shape
    (a, b) = np.indices((x, y))
    a = a.reshape(x*y)
    b = b.reshape(x*y)
    tensor_3 = np.bincount(np.ravel_multi_index((a, b, array.reshape(x*y)), (x, y, np.amax(array))))
    return tensor_3

如果你知道这里有什么问题或知道一个更好的方法来完成我的目标，无论是真正有用的，谢谢。

If you know what is wrong here or know an even better method to accomplish my goal, both would be really helpful, thanks.

推荐答案

您可以使用（A [：，：，np.newaxis] == np.arange（A.max（）+ 1 ））。astype（INT）。

下面是一个演示：

In [52]: A
Out[52]: 
array([[2, 0, 0, 2],
       [3, 1, 2, 3],
       [3, 2, 1, 0]])

In [53]: B = (A[:,:,np.newaxis] == np.arange(A.max()+1)).astype(int)

In [54]: B
Out[54]: 
array([[[0, 0, 1, 0],
        [1, 0, 0, 0],
        [1, 0, 0, 0],
        [0, 0, 1, 0]],

       [[0, 0, 0, 1],
        [0, 1, 0, 0],
        [0, 0, 1, 0],
        [0, 0, 0, 1]],

       [[0, 0, 0, 1],
        [0, 0, 1, 0],
        [0, 1, 0, 0],
        [1, 0, 0, 0]]])

检查几个单个元素 A ：

In [55]: A[0,0]
Out[55]: 2

In [56]: B[0,0,:]
Out[56]: array([0, 0, 1, 0])

In [57]: A[1,3]
Out[57]: 3

In [58]: B[1,3,:]
Out[58]: array([0, 0, 0, 1])

这位前pression A [：，：，np.newaxis] == np.arange（A.max（）+ 1）使用的广播来的每个元素比较 A 为 np.arange（A.max（）+ 1）。对于单个值，这看起来像：

The expression A[:,:,np.newaxis] == np.arange(A.max()+1) uses broadcasting to compare each element of A to np.arange(A.max()+1). For a single value, this looks like:

In [63]: 3 == np.arange(A.max()+1)
Out[63]: array([False, False, False,  True], dtype=bool)

In [64]: (3 == np.arange(A.max()+1)).astype(int)
Out[64]: array([0, 0, 0, 1])

A [：，：，np.newaxis] 是 A 与形状的三维视图（3,4,1）。的额外的维度被添加，使得比较 np.arange（A.max（）+ 1）将广播到每个元素，从而具有形状的结果（3，4，A.max（）+ 1）。

A[:,:,np.newaxis] is a three-dimensional view of A with shape (3,4,1). The extra dimension is added so that the comparison to np.arange(A.max()+1) will broadcast to each element, giving a result with shape (3, 4, A.max()+1).

通过一个微不足道的变化，这将对于n维阵列工作。索引的省略号 ... A numpy的数组是指所有其他方面。因此，

With a trivial change, this will work for an n-dimensional array. Indexing a numpy array with the ellipsis ... means "all the other dimensions". So

(A[..., np.newaxis] == np.arange(A.max()+1)).astype(int)

转换n维数组的（N + 1）维数组，其中最后一维是整数的 A 二元指标。这里有一个一维数组的例子：

converts an n-dimensional array to an (n+1)-dimensional array, where the last dimension is the binary indicator of the integer in A. Here's an example with a one-dimensional array:

In [6]: a = np.array([3, 4, 0, 1])

In [7]: (a[...,np.newaxis] == np.arange(a.max()+1)).astype(int)
Out[7]: 
array([[0, 0, 0, 1, 0],
       [0, 0, 0, 0, 1],
       [1, 0, 0, 0, 0],
       [0, 1, 0, 0, 0]])

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