将每行的最后一个非零元素设置为零-NumPy [英] Set last non-zero element of each row to zero - NumPy

问题描述

我有一个数组A:

A = array([[1, 2, 3,4], [5,6,7,0] , [8,9,0,0]])

我想将每行的最后一个非零更改为0

I want to change the last non-zero of each row to 0

A = array([[1, 2, 3,0], [5,6,0,0] , [8,0,0,0]])

如何为任何n * m numpy数组编写代码? 谢谢，S;-)

how to write the code for any n*m numpy array? Thanks, S ;-)

推荐答案

方法1

基于cumsum和argmax的一种方法-

A[np.arange(A.shape[0]),(A!=0).cumsum(1).argmax(1)] = 0

样品运行-

In [59]: A
Out[59]: 
array([[2, 0, 3, 4],
       [5, 6, 7, 0],
       [8, 9, 0, 0]])

In [60]: A[np.arange(A.shape[0]),(A!=0).cumsum(1).argmax(1)] = 0

In [61]: A
Out[61]: 
array([[2, 0, 3, 0],
       [5, 6, 0, 0],
       [8, 0, 0, 0]])

方法2

又一个基于argmax的代码，希望效率更高-

One more based on argmax and hopefully more efficient -

A[np.arange(A.shape[0]),A.shape[1] - 1 - (A[:,::-1]!=0).argmax(1)] = 0

---

说明

argmax的用途之一是获取max元素在数组中沿轴的 first 出现的ID.在第一种方法中，我们沿行获取总和，并获取第一个最大ID，该ID代表最后一个非零元素.这是因为剩余元素上的cumsum不会在最后一个非零元素之后增加总和值.

One of the uses of argmax is to get ID of the first occurence of the max element along an axis in an array . In the first approach we get the cumsum along the rows and get the first max ID, which represents the last non-zero elem. This is because cumsum on the leftover elements won't increase the sum value after that last non-zero element.

让我们以更详细的方式重新运行该案例-

Let's re-run that case in a bit more detailed manner -

In [105]: A
Out[105]: 
array([[2, 0, 3, 4],
       [5, 6, 7, 0],
       [8, 9, 0, 0]])

In [106]: (A!=0)
Out[106]: 
array([[ True, False,  True,  True],
       [ True,  True,  True, False],
       [ True,  True, False, False]], dtype=bool)

In [107]: (A!=0).cumsum(1)
Out[107]: 
array([[1, 1, 2, 3],
       [1, 2, 3, 3],
       [1, 2, 2, 2]])

In [108]: (A!=0).cumsum(1).argmax(1)
Out[108]: array([3, 2, 1])

最后，我们使用fancy-indexing将其用作列索引，以在A中设置适当的元素.

Finally, we use fancy-indexing to use those as the column indices to set appropriate elements in A.

在第二种方法中，当我们在布尔数组上使用argmax时，我们只得到了True的第一次出现，我们在输入数组的行翻转版本中使用了它.这样，我们将拥有原始顺序中的最后一个非零元素.剩下的想法是一样的.

In the second approach, when we use argmax on the boolean array, we simply got the first occurence of True, which we used on a row-flipped version of the input array. As such, we would have the last non-zero elem in the original order. Rest of the idea there, is the same.

这篇关于将每行的最后一个非零元素设置为零-NumPy的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！