没有fill_diagonal的inf的Python矩阵对角线 [英] Python matrix diagonal of inf without fill_diagonal

问题描述

我需要将矩阵的对角元素设置为Inf.

一种简单的方法是使用np.fill_diagonal.

np.fill_diagonal(my_matrix, float('inf')

但是fill_diagonal修改输入矩阵，而不是返回填充对角线的新矩阵. 这对我不起作用.我需要在不修改原始矩阵的情况下填充对角线.

当然我可以克隆原始矩阵，所以我将始终保留原始矩阵的副本.但是，我不太喜欢这种解决方案，因为我会经常更新原始矩阵，因此每次需要对角线为inf时，都必须对其进行复制.

是否有一个功能与fill_diagonal相同但不修改输入矩阵?像这样:

new_matrix = np.fill_diagonal(original_matrix, float('inf') 

为什么我需要这个:

我的矩阵是点之间的距离矩阵，我想在每一步中计算两个最近的点.当然，此矩阵的对角线为0(因为从点到自身的距离为0).因此，要确保我不会接受相同的观点，我的解决方案是将对角线设置为Inf.

但是，一旦找到两个点，我需要计算这两个点与其余点之间的平均距离，因此我实际上需要对角线为0而不是Inf.

当前我正在做的是

• 用Inf填充对角线
• 找到2个最接近的点
• 用0填充对角线

• 计算这两个点之间以及其余两个点之间的平均距离.

  # fill diagonal with Inf to avoid taking the diagonals
  np.fill_diagonal(data, float('inf'))  
  # find the minimum distance
  idx = np.argmin(data)
  # fill the diagonals back to 0
  np.fill_diagonal(data, 0.0) 
  # get the coordinates of the minimum distance
  row, col =  np.unravel_index(idx,data.shape)
  # compute the new node as the average distance between the two points
  new_node = np.mean((data[:,row],data[:,col]),0)
  # replace the first node (row) with the new node
  data[:,row] = new_node
  data[row,:] = new_node.T
  # delete the second node (col) from the matrix
  data = np.delete(data, col, 0)  # delete row
  data = np.delete(data, col, 1)  # delete column
  

但是，我不喜欢将对角线设置为Inf然后返回0的想法，我宁愿仅将函数传递给argmax，该函数返回用Inf填充对角线的数据，而无需实际修改矩阵数据.

类似的东西:

idx = np.argmin(return_filled_diagonals(data, float('Inf'))
# here I can operate with data as usual since it has not been modified.

解决方案

orig_mat = np.array([[1.2,2,3],[4,5,6],[7,8,9]])

#set diagonal to inf without making a copy of the array.
orig_mat + np.where(np.eye(orig_mat.shape[0])>0,np.inf,0)
array([[ inf,   2.,   3.],
       [  4.,  inf,   6.],
       [  7.,   8.,  inf]])

#the original array remains untorched.
print(orig_mat)
[[ 1.2  2.   3. ]
 [ 4.   5.   6. ]
 [ 7.   8.   9. ]]

I need to set the diagonal elements of a matrix to Inf.

An easy way to do it is to use np.fill_diagonal.

np.fill_diagonal(my_matrix, float('inf')

However fill_diagonal modifies the input matrix instead of returning a new matrix with the diagonal filled. This doesn't work for me. I need the diagonals filled WITHOUT modifying the original matrix.

Of course I could clone the original matrix, so I will always keep a copy of the original matrix. However I don't really like this solution, since I will update my original matrix often and therefore I'll have to make copies of it every time I need the diagonal to be inf.

Is there a function that will do the same that fill_diagonal but without modifying the input matrix? Something like:

new_matrix = np.fill_diagonal(original_matrix, float('inf') 

Why I need this:

My matrix is a distance matrix between points and I want to compute at each step the two closest points. Of course the diagonal of this matrix is 0 (since the distance from a point to itself is 0). So my solution to make sure I don't take the same point is to set the diagonals to Inf.

However once the two points are found, I need to compute the average of the distances between this two points and the rest of the points, so I actually need the diagonals to be 0 instead of Inf.

Currently what I'm doing is:

• Fill diagonals with Inf
• Find the 2 closest points
• Fill diagonals with 0

• Compute the average distance between this two points and the rest of them.

  # fill diagonal with Inf to avoid taking the diagonals
  np.fill_diagonal(data, float('inf'))  
  # find the minimum distance
  idx = np.argmin(data)
  # fill the diagonals back to 0
  np.fill_diagonal(data, 0.0) 
  # get the coordinates of the minimum distance
  row, col =  np.unravel_index(idx,data.shape)
  # compute the new node as the average distance between the two points
  new_node = np.mean((data[:,row],data[:,col]),0)
  # replace the first node (row) with the new node
  data[:,row] = new_node
  data[row,:] = new_node.T
  # delete the second node (col) from the matrix
  data = np.delete(data, col, 0)  # delete row
  data = np.delete(data, col, 1)  # delete column
  

However I don't like the idea of setting diagonals to Inf and then back to 0, I would prefer just passing a function to argmax that returns data with diagonal filled with Inf without actually modifying the matrix data.

Something like:

idx = np.argmin(return_filled_diagonals(data, float('Inf'))
# here I can operate with data as usual since it has not been modified.

解决方案

orig_mat = np.array([[1.2,2,3],[4,5,6],[7,8,9]])

#set diagonal to inf without making a copy of the array.
orig_mat + np.where(np.eye(orig_mat.shape[0])>0,np.inf,0)
array([[ inf,   2.,   3.],
       [  4.,  inf,   6.],
       [  7.,   8.,  inf]])

#the original array remains untorched.
print(orig_mat)
[[ 1.2  2.   3. ]
 [ 4.   5.   6. ]
 [ 7.   8.   9. ]]

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