使用另一个矩阵的索引引用矩阵中的行 [英] referencing rows in a matrix using index from another matrix

问题描述

您有一个原始的稀疏矩阵X:

You have an original sparse matrix X:

>>print type(X) 
>>print X.todense()
<class 'scipy.sparse.csr.csr_matrix'>
[[1,4,3]
 [3,4,1]
 [2,1,1]
 [3,6,3]]

您有第二个稀疏矩阵Z，它是从X的某些行派生而来的(例如，将值加倍，以便我们可以看到两个矩阵之间的差异).在pseudo-code:

You have a second sparse matrix Z, which is derived from some rows of X (say the values are doubled so we can see the difference between the two matrices). In pseudo-code:

>>Z = X[[0,2,3]]
>>print Z.todense()
[[1,4,3]
 [2,1,1]
 [3,6,3]]

>>Z = Z*2
>>print Z.todense()
[[2, 8, 6]
 [4, 2, 2]
 [6, 12,6]]

使用来自X的原始索引来检索Z中的行的最佳方法是什么.例如，以伪代码:

What's the best way of retrieving the rows in Z using the ORIGINAL indices from X. So for instance, in pseudo-code:

>>print Z[[0,3]]
[[2,8,6]              #0 from Z, and what would be row **0** from X)
 [6,12,6]]            #2 from Z, but what would be row **3** from X)

也就是说，如何使用引用原始矩阵X中原始行位置的索引从Z中检索行?为此，您无论如何都不能修改X(您不能向矩阵X添加索引列)，但是没有其他限制.

That is, how can you retrieve rows from Z, using indices that refer to the original rows position in the original matrix X? To do this, you can't modify X in anyway (you can't add an index column to the matrix X), but there are no other limits.

推荐答案

如果原始索引位于数组i中，并且i中的值按升序排列(如您的示例)，则可以使用numpy.searchsorted(i，[0，3])在Z中找到与原始X中的索引[0，3]对应的索引.这是IPython会话中的一个演示:

If you have the original indices in an array i, and the values in i are in increasing order (as in your example), you can use numpy.searchsorted(i, [0, 3]) to find the indices in Z that correspond to indices [0, 3] in the original X. Here's a demonstration in an IPython session:

In [39]: X = csr_matrix([[1,4,3],[3,4,1],[2,1,1],[3,6,3]])

In [40]: X.todense()
Out[40]: 
matrix([[1, 4, 3],
        [3, 4, 1],
        [2, 1, 1],
        [3, 6, 3]])

In [41]: i = array([0, 2, 3])

In [42]: Z = 2 * X[i]

In [43]: Z.todense()
Out[43]: 
matrix([[ 2,  8,  6],
        [ 4,  2,  2],
        [ 6, 12,  6]])

In [44]: Zsub = Z[searchsorted(i, [0, 3])]

In [45]: Zsub.todense()
Out[45]: 
matrix([[ 2,  8,  6],
        [ 6, 12,  6]])

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