如何展示随着数据增加而稀疏的矩阵

huangapple
go评论94阅读模式
英文:

how to show sparse matrix with increasing data

问题

如何使稀疏矩阵以值部分递增的方式显示？我认为默认情况下，列部分是递增的。

英文:

how can we have the sparse matrix to be displayed in a way that the value part of a sparse matrix is increasing? I think by default, column part is increasing.

答案1

得分: 1

一个直观的方法是使用 sorted(zip()) 方法通过值来“手动”对列/行索引进行排序：

  1. from scipy.sparse import csr_matrix
  2. m = csr_matrix([[0, 0, 7], [2, 0, 0], [0, 5, 0]])
  4. print("As returned by csr:")
  5. print(m)
  7. print("Sorted by value:")
  8. for d, r, c in sorted(zip(m.data, m.tocoo().row, m.tocoo().col)):
  9.     print(f"  ({r}, {c})\t{d}")

这将得到以下结果：

  1. As returned by csr:
  2.   (0, 2)	7
  3.   (1, 0)	2
  4.   (2, 1)	5
  5. Sorted by value:
  6.   (1, 0)	2
  7.   (2, 1)	5
  8.   (0, 2)	7
英文:

A straight-forward approach would be to "manually" sort the col/row indices by the value, using a sorted(zip()) approach:

  1. from scipy.sparse import csr_matrix
  2. m = csr_matrix([[0, 0, 7], [2, 0, 0], [0, 5, 0]])
  4. print("As returned by csr:")
  5. print(m)
  7. print("Sorted by value:")
  8. for d, r, c in sorted(zip(m.data, m.tocoo().row, m.tocoo().col)):
  9.     print(f"  ({r}, {c})\t{d}")

which gives the following result:

  1. As returned by csr:
  2.   (0, 2)	7
  3.   (1, 0)	2
  4.   (2, 1)	5
  5. Sorted by value:
  6.   (1, 0)	2
  7.   (2, 1)	5
  8.   (0, 2)	7

答案2

得分: 1

这是关于对data进行排序并以相同顺序显示坐标的基本方法的变体：

首先，创建一个带有整数值的示例随机矩阵（便于显示）。我使用coo格式，因为它显示了数据和坐标之间的最简单关系。

  1. M = (sparse.random(10, 10, 0.1, 'coo') * 10).astype(int)

在ipython中，基本显示是repr版本，而print显示str版本。

现在获取数据的排序顺序：

  1. idx = np.argsort(M.data)

然后创建一个新的矩阵，其中行和列以相同的顺序排列：

  1. M1 = sparse.coo_matrix((M.data[idx], (M.row[idx], M.col[idx])), shape=M.shape)

这是相同的矩阵，但data数组已排序。

  1. M1.A

将coo格式转换为csr格式需要排序（按行和列的词法顺序），因此两个coo矩阵以相同的方式转换：

  1. M.tocsr().data

csr格式的规范顺序按行和列排序：

  1. print(M1.tocsr())
英文:

Here's a variation on the basic approach of sorting the data, and displaying the coordinates in the same order

First make a sample random matrix with integer values (for ease of display). I'm using coo format because that's the one that shows the simplest relation between data and coordinates.

  1. In [9]: M = (sparse.random(10,10,.1, 'coo')*10).astype(int)
  2. In [10]: M
  3. Out[10]: 
  4. <10x10 sparse matrix of type '<class 'numpy.int32'>'
  5. 	with 10 stored elements in COOrdinate format>
  6. In [11]: print(M)
  7.   (1, 3)	1
  8.   (3, 4)	7
  9.   (5, 4)	1
  10.   (7, 4)	8
  11.   (5, 6)	9
  12.   (6, 6)	7
  13.   (5, 8)	4
  14.   (6, 8)	1
  15.   (4, 9)	5
  16.   (9, 9)	1

In ipython, the basic display is the repr version, while print shows the str version.

Now get the sort order of the data:

  1. In [12]: idx = np.argsort(M.data)

and make a new matrix with row and column in the same order:

  1. In [13]: M1 = sparse.coo_matrix((M.data[idx],(M.row[idx], M.col[idx])),shape=M.shape)    
  2. In [14]: M1
  3. Out[14]: 
  4. <10x10 sparse matrix of type '<class 'numpy.int32'>'
  5. 	with 10 stored elements in COOrdinate format>
  7. In [15]: print(M1)
  8.   (1, 3)	1
  9.   (5, 4)	1
  10.   (6, 8)	1
  11.   (9, 9)	1
  12.   (5, 8)	4
  13.   (4, 9)	5
  14.   (3, 4)	7
  15.   (6, 6)	7
  16.   (7, 4)	8
  17.   (5, 6)	9

It's the same matrix, but the data array is sorted

  1. In [16]: M.A
  2. Out[16]: 
  3. array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  4.        [0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
  5.        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  6.        [0, 0, 0, 0, 7, 0, 0, 0, 0, 0],
  7.        [0, 0, 0, 0, 0, 0, 0, 0, 0, 5],
  8.        [0, 0, 0, 0, 1, 0, 9, 0, 4, 0],
  9.        [0, 0, 0, 0, 0, 0, 7, 0, 1, 0],
  10.        [0, 0, 0, 0, 8, 0, 0, 0, 0, 0],
  11.        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  12.        [0, 0, 0, 0, 0, 0, 0, 0, 0, 1]])
  14. In [17]: M1.A
  15. Out[17]: 
  16. array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  17.        [0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
  18.        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  19.        [0, 0, 0, 0, 7, 0, 0, 0, 0, 0],
  20.        [0, 0, 0, 0, 0, 0, 0, 0, 0, 5],
  21.        [0, 0, 0, 0, 1, 0, 9, 0, 4, 0],
  22.        [0, 0, 0, 0, 0, 0, 7, 0, 1, 0],
  23.        [0, 0, 0, 0, 8, 0, 0, 0, 0, 0],
  24.        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  25.        [0, 0, 0, 0, 0, 0, 0, 0, 0, 1]])
  27. In [18]: M1.data
  28. Out[18]: array([1, 1, 1, 1, 4, 5, 7, 7, 8, 9])

Converting the coo to csr format requires sorting (lexical for row and col), so both coo matrices are converted in the same way:

  1. In [19]: M.tocsr().data
  2. Out[19]: array([1, 7, 5, 1, 9, 4, 7, 1, 8, 1], dtype=int32)
  3. In [20]: M1.tocsr().data
  4. Out[20]: array([1, 7, 5, 1, 9, 4, 7, 1, 8, 1], dtype=int32)

The cannonical csr order by rows and columns within those:

  1. In [23]: print(M1.tocsr())
  2.   (1, 3)	1
  3.   (3, 4)	7
  4.   (4, 9)	5
  5.   (5, 4)	1
  6.   (5, 6)	9
  7.   (5, 8)	4
  8.   (6, 6)	7
  9.   (6, 8)	1
  10.   (7, 4)	8
  11.   (9, 9)	1

huangapple
  • 本文由 发表于 2023年6月13日 14:51:15
  • 转载请务必保留本文链接：https://go.coder-hub.com/76462308.html
  • python
  • sparse-matrix
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定