python 稀疏矩阵问题整理

1. sparse.identity（N）： N*N的对角矩阵，对角元素为1
2. sparse.bmat(blocks[, format, dtype]）： 根据format的大小及其元素组织一个新的稀疏矩阵
   转：

https://www.it610.com/article/1344360099082219520.htm

        In [1]: A = np.arange(8).reshape(2, 4)
		
		In [2]: T = np.tri(5, 4)
		
		In [3]: L = [[8] * 4] * 2
		
		In [4]: I = sparse.identity(4)
		
		In [5]: Z = sparse.coo_matrix((2, 3))
		
		In [6]: sp.bmat([[   A,    Z,    L],
		    ...:          [None, None,    I],
		    ...:          [   T, None, None]], dtype=int)
		Out[7]:
		<11x11 sparse matrix of type 'numpy.int64'>'
		        with 33 stored elements in COOrdinate format>
		
		In [8]: _.toarray()  # ipython previous output
		Out[9]:
		array([[0, 1, 2, 3, 0, 0, 0, 8, 8, 8, 8],
		       [4, 5, 6, 7, 0, 0, 0, 8, 8, 8, 8],
		       [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
		       [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0],
		       [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0],
		       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
		       [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
		       [1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
		       [1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
		       [1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0],
		       [1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0]])