sparse_matrix

（1）ndarray 与 scipy.sparse.csr.csr_matrix 的互转

import numpy as np
from scipy import sparse

1.1 ndarry 转 csr_matrix

A = np.array([[1,2,0],[0,0,3],[1,0,4]])

array([[1, 2, 0],

[0, 0, 3],
[1, 0, 4]])

sA = sparse.csr_matrix(A) # Here's the initialization of the sparse matrix.

<3x3 sparse matrix of type '<type 'numpy.int32'>'
with 5 stored elements in Compressed Sparse Row format>

print sA

(0, 0) 1
(0, 1) 2
(1, 2) 3
(2, 0) 1
(2, 2) 4

1.2  csr_matrix转 ndarry 

my_matrix = scipy.sparse.csr_matrix((2,2))

my_array = my_matrix.A

type(my_array) numpy.ndarray

(2)在用python进行科学运算时，常常需要把一个稀疏的np.array压缩

按行压缩：sparse.csr_matrix(csr:Compressed Sparse Row marix)

按列压缩：sparse.csc_matric(csc:Compressed Sparse Column marix)

2.1 按row行来压缩

>>> indptr = np.array([0, 2, 3, 6])
>>> indices = np.array([0, 2, 2, 0, 1, 2])
>>> data = np.array([1, 2, 3, 4, 5, 6])
>>> csr_matrix((data, indices, indptr), shape=(3, 3)).toarray()
array([[1, 0, 2],
[0, 0, 3],
[4, 5, 6]])
# 对于第i行，非0数据列是indices[indptr[i]:indptr[i+1]] 数据是data[indptr[i]:indptr[i+1]]
# 第0行，有非0的数据列是indices[indptr[0]:indptr[1]] = indices[0:2] = [0,2]
# 数据是data[indptr[0]:indptr[1]] = data[0:2] = [1,2],所以在第0行第0列是1，第2列是2
# 第1行，有非0的数据列是indices[indptr[1]:indptr[2]] = indices[2:3] = [2]
# 数据是data[indptr[1]:indptr[2] = data[2:3] = [3],所以在第1行第2列是3
# 第2行，有非0的数据列是indices[indptr[2]:indptr[3]] = indices[3:6] = [0,1,2]
# 数据是data[indptr[2]:indptr[3]] = data[3:6] = [4,5,6],所以在第2行第0列是4，第1列是5,第2列是6

2.2 按col列来压缩
>>> indptr = np.array([0, 2, 3, 6])
>>> indices = np.array([0, 2, 2, 0, 1, 2])
>>> data = np.array([1, 2, 3, 4, 5, 6])
>>> csc_matrix((data, indices, indptr), shape=(3, 3)).toarray()
array([[1, 0, 4],
[0, 0, 5],
[2, 3, 6]])
# 对于第i列，非0数据行是indices[indptr[i]:indptr[i+1]] 数据是data[indptr[i]:indptr[i+1]]
# 第0列，有非0的数据行是indices[indptr[0]:indptr[1]] = indices[0:2] = [0,2]
# 数据是data[indptr[0]:indptr[1]] = data[0:2] = [1,2],所以在第0列第0行是1，第2行是2
# 第1行，有非0的数据行是indices[indptr[1]:indptr[2]] = indices[2:3] = [2]
# 数据是data[indptr[1]:indptr[2] = data[2:3] = [3],所以在第1列第2行是3
# 第2行，有非0的数据行是indices[indptr[2]:indptr[3]] = indices[3:6] = [0,1,2]
# 数据是data[indptr[2]:indptr[3]] = data[3:6] = [4,5,6],所以在第2列第0行是4，第1行是5,第2行是6

 2.3 初始化

row = np.array([0, 0, 1, 2, 2, 2])
col = np.array([0, 2, 2, 0, 1, 2])
data = np.array([1, 2, 3, 4, 5, 6])
a = csr_matrix((data, (row, col)), shape=(3, 3)).toarray()

array([[1, 0, 2],
       [0, 0, 3],
       [4, 5, 6]])