我试图采用稀疏矩阵及其转置的点积。我正在使用scipy.sparse库并发现结果不正确。见下文:
import numpy as np
import scipy.sparse
#Define the dense matrix
matrix_dense = np.zeros([100000,10])
for i in range(10):
i_0 = i*10000
i_1 = (i+1)*10000
matrix_dense[i_0:i_1,i] = 1
#Define the sparse matrix
cols = []
for i in range(10):
cols+=[i]*10000
dtype = np.uint8
rows = range(len(cols))
data_csc = np.ones(len(cols), dtype=dtype)
matrix_sparse = scipy.sparse.csc_matrix((data_csc, (rows, cols)), shape=(len(cols), 10), dtype=dtype)
#Check that the two matrices are identical
assert np.abs(matrix_sparse.todense() - matrix_dense).max() == 0
#Dot product of the dense matrix
dense_product = np.dot(matrix_dense.T,matrix_dense)
#Dot product of the sparse matrix
sparse_product = (matrix_sparse.T)*(matrix_sparse)
正确的答案(由dense_product给出)应该是对角矩阵,其中对角线项等于10,000。
print dense_product
[[ 10000. 0. 0. 0. 0. 0. 0. 0. 0.
0.]
[ 0. 10000. 0. 0. 0. 0. 0. 0. 0.
0.]
[ 0. 0. 10000. 0. 0. 0. 0. 0. 0.
0.]
[ 0. 0. 0. 10000. 0. 0. 0. 0. 0.
0.]
[ 0. 0. 0. 0. 10000. 0. 0. 0. 0.
0.]
[ 0. 0. 0. 0. 0. 10000. 0. 0. 0.
0.]
[ 0. 0. 0. 0. 0. 0. 10000. 0. 0.
0.]
[ 0. 0. 0. 0. 0. 0. 0. 10000. 0.
0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 10000.
0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0.
10000.]]
但是,无论我如何计算稀疏矩阵,结果都是错误的:
print sparse_product.todense()
[[16 0 0 0 0 0 0 0 0 0]
[ 0 16 0 0 0 0 0 0 0 0]
[ 0 0 16 0 0 0 0 0 0 0]
[ 0 0 0 16 0 0 0 0 0 0]
[ 0 0 0 0 16 0 0 0 0 0]
[ 0 0 0 0 0 16 0 0 0 0]
[ 0 0 0 0 0 0 16 0 0 0]
[ 0 0 0 0 0 0 0 16 0 0]
[ 0 0 0 0 0 0 0 0 16 0]
[ 0 0 0 0 0 0 0 0 0 16]]
我尝试过不同的方式来执行稀疏点积并获得完全相同的答案:
sparse_product_1 = np.dot(matrix_sparse.T,matrix_sparse)
sparse_product_2 = (matrix_sparse.T).dot(matrix_sparse)
sparse_product_3 = scipy.sparse.csr_matrix.dot((matrix_sparse.T),
matrix_sparse)
有什么想法吗?
答案 0 :(得分:2)
您似乎正在使用uint8
的数据类型,其最大值为256,并且可能是溢出的,最后是10000%256
,这样会给你16个。
以下是正在发生的事情的一个例子:
x = np.array(10000, dtype = np.uint8)
x
array(16, dtype=uint8)
将dtype更改为np.int64对我有用:
dtype = np.int64