1 Answer. You can use either todense() or toarray() function to convert a CSR matrix to a dense matrix.
A dense matrix stored in a NumPy array can be converted into a sparse matrix using the CSR representation by calling the csr_matrix() function.
The function csr_matrix() is used to create a sparse matrix of compressed sparse row format whereas csc_matrix() is used to create a sparse matrix of compressed sparse column format.
Transposing a sparse matrix is simple enough, we just have to swap the row and column values and then sort the rows in the sparse matrix.
from scipy.sparse import csr_matrix
A = csr_matrix([[1,0,2],[0,3,0]])
>>>A
<2x3 sparse matrix of type '<type 'numpy.int64'>'
with 3 stored elements in Compressed Sparse Row format>
>>> A.todense()
matrix([[1, 0, 2],
[0, 3, 0]])
>>> A.toarray()
array([[1, 0, 2],
[0, 3, 0]])
this is an example of how to convert a sparse matrix to a dense matrix taken from scipy
I solved this problem using Pandas. Because we want to keep the document ids and term ids.
from pandas import DataFrame
# A sparse matrix in dictionary form (can be a SQLite database). Tuples contains doc_id and term_id.
doc_term_dict={('d1','t1'):12, ('d2','t3'):10, ('d3','t2'):5}
#extract all unique documents and terms ids and intialize a empty dataframe.
rows = set([d for (d,t) in doc_term_dict.keys()])
cols = set([t for (d,t) in doc_term_dict.keys()])
df = DataFrame(index = rows, columns = cols )
df = df.fillna(0)
#assign all nonzero values in dataframe
for key, value in doc_term_dict.items():
df[key[1]][key[0]] = value
print df
Output:
t2 t3 t1
d2 0 10 0
d3 5 0 0
d1 0 0 12
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