将python稀疏矩阵dict转换为scipy稀疏矩阵

Question

I am using python scikit-learn for document clustering and I have a sparse matrix stored in a dict object:

For example:

doc_term_dict = { ('d1','t1'): 12,             \
                  ('d2','t3'): 10,             \
                  ('d3','t2'):  5              \
                  }                            # from mysql data table 
<type 'dict'>

I want to use scikit-learn to do the clustering where the input matrix type is scipy.sparse.csr.csr_matrix

Example:

(0, 2164)   0.245793088885
(0, 2076)   0.205702177467
(0, 2037)   0.193810934784
(0, 2005)   0.14547028437
(0, 1953)   0.153720023365
...
<class 'scipy.sparse.csr.csr_matrix'>

I can't find a way to convert dict to this csr-matrix (I have never used scipy .)

Answer 1

Pretty straightforward. First read the dictionary and convert the keys to the appropriate row and column. Scipy supports (and recommends for this purpose) the COO-rdinate format for sparse matrices.

Pass it data , row , and column , where A[row[k], column[k] = data[k] (for all k) defines the matrix. Then let Scipy do the conversion to CSR.

Please check, that I have rows and columns in the way you want them, I might have them transposed. I also assumed that the input would be 1-indexed.

My code below prints:

(0, 0)        12
(1, 2)        10
(2, 1)        5

Code:

#!/usr/bin/env python3
#http://stackoverflow.com/questions/26335059/converting-python-sparse-matrix-dict-to-scipy-sparse-matrix

from scipy.sparse import csr_matrix, coo_matrix

def convert(term_dict):
    ''' Convert a dictionary with elements of form ('d1', 't1'): 12 to a CSR type matrix.
    The element ('d1', 't1'): 12 becomes entry (0, 0) = 12.
    * Conversion from 1-indexed to 0-indexed.
    * d is row
    * t is column.
    '''
    # Create the appropriate format for the COO format.
    data = []
    row = []
    col = []
    for k, v in term_dict.items():
        r = int(k[0][1:])
        c = int(k[1][1:])
        data.append(v)
        row.append(r-1)
        col.append(c-1)
    # Create the COO-matrix
    coo = coo_matrix((data,(row,col)))
    # Let Scipy convert COO to CSR format and return
    return csr_matrix(coo)

if __name__=='__main__':
    doc_term_dict = { ('d1','t1'): 12,             \
                ('d2','t3'): 10,             \
                ('d3','t2'):  5              \
                }   
    print(convert(doc_term_dict))

Answer 2

We can make @Unapiedra's (excellent) answer a little more sparse:

from scipy.sparse import csr_matrix
def _dict_to_csr(term_dict):
    term_dict_v = list(term_dict.itervalues())
    term_dict_k = list(term_dict.iterkeys())
    shape = list(repeat(np.asarray(term_dict_k).max() + 1,2))
    csr = csr_matrix((term_dict_v, zip(*term_dict_k)), shape = shape)
    return csr

Answer 3

Same as @carsonc, but for Python 3.X :

from scipy.sparse import csr_matrix
def _dict_to_csr(term_dict):
    term_dict_v = term_dict.values()
    term_dict_k = term_dict.keys()
    term_dict_k_zip = zip(*term_dict_k)
    term_dict_k_zip_list = list(term_dict_k_zip)

    shape = (len(term_dict_k_zip_list[0]), len(term_dict_k_zip_list[1]))
    csr = csr_matrix((list(term_dict_v), list(map(list, zip(*term_dict_k)))), shape = shape)
    return csr