将 numpy float64 稀疏矩阵转换为 pandas 数据帧

Question

I have an nxn numpy float64 sparse matrix ( data , where n = 44 ), where the rows and columns are graph nodes and the values are edge weights:

>>> data
<44x44 sparse matrix of type '<class 'numpy.float64'>'
    with 668 stored elements in Compressed Sparse Row format>

>>> type(data)
<class 'scipy.sparse.csr.csr_matrix'>

>>> print(data)
  (0, 7)    0.11793236293516568
  (0, 9)    0.10992000939300195
  (0, 21)   0.7422196678913772
  (0, 23)   0.0630039712667936
  (0, 24)   0.027037442463504143
  (0, 27)   0.16908845414214152
  (0, 28)   0.6109227233402952
  (0, 32)   0.0514765253537568
  (0, 33)   0.016341754080557713
  (1, 6)    0.015070325434709386
  (1, 10)   9.346673769086203e-05
  (1, 11)   0.2471018034781923
  (1, 14)   0.0020684269551621776
  (1, 18)   0.015258704502643251
  (1, 20)   0.021798149289490358
  (1, 22)   0.0087026831764125
  (1, 24)   0.1454235884185166
  (1, 25)   0.022060777594183015
  (1, 29)   0.9117391202819067
  (1, 30)   0.018557883854566116
  (1, 31)   0.001876070225734826
  (1, 32)   0.025841354399637764
  (1, 33)   0.014766488228364438
  (1, 39)   0.002791226433410351
  (1, 43)   1.0
  : :
  (41, 7)   0.8922099840113696
  (41, 10)  0.015776226631920767
  (41, 12)  1.0
  (41, 15)  0.1839408706622038
  (41, 18)  0.5151025641025642
  (41, 20)  0.4599130036630037
  (41, 22)  0.29378473237788827
  (41, 33)  0.47474890700697153
  (41, 39)  1.0
  (42, 2)   1.0
  (42, 10)  0.023305789342610222
  (42, 11)  0.011349136164776494
  (42, 12)  1.0
  (42, 17)  0.886081346522542
  (42, 18)  1.0
  (42, 30)  1.0
  (42, 40)  1.0
  (43, 1)   1.0
  (43, 6)   1.0
  (43, 11)  0.039948959300013256
  (43, 13)  1.0
  (43, 14)  0.02669811947637717
  (43, 29)  1.0
  (43, 30)  1.0
  (43, 36)  0.3381986531986532

I'd like to convert it to a pandas data frame , in order to write it to a file, with the columns: node1, node2, edge_weight , which will therefore give:

node1, node2, edge_weight
0, 7, 0.11793236293516568
0, 9, 0.10992000939300195
:, :, :
43, 36, 0.3381986531986532

Any idea how to do that?

Note that:

>>> pandas.DataFrame(data)

gives:

                                                    0
0     (0, 7)\t0.11793236293516568\n  (0, 9)\t0.109...
1     (0, 6)\t0.015070325434709386\n  (0, 10)\t9.3...

And

>>> pandas.DataFrame(print(data))

Gives:

  (0, 7)    0.11793236293516568
  (0, 9)    0.10992000939300195

So I guess pandas.DataFrame(print(data)) is close to what I'm looking for.

Answer 1

你可以尝试toarray

pd.DataFrame(A.toarray())

Answer 2

This ipython session shows one way you could do it. The two steps are: convert the sparse matrix to COO format, and then create the Pandas DataFrame using the .row , .col and .data attributes of the COO matrix.

In [50]: data                                                                                                    
Out[50]: 
<15x15 sparse matrix of type '<class 'numpy.float64'>'
    with 11 stored elements in Compressed Sparse Row format>

In [51]: print(data)                                                                                             
  (1, 12)   0.8581958095588134
  (6, 12)   0.03828052946099181
  (6, 14)   0.7908634838351427
  (7, 1)    0.7995008873930302
  (7, 11)   0.48477191537121145
  (7, 13)   0.6226526443518743
  (9, 4)    0.37242576669669103
  (11, 1)   0.9604278557580955
  (11, 5)   0.13285436036287313
  (12, 11)  0.5631419223609928
  (13, 8)   0.16481624650723847

In [52]: import pandas as pd                                                                                     

In [53]: c = data.tocoo()                                                                                        

In [54]: df = pd.DataFrame({node1: c.row, node2: c.col, edge_weight: c.data})                                   

In [55]: df                                                                                                      
Out[55]: 
    node1  node2  edge_weight
0       1     12     0.858196
1       6     12     0.038281
2       6     14     0.790863
3       7      1     0.799501
4       7     11     0.484772
5       7     13     0.622653
6       9      4     0.372426
7      11      1     0.960428
8      11      5     0.132854
9      12     11     0.563142
10     13      8     0.164816

Answer 3

I ran into a similar problem when using OneHotEncoder I fixed it by changing sparse to False

enc = OneHotEncoder(sparse=False)