在Python中将json转换为coo稀疏矩阵答案

【问题标题】：Transforming json into coo sparse matrix in Python在Python中将json转换为coo稀疏矩阵
【发布时间】：2019-08-05 04:31:59
【问题描述】：

我正在尝试转换形状的 JSON 文件：

{"1": 
{"2": 0, "3": 0, "4": 0, "5": 1, "6": 0, "7": 1, "8": 0, "9": 0, "10": 0, "11": 1, "12": 1, "13": 0, "14": 1, "15": 1, "16": 0, "17": 0, "18": 0, "19": 0, "20": 0, "21": 0, "22": 0, "23": 0, "24": 0, "25": 0, "26": 0, "27": 0, "28": 0, "29": 0, "30": 0, "31": 1, "32": 0, "33": 0, "34": 1, "35": 0, "36": 0, "37": 0, "38": 0, "39": 0, "40": 0, "41": 0, "42": 0, "43": 0, "44": 0, "45": 0},
 "2": 
{"2": 0, "3": 0, "4": 0, "5": 0, "6": 1, "7": 0, "8": 1, "9": 1, "10": 1, "11": 0, "12": 0, "13": 1, "14": 1, "15": 1, "16": 0, "17": 0, "18": 1, "19": 0, "20": 1, "21": 1, "22": 0, "23": 0, "24": 0, "25": 1, "26": 0, "27": 0, "28": 0, "29": 1, "30": 0, "31": 1, "32": 1, "33": 0, "34": 0, "35": 0, "36": 0, "37": 1, "38": 0, "39": 0, "40": 1, "41": 1, "42": 0, "43": 0, "44": 1, "45": 1}, 
"3": 
{"2": 1, "3": 0, "4": 0, "5": 0, "6": 0, "7": 1, "8": 0, "9": 0, "10": 0, "11": 0, "12": 0, "13": 0, "14": 0, "15": 0, "16": 0, "17": 0, "18": 0, "19": 0, "20": 0, "21": 0, "22": 0, "23": 0, "24": 0, "25": 0, "26": 0, "27": 0, "28": 0, "29": 0, "30": 0, "31": 1, "32": 0, "33": 0, "34": 0, "35": 0, "36": 0, "37": 0, "38": 0, "39": 0, "40": 0, "41": 0, "42": 0, "43": 0, "44": 0, "45": 0}, 
"4": 
{"2": 1, "3": 1, "4": 1, "5": 1, "6": 0, "7": 0, "8": 0, "9": 0, "10": 0, "11": 1, "12": 1, "13": 0, "14": 0, "15": 0, "16": 1, "17": 1, "18": 0, "19": 0, "20": 0, "21": 0, "22": 1, "23": 1, "24": 1, "25": 0, "26": 1, "27": 1, "28": 1, "29": 0, "30": 1, "31": 0, "32": 0, "33": 0, "34": 1, "35": 0, "36": 1, "37": 0, "38": 1, "39": 0, "40": 0, "41": 0, "42": 1, "43": 1, "44": 0, "45": 0}}

进入一个 coo 稀疏矩阵，其中有一个坐标显示第一个键，然后是第二个键，然后是值，如下所示：

(1,2) 0
(1,3) 0
(1,4) 0
(1,5) 1
...
(4,44) 0
(4,45) 0

我尝试将 JSON 文件转换为如下所示的 pandas 数据框：

in  1   2   3   4
2   0   0   1   1
3   0   0   0   1
4   0   0   0   1
5   1   0   0   1
6   0   1   0   0
7   1   0   1   0
8   0   1   0   0
9   0   1   0   0
10  0   1   0   0
11  1   0   0   1
12  1   0   0   1
13  0   1   0   0
14  1   1   0   0
15  1   1   0   0
16  0   0   0   1
17  0   0   0   1
18  0   1   0   0
19  0   0   0   0
20  0   1   0   0
21  0   1   0   0
22  0   0   0   1
23  0   0   0   1
24  0   0   0   1
25  0   1   0   0
26  0   0   0   1
27  0   0   0   1
28  0   0   0   1
29  0   1   0   0
30  0   0   0   1
31  1   1   1   0
32  0   1   0   0
33  0   0   0   0
34  1   0   0   1
35  0   0   0   0
36  0   0   0   1
37  0   1   0   0
38  0   0   0   1
39  0   0   0   0
40  0   1   0   0
41  0   1   0   0
42  0   0   0   1
43  0   0   0   1
44  0   1   0   0
45  0   1   0   0

但我无法将其转换为稀疏矩阵，这将在放大时消除任何功能。

【问题讨论】：

只需遍历字典层，并通过列表追加创建标准coo 输入。有关这 3 个输入的详细信息，请参阅 coo 文档。

标签： python numpy scipy sparse-matrix

【解决方案1】：

当我将您的 json 复制粘贴到 Ipython 会话时，我会得到一个包含 4 个键的字典。

我可以将它解压到一个列表中：

In [466]: alist = [] 
     ...: for k,v in adict.items(): 
     ...:     for k1,v1 in v.items(): 
     ...:         alist.append((int(k),int(k1),v1)) 
     ...:

并制作一个数组：

In [467]: arr = np.array(alist)                                                                              
In [468]: arr.shape                                                                                          
Out[468]: (176, 3)

并使用数组的 3 列作为输入 sparse.coo_matrix:

In [469]: M = sparse.coo_matrix((arr[:,2],(arr[:,0],arr[:,1])))                                              
In [470]: M                                                                                                  
Out[470]: 
<5x46 sparse matrix of type '<class 'numpy.int64'>'
    with 176 stored elements in COOrdinate format>
In [471]: M.A                                                                                                
Out[471]: 
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0],
       [0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0],
       [0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1,
        0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0,
        1, 1],
       [0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0],
       [0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0,
        1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1,
        0, 0]])

变体：

In [472]: rows, cols, data = [],[],[] 
     ...: for k,v in adict.items(): 
     ...:     for k1,v1 in v.items(): 
     ...:         rows.append(int(k)) 
     ...:         cols.append(int(k1)) 
     ...:         data.append(v1) 
     ...:                                                                                                    
In [473]: len(rows)                                                                                          
Out[473]: 176
In [474]: M = sparse.coo_matrix((data,(rows,cols)))

【讨论】：

【解决方案2】：

我们可以通过字典理解将字典转换为“多索引数据帧”。例如：

pd.concat({k: pd.DataFrame.from_dict(v, orient='index') for k,v in data.items()})

对于给定的样本数据，这将产生一个包含 176 行和 1 列的数据框：

>>> pd.concat({k: pd.DataFrame.from_dict(v, orient='index') for k,v in data.items()})
      0
1 2   0
  3   0
  4   0
  5   1
  6   0
  7   1
  8   0
  9   0
  10  0
  11  1
  12  1
  13  0
  14  1
  15  1
  16  0
  17  0
  18  0
  19  0
  20  0
  21  0
  22  0
  23  0
  24  0
  25  0
  26  0
  27  0
  28  0
  29  0
  30  0
  31  1
...  ..
4 16  1
  17  1
  18  0
  19  0
  20  0
  21  0
  22  1
  23  1
  24  1
  25  0
  26  1
  27  1
  28  1
  29  0
  30  1
  31  0
  32  0
  33  0
  34  1
  35  0
  36  1
  37  0
  38  1
  39  0
  40  0
  41  0
  42  1
  43  1
  44  0
  45  0

[176 rows x 1 columns]

【讨论】：

感谢您的回复。这很棒！但它返回一个 DF。我正在尝试返回一个 coo 稀疏矩阵。