Python 将具有图形节点和边级别的数据帧转换为方形矩阵

我的Googlefu让我失望了

我有一个熊猫

数据帧

，格式如下：

Level 1   Level 2   Level 3   Level 4
-------------------------------------
A         B         C         NaN
A         B         D         E
A         B         D         F
G         H         NaN       NaN
G         I         J         K

     A   B   C   D   E   F   G   H   I   J   K
  ---------------------------------------------
A |  0   1   0   0   0   0   0   0   0   0   0
B |  0   0   1   1   0   0   0   0   0   0   0
C |  0   0   0   0   0   0   0   0   0   0   0
D |  0   0   0   0   1   1   0   0   0   0   0
E |  0   0   0   0   0   0   0   0   0   0   0
F |  0   0   0   0   0   0   0   0   0   0   0
G |  0   0   0   0   0   0   0   1   1   0   0
H |  0   0   0   0   0   0   0   0   0   0   0
I |  0   0   0   0   0   0   0   0   0   1   0
J |  0   0   0   0   0   0   0   0   0   0   1
K |  0   0   0   0   0   0   0   0   0   0   0

它基本上包含一个图的节点，其级别描述了从低阶级别到高阶级别的输出边。我想转换数据帧/创建表单的新数据帧：

Level 1   Level 2   Level 3   Level 4
-------------------------------------
A         B         C         NaN
A         B         D         E
A         B         D         F
G         H         NaN       NaN
G         I         J         K

     A   B   C   D   E   F   G   H   I   J   K
  ---------------------------------------------
A |  0   1   0   0   0   0   0   0   0   0   0
B |  0   0   1   1   0   0   0   0   0   0   0
C |  0   0   0   0   0   0   0   0   0   0   0
D |  0   0   0   0   1   1   0   0   0   0   0
E |  0   0   0   0   0   0   0   0   0   0   0
F |  0   0   0   0   0   0   0   0   0   0   0
G |  0   0   0   0   0   0   0   1   1   0   0
H |  0   0   0   0   0   0   0   0   0   0   0
I |  0   0   0   0   0   0   0   0   0   1   0
J |  0   0   0   0   0   0   0   0   0   0   1
K |  0   0   0   0   0   0   0   0   0   0   0

包含

1

的单元格表示从相应行到相应列的输出边。在Pandas中是否有一种不需要循环和条件的Pythonic方法来实现这一点？

试试下面的代码：

df = pd.DataFrame({'level_1':['A', 'A', 'A', 'G', 'G'], 'level_2':['B', 'B', 'B', 'H', 'I'],
    'level_3':['C', 'D', 'D', np.nan, 'J'], 'level_4':[np.nan, 'E', 'F', np.nan, 'K']})

您的输入数据帧是：

  level_1 level_2 level_3 level_4
0       A       B       C     NaN
1       A       B       D       E
2       A       B       D       F
3       G       H     NaN     NaN
4       G       I       J       K

解决办法是：

# Get unique values from input dataframe and filter out 'nan' values
list_nodes = []
for i_col in df.columns.tolist():
    list_nodes.extend(filter(lambda v: v==v, df[i_col].unique().tolist()))

# Initialize your result dataframe
df_res = pd.DataFrame(columns=sorted(list_nodes), index=sorted(list_nodes))
df_res = df_res.fillna(0)

# Get 'index-column' pairs from input dataframe ('nan's are exluded)
list_indexes = []
for i_col in range(df.shape[1]-1):
    list_indexes.extend(list(set([tuple(i) for i in df.iloc[:, i_col:i_col+2]\
        .dropna(axis=0).values.tolist()])))

# Use 'index-column' pairs to fill the result dataframe
for i_list_indexes in list_indexes:
    df_res.set_value(i_list_indexes[0], i_list_indexes[1], 1)

最终结果是：

   A  B  C  D  E  F  G  H  I  J  K
A  0  1  0  0  0  0  0  0  0  0  0
B  0  0  1  1  0  0  0  0  0  0  0
C  0  0  0  0  0  0  0  0  0  0  0
D  0  0  0  0  1  1  0  0  0  0  0
E  0  0  0  0  0  0  0  0  0  0  0
F  0  0  0  0  0  0  0  0  0  0  0
G  0  0  0  0  0  0  0  1  1  0  0
H  0  0  0  0  0  0  0  0  0  0  0
I  0  0  0  0  0  0  0  0  0  1  0
J  0  0  0  0  0  0  0  0  0  0  1
K  0  0  0  0  0  0  0  0  0  0  0