可以将生成器与itertools.permutations和pandas.crosstab一起使用：

from itertools import permutations

out = (pd.crosstab(*zip(*((a, b) for l in df.to_numpy().tolist()
                          for a,b in permutations(l, r=2))))
      ).rename_axis(index=None, columns=None)

输出：

Apple  Argentina  Bosch  China  Finance  John Fowler  Marketing  Messi  Peter Pan  UK  USA
Apple            0          0      0      1        1            2          2      0          1   1    1
Argentina        0          0      1      0        0            0          1      1          0   0    0
Bosch            0          1      0      0        0            0          1      1          0   0    0
China            1          0      0      0        1            1          0      0          0   0    0
Finance          1          0      0      1        0            1          0      0          0   0    0
John Fowler      2          0      0      1        1            0          1      0          0   0    1
Marketing        2          1      1      0        0            1          0      1          1   1    1
Messi            0          1      1      0        0            0          1      0          0   0    0
Peter Pan        1          0      0      0        0            0          1      0          0   1    0
UK               1          0      0      0        0            0          1      0          1   0    0
USA              1          0      0      0        0            1          1      0          0   0    0

工作原理：

• 对于每行，生成2个项目[(a,b), (b,a), ...]的所有组合
• 整形为2个数组[(a,b,...),(b,a,...)]
• 将数组传递给crosstab，扩展为前两个参数

对应图表：

可以使用sliding_window_view和networkx来获得加权邻接矩阵：

from numpy.lib.stride_tricks import sliding_window_view
import networkx as nx

edges = sliding_window_view(df1.values, (len(df1), 2)).reshape(-1, 2)
edges = pd.DataFrame(edges, columns=['source', 'target']).value_counts().reset_index()

G = nx.from_pandas_edgelist(edges, edge_attr='count')
adjmx = nx.to_pandas_adjacency(G, weight='count').astype(int)

输出：

>>> adjmx
             Apple  Marketing  Finance  Argentina  Bosch  China  John Fowler  Messi  Peter Pan  UK  USA
Apple            0          2        1          0      0      1            0      0          0   1    1
Marketing        2          0        0          0      1      0            1      1          1   0    0
Finance          1          0        0          0      0      0            1      0          0   0    0
Argentina        0          0        0          0      1      0            0      0          0   0    0
Bosch            0          1        0          1      0      0            0      0          0   0    0
China            1          0        0          0      0      0            0      0          0   0    0
John Fowler      0          1        1          0      0      0            0      0          0   0    0
Messi            0          1        0          0      0      0            0      0          0   0    0
Peter Pan        0          1        0          0      0      0            0      0          0   0    0
UK               1          0        0          0      0      0            0      0          0   0    0
USA              1          0        0          0      0      0            0      0          0   0    0