Python: Creating an adjacency matrix from a dataframe

Question

I have the following data frame:

Company Firm
125911  1
125911  2
32679   3
32679   5
32679   5
32679   8
32679   10
32679   12
43805   14
67734   8
67734   9
67734   10
67734   10
67734   11
67734   12
67734   13
74240   4
74240   6
74240   7

Where basically the firm makes an investment into the company at a specific year which in this case is the same year for all companies. What I want to do in python is to create a simple adjacency matrix with only 0's and 1's. 1 if two firms has made an investment into the same company. So even if firm 10 and 8 for example have invested in two different firms at the same it will still be a 1. The resulting matrix I am looking for looks like:

Firm 1  2   3   4   5   6   7   8   9   10  11  12  13  14
1   0   1   0   0   0   0   0   0   0   0   0   0   0   0
2   1   0   0   0   0   0   0   0   0   0   0   0   0   0
3   0   0   0   0   1   0   0   1   0   1   0   1   0   0
4   0   0   0   0   0   1   1   0   0   0   0   0   0   0
5   0   0   1   0   0   0   0   1   0   1   0   1   0   0
6   0   0   0   1   0   0   1   0   0   0   0   0   0   0
7   0   0   0   1   0   1   0   0   0   0   0   0   0   0
8   0   0   1   0   1   0   0   0   1   1   1   1   1   0
9   0   0   0   0   0   0   0   1   0   1   1   1   1   0
10  0   0   1   0   1   0   0   1   1   0   1   1   1   0
11  0   0   0   0   0   0   0   1   1   1   0   1   1   0
12  0   0   1   0   1   0   0   1   1   1   1   0   1   0
13  0   0   0   0   0   0   0   1   1   1   1   1   0   0
14  0   0   0   0   0   0   0   0   0   0   0   0   0   0

I have seen similar questions where you can use crosstab , however in that case each company will only have one row with all the firms in different columns instead. So I am wondering what the best and most efficient way to tackle this specific problem is? Any help is greatly appreciated.

Answer 1

dfs = []
for s in df.groupby("Company").agg(list).values:
    dfs.append(pd.DataFrame(index=set(s[0]), columns=set(s[0])).fillna(1))

out = pd.concat(dfs).groupby(level=0).sum().gt(0).astype(int)
np.fill_diagonal(out.values, 0)
print(out)

Prints:

    1   2   3   4   5   6   7   8   9   10  11  12  13  14
1    0   1   0   0   0   0   0   0   0   0   0   0   0   0
2    1   0   0   0   0   0   0   0   0   0   0   0   0   0
3    0   0   0   0   1   0   0   1   0   1   0   1   0   0
4    0   0   0   0   0   1   1   0   0   0   0   0   0   0
5    0   0   1   0   0   0   0   1   0   1   0   1   0   0
6    0   0   0   1   0   0   1   0   0   0   0   0   0   0
7    0   0   0   1   0   1   0   0   0   0   0   0   0   0
8    0   0   1   0   1   0   0   0   1   1   1   1   1   0
9    0   0   0   0   0   0   0   1   0   1   1   1   1   0
10   0   0   1   0   1   0   0   1   1   0   1   1   1   0
11   0   0   0   0   0   0   0   1   1   1   0   1   1   0
12   0   0   1   0   1   0   0   1   1   1   1   0   1   0
13   0   0   0   0   0   0   0   1   1   1   1   1   0   0
14   0   0   0   0   0   0   0   0   0   0   0   0   0   0

Answer 2

dfm = df.merge(df, on="Company").query("Firm_x != Firm_y")
out = pd.crosstab(dfm['Firm_x'], dfm['Firm_y'])

>>> out
Firm_y  1   2   3   4   5   6   7   8   9   10  11  12  13  14
Firm_x
1        1   0   0   0   0   0   0   0   0   0   0   0   0   0
2        0   1   0   0   0   0   0   0   0   0   0   0   0   0
3        0   0   1   0   0   0   0   0   0   0   0   0   0   0
4        0   0   0   1   0   0   0   0   0   0   0   0   0   0
5        0   0   0   0   4   0   0   0   0   0   0   0   0   0
6        0   0   0   0   0   1   0   0   0   0   0   0   0   0
7        0   0   0   0   0   0   1   0   0   0   0   0   0   0
8        0   0   0   0   0   0   0   2   0   0   0   0   0   0
9        0   0   0   0   0   0   0   0   1   0   0   0   0   0
10       0   0   0   0   0   0   0   0   0   5   0   0   0   0
11       0   0   0   0   0   0   0   0   0   0   1   0   0   0
12       0   0   0   0   0   0   0   0   0   0   0   2   0   0
13       0   0   0   0   0   0   0   0   0   0   0   0   1   0
14       0   0   0   0   0   0   0   0   0   0   0   0   0   1