获取Python中的层次 struct 数据

发布于02月19日

我有数据:

issue	father	son
1	33	34
1	34	35
2	33	34

以下是Python中的脚本，它按层次 struct 检索表:

G = nx.from_pandas_edgelist(df, source='father', target='son', create_using=nx.DiGraph ,edge_attr=True)
print(G.out_degree())
roots = [v for v, d in G.in_degree() if d == 0]
leaves = [v for v, d in G.out_degree() if d == 0]
all_paths = []
for root in roots:
    for leaf in leaves:
        paths = nx.all_simple_paths(G, root, leaf)
        all_paths.extend(paths)
for node in nx.nodes_with_selfloops(G):
    all_paths.append([node, node])
df_h=pd.DataFrame(sorted(all_paths)).add_prefix('Value_Depend_On_ID_').fillna('')
df_h=df_h.replace('',np.nan, regex=True)
df_h.astype('Int64')

输出:

Value_Depend_On_ID_0	Value_Depend_On_ID_1	Value_Depend_On_ID_2
33	34	35

所需的输出:

Value_Depend_On_ID_0	Value_Depend_On_ID_1	Value_Depend_On_ID_2	issue
33	34	35	1
33	34	null	2

如何添加问题？

推荐答案

如果我理解正确的话，您希望将数据帧按issue分组，然后搜索所有路径:

import networkx as nx


def group_func(g):
    G = nx.from_pandas_edgelist(
        g, source="father", target="son", create_using=nx.DiGraph
    )

    roots = [v for v, d in G.in_degree() if d == 0]
    leaves = [v for v, d in G.out_degree() if d == 0]

    all_paths = []
    for root in roots:
        for leaf in leaves:
            paths = nx.all_simple_paths(G, root, leaf)
            all_paths.extend(paths)

    out = []
    for p in all_paths:
        d = {}
        for i, v in enumerate(p):
            d[f"Value_Depend_On_ID_{i}"] = v
        out.append(d)

    return pd.DataFrame(out)


x = (
    df.groupby("issue")
    .apply(group_func, include_groups=False)
    .reset_index(1, drop=True)
    .reset_index()
)
print(x)

打印:

   issue  Value_Depend_On_ID_0  Value_Depend_On_ID_1  Value_Depend_On_ID_2
0      1                    33                    34                  35.0
1      2                    33                    34                   NaN