將父子字典串列轉換為嵌套字典

我嘗試并看到了上述問題的多種解決方案，但找不到適合我情況的解決方案。我有以下字典串列。

[
        {"id": "101", "logical_section": "ORGANIZATION", "parent_section_id": None},
        {"id": "102", "logical_section": "ORG_NAMES", "parent_section_id": "101"},
        {"id": "103", "logical_section": "ORG_ADDRESSES", "parent_section_id": "101"},
        {"id": "104", "logical_section": "SOURCEADDRESS", "parent_section_id": "103"},
        {"id": "105", "logical_section": "STANDARDIZEDADDRESS", "parent_section_id": "103"},
        {"id": "106", "logical_section": "ORG_EMPLOYES", "parent_section_id": "101"}
]

我需要通過以下方式將此串列轉換為嵌套字典串列

{
    "id": "101",
    "logical_section": "ORGANIZATION",
    "parent_section_id": None,
    "child": [
        {
            "id": "102",
            "logical_section": "ORG_NAMES",
            "parent_section_id": "101"
        },
        {
            "id": "103",
            "logical_section": "ORG_ADDRESSES",
            "parent_section_id": "101",
            "child": [
                {"id": "104", "logical_section": "SOURCEADDRESS", "parent_section_id": "103"},
                {"id": "105", "logical_section": "STANDARDIZEDADDRESS", "parent_section_id": "103"}
            ]
        },
        {
            "id": "106",
            "logical_section": "ORG_EMPLOYES",
            "parent_section_id": "101"}
    ]
}

我嘗試過的解決方案如下

levels = dict()
for n in test:
    levels.setdefault(n['parent_section_id'], []).append(n)

但這并沒有給出我正在尋找的輸出。

任何幫助，將不勝感激。

uj5u.com熱心網友回復：

我看了一下這個并想出了以下代碼。我絕對肯定這可以做得更好：我不建議將其用于大量資料。如果另一位更有經驗的程式員可以參與此解決方案，我將不勝感激。

test = [
    {"id": "101", "logical_section": "ORGANIZATION", "parent_section_id": None},
    {"id": "102", "logical_section": "ORG_NAMES", "parent_section_id": "101"},
    {"id": "103", "logical_section": "ORG_ADDRESSES", "parent_section_id": "101"},
    {"id": "104", "logical_section": "SOURCEADDRESS", "parent_section_id": "103"},
    {"id": "105", "logical_section": "STANDARDIZEDADDRESS", "parent_section_id": "103"},
    {"id": "106", "logical_section": "ORG_EMPLOYES", "parent_section_id": "101"}
]

ans_dict = test[0]


def insert(tree, node):
    if node["parent_section_id"] == tree["id"]:
        if "child" not in tree:
            tree["child"] = [node]
        else:
            tree["child"].append(node)
    elif "child" in tree:
        for c in tree["child"]:
            insert(c, node)


for record in test:
    insert(ans_dict, record)


print(ans_dict)

此代碼使用遞回二叉樹方法。使它如此低效的原因是，每個節點都insert()使用給定節點作為引數運行函式，每次插入新節點。

uj5u.com熱心網友回復：

這種方法使用廣度優先搜索演算法來構造一棵樹。請注意，由于字典是可變的，因此輸入串列中的所有元素都將被修改。您需要選擇“parent_section_id”為 None 的一個/s 來查找所有父母。

test = [
        {"id": "101", "logical_section": "ORGANIZATION", "parent_section_id": None},
        {"id": "102", "logical_section": "ORG_NAMES", "parent_section_id": "101"},
        {"id": "103", "logical_section": "ORG_ADDRESSES", "parent_section_id": "101"},
        {"id": "104", "logical_section": "SOURCEADDRESS", "parent_section_id": "103"},
        {"id": "105", "logical_section": "STANDARDIZEDADDRESS", "parent_section_id": "103"},
        {"id": "106", "logical_section": "ORG_EMPLOYES", "parent_section_id": "101"}
]

# No assumption about the position of the parent
nodes = [i for i in test if i["parent_section_id"] is None]
# Single parent
parent_idx = test.index(nodes[0])
# If more than 1 parent, then uncomment
#parent_idxs = [test.index(node) for node in nodes]

# Find all children
parent_id_vals = set([d["parent_section_id"] for d in test if d["parent_section_id"] is not None])

while nodes:
    focus_node = nodes[0]
    
    if focus_node["id"] in parent_id_vals:
        focus_node["child"] = []
    
    # Uncomment else clause if "test" is too large
    #else:
    #    nodes.remove(focus_node)
    #    continue
    
    for i in test:
        if i["parent_section_id"] == focus_node["id"]:
            focus_node["child"].append(i)
            nodes.append(i)
    
    nodes.remove(focus_node)
            
# If single parent node
ans = test[parent_idx]
# If multiple parents, then uncomment
#ans = [test[idx] for idx in parent_idxs]

print(ans)

我把它作為一個小挑戰來做。DCoxshall 的答案無論如何都適合您的需求。

標籤：Python 字典 树 嵌套的 等级制度

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