Python遍历Azure Document Intelligence嵌套Sections列表并扁平化输出

代码实现 #

def flatten_sections(sections):
    result = []
    # 栈元素：(当前section字典, 当前section的路径)
    stack = []
    # 初始化栈，加入所有根级sections
    for idx, section in enumerate(sections):
        stack.append( (section, f"/sections/{idx}") )
    
    while stack:
        current_section, current_path = stack.pop()
        # 处理当前section下的paragraphs
        if "paragraphs" in current_section:
            for p_idx, _ in enumerate(current_section["paragraphs"]):
                para_path = f"{current_path}/paragraphs/{p_idx}"
                result.append(para_path)
        # 处理当前section下的tables
        if "tables" in current_section:
            for t_idx, _ in enumerate(current_section["tables"]):
                table_path = f"{current_path}/tables/{t_idx}"
                result.append(table_path)
        # 处理嵌套的子sections，逆序压栈保证遍历顺序和原结构一致
        if "sections" in current_section:
            # 逆序遍历，因为栈是后进先出，这样弹出时顺序和原列表一致
            for s_idx in reversed(range(len(current_section["sections"]))):
                sub_section = current_section["sections"][s_idx]
                sub_path = f"{current_path}/sections/{s_idx}"
                stack.append( (sub_section, sub_path) )
    return result

代码说明 #

1. 栈初始化：将每个根级section与对应路径（/sections/索引）加入栈，作为遍历起点。
2. 循环处理栈元素：每次从栈顶弹出一个section及其路径，依次处理三类内容：
   • Paragraphs/Tables处理：若当前section包含paragraphs或tables字段，遍历每个元素，拼接完整路径后加入结果列表。
   • 嵌套Sections处理：若存在子sections，逆序压入栈（利用栈后进先出的特性，保证弹出时顺序与原列表一致），同时拼接子section的路径。
3. 结果返回：最终得到所有paragraph和table的完整路径列表，可直接用于后续关联提取内容。

使用示例 #

假设输入的sections简化结构如下：

sections = [
    {
        "paragraphs": [{"content": "段落1"}, {"content": "段落2"}],
        "sections": [
            {
                "paragraphs": [{"content": "嵌套段落1"}],
                "tables": [{"cells": []}]
            }
        ]
    },
    {
        "tables": [{"cells": []}]
    }
]

调用flatten_sections(sections)会得到：

[
    "/sections/0/paragraphs/0",
    "/sections/0/paragraphs/1",
    "/sections/0/sections/0/paragraphs/0",
    "/sections/0/sections/0/tables/0",
    "/sections/1/tables/0"
]

内容的提问来源于stack exchange，提问作者Shubham R