python - 如何在字典中查找短语？

Question

我有一个结构:

{
    "content": "Name 1", 
    "name": "directory", 
    "decendent": [
         {
            "content": "Name 2", 
            "name": "subdirectory", 
            "decendent": None
        }, 
        {
            "content": "Name 3", 
            "name": "subdirectory_two", 
            "decendent": [
                {
                    "content": "Name 4", 
                    "name": "subsubdirectory", 
                    "decendent": None
                }
            ]
        }
    ]
}

我必须查找在搜索字段中输入的单词序列:

 <form id="tfnewsearch" method="get" action="/help/search">
        <input type="text" class="tftextinput" name="q" size="21" maxlength="120">
        <input type="submit" value="Search" class="tfbutton">
 </form>

如果我找到了 - 将它们添加到

   [
        {
             "content": "The sought content",
             "phrase": "the sought phrase",
             "name": "unique name"
        }, ...
   ] 

对于每一个发现的巧合。

例如: 如果我寻找“我有”，我应该得到:

 [
       {
            "content": "I have a good day", 
            "phrase": "I have", 
            "name": "subdirectory", 

        },
        {
            "content": "While I have several ways to do it",
            "phrase": "I have", 
            "name": "subdirectory2"
        },
        {
            "content": "When I had it",
            "phrase": "I have",
            "name": "subdirectory3"
         },
 ]

如果我要在搜索过程中使用词法分析器(如 pymorhph2)更改此短语(例如:“have”、“had”)，如何使用递归在 Python 中实现它？

提前致谢!

Answer 1

我不知道 pymorph2，但你可以这样做:

def f1(a, s):
    if s in a["contents"]:
        a["phrase"] = s
    for b in a["descendent"]:
        f(b, s)

参数 a 是您的主要词典，s 要查找的短语。

但是，如果您再次调用 f，它将清除之前的值。如果这不是您想要的，您可以考虑替代，例如:

def f2(a, s):
    if s in a["contents"]:
        if "phrase" not in a:
            a["phrase"] = [s]
        else:
            a["phrase"].append(s)
    for b in a["descendent"]:
        f(b, s)

如果您还想返回匹配字典的列表:

def f3(a, s):
    r = []
    if s in a["contents"]:
        if "phrase" not in a:
            a["phrase"] = [s]
        else:
            a["phrase"].append(s)
        r.append(a)
    for b in a["descendent"]:
        r += f(b, s)
    return r

如果你想复制输出列表中的数据而不是只引用原始字典，你可以将 r.append(a) 替换为，例如

r.append({k: a[k] for k in ("content", "phrase", "name")})