遞歸搜索字典，返回完整的收入

Question

我的問題陳述是我有搜索查詢，我必須返回與查詢匹配的字典，以維護層次結構。

我能夠做到第一。 但是我想從一開始就返回完整的層次結構，如下所示

獲取此輸出：

{"Name":"google search","items":[],"properties":{"id":1,"process":123}

預期產量：

{
    "items":[
    {'Name':'chrome','items':
     [
       {"Name":"google search","items":[],"properties":{"id":1,"process":123}}
     ]
    },
    ]

}

這是我的示例輸入：

myinput = {
    "items":[
    {'Name':'firefox','items':[],"properties":{"one":1,"two":2}},
    {'Name':'chrome','items':[
                        {'Name':"stackoverflow","items":[],"properties":{"one":1,"two":2}},
                        {"Name":"google search","items":[],"properties":{"id":1,"process":123}}
                        ],
                        "properties":{"one":1,"two":2}},
    {'Name':'new','items':[],"properties":{"one":1,"two":2}},
    {'Name':'new','items':[],"properties":{"one":1,"two":2}},
    ]
}

這是我迄今為止嘗試過的

matched_items = []
def resursive_fun(Nodes, searchQuery):
    for key, value in Nodes.iteritems():
        if isinstance(value, list):
            for item in value:
                matchValue = match_items(searchQuery, item['properties'])
                if matchValue:
                    matched_items.append(item)
                resursive_fun(item, searchQuery)
    return matched_items

searchDict = {"id": 1, "process": 123}
resursive_fun(myinput, searchDict)

Answer 1

我認為您需要從任何成功的遞歸調用的返回值中構建返回值，而不是使用全局列表（如果您需要進行多次搜索，這將導致各種問題）。 如果沒有匹配項，則應返回具有特殊含義的內容（如None ）。

嘗試這樣的事情：

def recursive_search(data, query):
    if match_items(searchQuery, data["properties"]):            # base case
        return data

    if "items" in data and isinstance(data["items"], list):     # recursive case
        for item in data["items"]:
            x = recursive_search(item, query)
            if x is not None:                    # recursive search was successful!
                return {"Name": data["Name"], "items": [x]}

    return None                          # if search is not successful, return None

return None可以省略，因為None是不返回其他任何值的函數的默認返回值。 但是，我認為最好在“ None具有某些含義時要明確一些，就像在這里一樣。

如果您不僅要查找第一個匹配結果，而且要查找所有匹配結果，則需要用一些更細微的行為來替換早期return調用，這些行為只有在某處找到匹配項時才返回值：

def recursive_search(data, query):
    result = {}

    if if "properties" in data and match_items(searchQuery, data["properties"]):
        result["properties"] = data["properties"]

    if "items" in data and isinstance(data["items"], list):
        for item in data["items"]:
            result_items = []
            x = recursive_search(item, query)
            if x is not None:
                result_items.append(x)
        if result_items:         # recursive search found at least one match
            result["items"] = result_items

    if result:      # some part of the search found a match (either here or deeper)
        if "Name" in data:
            result["Name"] = data["Name"]
        return result
    else:
        return None

您也可以在失敗的匹配中返回一個空字典，而不是None 。 為此，將if result: if "Name" in data:行更改為單行if result and "Name" in data: if result: if "Name" in data:行的末尾，並無條件地執行return result （取消縮進並刪除else: return None ）。 將遞歸情況下的if x is not None更改為if x 。