Question

我试图从嵌套字典中获取用户输入定义的键的值。虽然我可以从函数本身中打印出值。 Return语句将值发送为None。

people = {1: {'name': 'John', 'age': '27', 'sex': 'Male'},
          2: {'name': 'Marie', 'age': '22', 'sex': 'Female'},
          3: 'James',
          4: 'lance',
          'change1': {5: 'place'},
          'country': {'Waterloo': 'Waterloo', 'Australia': 'Australia'},
          6: {'Position': 'GM'}
          }

def getfromdict(mydict, mykey):
    for i in mydict.keys():

        if type(mydict[i])==dict:
            newdict = mydict[i]
            getfromdict(newdict, mykey)

        elif i == mykey:
            print(mydict[i])
            valuefound = mydict[i]
            return valuefound

        else:
            continue

result = getfromdict(people,5)
print(result)

Answer 1

您的代码段的第三行应阅读

           return getfromdict(newdict, mykey)

要在树上找到一片叶子，就等距而言，

def leaf(tree, key):
    if key in tree: return tree[key]
    for subtree in filter(lambda s: isinstance(s, dict), tree.values()):
        ret_val = leaf(subtree, key)
        if ret_val is not None: return ret_val

此实现具有以下属性：如果在某个级别找到了密钥，则返回的键将返回叶或子树（不希望出现的情况是非常简单的）。

Answer 2

您可以尝试在嵌套JSON中搜索键

 def getfromdict(mydict, mykey):
    if type(mydict) == type({}):
        for k1 in mydict:
            if k1 == mykey:
                return mydict[k1]
            fromdict = getfromdict(mydict[k1], mykey)
            if fromdict is not None:
                return fromdict

从函数返回值无

2 个答案: