计算嵌套组合字典列表中相同键的函数

Question

我正在尝试计算这个嵌套组合字典列表中的“名称”键，我得到 3 而不是 6，我认为我的问题是递归函数 count_elem(tree) 中的基本情况def define_tree3():    tree3 ={'name':'GAS','grade':0.8,'children':[{'name':'CSA','grade':0.5,'children':[{'name':'MB','grade':0.1},{'name':'TA','grade':0.6}]},{'name':'IIW','grade':0.9,'children':[None,{'name':'IP','grade':0.99}]}]}    return tree3#this fuction is to delete the given key from the given dict and retur the new dictdef delkey(dict1,key):    d=dict(dict1)    del d[key]    return d#this function is to count the numbers of 'name'def count_elem(tree):    if len(tree)==0:        return 0    else:        for i in tree:            if i == None:                return 0            elif i == 'name':                return 1+ count_elem(delkey(tree,i))            elif i == 'grade':                return count_elem(delkey(tree,i))            elif i == 'children':                for j in tree[i]:                    if j == None:                        continue                    else:                        return count_elem(j)a=define_tree3()print(count_elem(a))

Answer 1

这不是您问题的答案。我只是尝试解决它并使用了不同的方法。作为参考可能有用。

tree3 = {'name':'GAS','grade':0.8,'children':[{'name':'CSA','grade':0.5,'children':[{'name':'MB','grade':0.1},{'name':'TA','grade':0.6}]},{'name':'IIW','grade':0.9,'children':[None,{'name':'IP','grade':0.99}]}]}

def count_name(entity):

    count = 0

    name = 'name'

    # print('\n')

    # print(count, entity)

    if type(entity) == dict:

        count += sum([key == name for key in entity.keys()])

    

        for value in entity.values():

            # print(count, value)

            if type(value) == list:

                count += sum([count_name(member) for member in value])

    return count

count_name(tree3)