Question

我有一个树形结构，每个节点都有一个文档列表（文档长度可以从5到500不等），每个文档包含许多单词。我想存储每个单词以字典形式出现的文档数量。例如：如果任何节点A = [['b','m','n'],['b'],['g'],['o','b','g'],['b','g']]处的文档计数应存储在每个节点中，如A.occurlist = {'b':4,'m':1,'n':1,'g':3,'o':1

我正在运行以下代码，但它无法产生递归并显示错误： TypeError：＆＃39; dict＆＃39;对象不可调用

代码：

def occurlist(node):
    wordSet = set()
    lis = []
    next_node = []
    child_nodes = node.children
    for child in child_nodes:
        next_node.append(child)
        lis += child.documents
    wordSet = set(itertools.chain.from_iterable(lis))
    occurDict = {word:0 for word in wordSet}
    for child in child_nodes:
        occurlist = {}
        occurlist.update(copy.deepcopy(occurDict))
        for doc in child.documents:
            for word in wordSet:
                if word in doc:
                    occurlist[word] +=1
        child.occurlist = occurlist
        print child.name
    print len(next_node)
    if next_node:
        for nn in next_node:
            if not nn.update:
                occurlist(nn)

occurlist(Savings_Accounts)

Savings_Accounts是根节点的名称。

Answer 1

如果您只希望每个元素的计数使用collections.Counter：

from collections import Counter

A = [['b','m','n'],['b'],['g'],['o','b','g'],['b','g']]

dict(Counter([ele for sub in A for ele in set(sub)]))
#{'b': 4, 'm': 1, 'g': 3, 'o': 1, 'n': 1}

遍历树时递归不起作用

1 个答案: