递归函数：'NoneType' 对象不可迭代

Question

我有一个嵌套字典，

d={
    "A":1, 
    "depth":0, 
    "chain":[
        {
            "A1":0.7, 
            "depth":1,
            "chain":[
                {
                    "A11":0.3,
                    "depth":2,
                    "key2":{"direct":{},"cumulative":{"B":0.3}}, 
                    "chain":[]
                }, 
                {
                    "A12":0.4, 
                    "depth":2,
                    "chain":[
                        {
                            "A121":0.4, 
                            "depth":3, 
                            "key2":{"direct": {}, "cumulative":{"C":0.2, "D": 0.2}}, 
                           "chain": []
                        }]}]},
        {
            "A2":0.3,
            "depth":1,
            "chain":[
                {
                    "A11":0.3, 
                    "depth":2, 
                    "key2":{"direct":{}, "cumulative":{"D":0.3}},
                    "chain":[]
                }]}]}

我想要return一个第一个键重复x次的列表。 x 是 "chain" 下的元素数。在这种情况下，它将 return:

["A", "A", "A1", "A1", "A2", "A12"]

我试过以下方法

def from_nodes(d):
    from_n=[list(d.keys())[0]]*len(d["chain"])
    for x in d["chain"]:
        if x is not None:
            from_n.extend(from_nodes(x))
            return from_n

并得到错误

TypeError                                 Traceback (most recent call last)
<ipython-input-196-6233463c604b>in <module>()
----> 1 from_nodes(test2)

<ipython-input-194-5b7ca4b6db75>in from_nodes(d)
  3     for x in d["chain"]:
  4         if x is not None:
----> 5             from_n.extend(from_nodes(x))
  6         return from_n

<ipython-input-194-5b7ca4b6db75> in from_nodes(d)
  3     for x in d["chain"]:
  4         if x is not None:
----> 5             from_n.extend(from_nodes(x))
  6         return from_n

Answer 1

正如我在 , your error is that the return statement is indented incorrectly. If d["chain"] is empty or None, your function will return None implicitly 中提到的那样。

将您的函数更改为以下内容：

def from_nodes(d):
    from_n=[list(d.keys())[0]]*len(d["chain"])
    for x in d["chain"]:
        if x is not None:
            from_n.extend(from_nodes(x))
    return from_n

将修复错误，我的计算机上的结果是：

print(from_nodes(d))
#['A', 'A', 'A1', 'A1', 'depth', 'depth']

这与您想要的输出不匹配——这是因为当您调用 .keys() 时不能保证获得确定性顺序。

为所需输出修改函数的一种方法是创建要忽略的键列表：

def from_nodes(d):
    ignore_keys = {"chain", "depth", "key2"}
    from_n=[list(k for k in d.keys() if k not in ignore_keys)[0]]*len(d["chain"])
    for x in d["chain"]:
        if x is not None:
            from_n.extend(from_nodes(x))
    return from_n

print(from_nodes(d))
#['A', 'A', 'A1', 'A1', 'A12', 'A2']

但是，这只是我推测您的要求。您需要为 "first" 键的含义定义正确的条件。

递归函数：'NoneType' 对象不可迭代

recursive function: 'NoneType' object is not iterable

python

recursion

nonetype