打印python嵌套字典的所有值的完整键路径

Question

如果下面是我的嵌套字典,我想通过递归方式解析并打印所有值以及嵌套键的完整路径.

my_dict = {'attr':{'types':{'tag':{'name':'Tom', 'gender':'male'},'category':'employee'}}}

预期产量:

Key structure : my_dict["attr"]["types"]["tag"]["name"]<br>
value : "Tom"<br>
Key structure : my_dict["attr"]["types"]["tag"]["gender"]<br>
value : "male"<br>
Key structure : my_dict["attr"]["types"]["category"]<br>
value : "employee"<br>

我写了一个递归函数,但运行到这个:

my_dict = {'attr':{'types':{'tag':{'name':'Tom','gender':'male'},'category':'employee'}}}

def dict_path(path,my_dict):
    for k,v in my_dict.iteritems():
        if isinstance(v,dict):
            path=path+"_"+k
            dict_path(path,v)
        else:
            path=path+"_"+k
            print path,"=>",v

    return
dict_path("",my_dict)

输出:

_attr_types_category => employee
_attr_types_category_tag_gender => male
_attr_types_category_tag_gender_name => Tom

在上面:对于男性,关键结构不应该包含"类别"如何保留正确的关键结构？

Answer 1

您不应该更改函数中的path变量dict_path():

def dict_path(path,my_dict):
    for k,v in my_dict.iteritems():
        if isinstance(v,dict):
            dict_path(path+"_"+k,v)
        else:
            print path+"_"+k,"=>",v
dict_path("",my_dict)

Answer 2

正如 catavaran 所提到的，您的问题是由将新路径组件添加到循环path内的变量引起的for。您需要将新路径放入调用中，以便将其传递到下一级递归，并且不会干扰for当前递归级别循环中后续项目的路径。

这是使用递归生成器的替代解决方案，而不是在dict_path函数内打印结果。（FWIW，我曾经print json.dumps(my_dict, indent=4)重新格式化字典）。

my_dict = {
    "attr": {
        "types": {
            "category": "employee", 
            "tag": {
                "gender": "male", 
                "name": "Tom"
            }
        }
    }
}

def dict_path(my_dict, path=None):
    if path is None:
        path = []
    for k,v in my_dict.iteritems():
        newpath = path + [k]
        if isinstance(v, dict):
            for u in dict_path(v, newpath):
                yield u
        else:
            yield newpath, v

for path, v in dict_path(my_dict):
    print '_'.join(path), "=>", v

输出

attr_types_category => employee
attr_types_tag_gender => male
attr_types_tag_name => Tom