Print complete key path for all the values of a python nested dictionary

Question

If below is my nested dictionary I want to parse through recursively and print all the values along with the complete path of the nested key.

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

Expected output:

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>

I wrote a recursive function, but running to this:

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)

Output:

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

In the above : For male, the key struct shouldnt contain “category”
How to retain the correct key structure?

Asked By: Suren

||

Source

Answer 1

You shouldn’t alter the path variable in the dict_path() function:

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)

Answered By: catavaran

Answer 2

As catavaran mentions, your problem is caused by adding the new path component to the path variable inside your for loop. You need to put the new path into the call so it gets passed to the next level of recursion and doesn’t interfere with the path of subsequent items in the for loop at the current recursion level.

Here’s an alternate solution that uses a recursive generator, rather than printing the results inside the dict_path function. (FWIW, I used print json.dumps(my_dict, indent=4) to reformat the dictionary).

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

output

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

Answered By: PM 2Ring

Answer 3

Just adding to above @catavaran code.
in case if the dict value is list, and if list may have dict or values itself, then this code could help.
I just modified delimiter as dot.

def dict_path(path,my_dict):
    for k,v in my_dict.iteritems():
        if isinstance(v,list):
            for i, item in enumerate(v):
                dict_path( path + "." + k + "." + str(i), item)
        elif isinstance(v,dict):
            dict_path(path+"."+k,v)
        else:
            print path+"."+k, "=>", v

Thanks @catavaran , your code helped me.

Answered By: Hara

Answer 4

Python 3 with f-strings

def dict_path(my_dict, path=None):
    seperator = ":"
    if path is None:
        path = "{"
    for k, v in my_dict.items():
        if isinstance(v, dict):
            dict_path(v, f"{path}{seperator if path != '{' else ''}{k}")
        else:
            print(f"{path}{seperator if path != '{' else ''}{k}:{v}" + "}")


dict_path(test_dict)

Answered By: user1023102

Print complete key path for all the values of a python nested dictionary

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