Access nested dictionary items via a list of keys?


I have a complex dictionary structure which I would like to access via a list of keys to address the correct item.

dataDict = {
        "r": 1,
        "s": 2,
        "t": 3
        "u": 1,
        "v": {
            "x": 1,
            "y": 2,
            "z": 3
        "w": 3

maplist = ["a", "r"]


maplist = ["b", "v", "y"]

I have made the following code which works but I’m sure there is a better and more efficient way to do this if anyone has an idea.

# Get a given data from a dictionary with position provided as a list
def getFromDict(dataDict, mapList):    
    for k in mapList: dataDict = dataDict[k]
    return dataDict

# Set a given data in a dictionary with position provided as a list
def setInDict(dataDict, mapList, value): 
    for k in mapList[:-1]: dataDict = dataDict[k]
    dataDict[mapList[-1]] = value
Asked By: kolergy



Use reduce() to traverse the dictionary:

from functools import reduce  # forward compatibility for Python 3
import operator

def getFromDict(dataDict, mapList):
    return reduce(operator.getitem, mapList, dataDict)

and reuse getFromDict to find the location to store the value for setInDict():

def setInDict(dataDict, mapList, value):
    getFromDict(dataDict, mapList[:-1])[mapList[-1]] = value

All but the last element in mapList is needed to find the ‘parent’ dictionary to add the value to, then use the last element to set the value to the right key.


>>> getFromDict(dataDict, ["a", "r"])
>>> getFromDict(dataDict, ["b", "v", "y"])
>>> setInDict(dataDict, ["b", "v", "w"], 4)
>>> import pprint
>>> pprint.pprint(dataDict)
{'a': {'r': 1, 's': 2, 't': 3},
 'b': {'u': 1, 'v': {'w': 4, 'x': 1, 'y': 2, 'z': 3}, 'w': 3}}

Note that the Python PEP8 style guide prescribes snake_case names for functions. The above works equally well for lists or a mix of dictionaries and lists, so the names should really be get_by_path() and set_by_path():

from functools import reduce  # forward compatibility for Python 3
import operator

def get_by_path(root, items):
    """Access a nested object in root by item sequence."""
    return reduce(operator.getitem, items, root)

def set_by_path(root, items, value):
    """Set a value in a nested object in root by item sequence."""
    get_by_path(root, items[:-1])[items[-1]] = value

And for completion’s sake, a function to delete a key:

def del_by_path(root, items):
    """Delete a key-value in a nested object in root by item sequence."""
    del get_by_path(root, items[:-1])[items[-1]]
Answered By: Martijn Pieters

This library may be helpful:

A python library for accessing and searching dictionaries via
/slashed/paths ala xpath

Basically it lets you glob over a dictionary as if it were a

Answered By: dmmfll

Using reduce is clever, but the OP’s set method may have issues if the parent keys do not pre-exist in the nested dictionary. Since this is the first SO post I saw for this subject in my google search, I would like to make it slightly better.

The set method in ( Setting a value in a nested python dictionary given a list of indices and value ) seems more robust to missing parental keys. To copy it over:

def nested_set(dic, keys, value):
    for key in keys[:-1]:
        dic = dic.setdefault(key, {})
    dic[keys[-1]] = value

Also, it can be convenient to have a method that traverses the key tree and get all the absolute key paths, for which I have created:

def keysInDict(dataDict, parent=[]):
    if not isinstance(dataDict, dict):
        return [tuple(parent)]
        return reduce(list.__add__, 
            [keysInDict(v,parent+[k]) for k,v in dataDict.items()], [])

One use of it is to convert the nested tree to a pandas DataFrame, using the following code (assuming that all leafs in the nested dictionary have the same depth).

def dict_to_df(dataDict):
    ret = []
    for k in keysInDict(dataDict):
        v = np.array( getFromDict(dataDict, k), )
        v = pd.DataFrame(v)
        v.columns = pd.MultiIndex.from_product(list(k) + [v.columns])
    return reduce(pd.DataFrame.join, ret)
Answered By: eafit

It seems more pythonic to use a for loop.
See the quote from What’s New In Python 3.0.

Removed reduce(). Use functools.reduce() if you really need it; however, 99 percent of the time an explicit for loop is more readable.

def nested_get(dic, keys):    
    for key in keys:
        dic = dic[key]
    return dic

Note that the accepted solution doesn’t set non-existing nested keys (it raises KeyError). Using the approach below will create non-existing nodes instead:

def nested_set(dic, keys, value):
    for key in keys[:-1]:
        dic = dic.setdefault(key, {})
    dic[keys[-1]] = value

The code works in both Python 2 and 3.

Answered By: DomTomCat

Instead of taking a performance hit each time you want to look up a value, how about you flatten the dictionary once then simply look up the key like b:v:y

def flatten(mydict,sep = ':'):
  new_dict = {}
  for key,value in mydict.items():
    if isinstance(value,dict):
      _dict = {sep.join([key, _key]):_value for _key, _value in flatten(value).items()}
  return new_dict

dataDict = {
    "r": 1,
    "s": 2,
    "t": 3
    "u": 1,
    "v": {
        "x": 1,
        "y": 2,
        "z": 3
    "w": 3

flat_dict = flatten(dataDict)
print flat_dict
{'b:w': 3, 'b:u': 1, 'b:v:y': 2, 'b:v:x': 1, 'b:v:z': 3, 'a:r': 1, 'a:s': 2, 'a:t': 3}

This way you can simply look up items using flat_dict['b:v:y'] which will give you 1.

And instead of traversing the dictionary on each lookup, you may be able to speed this up by flattening the dictionary and saving the output so that a lookup from cold start would mean loading up the flattened dictionary and simply performing a key/value lookup with no traversal.

Answered By: Okezie

How about using recursive functions?

To get a value:

def getFromDict(dataDict, maplist):
    first, rest = maplist[0], maplist[1:]

    if rest: 
        # if `rest` is not empty, run the function recursively
        return getFromDict(dataDict[first], rest)
        return dataDict[first]

And to set a value:

def setInDict(dataDict, maplist, value):
    first, rest = maplist[0], maplist[1:]

    if rest:
            if not isinstance(dataDict[first], dict):
                # if the key is not a dict, then make it a dict
                dataDict[first] = {}
        except KeyError:
            # if key doesn't exist, create one
            dataDict[first] = {}

        setInDict(dataDict[first], rest, value)
        dataDict[first] = value
Answered By: xyres

Pure Python style, without any import:

def nested_set(element, value, *keys):
    if type(element) is not dict:
        raise AttributeError('nested_set() expects dict as first argument.')
    if len(keys) < 2:
        raise AttributeError('nested_set() expects at least three arguments, not enough given.')

    _keys = keys[:-1]
    _element = element
    for key in _keys:
        _element = _element[key]
    _element[keys[-1]] = value

example = {"foo": { "bar": { "baz": "ok" } } }
keys = ['foo', 'bar']
nested_set(example, "yay", *keys)


{'foo': {'bar': 'yay'}}
Answered By: Arount

Solved this with recursion:

def get(d,l):
    if len(l)==1: return d[l[0]]
    return get(d[l[0]],l[1:])

Using your example:

dataDict = {
        "r": 1,
        "s": 2,
        "t": 3
        "u": 1,
        "v": {
            "x": 1,
            "y": 2,
            "z": 3
        "w": 3
maplist1 = ["a", "r"]
maplist2 = ["b", "v", "y"]
print(get(dataDict, maplist1)) # 1
print(get(dataDict, maplist2)) # 2
Answered By: Poh Zi How

If you also want the ability to work with arbitrary json including nested lists and dicts, and nicely handle invalid lookup paths, here’s my solution:

from functools import reduce

def get_furthest(s, path):
    Gets the furthest value along a given key path in a subscriptable structure.

    subscriptable, list -> any
    :param s: the subscriptable structure to examine
    :param path: the lookup path to follow
    :return: a tuple of the value at the furthest valid key, and whether the full path is valid

    def step_key(acc, key):
        s = acc[0]
        if isinstance(s, str):
            return (s, False)
            return (s[key], acc[1])
        except LookupError:
            return (s, False)

    return reduce(step_key, path, (s, True))

def get_val(s, path):
    val, successful = get_furthest(s, path)
    if successful:
        return val
        raise LookupError('Invalid lookup path: {}'.format(path))

def set_val(s, path, value):
    get_val(s, path[:-1])[path[-1]] = value
Answered By: Grant Palmer

An alternative way if you don’t want to raise errors if one of the keys is absent (so that your main code can run without interruption):

def get_value(self,your_dict,*keys):
    curr_dict_ = your_dict
    for k in keys:
        v = curr_dict.get(k,None)
        if v is None:
        if isinstance(v,dict):
            curr_dict = v
    return v

In this case, if any of the input keys is not present, None is returned, which can be used as a check in your main code to perform an alternative task.

Answered By: Pulkit

How about check and then set dict element without processing all indexes twice?


def nested_yield(nested, keys_list):
    Get current nested data by send(None) method. Allows change it to Value by calling send(Value) next time
    :param nested: list or dict of lists or dicts
    :param keys_list: list of indexes/keys
    if not len(keys_list):  # assign to 1st level list
        if isinstance(nested, list):
            while True:
                nested[:] = yield nested
            raise IndexError('Only lists can take element without key')

    last_key = keys_list.pop()
    for key in keys_list:
        nested = nested[key]

    while True:
            nested[last_key] = yield nested[last_key]
        except IndexError as e:
            print('no index {} in {}'.format(last_key, nested))
            yield None

Example workflow:

ny = nested_yield(nested_dict, nested_address)
data_element = ny.send(None)
if data_element:
    # process element
    # extend/update nested data


>>> cfg= {'Options': [[1,[0]],[2,[4,[8,16]]],[3,[9]]]}
    ny = nested_yield(cfg, ['Options',1,1,1])
[8, 16]
>>> ny.send('Hello!')
>>> cfg
{'Options': [[1, [0]], [2, [4, 'Hello!']], [3, [9]]]}
>>> ny.close()
Answered By: And0k

It’s satisfying to see these answers for having two static methods for setting & getting nested attributes. These solutions are way better than using nested trees

Here’s my implementation.


To set nested attribute call sattr(my_dict, 1, 2, 3, 5) is equal to my_dict[1][2][3][4]=5

To get a nested attribute call gattr(my_dict, 1, 2)

def gattr(d, *attrs):
    This method receives a dict and list of attributes to return the innermost value of the give dict       
        for at in attrs:
            d = d[at]
        return d
    except(KeyError, TypeError):
        return None

def sattr(d, *attrs):
    Adds "val" to dict in the hierarchy mentioned via *attrs
    For ex:
    sattr(animals, "cat", "leg","fingers", 4) is equivalent to animals["cat"]["leg"]["fingers"]=4
    This method creates necessary objects until it reaches the final depth
    This behaviour is also known as autovivification and plenty of implementation are around
    This implementation addresses the corner case of replacing existing primitives
    for attr in attrs[:-2]:
        if type(d.get(attr)) is not dict:
            d[attr] = {}
        d = d[attr]
    d[attrs[-2]] = attrs[-1]
Answered By: nehem

Very late to the party, but posting in case this may help someone in the future. For my use case, the following function worked the best. Works to pull any data type out of dictionary

dict is the dictionary containing our value

list is a list of “steps” towards our value

def getnestedvalue(dict, list):

    length = len(list)
        for depth, key in enumerate(list):
            if depth == length - 1:
                output = dict[key]
                return output
            dict = dict[key]
    except (KeyError, TypeError):
        return None

    return None
Answered By: Jack Casey

a method for concatenating strings:

def get_sub_object_from_path(dict_name, map_list):
    for i in map_list:
        _string = "['%s']" % i
        dict_name += _string
    value = eval(dict_name)
    return value
_dict = {'new': 'person', 'time': {'for': 'one'}}
map_list = ['time', 'for']
print get_sub_object_from_path("_dict",map_list)
Answered By: lucas

Extending @DomTomCat and others’ approach, these functional (ie, return modified data via deepcopy without affecting the input) setter and mapper works for nested dict and list.


def set_at_path(data0, keys, value):
    data = deepcopy(data0)
    if len(keys)>1:
        if isinstance(data,dict):
            return {k:(set_by_path(v,keys[1:],value) if k==keys[0] else v) for k,v in data.items()}
        if isinstance(data,list):
            return [set_by_path(x[1],keys[1:],value) if x[0]==keys[0] else x[1] for x in enumerate(data)]
        return data


def map_at_path(data0, keys, f):
    data = deepcopy(data0)
    if len(keys)>1:
        if isinstance(data,dict):
            return {k:(map_at_path(v,keys[1:],f) if k==keys[0] else v) for k,v in data.items()}
        if isinstance(data,list):
            return [map_at_path(x[1],keys[1:],f) if x[0]==keys[0] else x[1] for x in enumerate(data)]
        return data
Answered By: alancalvitti

You can make use of the eval function in python.

def nested_parse(nest, map_list):
    nestq = "nest['" + "']['".join(map_list) + "']"
    return eval(nestq, {'__builtins__':None}, {'nest':nest})


For your example query: maplist = ["b", "v", "y"]

nestq will be "nest['b']['v']['y']" where nest is the nested dictionary.

The eval builtin function executes the given string. However, it is important to be careful about possible vulnerabilities that arise from use of eval function. Discussion can be found here:


In the nested_parse() function, I have made sure that no __builtins__ globals are available and only local variable that is available is the nest dictionary.

Answered By: Abhirup Das

You can use pydash:

import pydash as _

_.get(dataDict, ["b", "v", "y"], default='Default')

Answered By: jamieroboto

I use this

def get_dictionary_value(dictionary_temp, variable_dictionary_keys):
          if(len(variable_dictionary_keys) == 0):
               return str(dictionary_temp)

          variable_dictionary_key = variable_dictionary_keys[0]

          return get_dictionary_value(dictionary_temp[variable_dictionary_key] , variable_dictionary_keys)

     except Exception as variable_exception:
          return ''

Answered By: Werner Venter

Check out NestedDict from the ndicts package (I am the author), it does exactly what you ask for.

from ndicts import NestedDict

data_dict = {
        "r": 1,
        "s": 2,
        "t": 3
        "u": 1,
        "v": {
            "x": 1,
            "y": 2,
            "z": 3
        "w": 3

nd = NestedDict(data_dict)

You can now access keys using comma separated values.

>>> nd["a", "r"]
>>> nd["b", "v"]
    {"x": 1, "y": 2, "z": 3}
Answered By: edd313

I’d rather use simple recursion function:

def get_value_by_path(data, maplist):
    if not maplist:
        return data
    for key in maplist:
        if key in data:
            return get_value_by_path(data[key], maplist[1:])
Answered By: funnydman
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