Update value of a nested dictionary of varying depth

Question

I'm looking for a way to update dict dictionary1 with the contents of dict update wihout overwriting levelA

dictionary1 = {
    "level1": {
        "level2": {"levelA": 0, "levelB": 1}
    }
}
update = {
    "level1": {
        "level2": {"levelB": 10}
    }
}
dictionary1.update(update)
print(dictionary1)

{
    "level1": {
        "level2": {"levelB": 10}
    }
}

I know that update deletes the values in level2 because it's updating the lowest key level1.

How could I tackle this, given that dictionary1 and update can have any length?

Answer 1

@FM's answer has the right general idea, i.e. a recursive solution, but somewhat peculiar coding and at least one bug. I'd recommend, instead:

Python 2:

import collections

def update(d, u):
    for k, v in u.iteritems():
        if isinstance(v, collections.Mapping):
            d[k] = update(d.get(k, {}), v)
        else:
            d[k] = v
    return d

Python 3:

import collections.abc

def update(d, u):
    for k, v in u.items():
        if isinstance(v, collections.abc.Mapping):
            d[k] = update(d.get(k, {}), v)
        else:
            d[k] = v
    return d

The bug shows up when the "update" has a k, v item where v is a dict and k is not originally a key in the dictionary being updated -- @FM's code "skips" this part of the update (because it performs it on an empty new dict which isn't saved or returned anywhere, just lost when the recursive call returns).

My other changes are minor: there is no reason for the if/else construct when .get does the same job faster and cleaner, and isinstance is best applied to abstract base classes (not concrete ones) for generality.

Answer 2

If you happen to be using pydantic (great lib, BTW), you can use one of its utility methods:

from pydantic.utils import deep_update

dictionary1 = deep_update(dictionary1, update)

UPDATE: reference to code, as pointed by @Jorgu. If installing pydantic is not desired, the code is short enough to be copied, provided adequate licenses compatibilities.

Answer 3

Took me a little bit on this one, but thanks to @Alex's post, he filled in the gap I was missing. However, I came across an issue if a value within the recursive dict happens to be a list, so I thought I'd share, and extend his answer.

import collections

def update(orig_dict, new_dict):
    for key, val in new_dict.iteritems():
        if isinstance(val, collections.Mapping):
            tmp = update(orig_dict.get(key, { }), val)
            orig_dict[key] = tmp
        elif isinstance(val, list):
            orig_dict[key] = (orig_dict.get(key, []) + val)
        else:
            orig_dict[key] = new_dict[key]
    return orig_dict

Answer 4

Same solution as the accepted one, but clearer variable naming, docstring, and fixed a bug where {} as a value would not override.

import collections


def deep_update(source, overrides):
    """
    Update a nested dictionary or similar mapping.
    Modify ``source`` in place.
    """
    for key, value in overrides.iteritems():
        if isinstance(value, collections.Mapping) and value:
            returned = deep_update(source.get(key, {}), value)
            source[key] = returned
        else:
            source[key] = overrides[key]
    return source

Here are a few test cases:

def test_deep_update():
    source = {'hello1': 1}
    overrides = {'hello2': 2}
    deep_update(source, overrides)
    assert source == {'hello1': 1, 'hello2': 2}

    source = {'hello': 'to_override'}
    overrides = {'hello': 'over'}
    deep_update(source, overrides)
    assert source == {'hello': 'over'}

    source = {'hello': {'value': 'to_override', 'no_change': 1}}
    overrides = {'hello': {'value': 'over'}}
    deep_update(source, overrides)
    assert source == {'hello': {'value': 'over', 'no_change': 1}}

    source = {'hello': {'value': 'to_override', 'no_change': 1}}
    overrides = {'hello': {'value': {}}}
    deep_update(source, overrides)
    assert source == {'hello': {'value': {}, 'no_change': 1}}

    source = {'hello': {'value': {}, 'no_change': 1}}
    overrides = {'hello': {'value': 2}}
    deep_update(source, overrides)
    assert source == {'hello': {'value': 2, 'no_change': 1}}

This functions is available in the charlatan package, in charlatan.utils.

Answer 5

@Alex's answer is good, but doesn't work when replacing an element such as an integer with a dictionary, such as update({'foo':0},{'foo':{'bar':1}}). This update addresses it:

import collections
def update(d, u):
    for k, v in u.iteritems():
        if isinstance(d, collections.Mapping):
            if isinstance(v, collections.Mapping):
                r = update(d.get(k, {}), v)
                d[k] = r
            else:
                d[k] = u[k]
        else:
            d = {k: u[k]}
    return d

update({'k1': 1}, {'k1': {'k2': {'k3': 3}}})

Answer 6

Here's an immutable version of recursive dictionary merge in case someone needs it.

Based upon @Alex Martelli's answer.

Python 3.x:

import collections
from copy import deepcopy


def merge(dict1, dict2):
    ''' Return a new dictionary by merging two dictionaries recursively. '''

    result = deepcopy(dict1)

    for key, value in dict2.items():
        if isinstance(value, collections.Mapping):
            result[key] = merge(result.get(key, {}), value)
        else:
            result[key] = deepcopy(dict2[key])

    return result

Python 2.x:

import collections
from copy import deepcopy


def merge(dict1, dict2):
    ''' Return a new dictionary by merging two dictionaries recursively. '''

    result = deepcopy(dict1)

    for key, value in dict2.iteritems():
        if isinstance(value, collections.Mapping):
            result[key] = merge(result.get(key, {}), value)
        else:
            result[key] = deepcopy(dict2[key])

    return result

Answer 7

This question is old, but I landed here when searching for a "deep merge" solution. The answers above inspired what follows. I ended up writing my own because there were bugs in all the versions I tested. The critical point missed was, at some arbitrary depth of the two input dicts, for some key, k, the decision tree when d[k] or u[k] is not a dict was faulty.

Also, this solution does not require recursion, which is more symmetric with how dict.update() works, and returns None.

import collections
def deep_merge(d, u):
   """Do a deep merge of one dict into another.

   This will update d with values in u, but will not delete keys in d
   not found in u at some arbitrary depth of d. That is, u is deeply
   merged into d.

   Args -
     d, u: dicts

   Note: this is destructive to d, but not u.

   Returns: None
   """
   stack = [(d,u)]
   while stack:
      d,u = stack.pop(0)
      for k,v in u.items():
         if not isinstance(v, collections.Mapping):
            # u[k] is not a dict, nothing to merge, so just set it,
            # regardless if d[k] *was* a dict
            d[k] = v

        else:
            # note: u[k] is a dict
            if k not in d:
                # add new key into d
                d[k] = v
            elif not isinstance(d[k], collections.Mapping):
                # d[k] is not a dict, so just set it to u[k],
                # overriding whatever it was
                d[k] = v
            else:
                # both d[k] and u[k] are dicts, push them on the stack
                # to merge
                stack.append((d[k], v))

Answer 8

Just use python-benedict (I did it), it has a merge (deepupdate) utility method and many others. It works with python 2 / python 3 and it is well tested.

from benedict import benedict

dictionary1=benedict({'level1':{'level2':{'levelA':0,'levelB':1}}})
update={'level1':{'level2':{'levelB':10}}}
dictionary1.merge(update)
print(dictionary1)
# >> {'level1':{'level2':{'levelA':0,'levelB':10}}}

Installation: pip install python-benedict

Documentation: https://github.com/fabiocaccamo/python-benedict

Note: I am the author of this project

Answer 9

Minor improvements to @Alex's answer that enables updating of dictionaries of differing depths as well as limiting the depth that the update dives into the original nested dictionary (but the updating dictionary depth is not limited). Only a few cases have been tested:

def update(d, u, depth=-1):
    """
    Recursively merge or update dict-like objects. 
    >>> update({'k1': {'k2': 2}}, {'k1': {'k2': {'k3': 3}}, 'k4': 4})
    {'k1': {'k2': {'k3': 3}}, 'k4': 4}
    """

    for k, v in u.iteritems():
        if isinstance(v, Mapping) and not depth == 0:
            r = update(d.get(k, {}), v, depth=max(depth - 1, -1))
            d[k] = r
        elif isinstance(d, Mapping):
            d[k] = u[k]
        else:
            d = {k: u[k]}
    return d

Answer 10

The code below should solve the update({'k1': 1}, {'k1': {'k2': 2}}) issue in @Alex Martelli's answer the right way.

def deepupdate(original, update):
    """Recursively update a dict.

    Subdict's won't be overwritten but also updated.
    """
    if not isinstance(original, abc.Mapping):
        return update
    for key, value in update.items():
        if isinstance(value, abc.Mapping):
            original[key] = deepupdate(original.get(key, {}), value)
        else:
            original[key] = value
    return original

Answer 11

I used the solution @Alex Martelli suggests, but it fails

TypeError 'bool' object does not support item assignment

when the two dictionaries differ in data type at some level.

In case at the same level the element of dictionary d is just a scalar (ie. Bool) while the element of dictionary u is still dictionary the reassignment fails as no dictionary assignment is possible into scalar (like True[k]).

One added condition fixes that:

from collections import Mapping

def update_deep(d, u):
    for k, v in u.items():
        # this condition handles the problem
        if not isinstance(d, Mapping):
            d = u
        elif isinstance(v, Mapping):
            r = update_deep(d.get(k, {}), v)
            d[k] = r
        else:
            d[k] = u[k]

    return d

Answer 12

It could be that you stumble over a non-standard-dictionary, like me today, which has no iteritems-Attribute. In this case it's easy to interpret this type of dictionary as a standard-dictionary. E.g.: Python 2.7:

    import collections
    def update(orig_dict, new_dict):
        for key, val in dict(new_dict).iteritems():
            if isinstance(val, collections.Mapping):
                tmp = update(orig_dict.get(key, { }), val)
                orig_dict[key] = tmp
            elif isinstance(val, list):
                orig_dict[key] = (orig_dict[key] + val)
            else:
                orig_dict[key] = new_dict[key]
        return orig_dict

    import multiprocessing
    d=multiprocessing.Manager().dict({'sample':'data'})
    u={'other': 1234}

    x=update(d, u)
    x.items()

Python 3.8:

    def update(orig_dict, new_dict):
        orig_dict=dict(orig_dict)
        for key, val in dict(new_dict).items():
            if isinstance(val, collections.abc.Mapping):
                tmp = update(orig_dict.get(key, { }), val)
                orig_dict[key] = tmp
            elif isinstance(val, list):
                orig_dict[key] = (orig_dict[key] + val)
            else:
                orig_dict[key] = new_dict[key]
        return orig_dict

    import collections
    import multiprocessing
    d=multiprocessing.Manager().dict({'sample':'data'})
    u={'other': 1234, "deeper": {'very': 'deep'}}

    x=update(d, u)
    x.items()

Answer 13

In neither of these answers the authors seem to understand the concept of updating an object stored in a dictionary nor even of iterating over dictionary items (as opposed to keys). So I had to write one which doesn't make pointless tautological dictionary stores and retrievals. The dicts are assumed to store other dicts or simple types.

def update_nested_dict(d, other):
    for k, v in other.items():
        if isinstance(v, collections.Mapping):
            d_v = d.get(k)
            if isinstance(d_v, collections.Mapping):
                update_nested_dict(d_v, v)
            else:
                d[k] = v.copy()
        else:
            d[k] = v

Or even simpler one working with any type:

def update_nested_dict(d, other):
    for k, v in other.items():
        d_v = d.get(k)
        if isinstance(v, collections.Mapping) and isinstance(d_v, collections.Mapping):
            update_nested_dict(d_v, v)
        else:
            d[k] = deepcopy(v) # or d[k] = v if you know what you're doing

Answer 14

Update to @Alex Martelli's answer to fix a bug in his code to make the solution more robust:

def update_dict(d, u):
    for k, v in u.items():
        if isinstance(v, collections.Mapping):
            default = v.copy()
            default.clear()
            r = update_dict(d.get(k, default), v)
            d[k] = r
        else:
            d[k] = v
    return d

The key is that we often want to create the same type at recursion, so here we use v.copy().clear() but not {}. And this is especially useful if the dict here is of type collections.defaultdict which can have different kinds of default_factorys.

Also notice that the u.iteritems() has been changed to u.items() in Python3.

Answer 15

def update(value, nvalue):
    if not isinstance(value, dict) or not isinstance(nvalue, dict):
        return nvalue
    for k, v in nvalue.items():
        value.setdefault(k, dict())
        if isinstance(v, dict):
            v = update(value[k], v)
        value[k] = v
    return value

use dict or collections.Mapping

Answer 16

I recommend to replace {} by type(v)() in order to propagate object type of any dict subclass stored in u but absent from d. For example, this would preserve types such as collections.OrderedDict:

Python 2:

import collections

def update(d, u):
    for k, v in u.iteritems():
        if isinstance(v, collections.Mapping):
            d[k] = update(d.get(k, type(v)()), v)
        else:
            d[k] = v
    return d

Python 3:

import collections.abc

def update(d, u):
    for k, v in u.items():
        if isinstance(v, collections.abc.Mapping):
            d[k] = update(d.get(k, type(v)()), v)
        else:
            d[k] = v
    return d

Answer 17

Thanks to hobs for his comment on Alex's answer. Indeed update({'k1': 1}, {'k1': {'k2': 2}}) will cause TypeError: 'int' object does not support item assignment.

We should check the types of the input values at the beginning of the function. So, I suggest the following function, which should solve this (and other) problem.

Python 3:

from collections.abc import Mapping


def deep_update(d1, d2):
    if all((isinstance(d, Mapping) for d in (d1, d2))):
        for k, v in d2.items():
            d1[k] = deep_update(d1.get(k), v)
        return d1
    return d2

Answer 18

I know this question is pretty old, but still posting what I do when I have to update a nested dictionary. We can use the fact that dicts are passed by reference in python Assuming that the path of the key is known and is dot separated. Forex if we have a dict named data:

{
"log_config_worker": {
    "version": 1, 
    "root": {
        "handlers": [
            "queue"
        ], 
        "level": "DEBUG"
    }, 
    "disable_existing_loggers": true, 
    "handlers": {
        "queue": {
            "queue": null, 
            "class": "myclass1.QueueHandler"
        }
    }
}, 
"number_of_archived_logs": 15, 
"log_max_size": "300M", 
"cron_job_dir": "/etc/cron.hourly/", 
"logs_dir": "/var/log/patternex/", 
"log_rotate_dir": "/etc/logrotate.d/"
}

And we want to update the queue class, the path of the key would be - log_config_worker.handlers.queue.class

We can use the following function to update the value:

def get_updated_dict(obj, path, value):
    key_list = path.split(".")

    for k in key_list[:-1]:
        obj = obj[k]

    obj[key_list[-1]] = value

get_updated_dict(data, "log_config_worker.handlers.queue.class", "myclass2.QueueHandler")

This would update the dictionary correctly.

Answer 19

I made a simple function, in which you give the key, the new value and the dictionary as input, and it recursively updates it with the value:

def update(key,value,dictionary):
    if key in dictionary.keys():
        dictionary[key] = value
        return
    dic_aux = []
    for val_aux in dictionary.values():
        if isinstance(val_aux,dict):
            dic_aux.append(val_aux)
    for i in dic_aux:
        update(key,value,i)
    for [key2,val_aux2] in dictionary.items():
        if isinstance(val_aux2,dict):
            dictionary[key2] = val_aux2

dictionary1={'level1':{'level2':{'levelA':0,'levelB':1}}}
update('levelB',10,dictionary1)
print(dictionary1)

#output: {'level1': {'level2': {'levelA': 0, 'levelB': 10}}}

Hope it answers.

Answer 20

If you want to replace a "full nested dictionary with arrays" you can use this snippet :

It will replace any "old_value" by "new_value". It's roughly doing a depth-first rebuilding of the dictionary. It can even work with List or Str/int given as input parameter of first level.

def update_values_dict(original_dict, future_dict, old_value, new_value):
    # Recursively updates values of a nested dict by performing recursive calls

    if isinstance(original_dict, Dict):
        # It's a dict
        tmp_dict = {}
        for key, value in original_dict.items():
            tmp_dict[key] = update_values_dict(value, future_dict, old_value, new_value)
        return tmp_dict
    elif isinstance(original_dict, List):
        # It's a List
        tmp_list = []
        for i in original_dict:
            tmp_list.append(update_values_dict(i, future_dict, old_value, new_value))
        return tmp_list
    else:
        # It's not a dict, maybe a int, a string, etc.
        return original_dict if original_dict != old_value else new_value

Answer 21

Yes! And another solution. My solution differs in the keys that are being checked. In all other solutions we only look at the keys in dict_b. But here we look in the union of both dictionaries.

Do with it as you please

def update_nested(dict_a, dict_b):
    set_keys = set(dict_a.keys()).union(set(dict_b.keys()))
    for k in set_keys:
        v = dict_a.get(k)
        if isinstance(v, dict):
            new_dict = dict_b.get(k, None)
            if new_dict:
                update_nested(v, new_dict)
        else:
            new_value = dict_b.get(k, None)
            if new_value:
                dict_a[k] = new_value

Answer 22

Another way of using recursion:

def updateDict(dict1,dict2):
    keys1 = list(dict1.keys())
    keys2= list(dict2.keys())
    keys2 = [x for x in keys2 if x in keys1]
    for x in keys2:
        if (x in keys1) & (type(dict1[x]) is dict) & (type(dict2[x]) is dict):
            updateDict(dict1[x],dict2[x])
        else:
            dict1.update({x:dict2[x]})
    return(dict1)

Answer 23

Credit to: @Gustavo Alves Casqueiro for original answer

I honestly would have preferred using a lib that could do the heavy lifting for me, but I just couldn't find something that did what I needed.

I have only added a couple of additional checks to this function.

I have included a check for lists within a dict and added a parameter for the name of a nested dict to correctly update the nested dict KEY when there may be another KEY within the OUTER dict with the same name.

Updated function:

def update(dictionary: dict[str, any], key: str, value: any, nested_dict_name: str = None) -> dict[str, any]:
    if not nested_dict_name:  # if current (outermost) dict should be updated
        if key in dictionary.keys():  # check if key exists in current dict
            dictionary[key] = value
            return dictionary
    else:  # if nested dict should be updated
        if nested_dict_name in dictionary.keys():  # check if dict is in next layer
            if isinstance(dictionary[nested_dict_name], dict):
                if key in dictionary[nested_dict_name].keys():  # check if key exists in current dict
                    dictionary[nested_dict_name][key] = value
                    return dictionary
            if isinstance(dictionary[nested_dict_name], list):
                list_index = random.choice(range(len(dictionary[nested_dict_name])))  # pick a random dict from the list

                if key in dictionary[nested_dict_name][list_index].keys():  # check if key exists in current dict
                    dictionary[nested_dict_name][list_index][key] = value
                    return dictionary

    dic_aux = []

    # this would only run IF the above if-statement was not able to identity and update a dict
    for val_aux in dictionary.values():
        if isinstance(val_aux, dict):
            dic_aux.append(val_aux)

    # call the update function again for recursion
    for i in dic_aux:
        return update(dictionary=i, key=key, value=value, nested_dict_name=nested_dict_name)

Original dict:

{
    "level1": {
        "level2": {
            "myBool": "Original",
            "myInt": "Original"
        },
        "myInt": "Original",
        "myBool": "Original"
    },
    "myStr": "Original",
    "level3": [
        {
            "myList": "Original",
            "myInt": "Original",
            "myBool": "Original"
        }
    ],
    "level4": [
        {
            "myList": "Original",
            "myInt": "UPDATED",
            "myBool": "Original"
        }
    ],
    "level5": {
        "level6": {
            "myBool": "Original",
            "myInt": "Original"
        },
        "myInt": "Original",
        "myBool": "Original"
    }
}

Data for updating (using pytest):

@pytest.fixture(params=[(None, 'myStr', 'UPDATED'),
                        ('level1', 'myInt', 'UPDATED'),
                        ('level2', 'myBool', 'UPDATED'),
                        ('level3', 'myList', 'UPDATED'),
                        ('level4', 'myInt', 'UPDATED'),
                        ('level5', 'myBool', 'UPDATED')])
def sample_data(request):
    return request.param

The 'UPDATED' parameter doesn't make sense in this smaller use case (since I could just hard-code it), but for simplicity when reading the logs, I didn't want to see multiple data-types and just made it show me an 'UPDATED' string.

Test:

@pytest.mark.usefixtures('sample_data')
def test_this(sample_data):
    nested_dict, param, update_value = sample_data

    if nested_dict is None:
        print(f'\nDict Value: Level0\nParam: {param}\nUpdate Value: {update_value}')
    else:
        print(f'\nDict Value: {nested_dict}\nParam: {param}\nUpdate Value: {update_value}')

    # initialise data dict
    data_object = # insert data here (see example dict above)

    # first print as is
    print(f'\nOriginal Dict:\n{data_object}')

    update(dictionary=data_object,
           key=param,
           value=update_value,
           nested_dict_name=nested_dict)

    # print updated
    print(f'\nUpdated Dict:\n{data_object}')

There is one caveat, when you have a dict like this:

{
    "level1": {
        "level2": {
            "myBool": "Original"
        },
        "myBool": "Original"
    },
    "level3": {
        "level2": {
            "myBool": "Original"
        },
        "myInt": "Original"
    }
}

Where level2 is under level1 AND level3. This would require making using of a list or something with the nested_dict_name and passing in the name of the outer dict AND inner dict (['level5', 'level2']) and then somehow looping through the values to find that dict.

However, since I haven't yet ran into this issue for the data objects I use, I haven't spent the time trying to solve this "issue".

Answer 24

Convert your dictionaries into NestedDict

from ndicts.ndicts import NestedDict

dictionary1 = {'level1': {'level2': {'levelA': 0, 'levelB': 1}}}
update = {'level1': {'level2': {'levelB': 10}}}

nd, nd_update = NestedDict(dictionary1), NestedDict(update)

Then just use update

>>> nd.update(nd_update)
>>> nd
NestedDict({'level1': {'level2': {'levelA': 0, 'levelB': 10}}})

If you need the result as a dictionary nd.to_dict()

To install ndicts pip install ndicts

Answer 25

d is dict to update, u is dict-updater.

def recursively_update_dict(d: dict, u: dict):
    for k, v in u.items():
        if isinstance(v, dict):
            d.setdefault(k, {})
            recursively_update_dict(d[k], v)
        else:
            d[k] = v

Or for defaultdict

from collections import defaultdict

def recursively_update_defaultdict(d: defaultdict[dict], u: dict):
    for k, v in u.items():
        if isinstance(v, dict):
            recursively_update_dict(d[k], v)
        else:
            d[k] = v

Answer 26

Basic solution (in case someone arrives here looking for that)

dictionary1 = {
    "level1": {
        "level2": {"levelA": 0, "levelB": 1}
    }
}

dictionary1.update({
    "level1": {
        "level2": {
            **dictionary1["level1"]["level2"],
            "levelB": 10
        }
    }
})

Result:

{'level1': {'level2': {'levelA': 0, 'levelB': 10}}}

Answer 27

Here's what I use:

from collections.abc import MutableMapping as Map

def merge(d, v):
    """
    Merge two dictionaries.

    Merge dict-like `v` into dict-like `d`. In case keys between them
    are the same, merge their sub-dictionaries. Otherwise, values in  
    `v` overwrite `d`.
    """
    for key in v:
        if key in d and isinstance(d[key], Map) and isinstance(v[key], Map):
            d[key] = merge(d[key], v[key])
        else:
            d[key] = v[key]
    return d

merge(dictionary1, update)
print(dictionary1)

>>> {'level1': {'level2': {'levelA': 0, 'levelB': 10}}}

Answer 28

The answer from @kepler helps me, but when I run

from pydantic.utils import deep_update
dict1 = deep_update(dict1, dict2)

a warning comes out:

UserWarning: `pydantic.utils:deep_update` has been removed. We are importing from `pydantic.v1.utils:deep_update` instead.See the migration guide for more details: https://docs.pydantic.dev/latest/migration/
  f'`{import_path}` has been removed. We are importing from `{new_location}` instead.'

So I change the import command to

from pydantic.v1.utils import deep_update

Answer 29

a new Q how to By a keys chain

dictionary1={'level1':{'level2':{'levelA':0,'levelB':1}},'anotherLevel1':{'anotherLevel2':{'anotherLevelA':0,'anotherLevelB':1}}}
update={'anotherLevel1':{'anotherLevel2':1014}}
dictionary1.update(update)
print dictionary1
{'level1':{'level2':{'levelA':0,'levelB':1}},'anotherLevel1':{'anotherLevel2':1014}}

Answer 30

you could try this, it works with lists and is pure:

def update_keys(newd, dic, mapping):
  def upsingle(d,k,v):
    if k in mapping:
      d[mapping[k]] = v
    else:
      d[k] = v
  for ekey, evalue in dic.items():
    upsingle(newd, ekey, evalue)
    if type(evalue) is dict:
      update_keys(newd, evalue, mapping)
    if type(evalue) is list:
      upsingle(newd, ekey, [update_keys({}, i, mapping) for i in evalue])
  return newd

Answer 31

That's a bit to the side but do you really need nested dictionaries? Depending on the problem, sometimes flat dictionary may suffice... and look good at it:

>>> dict1 = {('level1','level2','levelA'): 0}
>>> dict1['level1','level2','levelB'] = 1
>>> update = {('level1','level2','levelB'): 10}
>>> dict1.update(update)
>>> print dict1
{('level1', 'level2', 'levelB'): 10, ('level1', 'level2', 'levelA'): 0}