Deep merge dictionaries of dictionaries in Python

Question

I need to merge multiple dictionaries, here's what I have for instance:

dict1 = {1:{"a":{"A"}}, 2:{"b":{"B"}}}

dict2 = {2:{"c":{"C"}}, 3:{"d":{"D"}}}

With A B C and D being leaves of the tree, like {"info1":"value", "info2":"value2"}

There is an unknown level(depth) of dictionaries, it could be {2:{"c":{"z":{"y":{C}}}}}

In my case it represents a directory/files structure with nodes being docs and leaves being files.

I want to merge them to obtain:

 dict3 = {1:{"a":{"A"}}, 2:{"b":{"B"},"c":{"C"}}, 3:{"d":{"D"}}}

I'm not sure how I could do that easily with Python.

Answer 1

This is actually quite tricky - particularly if you want a useful error message when things are inconsistent, while correctly accepting duplicate but consistent entries (something no other answer here does..)

Assuming you don't have huge numbers of entries, a recursive function is easiest:

def merge(a: dict, b: dict, path=[]):
    for key in b:
        if key in a:
            if isinstance(a[key], dict) and isinstance(b[key], dict):
                merge(a[key], b[key], path + [str(key)])
            elif a[key] != b[key]:
                raise Exception('Conflict at ' + '.'.join(path + [str(key)]))
        else:
            a[key] = b[key]
    return a

# works
print(merge({1:{"a":"A"},2:{"b":"B"}}, {2:{"c":"C"},3:{"d":"D"}}))
# has conflict
merge({1:{"a":"A"},2:{"b":"B"}}, {1:{"a":"A"},2:{"b":"C"}})

note that this mutates a - the contents of b are added to a (which is also returned). If you want to keep a you could call it like merge(dict(a), b).

agf pointed out (below) that you may have more than two dicts, in which case you can use:

from functools import reduce
reduce(merge, [dict1, dict2, dict3...])

where everything will be added to dict1.

Note: I edited my initial answer to mutate the first argument; that makes the "reduce" easier to explain

Answer 2

You could try mergedeep.

---

Installation

$ pip3 install mergedeep

Usage

from mergedeep import merge

a = {"keyA": 1}
b = {"keyB": {"sub1": 10}}
c = {"keyB": {"sub2": 20}}

merge(a, b, c) 

print(a)
# {"keyA": 1, "keyB": {"sub1": 10, "sub2": 20}}

For a full list of options, check out the docs!

Answer 3

Here's an easy way to do it using generators:

def mergedicts(dict1, dict2):
    for k in set(dict1.keys()).union(dict2.keys()):
        if k in dict1 and k in dict2:
            if isinstance(dict1[k], dict) and isinstance(dict2[k], dict):
                yield (k, dict(mergedicts(dict1[k], dict2[k])))
            else:
                # If one of the values is not a dict, you can't continue merging it.
                # Value from second dict overrides one in first and we move on.
                yield (k, dict2[k])
                # Alternatively, replace this with exception raiser to alert you of value conflicts
        elif k in dict1:
            yield (k, dict1[k])
        else:
            yield (k, dict2[k])

dict1 = {1:{"a":"A"},2:{"b":"B"}}
dict2 = {2:{"c":"C"},3:{"d":"D"}}

print dict(mergedicts(dict1,dict2))

This prints:

{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}

Answer 4

One issue with this question is that the values of the dict can be arbitrarily complex pieces of data. Based upon these and other answers I came up with this code:

class YamlReaderError(Exception):
    pass

def data_merge(a, b):
    """merges b into a and return merged result

    NOTE: tuples and arbitrary objects are not handled as it is totally ambiguous what should happen"""
    key = None
    # ## debug output
    # sys.stderr.write("DEBUG: %s to %s\n" %(b,a))
    try:
        if a is None or isinstance(a, str) or isinstance(a, unicode) or isinstance(a, int) or isinstance(a, long) or isinstance(a, float):
            # border case for first run or if a is a primitive
            a = b
        elif isinstance(a, list):
            # lists can be only appended
            if isinstance(b, list):
                # merge lists
                a.extend(b)
            else:
                # append to list
                a.append(b)
        elif isinstance(a, dict):
            # dicts must be merged
            if isinstance(b, dict):
                for key in b:
                    if key in a:
                        a[key] = data_merge(a[key], b[key])
                    else:
                        a[key] = b[key]
            else:
                raise YamlReaderError('Cannot merge non-dict "%s" into dict "%s"' % (b, a))
        else:
            raise YamlReaderError('NOT IMPLEMENTED "%s" into "%s"' % (b, a))
    except TypeError, e:
        raise YamlReaderError('TypeError "%s" in key "%s" when merging "%s" into "%s"' % (e, key, b, a))
    return a

My use case is merging YAML files where I only have to deal with a subset of possible data types. Hence I can ignore tuples and other objects. For me a sensible merge logic means

• replace scalars
• append lists
• merge dicts by adding missing keys and updating existing keys

Everything else and the unforeseens results in an error.

Answer 5

Dictionaries of dictionaries merge

As this is the canonical question (in spite of certain non-generalities) I'm providing the canonical Pythonic approach to solving this issue.

Simplest Case: "leaves are nested dicts that end in empty dicts":

d1 = {'a': {1: {'foo': {}}, 2: {}}}
d2 = {'a': {1: {}, 2: {'bar': {}}}}
d3 = {'b': {3: {'baz': {}}}}
d4 = {'a': {1: {'quux': {}}}}

This is the simplest case for recursion, and I would recommend two naive approaches:

def rec_merge1(d1, d2):
    '''return new merged dict of dicts'''
    for k, v in d1.items(): # in Python 2, use .iteritems()!
        if k in d2:
            d2[k] = rec_merge1(v, d2[k])
    d3 = d1.copy()
    d3.update(d2)
    return d3

def rec_merge2(d1, d2):
    '''update first dict with second recursively'''
    for k, v in d1.items(): # in Python 2, use .iteritems()!
        if k in d2:
            d2[k] = rec_merge2(v, d2[k])
    d1.update(d2)
    return d1

I believe I would prefer the second to the first, but keep in mind that the original state of the first would have to be rebuilt from its origin. Here's the usage:

>>> from functools import reduce # only required for Python 3.
>>> reduce(rec_merge1, (d1, d2, d3, d4))
{'a': {1: {'quux': {}, 'foo': {}}, 2: {'bar': {}}}, 'b': {3: {'baz': {}}}}
>>> reduce(rec_merge2, (d1, d2, d3, d4))
{'a': {1: {'quux': {}, 'foo': {}}, 2: {'bar': {}}}, 'b': {3: {'baz': {}}}}

Complex Case: "leaves are of any other type:"

So if they end in dicts, it's a simple case of merging the end empty dicts. If not, it's not so trivial. If strings, how do you merge them? Sets can be updated similarly, so we could give that treatment, but we lose the order in which they were merged. So does order matter?

So in lieu of more information, the simplest approach will be to give them the standard update treatment if both values are not dicts: i.e. the second dict's value will overwrite the first, even if the second dict's value is None and the first's value is a dict with a lot of info.

d1 = {'a': {1: 'foo', 2: None}}
d2 = {'a': {1: None, 2: 'bar'}}
d3 = {'b': {3: 'baz'}}
d4 = {'a': {1: 'quux'}}

from collections.abc import MutableMapping

def rec_merge(d1, d2):
    '''
    Update two dicts of dicts recursively, 
    if either mapping has leaves that are non-dicts, 
    the second's leaf overwrites the first's.
    '''
    for k, v in d1.items():
        if k in d2:
            # this next check is the only difference!
            if all(isinstance(e, MutableMapping) for e in (v, d2[k])):
                d2[k] = rec_merge(v, d2[k])
            # we could further check types and merge as appropriate here.
    d3 = d1.copy()
    d3.update(d2)
    return d3

And now

from functools import reduce
reduce(rec_merge, (d1, d2, d3, d4))

returns

{'a': {1: 'quux', 2: 'bar'}, 'b': {3: 'baz'}}

Application to the original question:

I've had to remove the curly braces around the letters and put them in single quotes for this to be legit Python (else they would be set literals in Python 2.7+) as well as append a missing brace:

dict1 = {1:{"a":'A'}, 2:{"b":'B'}}
dict2 = {2:{"c":'C'}, 3:{"d":'D'}}

and rec_merge(dict1, dict2) now returns:

{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}

Which matches the desired outcome of the original question (after changing, e.g. the {A} to 'A'.)

Answer 6

Based on @andrew cooke. This version handles nested lists of dicts and also allows the option to update the values

def merge(a, b, path=None, update=True):
    "http://stackoverflow.com/questions/7204805/python-dictionaries-of-dictionaries-merge"
    "merges b into a"
    if path is None: path = []
    for key in b:
        if key in a:
            if isinstance(a[key], dict) and isinstance(b[key], dict):
                merge(a[key], b[key], path + [str(key)])
            elif a[key] == b[key]:
                pass # same leaf value
            elif isinstance(a[key], list) and isinstance(b[key], list):
                for idx, val in enumerate(b[key]):
                    a[key][idx] = merge(a[key][idx], b[key][idx], path + [str(key), str(idx)], update=update)
            elif update:
                a[key] = b[key]
            else:
                raise Exception('Conflict at %s' % '.'.join(path + [str(key)]))
        else:
            a[key] = b[key]
    return a

Answer 7

This simple recursive procedure will merge one dictionary into another while overriding conflicting keys:

#!/usr/bin/env python2.7

def merge_dicts(dict1, dict2):
    """ Recursively merges dict2 into dict1 """
    if not isinstance(dict1, dict) or not isinstance(dict2, dict):
        return dict2
    for k in dict2:
        if k in dict1:
            dict1[k] = merge_dicts(dict1[k], dict2[k])
        else:
            dict1[k] = dict2[k]
    return dict1

print (merge_dicts({1:{"a":"A"}, 2:{"b":"B"}}, {2:{"c":"C"}, 3:{"d":"D"}}))
print (merge_dicts({1:{"a":"A"}, 2:{"b":"B"}}, {1:{"a":"A"}, 2:{"b":"C"}}))

Output:

{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}
{1: {'a': 'A'}, 2: {'b': 'C'}}

Answer 8

In case someone wants yet another approach to this problem, here's my solution.

Virtues: short, declarative, and functional in style (recursive, does no mutation).

Potential Drawback: This might not be the merge you're looking for. Consult the docstring for semantics.

def deep_merge(a, b):
    """
    Merge two values, with `b` taking precedence over `a`.

    Semantics:
    - If either `a` or `b` is not a dictionary, `a` will be returned only if
      `b` is `None`. Otherwise `b` will be returned.
    - If both values are dictionaries, they are merged as follows:
        * Each key that is found only in `a` or only in `b` will be included in
          the output collection with its value intact.
        * For any key in common between `a` and `b`, the corresponding values
          will be merged with the same semantics.
    """
    if not isinstance(a, dict) or not isinstance(b, dict):
        return a if b is None else b
    else:
        # If we're here, both a and b must be dictionaries or subtypes thereof.

        # Compute set of all keys in both dictionaries.
        keys = set(a.keys()) | set(b.keys())

        # Build output dictionary, merging recursively values with common keys,
        # where `None` is used to mean the absence of a value.
        return {
            key: deep_merge(a.get(key), b.get(key))
            for key in keys
        }

Answer 9

Based on answers from @andrew cooke. It takes care of nested lists in a better way.

def deep_merge_lists(original, incoming):
    """
    Deep merge two lists. Modifies original.
    Recursively call deep merge on each correlated element of list. 
    If item type in both elements are
     a. dict: Call deep_merge_dicts on both values.
     b. list: Recursively call deep_merge_lists on both values.
     c. any other type: Value is overridden.
     d. conflicting types: Value is overridden.

    If length of incoming list is more that of original then extra values are appended.
    """
    common_length = min(len(original), len(incoming))
    for idx in range(common_length):
        if isinstance(original[idx], dict) and isinstance(incoming[idx], dict):
            deep_merge_dicts(original[idx], incoming[idx])

        elif isinstance(original[idx], list) and isinstance(incoming[idx], list):
            deep_merge_lists(original[idx], incoming[idx])

        else:
            original[idx] = incoming[idx]

    for idx in range(common_length, len(incoming)):
        original.append(incoming[idx])


def deep_merge_dicts(original, incoming):
    """
    Deep merge two dictionaries. Modifies original.
    For key conflicts if both values are:
     a. dict: Recursively call deep_merge_dicts on both values.
     b. list: Call deep_merge_lists on both values.
     c. any other type: Value is overridden.
     d. conflicting types: Value is overridden.

    """
    for key in incoming:
        if key in original:
            if isinstance(original[key], dict) and isinstance(incoming[key], dict):
                deep_merge_dicts(original[key], incoming[key])

            elif isinstance(original[key], list) and isinstance(incoming[key], list):
                deep_merge_lists(original[key], incoming[key])

            else:
                original[key] = incoming[key]
        else:
            original[key] = incoming[key]

Answer 10

Short-n-sweet:

from collections.abc import MutableMapping as Map

def nested_update(d, v):
    """
    Nested update of dict-like 'd' with dict-like 'v'.
    """

    for key in v:
        if key in d and isinstance(d[key], Map) and isinstance(v[key], Map):
            nested_update(d[key], v[key])
        else:
            d[key] = v[key]

This works like (and is build on) Python's dict.update method. It returns None (you can always add return d if you prefer) as it updates dict d in-place. Keys in v will overwrite any existing keys in d (it does not try to interpret the dict's contents).

It will also work for other ("dict-like") mappings.

Example:

people = {'pete': {'gender': 'male'}, 'mary': {'age': 34}}
nested_update(people, {'pete': {'age': 41}})

# Pete's age was merged in
print(people)
{'pete': {'gender': 'male', 'age': 41}, 'mary': {'age': 34}}

Where Python's regular dict.update method yields:

people = {'pete': {'gender': 'male'}, 'mary': {'age': 34}}
people.update({'pete': {'age': 41}})

# We lost Pete's gender here!
print(people) 
{'pete': {'age': 41}, 'mary': {'age': 34}}

Answer 11

If you have an unknown level of dictionaries, then I would suggest a recursive function:

def combineDicts(dictionary1, dictionary2):
    output = {}
    for item, value in dictionary1.iteritems():
        if dictionary2.has_key(item):
            if isinstance(dictionary2[item], dict):
                output[item] = combineDicts(value, dictionary2.pop(item))
        else:
            output[item] = value
    for item, value in dictionary2.iteritems():
         output[item] = value
    return output

Answer 12

Overview

The following approach subdivides the problem of a deep merge of dicts into:

1. A parameterized shallow merge function merge(f)(a,b) that uses a function f to merge two dicts a and b

2. A recursive merger function f to be used together with merge

---

Implementation

A function for merging two (non nested) dicts can be written in a lot of ways. I personally like

def merge(f):
    def merge(a,b): 
        keys = a.keys() | b.keys()
        return {key:f(a.get(key), b.get(key)) for key in keys}
    return merge

---

A nice way of defining an appropriate recursive merger function f is using multipledispatch which allows to define functions that evaluate along different paths depending on the type of their arguments.

from multipledispatch import dispatch

#for anything that is not a dict return
@dispatch(object, object)
def f(a, b):
    return b if b is not None else a

#for dicts recurse 
@dispatch(dict, dict)
def f(a,b):
    return merge(f)(a,b)

---

Example

To merge two nested dicts simply use merge(f) e.g.:

dict1 = {1:{"a":"A"},2:{"b":"B"}}
dict2 = {2:{"c":"C"},3:{"d":"D"}}
merge(f)(dict1, dict2)
#returns {1: {'a': 'A'}, 2: {'b': 'B', 'c': 'C'}, 3: {'d': 'D'}} 

---

Notes:

The advantages of this approach are:

• The function is build from smaller functions that each do a single thing which makes the code simpler to reason about and test

• The behaviour is not hard-coded but can be changed and extended as needed which improves code reuse (see example below).

---

Customization

Some answers also considered dicts that contain lists e.g. of other (potentially nested) dicts. In this case one might want map over the lists and merge them based on position. This can be done by adding another definition to the merger function f:

import itertools
@dispatch(list, list)
def f(a,b):
    return [merge(f)(*arg) for arg in itertools.zip_longest(a, b)]

Answer 13

There's a slight problem with andrew cookes answer: In some cases it modifies the second argument b when you modify the returned dict. Specifically it's because of this line:

if key in a:
    ...
else:
    a[key] = b[key]

If b[key] is a dict, it will simply be assigned to a, meaning any subsequent modifications to that dict will affect both a and b.

a={}
b={'1':{'2':'b'}}
c={'1':{'3':'c'}}
merge(merge(a,b), c) # {'1': {'3': 'c', '2': 'b'}}
a # {'1': {'3': 'c', '2': 'b'}} (as expected)
b # {'1': {'3': 'c', '2': 'b'}} <----
c # {'1': {'3': 'c'}} (unmodified)

To fix this, the line would have to be substituted with this:

if isinstance(b[key], dict):
    a[key] = clone_dict(b[key])
else:
    a[key] = b[key]

Where clone_dict is:

def clone_dict(obj):
    clone = {}
    for key, value in obj.iteritems():
        if isinstance(value, dict):
            clone[key] = clone_dict(value)
        else:
            clone[key] = value
    return

Still. This obviously doesn't account for list, set and other stuff, but I hope it illustrates the pitfalls when trying to merge dicts.

And for completeness sake, here is my version, where you can pass it multiple dicts:

def merge_dicts(*args):
    def clone_dict(obj):
        clone = {}
        for key, value in obj.iteritems():
            if isinstance(value, dict):
                clone[key] = clone_dict(value)
            else:
                clone[key] = value
        return

    def merge(a, b, path=[]):
        for key in b:
            if key in a:
                if isinstance(a[key], dict) and isinstance(b[key], dict):
                    merge(a[key], b[key], path + [str(key)])
                elif a[key] == b[key]:
                    pass
                else:
                    raise Exception('Conflict at `{path}\''.format(path='.'.join(path + [str(key)])))
            else:
                if isinstance(b[key], dict):
                    a[key] = clone_dict(b[key])
                else:
                    a[key] = b[key]
        return a
    return reduce(merge, args, {})

Answer 14

I have an iterative solution - works much much better with big dicts & a lot of them (for example jsons etc):

import collections


def merge_dict_with_subdicts(dict1: dict, dict2: dict) -> dict:
    """
    similar behaviour to builtin dict.update - but knows how to handle nested dicts
    """
    q = collections.deque([(dict1, dict2)])
    while len(q) > 0:
        d1, d2 = q.pop()
        for k, v in d2.items():
            if k in d1 and isinstance(d1[k], dict) and isinstance(v, dict):
                q.append((d1[k], v))
            else:
                d1[k] = v

    return dict1

note that this will use the value in d2 to override d1, in case they are not both dicts. (same as python's dict.update())

some tests:

def test_deep_update():
    d = dict()
    merge_dict_with_subdicts(d, {"a": 4})
    assert d == {"a": 4}

    new_dict = {
        "b": {
            "c": {
                "d": 6
            }
        }
    }
    merge_dict_with_subdicts(d, new_dict)
    assert d == {
        "a": 4,
        "b": {
            "c": {
                "d": 6
            }
        }
    }

    new_dict = {
        "a": 3,
        "b": {
            "f": 7
        }
    }
    merge_dict_with_subdicts(d, new_dict)
    assert d == {
        "a": 3,
        "b": {
            "c": {
                "d": 6
            },
            "f": 7
        }
    }

    # test a case where one of the dicts has dict as value and the other has something else
    new_dict = {
        'a': {
            'b': 4
        }
    }
    merge_dict_with_subdicts(d, new_dict)
    assert d['a']['b'] == 4

I've tested with around ~1200 dicts - this method took 0.4 seconds, while the recursive solution took ~2.5 seconds.

Answer 15

As noted in many other answers, a recursive algorithm makes the most sense here. In general, when working with recursion, it is preferable to create new values rather than trying to modify any input data structure.

We need to define what happens at each merge step. If both inputs are dictionaries, this is easy: we copy across unique keys from each side, and recursively merge the values of the duplicated keys. It's the base cases that cause a problem. It will be easier to understand the logic if we pull out a separate function for that. As a placeholder, we could just wrap the two values in a tuple:

def merge_leaves(x, y):
    return (x, y)

Now the core of our logic looks like:

def merge(x, y):
    if not(isinstance(x, dict) and isinstance(y, dict)):
        return merge_leaves(x, y)
    x_keys, y_keys = x.keys(), y.keys()
    result = { k: merge(x[k], y[k]) for k in x_keys & y_keys }
    result.update({k: x[k] for k in x_keys - y_keys})
    result.update({k: y[k] for k in y_keys - x_keys})
    return result

Let's test it:

>>> x = {'a': {'b': 'c', 'd': 'e'}, 'f': 1, 'g': {'h', 'i'}, 'j': None}
>>> y = {'a': {'d': 'e', 'h': 'i'}, 'f': {'b': 'c'}, 'g': 1, 'k': None}
>>> merge(x, y)
{'f': (1, {'b': 'c'}), 'g': ({'h', 'i'}, 1), 'a': {'d': ('e', 'e'), 'b': 'c', 'h': 'i'}, 'j': None, 'k': None}
>>> x # The originals are unmodified.
{'a': {'b': 'c', 'd': 'e'}, 'f': 1, 'g': {'h', 'i'}, 'j': None}
>>> y
{'a': {'d': 'e', 'h': 'i'}, 'f': {'b': 'c'}, 'g': 1, 'k': None}

We can easily modify the leaf-merging rule, for example:

def merge_leaves(x, y):
    try:
        return x + y
    except TypeError:
        return Ellipsis

and observe the effects:

>>> merge(x, y)
{'f': Ellipsis, 'g': Ellipsis, 'a': {'d': 'ee', 'b': 'c', 'h': 'i'}, 'j': None, 'k': None}

We could also potentially clean this up by using a third-party library to dispatch based on the type of the inputs. For example, using multipledispatch, we could do things like:

@dispatch(dict, dict)
def merge(x, y):
    x_keys, y_keys = x.keys(), y.keys()
    result = { k: merge(x[k], y[k]) for k in x_keys & y_keys }
    result.update({k: x[k] for k in x_keys - y_keys})
    result.update({k: y[k] for k in y_keys - x_keys})
    return result

@dispatch(str, str)
def merge(x, y):
    return x + y

@dispatch(tuple, tuple)
def merge(x, y):
    return x + y

@dispatch(list, list)
def merge(x, y):
    return x + y

@dispatch(int, int):
def merge(x, y):
    raise ValueError("integer value conflict")

@dispatch(object, object):
    return (x, y)

This allows us to handle various combinations of leaf-type special cases without writing our own type checking, and also replaces the type check in the main recursive function.

Answer 16

You can use the merge function from the toolz package, for example:

>>> import toolz
>>> dict1 = {1: {'a': 'A'}, 2: {'b': 'B'}}
>>> dict2 = {2: {'c': 'C'}, 3: {'d': 'D'}}
>>> toolz.merge_with(toolz.merge, dict1, dict2)
{1: {'a': 'A'}, 2: {'c': 'C'}, 3: {'d': 'D'}}

Answer 17

This version of the function will account for N number of dictionaries, and only dictionaries -- no improper parameters can be passed, or it will raise a TypeError. The merge itself accounts for key conflicts, and instead of overwriting data from a dictionary further down the merge chain, it creates a set of values and appends to that; no data is lost.

It might not be the most effecient on the page, but it's the most thorough and you're not going to lose any information when you merge your 2 to N dicts.

def merge_dicts(*dicts):
    if not reduce(lambda x, y: isinstance(y, dict) and x, dicts, True):
        raise TypeError, "Object in *dicts not of type dict"
    if len(dicts) < 2:
        raise ValueError, "Requires 2 or more dict objects"


    def merge(a, b):
        for d in set(a.keys()).union(b.keys()):
            if d in a and d in b:
                if type(a[d]) == type(b[d]):
                    if not isinstance(a[d], dict):
                        ret = list({a[d], b[d]})
                        if len(ret) == 1: ret = ret[0]
                        yield (d, sorted(ret))
                    else:
                        yield (d, dict(merge(a[d], b[d])))
                else:
                    raise TypeError, "Conflicting key:value type assignment"
            elif d in a:
                yield (d, a[d])
            elif d in b:
                yield (d, b[d])
            else:
                raise KeyError

    return reduce(lambda x, y: dict(merge(x, y)), dicts[1:], dicts[0])

print merge_dicts({1:1,2:{1:2}},{1:2,2:{3:1}},{4:4})

output: {1: [1, 2], 2: {1: 2, 3: 1}, 4: 4}

Answer 18

Since dictviews support set operations, I was able to greatly simplify jterrace's answer.

def merge(dict1, dict2):
    for k in dict1.keys() - dict2.keys():
        yield (k, dict1[k])

    for k in dict2.keys() - dict1.keys():
        yield (k, dict2[k])

    for k in dict1.keys() & dict2.keys():
        yield (k, dict(merge(dict1[k], dict2[k])))

Any attempt to combine a dict with a non dict (technically, an object with a 'keys' method and an object without a 'keys' method) will raise an AttributeError. This includes both the initial call to the function and recursive calls. This is exactly what I wanted so I left it. You could easily catch an AttributeErrors thrown by the recursive call and then yield any value you please.

Answer 19

The following function merges b into a.

def mergedicts(a, b):
    for key in b:
        if isinstance(a.get(key), dict) or isinstance(b.get(key), dict):
            mergedicts(a[key], b[key])
        else:
            a[key] = b[key]

Answer 20

The code will depend on your rules for resolving merge conflicts, of course. Here's a version which can take an arbitrary number of arguments and merges them recursively to an arbitrary depth, without using any object mutation. It uses the following rules to resolve merge conflicts:

• dictionaries take precedence over non-dict values ({"foo": {...}} takes precedence over {"foo": "bar"})
• later arguments take precedence over earlier arguments (if you merge {"a": 1}, {"a", 2}, and {"a": 3} in order, the result will be {"a": 3})

try:
    from collections import Mapping
except ImportError:
    Mapping = dict

def merge_dicts(*dicts):                                                            
    """                                                                             
    Return a new dictionary that is the result of merging the arguments together.   
    In case of conflicts, later arguments take precedence over earlier arguments.   
    """                                                                             
    updated = {}                                                                    
    # grab all keys                                                                 
    keys = set()                                                                    
    for d in dicts:                                                                 
        keys = keys.union(set(d))                                                   

    for key in keys:                                                                
        values = [d[key] for d in dicts if key in d]                                
        # which ones are mapping types? (aka dict)                                  
        maps = [value for value in values if isinstance(value, Mapping)]            
        if maps:                                                                    
            # if we have any mapping types, call recursively to merge them          
            updated[key] = merge_dicts(*maps)                                       
        else:                                                                       
            # otherwise, just grab the last value we have, since later arguments    
            # take precedence over earlier arguments                                
            updated[key] = values[-1]                                               
    return updated  

Answer 21

And just another slight variation:

Here is a pure python3 set based deep update function. It updates nested dictionaries by looping through one level at a time and calls itself to update each next level of dictionary values:

def deep_update(dict_original, dict_update):
    if isinstance(dict_original, dict) and isinstance(dict_update, dict):
        output=dict(dict_original)
        keys_original=set(dict_original.keys())
        keys_update=set(dict_update.keys())
        similar_keys=keys_original.intersection(keys_update)
        similar_dict={key:deep_update(dict_original[key], dict_update[key]) for key in similar_keys}
        new_keys=keys_update.difference(keys_original)
        new_dict={key:dict_update[key] for key in new_keys}
        output.update(similar_dict)
        output.update(new_dict)
        return output
    else:
        return dict_update

A simple example:

x={'a':{'b':{'c':1, 'd':1}}}
y={'a':{'b':{'d':2, 'e':2}}, 'f':2}

print(deep_update(x, y))
>>> {'a': {'b': {'c': 1, 'd': 2, 'e': 2}}, 'f': 2}

Answer 22

How about another answer?!? This one also avoids mutation/side effects:

def merge(dict1, dict2):
    output = {}

    # adds keys from `dict1` if they do not exist in `dict2` and vice-versa
    intersection = {**dict2, **dict1}

    for k_intersect, v_intersect in intersection.items():
        if k_intersect not in dict1:
            v_dict2 = dict2[k_intersect]
            output[k_intersect] = v_dict2

        elif k_intersect not in dict2:
            output[k_intersect] = v_intersect

        elif isinstance(v_intersect, dict):
            v_dict2 = dict2[k_intersect]
            output[k_intersect] = merge(v_intersect, v_dict2)

        else:
            output[k_intersect] = v_intersect

    return output

dict1 = {1:{"a":{"A"}}, 2:{"b":{"B"}}}
dict2 = {2:{"c":{"C"}}, 3:{"d":{"D"}}}
dict3 = {1:{"a":{"A"}}, 2:{"b":{"B"},"c":{"C"}}, 3:{"d":{"D"}}}

assert dict3 == merge(dict1, dict2)

Answer 23

I had two dictionaries (a and b) which could each contain any number of nested dictionaries. I wanted to recursively merge them, with b taking precedence over a.

Considering the nested dictionaries as trees, what I wanted was:

• To update a so that every path to every leaf in b would be represented in a
• To overwrite subtrees of a if a leaf is found in the corresponding path in b
  • Maintain the invariant that all b leaf nodes remain leafs.

The existing answers were a little complicated for my taste and left some details on the shelf. I implemented the following, which passes unit tests for my data set.

  def merge_map(a, b):
    if not isinstance(a, dict) or not isinstance(b, dict):
      return b

    for key in b:
      a[key] = merge_map(a[key], b[key]) if key in a else b[key]
    return a

Example (formatted for clarity):

 a = {
    1 : {'a': 'red', 
         'b': {'blue': 'fish', 'yellow': 'bear' },
         'c': { 'orange': 'dog'},
    },
    2 : {'d': 'green'},
    3: 'e'
  }

  b = {
    1 : {'b': 'white'},
    2 : {'d': 'black'},
    3: 'e'
  }


  >>> merge_map(a, b)
  {1: {'a': 'red', 
       'b': 'white',
       'c': {'orange': 'dog'},},
   2: {'d': 'black'},
   3: 'e'}

The paths in b that needed to be maintained were:

• 1 -> 'b' -> 'white'
• 2 -> 'd' -> 'black'
• 3 -> 'e'.

a had the unique and non-conflicting paths of:

• 1 -> 'a' -> 'red'
• 1 -> 'c' -> 'orange' -> 'dog'

so they are still represented in the merged map.

Answer 24

This is a solution I made that recursively merges dictionaries. The first dictionary passed to the function is the master dictionary - values in it will overwrite the values in the same key in the second dictionary.

def merge(dict1: dict, dict2: dict) -> dict:
    merged = dict1

    for key in dict2:
        if type(dict2[key]) == dict:
            merged[key] = merge(dict1[key] if key in dict1 else {}, dict2[key])
        else:
            if key not in dict1.keys():
                merged[key] = dict2[key]

    return merged

Answer 25

This is a recursive solution, similar to the ones written above, but it doesn't raise any exception and simply merge the two dicts.

def merge(ans: dict,  a:dict ):
keys = a.keys()
for key in keys:
    if key in ans:
        ans[key] = merge(ans[key], a[key])
        return ans
    else:
        ans[key] = a[key]
return ans

Answer 26

I've been testing your solutions and decided to use this one in my project:

def mergedicts(dict1, dict2, conflict, no_conflict):
    for k in set(dict1.keys()).union(dict2.keys()):
        if k in dict1 and k in dict2:
            yield (k, conflict(dict1[k], dict2[k]))
        elif k in dict1:
            yield (k, no_conflict(dict1[k]))
        else:
            yield (k, no_conflict(dict2[k]))

dict1 = {1:{"a":"A"}, 2:{"b":"B"}}
dict2 = {2:{"c":"C"}, 3:{"d":"D"}}

#this helper function allows for recursion and the use of reduce
def f2(x, y):
    return dict(mergedicts(x, y, f2, lambda x: x))

print dict(mergedicts(dict1, dict2, f2, lambda x: x))
print dict(reduce(f2, [dict1, dict2]))

Passing functions as parameteres is key to extend jterrace solution to behave as all the other recursive solutions.

Answer 27

Easiest way i can think of is :

#!/usr/bin/python

from copy import deepcopy
def dict_merge(a, b):
    if not isinstance(b, dict):
        return b
    result = deepcopy(a)
    for k, v in b.iteritems():
        if k in result and isinstance(result[k], dict):
                result[k] = dict_merge(result[k], v)
        else:
            result[k] = deepcopy(v)
    return result

a = {1:{"a":'A'}, 2:{"b":'B'}}
b = {2:{"c":'C'}, 3:{"d":'D'}}

print dict_merge(a,b)

Output:

{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}

Answer 28

I have another slightly different solution here:

def deepMerge(d1, d2, inconflict = lambda v1,v2 : v2) :
''' merge d2 into d1. using inconflict function to resolve the leaf conflicts '''
    for k in d2:
        if k in d1 : 
            if isinstance(d1[k], dict) and isinstance(d2[k], dict) :
                deepMerge(d1[k], d2[k], inconflict)
            elif d1[k] != d2[k] :
                d1[k] = inconflict(d1[k], d2[k])
        else :
            d1[k] = d2[k]
    return d1

By default it resolves conflicts in favor of values from the second dict, but you can easily override this, with some witchery you may be able to even throw exceptions out of it. :).

Answer 29

hey there I also had the same problem but I though of a solution and I will post it here, in case it is also useful for others, basically merging nested dictionaries and also adding the values, for me I needed to calculate some probabilities so this one worked great:

#used to copy a nested dict to a nested dict
def deepupdate(target, src):
    for k, v in src.items():
        if k in target:
            for k2, v2 in src[k].items():
                if k2 in target[k]:
                    target[k][k2]+=v2
                else:
                    target[k][k2] = v2
        else:
            target[k] = copy.deepcopy(v)

by using the above method we can merge:

target = {'6,6': {'6,63': 1}, '63,4': {'4,4': 1}, '4,4': {'4,3': 1}, '6,63': {'63,4': 1}}

src = {'5,4': {'4,4': 1}, '5,5': {'5,4': 1}, '4,4': {'4,3': 1}}

and this will become: {'5,5': {'5,4': 1}, '5,4': {'4,4': 1}, '6,6': {'6,63': 1}, '63,4': {'4,4': 1}, '4,4': {'4,3': 2}, '6,63': {'63,4': 1}}

also notice the changes here:

target = {'6,6': {'6,63': 1}, '6,63': {'63,4': 1}, '4,4': {'4,3': 1}, '63,4': {'4,4': 1}}

src = {'5,4': {'4,4': 1}, '4,3': {'3,4': 1}, '4,4': {'4,9': 1}, '3,4': {'4,4': 1}, '5,5': {'5,4': 1}}

merge = {'5,4': {'4,4': 1}, '4,3': {'3,4': 1}, '6,63': {'63,4': 1}, '5,5': {'5,4': 1}, '6,6': {'6,63': 1}, '3,4': {'4,4': 1}, '63,4': {'4,4': 1}, '4,4': {'4,3': 1, '4,9': 1}}

dont forget to also add the import for copy:

import copy

Answer 30

from collections import defaultdict
from itertools import chain

class DictHelper:

@staticmethod
def merge_dictionaries(*dictionaries, override=True):
    merged_dict = defaultdict(set)
    all_unique_keys = set(chain(*[list(dictionary.keys()) for dictionary in dictionaries]))  # Build a set using all dict keys
    for key in all_unique_keys:
        keys_value_type = list(set(filter(lambda obj_type: obj_type != type(None), [type(dictionary.get(key, None)) for dictionary in dictionaries])))
        # Establish the object type for each key, return None if key is not present in dict and remove None from final result
        if len(keys_value_type) != 1:
            raise Exception("Different objects type for same key: {keys_value_type}".format(keys_value_type=keys_value_type))

        if keys_value_type[0] == list:
            values = list(chain(*[dictionary.get(key, []) for dictionary in dictionaries]))  # Extract the value for each key
            merged_dict[key].update(values)

        elif keys_value_type[0] == dict:
            # Extract all dictionaries by key and enter in recursion
            dicts_to_merge = list(filter(lambda obj: obj != None, [dictionary.get(key, None) for dictionary in dictionaries]))
            merged_dict[key] = DictHelper.merge_dictionaries(*dicts_to_merge)

        else:
            # if override => get value from last dictionary else make a list of all values
            values = list(filter(lambda obj: obj != None, [dictionary.get(key, None) for dictionary in dictionaries]))
            merged_dict[key] = values[-1] if override else values

    return dict(merged_dict)



if __name__ == '__main__':
  d1 = {'aaaaaaaaa': ['to short', 'to long'], 'bbbbb': ['to short', 'to long'], "cccccc": ["the is a test"]}
  d2 = {'aaaaaaaaa': ['field is not a bool'], 'bbbbb': ['field is not a bool']}
  d3 = {'aaaaaaaaa': ['filed is not a string', "to short"], 'bbbbb': ['field is not an integer']}
  print(DictHelper.merge_dictionaries(d1, d2, d3))

  d4 = {"a": {"x": 1, "y": 2, "z": 3, "d": {"x1": 10}}}
  d5 = {"a": {"x": 10, "y": 20, "d": {"x2": 20}}}
  print(DictHelper.merge_dictionaries(d4, d5))

Output:

{'bbbbb': {'to long', 'field is not an integer', 'to short', 'field is not a bool'}, 
'aaaaaaaaa': {'to long', 'to short', 'filed is not a string', 'field is not a bool'}, 
'cccccc': {'the is a test'}}

{'a': {'y': 20, 'd': {'x1': 10, 'x2': 20}, 'z': 3, 'x': 10}}