How to merge dictionaries of dictionaries?

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, b, path=None):
    "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
            else:
                raise Exception('Conflict at %s' % '.'.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:

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]

ps in python 3, you will also need from functools import reduce

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'}}

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!

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.

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'.)