By Carl Meyer


2008-09-02 07:44:30 8 Comments

I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged. The update() method would be what I need, if it returned its result instead of modifying a dict in-place.

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> z = x.update(y)
>>> print(z)
None
>>> x
{'a': 1, 'b': 10, 'c': 11}

How can I get that final merged dict in z, not x?

(To be extra-clear, the last-one-wins conflict-handling of dict.update() is what I'm looking for as well.)

30 comments

@Surya Kannan KMS 2019-03-19 03:40:21

I would say that you should do the following:

z = dict(x.items() + y.items())

@ShadowRanger 2019-02-28 17:16:22

There will be a new option when Python 3.8 releases (scheduled for 20 October, 2019), thanks to PEP 572: Assignment Expressions. The new assignment expression operator := allows you to assign the result of the copy and still use it to call update, leaving the combined code a single expression, rather than two statements, changing:

newdict = dict1.copy()
newdict.update(dict2)

to:

(newdict := dict1.copy()).update(dict2)

while behaving identically in every way. If you must also return the resulting dict (you asked for an expression returning the dict; the above creates and assigns to newdict, but doesn't return it, so you couldn't use it to pass an argument to a function as is, a la myfunc((newdict := dict1.copy()).update(dict2))), then just add or newdict to the end (since update returns None, which is falsy, it will then evaluate and return newdict as the result of the expression):

(newdict := dict1.copy()).update(dict2) or newdict

Important caveat: In general, I'd discourage this approach in favor of:

newdict = {**dict1, **dict2}

The unpacking approach is clearer (to anyone who knows about generalized unpacking in the first place, which you should), doesn't require a name for the result at all (so it's much more concise when constructing a temporary that is immediately passed to a function or included in a list/tuple literal or the like), and is almost certainly faster as well, being (on CPython) roughly equivalent to:

newdict = {}
newdict.update(dict1)
newdict.update(dict2)

but done at the C layer, using the concrete dict API, so no dynamic method lookup/binding or function call dispatch overhead is involved (where (newdict := dict1.copy()).update(dict2) is unavoidably identical to the original two-liner in behavior, performing the work in discrete steps, with dynamic lookup/binding/invocation of methods.

It's also more extensible, as merging three dicts is obvious:

 newdict = {**dict1, **dict2, **dict3}

where using assignment expressions won't scale like that; the closest you could get would be:

 (newdict := dict1.copy()).update(dict2), newdict.update(dict3)

or without the temporary tuple of Nones, but with truthiness testing of each None result:

 (newdict := dict1.copy()).update(dict2) or newdict.update(dict3)

either of which is obviously much uglier, and includes further inefficiencies (either a wasted temporary tuple of Nones for comma separation, or pointless truthiness testing of each update's None return for or separation).

The only real advantage to the assignment expression approach occurs if:

  1. You have generic code that needs handle both sets and dicts (both of them support copy and update, so the code works roughly as you'd expect it to)
  2. You expect to receive arbitrary dict-like objects, not just dict itself, and must preserve the type and semantics of the left hand side (rather than ending up with a plain dict). While myspecialdict({**speciala, **specialb}) might work, it would involve an extra temporary dict, and if myspecialdict has features plain dict can't preserve (e.g. regular dicts now preserve order based on the first appearance of a key, and value based on the last appearance of a key; you might want one that preserves order based on the last appearance of a key so updating a value also moves it to the end), then the semantics would be wrong. Since the assignment expression version uses the named methods (which are presumably overloaded to behave appropriately), it never creates a dict at all (unless dict1 was already a dict), preserving the original type (and original type's semantics), all while avoiding any temporaries.

@Tigran Saluev 2018-05-11 10:00:39

I think my ugly one-liners are just necessary here.

z = next(z.update(y) or z for z in [x.copy()])
# or
z = (lambda z: z.update(y) or z)(x.copy())
  1. Dicts are merged.
  2. Single expression.
  3. Don't ever dare to use it.

P.S. This is a solution working in both versions of Python. I know that Python 3 has this {**x, **y} thing and it is the right thing to use (as well as moving to Python 3 if you still have Python 2 is the right thing to do).

@Aaron Hall 2014-11-10 22:11:48

How can I merge two Python dictionaries in a single expression?

For dictionaries x and y, z becomes a shallowly merged dictionary with values from y replacing those from x.

  • In Python 3.5 or greater:

    z = {**x, **y}
    
  • In Python 2, (or 3.4 or lower) write a function:

    def merge_two_dicts(x, y):
        z = x.copy()   # start with x's keys and values
        z.update(y)    # modifies z with y's keys and values & returns None
        return z
    

    and now:

    z = merge_two_dicts(x, y)
    

Explanation

Say you have two dicts and you want to merge them into a new dict without altering the original dicts:

x = {'a': 1, 'b': 2}
y = {'b': 3, 'c': 4}

The desired result is to get a new dictionary (z) with the values merged, and the second dict's values overwriting those from the first.

>>> z
{'a': 1, 'b': 3, 'c': 4}

A new syntax for this, proposed in PEP 448 and available as of Python 3.5, is

z = {**x, **y}

And it is indeed a single expression.

Note that we can merge in with literal notation as well:

z = {**x, 'foo': 1, 'bar': 2, **y}

and now:

>>> z
{'a': 1, 'b': 3, 'foo': 1, 'bar': 2, 'c': 4}

It is now showing as implemented in the release schedule for 3.5, PEP 478, and it has now made its way into What's New in Python 3.5 document.

However, since many organizations are still on Python 2, you may wish to do this in a backwards compatible way. The classically Pythonic way, available in Python 2 and Python 3.0-3.4, is to do this as a two-step process:

z = x.copy()
z.update(y) # which returns None since it mutates z

In both approaches, y will come second and its values will replace x's values, thus 'b' will point to 3 in our final result.

Not yet on Python 3.5, but want a single expression

If you are not yet on Python 3.5, or need to write backward-compatible code, and you want this in a single expression, the most performant while correct approach is to put it in a function:

def merge_two_dicts(x, y):
    """Given two dicts, merge them into a new dict as a shallow copy."""
    z = x.copy()
    z.update(y)
    return z

and then you have a single expression:

z = merge_two_dicts(x, y)

You can also make a function to merge an undefined number of dicts, from zero to a very large number:

def merge_dicts(*dict_args):
    """
    Given any number of dicts, shallow copy and merge into a new dict,
    precedence goes to key value pairs in latter dicts.
    """
    result = {}
    for dictionary in dict_args:
        result.update(dictionary)
    return result

This function will work in Python 2 and 3 for all dicts. e.g. given dicts a to g:

z = merge_dicts(a, b, c, d, e, f, g) 

and key value pairs in g will take precedence over dicts a to f, and so on.

Critiques of Other Answers

Don't use what you see in the formerly accepted answer:

z = dict(x.items() + y.items())

In Python 2, you create two lists in memory for each dict, create a third list in memory with length equal to the length of the first two put together, and then discard all three lists to create the dict. In Python 3, this will fail because you're adding two dict_items objects together, not two lists -

>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: 'dict_items' and 'dict_items'

and you would have to explicitly create them as lists, e.g. z = dict(list(x.items()) + list(y.items())). This is a waste of resources and computation power.

Similarly, taking the union of items() in Python 3 (viewitems() in Python 2.7) will also fail when values are unhashable objects (like lists, for example). Even if your values are hashable, since sets are semantically unordered, the behavior is undefined in regards to precedence. So don't do this:

>>> c = dict(a.items() | b.items())

This example demonstrates what happens when values are unhashable:

>>> x = {'a': []}
>>> y = {'b': []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: 'list'

Here's an example where y should have precedence, but instead the value from x is retained due to the arbitrary order of sets:

>>> x = {'a': 2}
>>> y = {'a': 1}
>>> dict(x.items() | y.items())
{'a': 2}

Another hack you should not use:

z = dict(x, **y)

This uses the dict constructor, and is very fast and memory efficient (even slightly more-so than our two-step process) but unless you know precisely what is happening here (that is, the second dict is being passed as keyword arguments to the dict constructor), it's difficult to read, it's not the intended usage, and so it is not Pythonic.

Here's an example of the usage being remediated in django.

Dicts are intended to take hashable keys (e.g. frozensets or tuples), but this method fails in Python 3 when keys are not strings.

>>> c = dict(a, **b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings

From the mailing list, Guido van Rossum, the creator of the language, wrote:

I am fine with declaring dict({}, **{1:3}) illegal, since after all it is abuse of the ** mechanism.

and

Apparently dict(x, **y) is going around as "cool hack" for "call x.update(y) and return x". Personally I find it more despicable than cool.

It is my understanding (as well as the understanding of the creator of the language) that the intended usage for dict(**y) is for creating dicts for readability purposes, e.g.:

dict(a=1, b=10, c=11)

instead of

{'a': 1, 'b': 10, 'c': 11}

Response to comments

Despite what Guido says, dict(x, **y) is in line with the dict specification, which btw. works for both Python 2 and 3. The fact that this only works for string keys is a direct consequence of how keyword parameters work and not a short-comming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact ** was designed precisely to pass dicts as keywords.

Again, it doesn't work for 3 when keys are non-strings. The implicit calling contract is that namespaces take ordinary dicts, while users must only pass keyword arguments that are strings. All other callables enforced it. dict broke this consistency in Python 2:

>>> foo(**{('a', 'b'): None})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() keywords must be strings
>>> dict(**{('a', 'b'): None})
{('a', 'b'): None}

This inconsistency was bad given other implementations of Python (Pypy, Jython, IronPython). Thus it was fixed in Python 3, as this usage could be a breaking change.

I submit to you that it is malicious incompetence to intentionally write code that only works in one version of a language or that only works given certain arbitrary constraints.

More comments:

dict(x.items() + y.items()) is still the most readable solution for Python 2. Readability counts.

My response: merge_two_dicts(x, y) actually seems much clearer to me, if we're actually concerned about readability. And it is not forward compatible, as Python 2 is increasingly deprecated.

{**x, **y} does not seem to handle nested dictionaries. the contents of nested keys are simply overwritten, not merged [...] I ended up being burnt by these answers that do not merge recursively and I was surprised no one mentioned it. In my interpretation of the word "merging" these answers describe "updating one dict with another", and not merging.

Yes. I must refer you back to the question, which is asking for a shallow merge of two dictionaries, with the first's values being overwritten by the second's - in a single expression.

Assuming two dictionary of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dicts from either source, and the surest way to avoid that is to make a copy when assigning values. As keys must be hashable and are usually therefore immutable, it is pointless to copy them:

from copy import deepcopy

def dict_of_dicts_merge(x, y):
    z = {}
    overlapping_keys = x.keys() & y.keys()
    for key in overlapping_keys:
        z[key] = dict_of_dicts_merge(x[key], y[key])
    for key in x.keys() - overlapping_keys:
        z[key] = deepcopy(x[key])
    for key in y.keys() - overlapping_keys:
        z[key] = deepcopy(y[key])
    return z

Usage:

>>> x = {'a':{1:{}}, 'b': {2:{}}}
>>> y = {'b':{10:{}}, 'c': {11:{}}}
>>> dict_of_dicts_merge(x, y)
{'b': {2: {}, 10: {}}, 'a': {1: {}}, 'c': {11: {}}}

Coming up with contingencies for other value types is far beyond the scope of this question, so I will point you at my answer to the canonical question on a "Dictionaries of dictionaries merge".

Less Performant But Correct Ad-hocs

These approaches are less performant, but they will provide correct behavior. They will be much less performant than copy and update or the new unpacking because they iterate through each key-value pair at a higher level of abstraction, but they do respect the order of precedence (latter dicts have precedence)

You can also chain the dicts manually inside a dict comprehension:

{k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7

or in python 2.6 (and perhaps as early as 2.4 when generator expressions were introduced):

dict((k, v) for d in dicts for k, v in d.items())

itertools.chain will chain the iterators over the key-value pairs in the correct order:

import itertools
z = dict(itertools.chain(x.iteritems(), y.iteritems()))

Performance Analysis

I'm only going to do the performance analysis of the usages known to behave correctly.

import timeit

The following is done on Ubuntu 14.04

In Python 2.7 (system Python):

>>> min(timeit.repeat(lambda: merge_two_dicts(x, y)))
0.5726828575134277
>>> min(timeit.repeat(lambda: {k: v for d in (x, y) for k, v in d.items()} ))
1.163769006729126
>>> min(timeit.repeat(lambda: dict(itertools.chain(x.iteritems(), y.iteritems()))))
1.1614501476287842
>>> min(timeit.repeat(lambda: dict((k, v) for d in (x, y) for k, v in d.items())))
2.2345519065856934

In Python 3.5 (deadsnakes PPA):

>>> min(timeit.repeat(lambda: {**x, **y}))
0.4094954460160807
>>> min(timeit.repeat(lambda: merge_two_dicts(x, y)))
0.7881555100320838
>>> min(timeit.repeat(lambda: {k: v for d in (x, y) for k, v in d.items()} ))
1.4525277839857154
>>> min(timeit.repeat(lambda: dict(itertools.chain(x.items(), y.items()))))
2.3143140770262107
>>> min(timeit.repeat(lambda: dict((k, v) for d in (x, y) for k, v in d.items())))
3.2069112799945287

Resources on Dictionaries

@Thomas Vander Stichele 2008-09-02 07:50:12

In your case, what you can do is:

z = dict(x.items() + y.items())

This will, as you want it, put the final dict in z, and make the value for key b be properly overridden by the second (y) dict's value:

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> z = dict(x.items() + y.items())
>>> z
{'a': 1, 'c': 11, 'b': 10}

If you use Python 3, it is only a little more complicated. To create z:

>>> z = dict(list(x.items()) + list(y.items()))
>>> z
{'a': 1, 'c': 11, 'b': 10}

@Josh Bode 2018-04-15 23:02:55

This is an expression for Python 3.5 or greater that merges dictionaries using reduce:

>>> from functools import reduce
>>> l = [{'a': 1}, {'b': 2}, {'a': 100, 'c': 3}]
>>> reduce(lambda x, y: {**x, **y}, l, {})
{'a': 100, 'b': 2, 'c': 3}

Note: this works even if the dictionary list is empty or contains only one element.

@litepresence 2018-03-22 04:08:47

I was curious if I could beat the accepted answer's time with a one line stringify approach:

I tried 5 methods, none previously mentioned - all one liner - all producing correct answers - and I couldn't come close.

So... to save you the trouble and perhaps fulfill curiosity:

import json
import yaml
import time
from ast import literal_eval as literal

def merge_two_dicts(x, y):
    z = x.copy()   # start with x's keys and values
    z.update(y)    # modifies z with y's keys and values & returns None
    return z

x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}

start = time.time()
for i in range(10000):
    z = yaml.load((str(x)+str(y)).replace('}{',', '))
elapsed = (time.time()-start)
print (elapsed, z, 'stringify yaml')

start = time.time()
for i in range(10000):
    z = literal((str(x)+str(y)).replace('}{',', '))
elapsed = (time.time()-start)
print (elapsed, z, 'stringify literal')

start = time.time()
for i in range(10000):
    z = eval((str(x)+str(y)).replace('}{',', '))
elapsed = (time.time()-start)
print (elapsed, z, 'stringify eval')

start = time.time()
for i in range(10000):
    z = {k:int(v) for k,v in (dict(zip(
            ((str(x)+str(y))
            .replace('}',' ')
            .replace('{',' ')
            .replace(':',' ')
            .replace(',',' ')
            .replace("'",'')
            .strip()
            .split('  '))[::2], 
            ((str(x)+str(y))
            .replace('}',' ')
            .replace('{',' ').replace(':',' ')
            .replace(',',' ')
            .replace("'",'')
            .strip()
            .split('  '))[1::2]
             ))).items()}
elapsed = (time.time()-start)
print (elapsed, z, 'stringify replace')

start = time.time()
for i in range(10000):
    z = json.loads(str((str(x)+str(y)).replace('}{',', ').replace("'",'"')))
elapsed = (time.time()-start)
print (elapsed, z, 'stringify json')

start = time.time()
for i in range(10000):
    z = merge_two_dicts(x, y)
elapsed = (time.time()-start)
print (elapsed, z, 'accepted')

results:

7.693928956985474 {'c': 11, 'b': 10, 'a': 1} stringify yaml
0.29134678840637207 {'c': 11, 'b': 10, 'a': 1} stringify literal
0.2208399772644043 {'c': 11, 'b': 10, 'a': 1} stringify eval
0.1106564998626709 {'c': 11, 'b': 10, 'a': 1} stringify replace
0.07989692687988281 {'c': 11, 'b': 10, 'a': 1} stringify json
0.005082368850708008 {'c': 11, 'b': 10, 'a': 1} accepted

what I did learn from this is that json approach is the fastest way (of those attempted) to return a dictionary from string-of-dictionary; much faster (about 1/4th of the time) of what I considered to be the normal method using ast. I also learned that, the yaml approach should be avoided at all cost.

Yes, I get that this is not the best/correct way so please don't downvote to negative oblivion, zero is just fine. I was curious if it was faster, which it isn't; I posted to prove it so.

@ShadowRanger 2019-02-28 17:29:29

Note that the json approach is faster than ast.literal_eval, but it's also not as comprehensive. It can't handle Python literals not in the JSON spec, so no tuples, sets, frozensets, bools (it can handle JSON bools, but not the result of stringifying a Python bool directly), etc. ast.literal_eval is slower, but at least some of that is a consequence of handling more complex inputs. That said, I'm pretty sure it could be faster if they bothered to optimize it, it's just pretty rare that evaluating strings of Python literals is the chokepoint in code.

@gilch 2017-09-22 02:57:15

If you don't mind mutating x,

x.update(y) or x

Simple, readable, performant. You know update() always returns None, which is a false value. So it will always evaluate to x.

Mutating methods in the standard library, like update, return None by convention, so this trick will work on those too.

If you're using a library that doesn't follow this convention, you can use a tuple display and index to make it a single expression, instead of or, but it's not as readable.

(x.update(y), x)[-1]

If you don't have x in a variable yet, you can use lambda to make a local without using an assignment statement. This amounts to using lambda as a let expression, which is a common technique in functional languages, but rather unpythonic.

(lambda x: x.update(y) or x)({'a':1, 'b': 2})

If you do want a copy, PEP 448 is best {**x, **y}. But if that's not available, let works here too.

(lambda z: z.update(y) or z)(x.copy())

@gboffi 2017-05-30 12:28:26

The question is tagged python-3x but, taking into account that it's a relatively recent addition and that the most voted, accepted answer deals extensively with a Python 2.x solution, I dare add a one liner that draws on an irritating feature of Python 2.x list comprehension, that is name leaking...

$ python2
Python 2.7.13 (default, Jan 19 2017, 14:48:08) 
[GCC 6.3.0 20170118] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> [z.update(d) for z in [{}] for d in (x, y)]
[None, None]
>>> z
{'a': 1, 'c': 11, 'b': 10}
>>> ...

I'm happy to say that the above doesn't work any more on any version of Python 3.

@gboffi 2018-04-11 08:42:22

Dear future down-voter, could you please: 1. consider that my answer is correct w/r to the original question; 2. if, otoh, you indeed consider my answer not appropriate for the site down voting is not the right action to perform, you should rather flag my answer as "in need of moderator intervention" and explain why you think that my answer isn't appropriate — if you were convincing, then a moderator could remove my answer leaving you fully satisfied; 3. if you think my answer is not appropriate but you are lazy, you can of course just down vote... it's your right, isn't it?

@gboffi 2018-04-11 08:52:08

@clacke I'd like to thank you for the comment. It made me smile! If I had to vote the single feature of Python 2 that made the case for breaking backwards compatibility, for me it's name leaking in comprehensions

@Skyduy 2017-05-24 07:24:50

In python 3:

import collections
a = {1: 1, 2: 2}
b = {2: 3, 3: 4}
c = {3: 5}

r = dict(collections.ChainMap(a, b, c))
print(r)

Out:

{1: 1, 2: 2, 3: 4}

Docs: https://docs.python.org/3/library/collections.html#collections.ChainMap:

@Mike Graham 2016-11-18 12:53:29

You can use toolz.merge([x, y]) for this.

@Mathieu Larose 2012-10-17 02:09:45

Two dictionaries

def union2(dict1, dict2):
    return dict(list(dict1.items()) + list(dict2.items()))

n dictionaries

def union(*dicts):
    return dict(itertools.chain.from_iterable(dct.items() for dct in dicts))

sum has bad performance. See https://mathieularose.com/how-not-to-flatten-a-list-of-lists-in-python/

@Robino 2016-01-20 11:46:22

Be pythonic. Use a comprehension:

z={i:d[i] for d in [x,y] for i in d}

>>> print z
{'a': 1, 'c': 11, 'b': 10}

@Jesse Knight 2018-06-06 18:27:21

As a function: def dictmerge(*args): return {i:d[i] for d in args for i in d}

@ShadowRanger 2019-03-05 19:33:32

Save a lookup by iterating the key/value pairs directly: z={k: v for d in (x, y) for k, v in d.items()}

@levi 2016-09-28 00:33:55

In Python 3.5 you can use unpack ** in order to create new dictionary. This method has no been showed in past answers. Also, it's better to use {} instead of dict(). Because {} is a python literal and dict() involves a function call.

dict1 = {'a':1}
dict2 = {'b':2}
new_dict = {**dict1, **dict2}
>>>new_dict
{'a':1, 'a':2}

@Yeo 2016-10-10 20:51:33

Duplicate answer posted earlier. stackoverflow.com/a/28753078/764592

@reubano 2015-08-04 14:54:58

Simple solution using itertools that preserves order (latter dicts have precedence)

import itertools as it
merge = lambda *args: dict(it.chain.from_iterable(it.imap(dict.iteritems, args)))

And it's usage:

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> merge(x, y)
{'a': 1, 'b': 10, 'c': 11}

>>> z = {'c': 3, 'd': 4}
>>> merge(x, y, z)
{'a': 1, 'b': 10, 'c': 3, 'd': 4}

@Kalpesh Dusane 2016-08-31 13:53:20

For Python 2 :

x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}
z = dict(x.items()+y.items())
print(z)

For Python 3:

x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}
z = dict(x.items()|y.items())
print(z)

It gives output:{'a': 1, 'c': 11, 'b': 10}

@ShadowRanger 2016-09-23 02:56:43

For the record, on 2.7, dict(x.viewitems()|y.viewitems()) would be directly equivalent to your Py3.x code. That said, this answer has ordering issues; if the same key appears in x and y, it's arbitrary which value survives in z; better solutions avoid this arbitrariness (usually by saying that the last appearance of the key, in this case, the value from y, is the one which survives).

@Alfe 2016-05-18 15:57:32

I know this does not really fit the specifics of the questions ("one liner"), but since none of the answers above went into this direction while lots and lots of answers addressed the performance issue, I felt I should contribute my thoughts.

Depending on the use case it might not be necessary to create a "real" merged dictionary of the given input dictionaries. A view which does this might be sufficient in many cases, i. e. an object which acts like the merged dictionary would without computing it completely. A lazy version of the merged dictionary, so to speak.

In Python, this is rather simple and can be done with the code shown at the end of my post. This given, the answer to the original question would be:

z = MergeDict(x, y)

When using this new object, it will behave like a merged dictionary but it will have constant creation time and constant memory footprint while leaving the original dictionaries untouched. Creating it is way cheaper than in the other solutions proposed.

Of course, if you use the result a lot, then you will at some point reach the limit where creating a real merged dictionary would have been the faster solution. As I said, it depends on your use case.

If you ever felt you would prefer to have a real merged dict, then calling dict(z) would produce it (but way more costly than the other solutions of course, so this is just worth mentioning).

You can also use this class to make a kind of copy-on-write dictionary:

a = { 'x': 3, 'y': 4 }
b = MergeDict(a)  # we merge just one dict
b['x'] = 5
print b  # will print {'x': 5, 'y': 4}
print a  # will print {'y': 4, 'x': 3}

Here's the straight-forward code of MergeDict:

class MergeDict(object):
  def __init__(self, *originals):
    self.originals = ({},) + originals[::-1]  # reversed

  def __getitem__(self, key):
    for original in self.originals:
      try:
        return original[key]
      except KeyError:
        pass
    raise KeyError(key)

  def __setitem__(self, key, value):
    self.originals[0][key] = value

  def __iter__(self):
    return iter(self.keys())

  def __repr__(self):
    return '%s(%s)' % (
      self.__class__.__name__,
      ', '.join(repr(original)
          for original in reversed(self.originals)))

  def __str__(self):
    return '{%s}' % ', '.join(
        '%r: %r' % i for i in self.iteritems())

  def iteritems(self):
    found = set()
    for original in self.originals:
      for k, v in original.iteritems():
        if k not in found:
          yield k, v
          found.add(k)

  def items(self):
    return list(self.iteritems())

  def keys(self):
    return list(k for k, _ in self.iteritems())

  def values(self):
    return list(v for _, v in self.iteritems())

@Alfe 2016-05-18 16:10:50

I saw by now that some answers refer to a class called ChainMap which is available in Python 3 only and which does more or less what my code does. So shame on me for not reading everything carefully enough. But given that this only exists for Python 3, please take my answer as a contribution for the Python 2 users ;-)

@clacke 2016-07-28 11:19:43

ChainMap was backported for earlier Pythons: pypi.python.org/pypi/chainmap

@Alfe 2018-04-06 09:50:20

downvoter: care to state why you downvoted?

@clacke 2018-04-07 12:16:31

I didn't. But the question is pretty clear and your answer is "[I know you didn't ask this other question, but let me answer that anyway]". Which can be kind of fine, but maybe better as a comment. Your class doesn't fit in a comment, but then the asker also specifically asked for a one-liner. Finally the question is marked py3, so there's ChainMap in the stdlib. So I'm not surprised someone did, but personally I find it a bit harsh. I just didn't upvote. I get why you wrote it, I get easily carried away too when I have an idea or perspective. I've been voted down for lesser infractions. :-)

@clacke 2018-04-07 12:22:15

Apparently the py3 tag may be more recent than your answer, but the downvoter might not know that.

@Alfe 2018-04-07 20:42:09

Thank you for the work you put into writing these possible explanations, but I already was aware of them. I am still not aware of the reasons of the downvoter, and maybe they have ever better reasons. I would really like to know them. And in any case, a downvote without a clear reason is not very helpful and should be declared as such.

@kjo 2016-03-28 13:13:27

(For Python2.7* only; there are simpler solutions for Python3*.)

If you're not averse to importing a standard library module, you can do

from functools import reduce

def merge_dicts(*dicts):
    return reduce(lambda a, d: a.update(d) or a, dicts, {})

(The or a bit in the lambda is necessary because dict.update always returns None on success.)

@Carl Meyer 2008-09-02 15:52:07

Another, more concise, option:

z = dict(x, **y)

Note: this has become a popular answer, but it is important to point out that if y has any non-string keys, the fact that this works at all is an abuse of a CPython implementation detail, and it does not work in Python 3, or in PyPy, IronPython, or Jython. Also, Guido is not a fan. So I can't recommend this technique for forward-compatible or cross-implementation portable code, which really means it should be avoided entirely.

@Stan 2011-11-29 11:52:15

Recursively/deep update a dict

def deepupdate(original, update):
    """
    Recursively update a dict.
    Subdict's won't be overwritten but also updated.
    """
    for key, value in original.iteritems(): 
        if key not in update:
            update[key] = value
        elif isinstance(value, dict):
            deepupdate(value, update[key]) 
    return update

Demonstration:

pluto_original = {
    'name': 'Pluto',
    'details': {
        'tail': True,
        'color': 'orange'
    }
}

pluto_update = {
    'name': 'Pluutoo',
    'details': {
        'color': 'blue'
    }
}

print deepupdate(pluto_original, pluto_update)

Outputs:

{
    'name': 'Pluutoo',
    'details': {
        'color': 'blue',
        'tail': True
    }
}

Thanks rednaw for edits.

@Aaron Hall 2018-11-09 02:14:04

This does not answer the question. The question clearly asks for a new dictionary, z, from original dictionaries, x and y, with values from y replacing those of x - not an updated dictionary. This answer modifies y in-place by adding values from x. Worse, it does not copy these values, so one could further modify the modified dictionary, y, and modifications could be reflected in dictionary x. @Jérôme I hope this code is not causing any bugs for your application - at least consider using deepcopy to copy the values.

@Jérôme 2018-11-09 13:24:32

@AaronHall agreed this does not answer the question. But it answers my need. I understand those limitations, but that's not an issue in my case. Thinking of it, maybe the name is misleading, as it might evoke a deepcopy, which it does not provide. But it addresses deep nesting. Here's another implementation from the Martellibot: stackoverflow.com/questions/3232943/….

@reetesh11 2015-11-30 13:04:00

from collections import Counter
dict1 = {'a':1, 'b': 2}
dict2 = {'b':10, 'c': 11}
result = dict(Counter(dict1) + Counter(dict2))

This should solve your problem.

@RemcoGerlich 2015-07-17 14:47:23

This can be done with a single dict comprehension:

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> { key: y[key] if key in y else x[key]
      for key in set(x) + set(y)
    }

In my view the best answer for the 'single expression' part as no extra functions are needed, and it is short.

@Breezer 2017-02-16 14:57:42

I suspect performance will not be very good though; creating a set out of each dict then only iterating through the keys means another lookup for the value each time (though relatively fast, still increases the order of the function for scaling)

@Rashid Mv 2017-12-23 15:50:00

it all depends on the version of the python we are using. In 3.5 and above {**x,**y} gives the concatenated dictionary

@Bilal Syed Hussain 2015-02-26 21:27:52

Python 3.5 (PEP 448) allows a nicer syntax option:

x = {'a': 1, 'b': 1}
y = {'a': 2, 'c': 2}
final = {**x, **y} 
final
# {'a': 2, 'b': 1, 'c': 2}

Or even

final = {'a': 1, 'b': 1, **x, **y}

@Blackeagle52 2015-03-04 11:09:13

In what way is this solution better than the dict(x, **y)-solution? As you (@CarlMeyer) mentioned within the note of your own answer (stackoverflow.com/a/39858/2798610) Guido considers that solution illegal.

@Carl Meyer 2015-03-04 22:24:44

Guido dislikes dict(x, **y) for the (very good) reason that it relies on y only having keys which are valid keyword argument names (unless you are using CPython 2.7, where the dict constructor cheats). This objection/restriction does not apply to PEP 448, which generalizes the ** unpacking syntax to dict literals. So this solution has the same concision as dict(x, **y), without the downside.

@zaphod 2008-10-23 02:38:56

In a follow-up answer, you asked about the relative performance of these two alternatives:

z1 = dict(x.items() + y.items())
z2 = dict(x, **y)

On my machine, at least (a fairly ordinary x86_64 running Python 2.5.2), alternative z2 is not only shorter and simpler but also significantly faster. You can verify this for yourself using the timeit module that comes with Python.

Example 1: identical dictionaries mapping 20 consecutive integers to themselves:

% python -m timeit -s 'x=y=dict((i,i) for i in range(20))' 'z1=dict(x.items() + y.items())'
100000 loops, best of 3: 5.67 usec per loop
% python -m timeit -s 'x=y=dict((i,i) for i in range(20))' 'z2=dict(x, **y)' 
100000 loops, best of 3: 1.53 usec per loop

z2 wins by a factor of 3.5 or so. Different dictionaries seem to yield quite different results, but z2 always seems to come out ahead. (If you get inconsistent results for the same test, try passing in -r with a number larger than the default 3.)

Example 2: non-overlapping dictionaries mapping 252 short strings to integers and vice versa:

% python -m timeit -s 'from htmlentitydefs import codepoint2name as x, name2codepoint as y' 'z1=dict(x.items() + y.items())'
1000 loops, best of 3: 260 usec per loop
% python -m timeit -s 'from htmlentitydefs import codepoint2name as x, name2codepoint as y' 'z2=dict(x, **y)'               
10000 loops, best of 3: 26.9 usec per loop

z2 wins by about a factor of 10. That's a pretty big win in my book!

After comparing those two, I wondered if z1's poor performance could be attributed to the overhead of constructing the two item lists, which in turn led me to wonder if this variation might work better:

from itertools import chain
z3 = dict(chain(x.iteritems(), y.iteritems()))

A few quick tests, e.g.

% python -m timeit -s 'from itertools import chain; from htmlentitydefs import codepoint2name as x, name2codepoint as y' 'z3=dict(chain(x.iteritems(), y.iteritems()))'
10000 loops, best of 3: 66 usec per loop

lead me to conclude that z3 is somewhat faster than z1, but not nearly as fast as z2. Definitely not worth all the extra typing.

This discussion is still missing something important, which is a performance comparison of these alternatives with the "obvious" way of merging two lists: using the update method. To try to keep things on an equal footing with the expressions, none of which modify x or y, I'm going to make a copy of x instead of modifying it in-place, as follows:

z0 = dict(x)
z0.update(y)

A typical result:

% python -m timeit -s 'from htmlentitydefs import codepoint2name as x, name2codepoint as y' 'z0=dict(x); z0.update(y)'
10000 loops, best of 3: 26.9 usec per loop

In other words, z0 and z2 seem to have essentially identical performance. Do you think this might be a coincidence? I don't....

In fact, I'd go so far as to claim that it's impossible for pure Python code to do any better than this. And if you can do significantly better in a C extension module, I imagine the Python folks might well be interested in incorporating your code (or a variation on your approach) into the Python core. Python uses dict in lots of places; optimizing its operations is a big deal.

You could also write this as

z0 = x.copy()
z0.update(y)

as Tony does, but (not surprisingly) the difference in notation turns out not to have any measurable effect on performance. Use whichever looks right to you. Of course, he's absolutely correct to point out that the two-statement version is much easier to understand.

@Antti Haapala 2015-03-16 05:50:54

This does not work in Python 3; items() is not catenable, and iteritems does not exist.

@bassounds 2014-09-30 02:36:19

A union of the OP's two dictionaries would be something like:

{'a': 1, 'b': 2, 10, 'c': 11}

Specifically, the union of two entities(x and y) contains all the elements of x and/or y. Unfortunately, what the OP asks for is not a union, despite the title of the post.

My code below is neither elegant nor a one-liner, but I believe it is consistent with the meaning of union.

From the OP's example:

x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}

z = {}
for k, v in x.items():
    if not k in z:
        z[k] = [(v)]
    else:
        z[k].append((v))
for k, v in y.items():
    if not k in z:
        z[k] = [(v)]
    else:
        z[k].append((v))

{'a': [1], 'b': [2, 10], 'c': [11]}

Whether one wants lists could be changed, but the above will work if a dictionary contains lists (and nested lists) as values in either dictionary.

@Carl Meyer 2014-09-30 15:49:34

I've edited the question to not use the word union, for clarity.

@Alfe 2016-05-18 16:07:25

Perhaps you mean {'a': 1, 'b': (2, 10), 'c': 11} …?

@Raymond Hettinger 2013-04-28 03:15:38

In Python 3, you can use collections.ChainMap which groups multiple dicts or other mappings together to create a single, updateable view:

>>> from collections import ChainMap
>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> z = ChainMap({}, y, x)
>>> for k, v in z.items():
        print(k, '-->', v)

a --> 1
b --> 10
c --> 11

@Prerit 2017-02-14 05:14:37

But one should be cautious while using ChainMap there's a catch that if you have duplicate keys the values from first mapping get used and when you call a del on say a ChainMap c will delete the first mapping of that key.

@Raymond Hettinger 2017-02-15 07:24:40

@Prerit What else would you expect it to do? That's the normal way chained namespaces work. Consider how $PATH works in bash. Deleting an executable on the path doesn't preclude another executable with the same name further upstream.

@Prerit 2017-02-15 15:15:26

@Raymond Hettinger I agree, just added a caution. Most people may not know about it. :D

@rcreswick 2008-09-04 19:08:25

I wanted something similar, but with the ability to specify how the values on duplicate keys were merged, so I hacked this out (but did not heavily test it). Obviously this is not a single expression, but it is a single function call.

def merge(d1, d2, merge_fn=lambda x,y:y):
    """
    Merges two dictionaries, non-destructively, combining 
    values on duplicate keys as defined by the optional merge
    function.  The default behavior replaces the values in d1
    with corresponding values in d2.  (There is no other generally
    applicable merge strategy, but often you'll have homogeneous 
    types in your dicts, so specifying a merge technique can be 
    valuable.)

    Examples:

    >>> d1
    {'a': 1, 'c': 3, 'b': 2}
    >>> merge(d1, d1)
    {'a': 1, 'c': 3, 'b': 2}
    >>> merge(d1, d1, lambda x,y: x+y)
    {'a': 2, 'c': 6, 'b': 4}

    """
    result = dict(d1)
    for k,v in d2.iteritems():
        if k in result:
            result[k] = merge_fn(result[k], v)
        else:
            result[k] = v
    return result

@Claudiu 2013-08-07 21:23:08

Abuse leading to a one-expression solution for Matthew's answer:

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> z = (lambda f=x.copy(): (f.update(y), f)[1])()
>>> z
{'a': 1, 'c': 11, 'b': 10}

You said you wanted one expression, so I abused lambda to bind a name, and tuples to override lambda's one-expression limit. Feel free to cringe.

You could also do this of course if you don't care about copying it:

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> z = (x.update(y), x)[1]
>>> z
{'a': 1, 'b': 10, 'c': 11}

@Tony Meyer 2008-09-08 11:16:54

This probably won't be a popular answer, but you almost certainly do not want to do this. If you want a copy that's a merge, then use copy (or deepcopy, depending on what you want) and then update. The two lines of code are much more readable - more Pythonic - than the single line creation with .items() + .items(). Explicit is better than implicit.

In addition, when you use .items() (pre Python 3.0), you're creating a new list that contains the items from the dict. If your dictionaries are large, then that is quite a lot of overhead (two large lists that will be thrown away as soon as the merged dict is created). update() can work more efficiently, because it can run through the second dict item-by-item.

In terms of time:

>>> timeit.Timer("dict(x, **y)", "x = dict(zip(range(1000), range(1000)))\ny=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
15.52571702003479
>>> timeit.Timer("temp = x.copy()\ntemp.update(y)", "x = dict(zip(range(1000), range(1000)))\ny=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
15.694622993469238
>>> timeit.Timer("dict(x.items() + y.items())", "x = dict(zip(range(1000), range(1000)))\ny=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
41.484580039978027

IMO the tiny slowdown between the first two is worth it for the readability. In addition, keyword arguments for dictionary creation was only added in Python 2.3, whereas copy() and update() will work in older versions.

@thiruvenkadam 2013-12-05 08:02:05

I have a solution which is not specified here(Man I LOVE python) :-)

z = {}
z.update(x) or z.update(y)

This will not update x as well as y. Performance? I don't think it will be terribly slow :-)

NOTE: It is supposed to be 'or' operation and not 'and' operation. Edited to correct the code.

@Navin 2013-12-11 09:50:20

Noooo, it's still 2 lines and also longer than z = x.copy(); z.update(y)

@clacke 2018-04-07 12:20:52

... or z. Otherwise it returns None.

@thiruvenkadam 2018-06-27 07:43:46

I am not assigning th update operations to z :-) But got an idea to make them single liner.. Thanks.

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