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

Question

I want to merge two dictionaries into a new dictionary.

x = {'a': 1, 'b': 2}
y = {'b': 3, 'c': 4}
z = merge(x, y)

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

Whenever a key k is present in both dictionaries, only the value y[k] should be kept.

Answer 1

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

For dictionaries x and y, their shallowly-merged dictionary z takes values from y, replacing those from x.

• In Python 3.9.0 or greater (released 17 October 2020, PEP-584, discussed here):

  z = x | y
  

• 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 keys and values of x
      z.update(y)    # modifies z with keys and values of y
      return z
  

  and now:

  z = merge_two_dicts(x, y)
  

Explanation

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

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 dictionary'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 the 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 backward-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 the correct approach is to put it in a function:

def merge_two_dicts(x, y):
    """Given two dictionaries, 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 arbitrary number of dictionaries, from zero to a very large number:

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

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

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

and key-value pairs in g will take precedence over dictionaries 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.

Dictionaries 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 dictionaries 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-coming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact, ** was designed precisely to pass dictionaries as keywords.

Again, it doesn't work for 3 when keys are not strings. The implicit calling contract is that namespaces take ordinary dictionaries, 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 dictionaries of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dictionaries 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 dictionaries have precedence)

You can also chain the dictionaries 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()) # iteritems in Python 2

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

from itertools import chain
z = dict(chain(x.items(), y.items())) # iteritems in Python 2

Performance Analysis

I'm only going to do the performance analysis of the usages known to behave correctly. (Self-contained so you can copy and paste yourself.)

from timeit import repeat
from itertools import chain

x = dict.fromkeys('abcdefg')
y = dict.fromkeys('efghijk')

def merge_two_dicts(x, y):
    z = x.copy()
    z.update(y)
    return z

min(repeat(lambda: {**x, **y}))
min(repeat(lambda: merge_two_dicts(x, y)))
min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
min(repeat(lambda: dict(chain(x.items(), y.items()))))
min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))

In Python 3.8.1, NixOS:

>>> min(repeat(lambda: {**x, **y}))
1.0804965235292912
>>> min(repeat(lambda: merge_two_dicts(x, y)))
1.636518670246005
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
3.1779992282390594
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
2.740647904574871
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
4.266070580109954

$ uname -a
Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux

Resources on Dictionaries

• My explanation of Python's dictionary implementation, updated for 3.6.
• Answer on how to add new keys to a dictionary
• Mapping two lists into a dictionary
• The official Python docs on dictionaries
• The Dictionary Even Mightier - talk by Brandon Rhodes at Pycon 2017
• Modern Python Dictionaries, A Confluence of Great Ideas - talk by Raymond Hettinger at Pycon 2017

Answer 2

In your case, you can do:

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

If you use Python 2, you can even remove the list() calls. To create z:

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

If you use Python version 3.9.0a4 or greater, you can directly use:

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

Answer 3

An alternative:

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

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

In Python 3.0 and later, 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 = dict(ChainMap({}, y, x))
>>> for k, v in z.items():
        print(k, '-->', v)
    
a --> 1
b --> 10
c --> 11

Update for Python 3.5 and later: You can use PEP 448 extended dictionary packing and unpacking. This is fast and easy:

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

Update for Python 3.9 and later: You can use the PEP 584 union operator:

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

Answer 8

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

Answer 9

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.

Answer 10

I benchmarked the suggested with perfplot and found that

x | y   # Python 3.9+

is the fastest solution together with the good old

{**x, **y}

and

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

---

Code to reproduce the plot:

from collections import ChainMap
from itertools import chain
import perfplot


def setup(n):
    x = dict(zip(range(n), range(n)))
    y = dict(zip(range(n, 2 * n), range(n, 2 * n)))
    return x, y


def copy_update(x, y):
    temp = x.copy()
    temp.update(y)
    return temp


def add_items(x, y):
    return dict(list(x.items()) + list(y.items()))


def curly_star(x, y):
    return {**x, **y}


def chain_map(x, y):
    return dict(ChainMap({}, y, x))


def itertools_chain(x, y):
    return dict(chain(x.items(), y.items()))


def python39_concat(x, y):
    return x | y


b = perfplot.bench(
    setup=setup,
    kernels=[
        copy_update,
        add_items,
        curly_star,
        chain_map,
        itertools_chain,
        python39_concat,
    ],
    labels=[
        "copy_update",
        "dict(list(x.items()) + list(y.items()))",
        "{**x, **y}",
        "chain_map",
        "itertools.chain",
        "x | y",
    ],
    n_range=[2 ** k for k in range(18)],
    xlabel="len(x), len(y)",
    equality_check=None,
)
b.save("out.png")
b.show()

Answer 11

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}

In Python 3.9 you also use | and |= with the below example from PEP 584

d = {'spam': 1, 'eggs': 2, 'cheese': 3}
e = {'cheese': 'cheddar', 'aardvark': 'Ethel'}
d | e
# {'spam': 1, 'eggs': 2, 'cheese': 'cheddar', 'aardvark': 'Ethel'}

Answer 12

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

For items with keys in both dictionaries ('b'), you can control which one ends up in the output by putting that one last.

Answer 13

The best version I could think while not using copy would be:

from itertools import chain
x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}
dict(chain(x.iteritems(), y.iteritems()))

It's faster than dict(x.items() + y.items()) but not as fast as n = copy(a); n.update(b), at least on CPython. This version also works in Python 3 if you change iteritems() to items(), which is automatically done by the 2to3 tool.

Personally I like this version best because it describes fairly good what I want in a single functional syntax. The only minor problem is that it doesn't make completely obvious that values from y takes precedence over values from x, but I don't believe it's difficult to figure that out.

Answer 14

While the question has already been answered several times, this simple solution to the problem has not been listed yet.

x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}
z4 = {}
z4.update(x)
z4.update(y)

It is as fast as z0 and the evil z2 mentioned above, but easy to understand and change.

Answer 15

def dict_merge(a, b):
  c = a.copy()
  c.update(b)
  return c

new = dict_merge(old, extras)

Among such shady and dubious answers, this shining example is the one and only good way to merge dicts in Python, endorsed by dictator for life Guido van Rossum himself! Someone else suggested half of this, but did not put it in a function.

print dict_merge(
      {'color':'red', 'model':'Mini'},
      {'model':'Ferrari', 'owner':'Carl'})

gives:

{'color': 'red', 'owner': 'Carl', 'model': 'Ferrari'}

Answer 16

Be Pythonic. Use a comprehension:

z={k: v for d in [x,y] for k, v in d.items()}

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

Answer 17

If you think lambdas are evil then read no further. As requested, you can write the fast and memory-efficient solution with one expression:

x = {'a':1, 'b':2}
y = {'b':10, 'c':11}
z = (lambda a, b: (lambda a_copy: a_copy.update(b) or a_copy)(a.copy()))(x, y)
print z
{'a': 1, 'c': 11, 'b': 10}
print x
{'a': 1, 'b': 2}

As suggested above, using two lines or writing a function is probably a better way to go.

Answer 18

In python3, the items method no longer returns a list, but rather a view, which acts like a set. In this case you'll need to take the set union since concatenating with + won't work:

dict(x.items() | y.items())

For python3-like behavior in version 2.7, the viewitems method should work in place of items:

dict(x.viewitems() | y.viewitems())

I prefer this notation anyways since it seems more natural to think of it as a set union operation rather than concatenation (as the title shows).

Edit:

A couple more points for python 3. First, note that the dict(x, **y) trick won't work in python 3 unless the keys in y are strings.

Also, Raymond Hettinger's Chainmap answer is pretty elegant, since it can take an arbitrary number of dicts as arguments, but from the docs it looks like it sequentially looks through a list of all the dicts for each lookup:

Lookups search the underlying mappings successively until a key is found.

This can slow you down if you have a lot of lookups in your application:

In [1]: from collections import ChainMap
In [2]: from string import ascii_uppercase as up, ascii_lowercase as lo; x = dict(zip(lo, up)); y = dict(zip(up, lo))
In [3]: chainmap_dict = ChainMap(y, x)
In [4]: union_dict = dict(x.items() | y.items())
In [5]: timeit for k in union_dict: union_dict[k]
100000 loops, best of 3: 2.15 µs per loop
In [6]: timeit for k in chainmap_dict: chainmap_dict[k]
10000 loops, best of 3: 27.1 µs per loop

So about an order of magnitude slower for lookups. I'm a fan of Chainmap, but looks less practical where there may be many lookups.

Answer 19

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/

Answer 20

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

# py2
from itertools import chain, imap
merge = lambda *args: dict(chain.from_iterable(imap(dict.iteritems, args)))

# py3
from itertools import chain
merge = lambda *args: dict(chain.from_iterable(map(dict.items, 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}

Answer 21

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}

Answer 22

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 the above expression will always evaluate to x, after updating it.

Most mutating methods in the standard library (like .update()) return None by convention, so this kind of pattern will work on those too. However, if you're using a dict subclass or some other method that doesn't follow this convention, then or may return its left operand, which may not be what you want. Instead, you can use a tuple display and index, which works regardless of what the first element evaluates to (although it's not quite as pretty):

(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 is maybe unpythonic.

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

Although it's not that different from the following use of the new walrus operator (Python 3.8+ only),

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

especially if you use a default argument:

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

If you do want a copy, PEP 584 style x | y is the most Pythonic on 3.9+. If you must support older versions, PEP 448 style {**x, **y} is easiest for 3.5+. But if that's not available in your (even older) Python version, the let expression pattern works here too.

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

(That is, of course, nearly equivalent to (z := x.copy()).update(y) or z, but if your Python version is new enough for that, then the PEP 448 style will be available.)

Answer 23

New in Python 3.9: Use the union operator (|) to merge dicts similar to sets:

>>> d = {'a': 1, 'b': 2}
>>> e = {'a': 9, 'c': 3}
>>> d | e
{'a': 9, 'b': 2, 'c': 3}

For matching keys, the right dict takes precedence.

This also works for |= to modify a dict in-place:

>>> e |= d    # e = e | d
>>> e
{'a': 1, 'c': 3, 'b': 2}

Answer 24

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.

Answer 25

Drawing on ideas here and elsewhere I've comprehended a function:

def merge(*dicts, **kv): 
      return { k:v for d in list(dicts) + [kv] for k,v in d.items() }

Usage (tested in python 3):

assert (merge({1:11,'a':'aaa'},{1:99, 'b':'bbb'},foo='bar')==\
    {1: 99, 'foo': 'bar', 'b': 'bbb', 'a': 'aaa'})

assert (merge(foo='bar')=={'foo': 'bar'})

assert (merge({1:11},{1:99},foo='bar',baz='quux')==\
    {1: 99, 'foo': 'bar', 'baz':'quux'})

assert (merge({1:11},{1:99})=={1: 99})

You could use a lambda instead.

Answer 26

(For Python 2.7* only; there are simpler solutions for Python 3*.)

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.)

Answer 27

The problem I have with solutions listed to date is that, in the merged dictionary, the value for key "b" is 10 but, to my way of thinking, it should be 12. In that light, I present the following:

import timeit

n=100000
su = """
x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}
"""

def timeMerge(f,su,niter):
    print "{:4f} sec for: {:30s}".format(timeit.Timer(f,setup=su).timeit(n),f)

timeMerge("dict(x, **y)",su,n)
timeMerge("x.update(y)",su,n)
timeMerge("dict(x.items() + y.items())",su,n)
timeMerge("for k in y.keys(): x[k] = k in x and x[k]+y[k] or y[k] ",su,n)

#confirm for loop adds b entries together
x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}
for k in y.keys(): x[k] = k in x and x[k]+y[k] or y[k]
print "confirm b elements are added:",x

Results:

0.049465 sec for: dict(x, **y)
0.033729 sec for: x.update(y)                   
0.150380 sec for: dict(x.items() + y.items())   
0.083120 sec for: for k in y.keys(): x[k] = k in x and x[k]+y[k] or y[k]

confirm b elements are added: {'a': 1, 'c': 11, 'b': 12}

Answer 28

It's so silly that .update returns nothing.
I just use a simple helper function to solve the problem:

def merge(dict1,*dicts):
    for dict2 in dicts:
        dict1.update(dict2)
    return dict1

Examples:

merge(dict1,dict2)
merge(dict1,dict2,dict3)
merge(dict1,dict2,dict3,dict4)
merge({},dict1,dict2)  # this one returns a new copy

Answer 29

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.

Answer 30

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.