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 3.9.0 or greater (released 17 October 2020): PEP-584, discussed here, was implemented and provides the simplest method:

  z = x | y          # NOTE: 3.9+ ONLY
  

• 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

To understand what yield does, you must understand what generators are. And before you can understand generators, you must understand iterables.

Iterables

When you create a list, you can read its items one by one. Reading its items one by one is called iteration:

>>> mylist = [1, 2, 3]
>>> for i in mylist:
...    print(i)
1
2
3

mylist is an iterable. When you use a list comprehension, you create a list, and so an iterable:

>>> mylist = [x*x for x in range(3)]
>>> for i in mylist:
...    print(i)
0
1
4

Everything you can use "for... in..." on is an iterable; lists, strings, files...

These iterables are handy because you can read them as much as you wish, but you store all the values in memory and this is not always what you want when you have a lot of values.

Generators

Generators are iterators, a kind of iterable you can only iterate over once. Generators do not store all the values in memory, they generate the values on the fly:

>>> mygenerator = (x*x for x in range(3))
>>> for i in mygenerator:
...    print(i)
0
1
4

It is just the same except you used () instead of []. BUT, you cannot perform for i in mygenerator a second time since generators can only be used once: they calculate 0, then forget about it and calculate 1, and end calculating 4, one by one.

Yield

yield is a keyword that is used like return, except the function will return a generator.

>>> def create_generator():
...    mylist = range(3)
...    for i in mylist:
...        yield i*i
...
>>> mygenerator = create_generator() # create a generator
>>> print(mygenerator) # mygenerator is an object!
<generator object create_generator at 0xb7555c34>
>>> for i in mygenerator:
...     print(i)
0
1
4

Here it's a useless example, but it's handy when you know your function will return a huge set of values that you will only need to read once.

To master yield, you must understand that when you call the function, the code you have written in the function body does not run. The function only returns the generator object, this is a bit tricky.

Then, your code will continue from where it left off each time for uses the generator.

Now the hard part:

The first time the for calls the generator object created from your function, it will run the code in your function from the beginning until it hits yield, then it'll return the first value of the loop. Then, each subsequent call will run another iteration of the loop you have written in the function and return the next value. This will continue until the generator is considered empty, which happens when the function runs without hitting yield. That can be because the loop has come to an end, or because you no longer satisfy an "if/else".

Your code explained

Generator:

# Here you create the method of the node object that will return the generator
def _get_child_candidates(self, distance, min_dist, max_dist):

    # Here is the code that will be called each time you use the generator object:

    # If there is still a child of the node object on its left
    # AND if the distance is ok, return the next child
    if self._leftchild and distance - max_dist < self._median:
        yield self._leftchild

    # If there is still a child of the node object on its right
    # AND if the distance is ok, return the next child
    if self._rightchild and distance + max_dist >= self._median:
        yield self._rightchild

    # If the function arrives here, the generator will be considered empty
    # there is no more than two values: the left and the right children

Caller:

# Create an empty list and a list with the current object reference
result, candidates = list(), [self]

# Loop on candidates (they contain only one element at the beginning)
while candidates:

    # Get the last candidate and remove it from the list
    node = candidates.pop()

    # Get the distance between obj and the candidate
    distance = node._get_dist(obj)

    # If distance is ok, then you can fill the result
    if distance <= max_dist and distance >= min_dist:
        result.extend(node._values)

    # Add the children of the candidate in the candidate's list
    # so the loop will keep running until it will have looked
    # at all the children of the children of the children, etc. of the candidate
    candidates.extend(node._get_child_candidates(distance, min_dist, max_dist))

return result

This code contains several smart parts:

• The loop iterates on a list, but the list expands while the loop is being iterated. It's a concise way to go through all these nested data even if it's a bit dangerous since you can end up with an infinite loop. In this case, candidates.extend(node._get_child_candidates(distance, min_dist, max_dist)) exhaust all the values of the generator, but while keeps creating new generator objects which will produce different values from the previous ones since it's not applied on the same node.

• The extend() method is a list object method that expects an iterable and adds its values to the list.

Usually we pass a list to it:

>>> a = [1, 2]
>>> b = [3, 4]
>>> a.extend(b)
>>> print(a)
[1, 2, 3, 4]

But in your code, it gets a generator, which is good because:

1. You don't need to read the values twice.
2. You may have a lot of children and you don't want them all stored in memory.

And it works because Python does not care if the argument of a method is a list or not. Python expects iterables so it will work with strings, lists, tuples, and generators! This is called duck typing and is one of the reasons why Python is so cool. But this is another story, for another question...

You can stop here, or read a little bit to see an advanced use of a generator:

Controlling a generator exhaustion

>>> class Bank(): # Let's create a bank, building ATMs
...    crisis = False
...    def create_atm(self):
...        while not self.crisis:
...            yield "$100"
>>> hsbc = Bank() # When everything's ok the ATM gives you as much as you want
>>> corner_street_atm = hsbc.create_atm()
>>> print(corner_street_atm.next())
$100
>>> print(corner_street_atm.next())
$100
>>> print([corner_street_atm.next() for cash in range(5)])
['$100', '$100', '$100', '$100', '$100']
>>> hsbc.crisis = True # Crisis is coming, no more money!
>>> print(corner_street_atm.next())
<type 'exceptions.StopIteration'>
>>> wall_street_atm = hsbc.create_atm() # It's even true for new ATMs
>>> print(wall_street_atm.next())
<type 'exceptions.StopIteration'>
>>> hsbc.crisis = False # The trouble is, even post-crisis the ATM remains empty
>>> print(corner_street_atm.next())
<type 'exceptions.StopIteration'>
>>> brand_new_atm = hsbc.create_atm() # Build a new one to get back in business
>>> for cash in brand_new_atm:
...    print cash
$100
$100
$100
$100
$100
$100
$100
$100
$100
...

Note: For Python 3, useprint(corner_street_atm.__next__()) or print(next(corner_street_atm))

It can be useful for various things like controlling access to a resource.

Itertools, your best friend

The itertools module contains special functions to manipulate iterables. Ever wish to duplicate a generator? Chain two generators? Group values in a nested list with a one-liner? Map / Zip without creating another list?

Then just import itertools.

An example? Let's see the possible orders of arrival for a four-horse race:

>>> horses = [1, 2, 3, 4]
>>> races = itertools.permutations(horses)
>>> print(races)
<itertools.permutations object at 0xb754f1dc>
>>> print(list(itertools.permutations(horses)))
[(1, 2, 3, 4),
 (1, 2, 4, 3),
 (1, 3, 2, 4),
 (1, 3, 4, 2),
 (1, 4, 2, 3),
 (1, 4, 3, 2),
 (2, 1, 3, 4),
 (2, 1, 4, 3),
 (2, 3, 1, 4),
 (2, 3, 4, 1),
 (2, 4, 1, 3),
 (2, 4, 3, 1),
 (3, 1, 2, 4),
 (3, 1, 4, 2),
 (3, 2, 1, 4),
 (3, 2, 4, 1),
 (3, 4, 1, 2),
 (3, 4, 2, 1),
 (4, 1, 2, 3),
 (4, 1, 3, 2),
 (4, 2, 1, 3),
 (4, 2, 3, 1),
 (4, 3, 1, 2),
 (4, 3, 2, 1)]

Understanding the inner mechanisms of iteration

Iteration is a process implying iterables (implementing the __iter__() method) and iterators (implementing the __next__() method). Iterables are any objects you can get an iterator from. Iterators are objects that let you iterate on iterables.

There is more about it in this article about how for loops work.