671

What is the best way to extend a dictionary with another one while avoiding the use of a for loop? For instance:

>>> a = { "a" : 1, "b" : 2 }
>>> b = { "c" : 3, "d" : 4 }
>>> a
{'a': 1, 'b': 2}
>>> b
{'c': 3, 'd': 4}

Result:

{ "a" : 1, "b" : 2, "c" : 3, "d" : 4 }

Something like:

a.extend(b)  # This does not work
Salvatore
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FerranB
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    I knew for lists [], then i supose may be work for others, not extrange ;-) – FerranB Feb 23 '09 at 13:05
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    This question is not uniquely defined as the example should cover the case where a and b have common keys – ntg Jun 25 '20 at 12:17

8 Answers8

970
a.update(b)

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phoenix
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Nick Fortescue
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    keep in mind that update() directly modifies the dict and returns None. – e18r Dec 05 '15 at 23:15
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    But what if you only want to extend the dict with values that are _not_ defined already (i.e. not overwriting existing values)? For instance, update a dict `{"a":2, "b":3}` with dict `{"b":4, "c":5}` to dict `{"a":2, "b":3,"c":5}`? Of course it's possible using `update()` by moving some stuff around, but it would be nicer if it could be accomplished in only one line... – Nearoo Jan 06 '16 at 16:28
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    @Nearoo - simply update the opposite way; instead of x.update(y) [which would overwrite x values with y], use y.update(x) [which overwrites y values with x] and use y as your chosen dict for further operations – jonathanl Jan 27 '16 at 02:20
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    keep in mind that `update` silently overwrites existing entries. Ie, any values in `b` overwrite those in `a` for overlapping keys. – stefanbschneider Apr 01 '20 at 11:43
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    @e18r Yes, please dont do a = a.update(b), this will turn a to None, a very common mistake. – c8999c 3f964f64 May 30 '23 at 12:07
251

A beautiful gem in this closed question:

The "oneliner way", altering neither of the input dicts, is

basket = dict(basket_one, **basket_two)

Learn what **basket_two (the **) means here.

In case of conflict, the items from basket_two will override the ones from basket_one. As one-liners go, this is pretty readable and transparent, and I have no compunction against using it any time a dict that's a mix of two others comes in handy (any reader who has trouble understanding it will in fact be very well served by the way this prompts him or her towards learning about dict and the ** form;-). So, for example, uses like:

x = mungesomedict(dict(adict, **anotherdict))

are reasonably frequent occurrences in my code.

Originally submitted by Alex Martelli

Note: In Python 3, this will only work if every key in basket_two is a string.

Community
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Tom Leys
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    Documentation for [`dict`](http://docs.python.org/2/library/stdtypes.html#dict) is easy to find while [`**`](http://www.saltycrane.com/blog/2008/01/how-to-use-args-and-kwargs-in-python/) is a bit more tricky (keyword is *kwargs*). Here is a nice explanation: http://www.saltycrane.com/blog/2008/01/how-to-use-args-and-kwargs-in-python/ – johndodo Mar 27 '13 at 10:37
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    this may be used to generate a second variable with a single command, whereas `basket_one.update()` as the name said, updates an existing dictionary (or a cloned one). – furins Jan 17 '14 at 16:08
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    Note that in Python 3, argument names, and thus the keys of `**anotherdict`, must be strings. – Petr Viktorin Aug 19 '14 at 18:56
  • Thanks @johndodo - I've integrated your suggestions into the post. – Tom Leys Jun 30 '16 at 22:27
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    A nice functional alternative to the accepted answer. This approach is desirable if you need to directly use the newly combined dict. (also see @e18r's comment on the accepted answer). – chinnychinchin Apr 08 '19 at 19:47
100

Have you tried using dictionary comprehension with dictionary mapping:

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

c = {**a, **b}
# c = {"a": 1, "b": 2, "c": 3, "d": 4}

Another way of doing is by Using dict(iterable, **kwarg)

c = dict(a, **b)
# c = {'a': 1, 'b': 2, 'c': 3, 'd': 4}

In Python 3.9 you can add two dict using union | operator

# use the merging operator |
c = a | b
# c = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
Vlad Bezden
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32
a.update(b)

Will add keys and values from b to a, overwriting if there's already a value for a key.

vartec
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22

As others have mentioned, a.update(b) for some dicts a and b will achieve the result you've asked for in your question. However, I want to point out that many times I have seen the extend method of mapping/set objects desire that in the syntax a.extend(b), a's values should NOT be overwritten by b's values. a.update(b) overwrites a's values, and so isn't a good choice for extend.

Note that some languages call this method defaults or inject, as it can be thought of as a way of injecting b's values (which might be a set of default values) in to a dictionary without overwriting values that might already exist.

Of course, you could simple note that a.extend(b) is nearly the same as b.update(a); a=b. To remove the assignment, you could do it thus:

def extend(a,b):
    """Create a new dictionary with a's properties extended by b,
    without overwriting.

    >>> extend({'a':1,'b':2},{'b':3,'c':4})
    {'a': 1, 'c': 4, 'b': 2}
    """
    return dict(b,**a)

Thanks to Tom Leys for that smart idea using a side-effect-less dict constructor for extend.

eblume
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17

Notice that since Python 3.9 a much easier syntax was introduced (Union Operators):

d1 = {'a': 1}
d2 = {'b': 2}

extended_dict = d1 | d2
>> {'a':1, 'b': 2}

Pay attention: in case first dict shared keys with second dict, position matters!

d1 = {'b': 1}
d2 = {'b': 2}
d1 | d2 
>> {'b': 2} 

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Nuno André
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Aaron_ab
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5

You can also use python's collections.ChainMap which was introduced in python 3.3.

from collections import ChainMap
c = ChainMap(a, b)
c['a'] # returns 1

This has a few possible advantages, depending on your use-case. They are explained in more detail here, but I'll give a brief overview:

  • A chainmap only uses views of the dictionaries, so no data is actually copied. This results in faster chaining (but slower lookup)
  • No keys are actually overwritten so, if necessary, you know whether the data comes from a or b.

This mainly makes it useful for things like configuration dictionaries.

Ivo Merchiers
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1

In terms of efficiency, it seems faster to use the unpack operation, compared with the update method.

Here an image of a test I did: enter image description here