How do I merge a list of dicts into a single dict?


How can I turn a list of dicts like [{'a':1}, {'b':2}, {'c':1}, {'d':2}], into a single dict like {'a':1, 'b':2, 'c':1, 'd':2}?

Answers here will overwrite keys that match between two of the input dicts, because a dict cannot have duplicate keys. If you want to collect multiple values from matching keys, see How to merge dicts, collecting values from matching keys?.

Asked By: killown



dict1.update( dict2 )

This is asymmetrical because you need to choose what to do with duplicate keys; in this case, dict2 will overwrite dict1. Exchange them for the other way.

EDIT: Ah, sorry, didn’t see that.

It is possible to do this in a single expression:

>>> from itertools import chain
>>> dict( chain( *map( dict.items, theDicts ) ) )
{'a': 1, 'c': 1, 'b': 2, 'd': 2}

No credit to me for this last!

However, I’d argue that it might be more Pythonic (explicit > implicit, flat > nested ) to do this with a simple for loop. YMMV.

Answered By: Katriel
>>> L=[{'a': 1}, {'b': 2}, {'c': 1}, {'d': 2}]    
>>> dict(i.items()[0] for i in L)
{'a': 1, 'c': 1, 'b': 2, 'd': 2}

Note: the order of ‘b’ and ‘c’ doesn’t match your output because dicts are unordered

if the dicts can have more than one key/value

>>> dict(j for i in L for j in i.items())
Answered By: John La Rooy

This works for dictionaries of any length:

>>> result = {}
>>> for d in L:
...    result.update(d)
>>> result

As a comprehension:

# Python >= 2.7
{k: v for d in L for k, v in d.items()}

# Python < 2.7
dict(pair for d in L for pair in d.items())
Answered By: user395760
>>> dictlist = [{'a':1},{'b':2},{'c':1},{'d':2, 'e':3}]
>>> dict(kv for d in dictlist for kv in d.iteritems())
{'a': 1, 'c': 1, 'b': 2, 'e': 3, 'd': 2}

Note I added a second key/value pair to the last dictionary to show it works with multiple entries.
Also keys from dicts later in the list will overwrite the same key from an earlier dict.

Answered By: Dave Kirby

For flat dictionaries you can do this:

from functools import reduce
reduce(lambda a, b: dict(a, **b), list_of_dicts)
Answered By: dietbuddha

You can use join function from funcy library:

from funcy import join
Answered By: Suor

In case of Python 3.3+, there is a ChainMap collection:

>>> from collections import ChainMap
>>> a = [{'a':1},{'b':2},{'c':1},{'d':2}]
>>> dict(ChainMap(*a))
{'b': 2, 'c': 1, 'a': 1, 'd': 2}

Also see:

Answered By: alecxe

This is similar to @delnan but offers the option to modify the k/v (key/value) items and I believe is more readable:

new_dict = {k:v for list_item in list_of_dicts for (k,v) in list_item.items()}

for instance, replace k/v elems as follows:

new_dict = {str(k).replace(" ","_"):v for list_item in list_of_dicts for (k,v) in list_item.items()}

unpacks the k,v tuple from the dictionary .items() generator after pulling the dict object out of the list

Answered By: Schalton

Little improvement for @dietbuddha answer with dictionary unpacking from PEP 448, for me, it`s more readable this way, also, it is faster as well:

from functools import reduce
result_dict = reduce(lambda a, b: {**a, **b}, list_of_dicts)

But keep in mind, this works only with Python 3.5+ versions.

Answered By: Insomniac631

If you don’t need the singleton dicts anymore:

>>> L = [{'a':1}, {'b':2}, {'c':1}, {'d':2}]
>>> dict(map(dict.popitem, L))
{'a': 1, 'b': 2, 'c': 1, 'd': 2}
Answered By: no comment
Categories: questions Tags: , ,
Answers are sorted by their score. The answer accepted by the question owner as the best is marked with
at the top-right corner.