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

Question

This works for dictionaries of any length:

>>> result = {}
>>> for d in L:
...    result.update(d)
... 
>>> result
{'a':1,'c':1,'b':2,'d':2}

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())

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:

• What is the purpose of collections.ChainMap?

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

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.

For flat dictionaries you can do this:

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

>>> 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())

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.