Python merge two lists of dictionaries by key

In python 3.5 or higher, you can merge dictionaries in a single statement.

So for python 3.5 or higher, a quick solution would be:

from itertools import zip_longest

l3 = [{**u, **v} for u, v in zip_longest[l1, l2, fillvalue={}]]

print[l3]
#[
#    {'index': 1, 'b': 2, 'c': 4}, 
#    {'index': 2, 'b': 3, 'c': 5}, 
#    {'index': 3, 'green': 'eggs'}
#]

However if the two lists were the same size, you could simply use zip:

l3 = [{**u, **v} for u, v in zip[l1, l2]]

Note: This assumes that the lists are sorted the same way by index, which is stated by OP to not be the case in general.

In order to generalize for that case, one way is to create a custom zip-longest type function which yields values from the two lists only if they match on a key.

For instance:

def sortedZipLongest[l1, l2, key, fillvalue={}]:  
    l1 = iter[sorted[l1, key=lambda x: x[key]]]
    l2 = iter[sorted[l2, key=lambda x: x[key]]]
    u = next[l1, None]
    v = next[l2, None]

    while [u is not None] or [v is not None]:  
        if u is None:
            yield fillvalue, v
            v = next[l2, None]
        elif v is None:
            yield u, fillvalue
            u = next[l1, None]
        elif u.get[key] == v.get[key]:
            yield u, v
            u = next[l1, None]
            v = next[l2, None]
        elif u.get[key] < v.get[key]:
            yield u, fillvalue
            u = next[l1, None]
        else:
            yield fillvalue, v
            v = next[l2, None]

Now if you had the following out of order lists:

l1 = [{"index":1, "b":2}, {"index":2, "b":3}, {"index":3, "green":"eggs"}, 
      {"index":4, "b": 4}]
l2 = [{"index":1, "c":4}, {"index":2, "c":5}, {"index":0, "green": "ham"}, 
      {"index":4, "green": "ham"}]

Using the sortedZipLongest function instead of itertools.zip_longest:

l3 = [{**u, **v} for u, v in sortedZipLongest[l1, l2, key="index", fillvalue={}]]
print[l3]
#[{'index': 0, 'green': 'ham'},
# {'index': 1, 'b': 2, 'c': 4},
# {'index': 2, 'b': 3, 'c': 5},
# {'index': 3, 'green': 'eggs'},
# {'index': 4, 'b': 4, 'green': 'ham'}]

Whereas original method would produce the incorrect answer:

l3 = [{**u, **v} for u, v in zip_longest[l1, l2, fillvalue={}]]
print[l3]
#[{'index': 1, 'b': 2, 'c': 4},
# {'index': 2, 'b': 3, 'c': 5},
# {'index': 0, 'green': 'ham'},
# {'index': 4, 'b': 4, 'green': 'ham'}]

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