Remove duplicate dict in list in Python

Another one-liner based on list comprehensions:

>>> d = [{'a': 123}, {'b': 123}, {'a': 123}]
>>> [i for n, i in enumerate(d) if i not in d[n + 1:]]
[{'b': 123}, {'a': 123}]

Here since we can use dict comparison, we only keep the elements that are not in the rest of the initial list (this notion is only accessible through the index n, hence the use of enumerate).

Try this:

[dict(t) for t in {tuple(d.items()) for d in l}]

The strategy is to convert the list of dictionaries to a list of tuples where the tuples contain the items of the dictionary. Since the tuples can be hashed, you can remove duplicates using set (using a set comprehension here, older python alternative would be set(tuple(d.items()) for d in l)) and, after that, re-create the dictionaries from tuples with dict.

where:

• l is the original list
• d is one of the dictionaries in the list
• t is one of the tuples created from a dictionary

Edit: If you want to preserve ordering, the one-liner above won't work since set won't do that. However, with a few lines of code, you can also do that:

l = [{'a': 123, 'b': 1234},
        {'a': 3222, 'b': 1234},
        {'a': 123, 'b': 1234}]

seen = set()
new_l = []
for d in l:
    t = tuple(d.items())
    if t not in seen:
        seen.add(t)
        new_l.append(d)

print new_l

Example output:

[{'a': 123, 'b': 1234}, {'a': 3222, 'b': 1234}]

Note: As pointed out by @alexis it might happen that two dictionaries with the same keys and values, don't result in the same tuple. That could happen if they go through a different adding/removing keys history. If that's the case for your problem, then consider sorting d.items() as he suggests.