Asked  7 Months ago    Answers:  5   Viewed   37 times

I have a list of dicts, and I'd like to remove the dicts with identical key and value pairs.

For this list: [{'a': 123}, {'b': 123}, {'a': 123}]

I'd like to return this: [{'a': 123}, {'b': 123}]

Another example:

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

I'd like to return this: [{'a': 123, 'b': 1234}, {'a': 3222, 'b': 1234}]



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.


  • 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:

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.

Tuesday, June 1, 2021
answered 7 Months ago

If it weren't for the pretty printing, the reprlib module would be the way to go: Safe, elegant and customizable handling of deeply nested and recursive / self-referencing data structures is what it has been made for.

However, it turns out combining the reprlib and pprint modules isn't trivial, at least I couldn't come up with a clean way without breaking (some) of the pretty printing aspects.

So instead, here's a solution that just subclasses PrettyPrinter to crop / abbreviate lists as necessary:

from pprint import PrettyPrinter

obj = {
    'key_1': [
        'EG8XYD9FVN', 'S2WARDCVAO', 'J00YCU55DP', 'R07BUIF2F7', 'VGPS1JD0UM',
        'WL3TWSDP8E', 'LD8QY7DMJ3', 'J36U3Z9KOQ', 'KU2FUGYB2U', 'JF3RQ315BY',
    'key_2': [
        '162LO154PM', '3ROAV881V2', 'I4T79LP18J', 'WBD36EM6QL', 'DEIODVQU46',
        'KWSJA5WDKQ', 'WX9SVRFO0G', '6UN63WU64G', '3Z89U7XM60', '167CYON6YN',
    # Test case to make sure we didn't break handling of recursive structures
    'key_3': [
        '162LO154PM', '3ROAV881V2', [1, 2, ['a', 'b', 'c'], 3, 4, 5, 6, 7],
        'KWSJA5WDKQ', 'WX9SVRFO0G', '6UN63WU64G', '3Z89U7XM60', '167CYON6YN',

class CroppingPrettyPrinter(PrettyPrinter):

    def __init__(self, *args, **kwargs):
        self.maxlist = kwargs.pop('maxlist', 6)
        return PrettyPrinter.__init__(self, *args, **kwargs)

    def _format(self, obj, stream, indent, allowance, context, level):
        if isinstance(obj, list):
            # If object is a list, crop a copy of it according to self.maxlist
            # and append an ellipsis
            if len(obj) > self.maxlist:
                cropped_obj = obj[:self.maxlist] + ['...']
                return PrettyPrinter._format(
                    self, cropped_obj, stream, indent,
                    allowance, context, level)

        # Let the original implementation handle anything else
        # Note: No use of super() because PrettyPrinter is an old-style class
        return PrettyPrinter._format(
            self, obj, stream, indent, allowance, context, level)

p = CroppingPrettyPrinter(maxlist=3)

Output with maxlist=3:

{'key_1': ['EG8XYD9FVN', 'S2WARDCVAO', 'J00YCU55DP', '...'],
 'key_2': ['162LO154PM',
           [1, 2, ['a', 'b', 'c'], '...'],

Output with maxlist=5 (triggers splitting the lists on separate lines):

{'key_1': ['EG8XYD9FVN',
 'key_2': ['162LO154PM',
 'key_3': ['162LO154PM',
           [1, 2, ['a', 'b', 'c'], 3, 4, '...'],


  • This will create copies of lists. Depending on the size of the data structures, this can be very expensive in terms of memory use.
  • This only deals with the special case of lists. Equivalent behavior would have to be implemented for dicts, tuples, sets, frozensets, ... for this class to be of general use.
Thursday, July 29, 2021
answered 4 Months ago

(ab)using side-effects version of a list comp:

seen = set()

[x for x in g if frozenset(x) not in seen and not seen.add(frozenset(x))]
Out[4]: [[1, 2, 3], [9, 0, 1], [4, 3, 2]]

For those (unlike myself) who don't like using side-effects in this manner:

res = []
seen = set()

for x in g:
    x_set = frozenset(x)
    if x_set not in seen:

The reason that you add frozensets to the set is that you can only add hashable objects to a set, and vanilla sets are not hashable.

Tuesday, August 17, 2021
answered 4 Months ago
for key, value in my_dict.iteritems()

This will iterate through the dictionary, storing each key in key and each value in value. See the docs.

Sunday, August 22, 2021
answered 4 Months ago

Try this :

list_ = [["A"], ["B"], ["A","B"], ["B","A"], ["A","B","C"], ["B", "A", "C"]]
l = list(map(list, set(map(tuple, map(set, list_)))))

Output :

[['A', 'B'], ['B'], ['A', 'B', 'C'], ['A']]

This process goes through like :

  1. First convert each sub-list into a set. Thus ['A', 'B'] and ['B', 'A'] both are converted to {'A', 'B'}.
  2. Now convert each of them to a tuple for removing duplicate items as set() operation can not be done with set sub-items in the list.
  3. With set() operation make a list of unique tuples.
  4. Now convert each tuple items in the list into list type.

This is equivalent to :

list_ = [['A'], ['B'], ['A', 'B'], ['B', 'A'], ['A', 'B', 'C'], ['B', 'A', 'C']]
l0 = [set(i) for i in list_]
# l0 = [{'A'}, {'B'}, {'A', 'B'}, {'A', 'B'}, {'A', 'B', 'C'}, {'A', 'B', 'C'}]
l1 = [tuple(i) for i in l0]
# l1 = [('A',), ('B',), ('A', 'B'), ('A', 'B'), ('A', 'B', 'C'), ('A', 'B', 'C')]
l2 = set(l1)
# l2 = {('A', 'B'), ('A',), ('B',), ('A', 'B', 'C')}
l = [list(i) for i in l2]
# l = [['A', 'B'], ['A'], ['B'], ['A', 'B', 'C']]
Wednesday, September 1, 2021
Morrison Chang
answered 3 Months ago
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