How to get count dict of items but maintain the order in which they appear?

Question

For example, I need to count how many times a word appears in a list, not sorted by frequency but with the order in which the words appear, i.e. insertion order.

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Answer 1

In Python 3.6, dictionaries are insertion ordered, but this is an implementation detail.

In Python 3.7+, insertion order is guaranteed and can be relied upon. See Are dictionaries ordered in Python 3.6+? for more details.

So, depending on your Python version, you may wish to just use Counter as is, without creating an OrderedCounter class as described in the documentation. This works because Counter is a subclass of dict, i.e. issubclass(Counter, dict) returns True, and therefore inherits the insertion ordering behaviour of dict.

String representation

It is worth noting the the string representation for Counter, as defined in the repr method, has not been updated to reflect the change in 3.6 / 3.7, i.e. print(Counter(some_iterable)) still returns items from largest counts descending. You can trivially return the insertion order via list(Counter(some_iterable)).

Here are some examples demonstrating the behaviour:

x = 'xyyxy'
print(Counter(x))         # Counter({'y': 3, 'x': 2}), i.e. most common first
print(list(Counter(x)))   # ['x', 'y'], i.e. insertion ordered
print(OrderedCounter(x))  # OC(OD([('x', 2), ('y', 3)])), i.e. insertion ordered

Exceptions

You should not use a regular Counter if additional or overwritten methods available to OrderedCounter are important to you. Of particular note:

1. OrderedDict and consequently OrderedCounter offer popitem and move_to_end methods.
2. Equality tests between OrderedCounter objects are order-sensitive and are implemented as list(oc1.items()) == list(oc2.items()).

For example, equality tests will yield different results:

Counter('xy') == Counter('yx')                # True
OrderedCounter('xy') == OrderedCounter('yx')  # False

Answer 2

On Python 3.6+, dict will now maintain insertion order.

So you can do:

words = ["oranges", "apples", "apples", "bananas", "kiwis", "kiwis", "apples"]
counter={}
for w in words: counter[w]=counter.get(w, 0)+1
>>> counter
{'oranges': 1, 'apples': 3, 'bananas': 1, 'kiwis': 2}

Unfortunately, the Counter in Python 3.6 and 3.7 does not display the insertion order that it maintains; instead, __repr__ sorts the return by the most to least common.

But you can use the same OrderedDict recipe but just use the Python 3.6+ dict instead:

from collections import Counter

class OrderedCounter(Counter, dict):
    'Counter that remembers the order elements are first encountered'
    def __repr__(self):
        return '%s(%r)' % (self.__class__.__name__, dict(self))

    def __reduce__(self):
        return self.__class__, (dict(self),)

>>> OrderedCounter(words)
OrderedCounter({'oranges': 1, 'apples': 3, 'bananas': 1, 'kiwis': 2})

Or, since Counter is a subclass of dict that maintains order in Python 3.6+, you can just avoid using Counter's __repr__ by either calling .items() on the counter or turning the counter back into a dict:

>>> c=Counter(words)

This presentation of that Counter is sorted by most common element to least and uses Counters __repr__ method:

>>> c
Counter({'apples': 3, 'kiwis': 2, 'oranges': 1, 'bananas': 1})

This presentation is as encountered, or insertion order:

>>> c.items()
dict_items([('oranges', 1), ('apples', 3), ('bananas', 1), ('kiwis', 2)])

Or,

>>> dict(c)
{'oranges': 1, 'apples': 3, 'bananas': 1, 'kiwis': 2}

Answer 3

You can use the recipe that uses collections.Counter and collections.OrderedDict:

from collections import Counter, OrderedDict

class OrderedCounter(Counter, OrderedDict):
    'Counter that remembers the order elements are first encountered'

    def __repr__(self):
        return '%s(%r)' % (self.__class__.__name__, OrderedDict(self))

    def __reduce__(self):
        return self.__class__, (OrderedDict(self),)

words = ["oranges", "apples", "apples", "bananas", "kiwis", "kiwis", "apples"]
c = OrderedCounter(words)
print(c)
# OrderedCounter(OrderedDict([('oranges', 1), ('apples', 3), ('bananas', 1), ('kiwis', 2)]))