sorting a counter in python by keys

Question

I have a counter that looks a bit like this:

Counter: {('A': 10), ('C':5), ('H':4)}

I want to sort on keys specifically in an alphabetical order, NOT by counter.most_common()

is there any way to achieve this?

Asked By: corvid

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Source

Answer 1

Just use sorted:

>>> from collections import Counter
>>> counter = Counter({'A': 10, 'C': 5, 'H': 7})
>>> counter.most_common()
[('A', 10), ('H', 7), ('C', 5)]
>>> sorted(counter.items())
[('A', 10), ('C', 5), ('H', 7)]

Answered By: falsetru

Answer 2

>>> from operator import itemgetter
>>> from collections import Counter
>>> c = Counter({'A': 10, 'C':5, 'H':4})
>>> sorted(c.items(), key=itemgetter(0))
[('A', 10), ('C', 5), ('H', 4)]

Answered By: Roman Pekar

Answer 3

In Python 3, you can use the most_common function of collections.Counter:

x = ['a', 'b', 'c', 'c', 'c', 'd', 'd']
counts = collections.Counter(x)
counts.most_common(len(counts))

This uses the most_common function available in collections.Counter, which allows you to find the keys and counts of n most common keys.

Answered By: wwwilliam

Answer 4

To get values as list in sorted order

array              = [1, 2, 3, 4, 5]
counter            = collections.Counter(array)
sorted_occurrences = list(dict(sorted(counter.items())).values())

Answered By: Redzhep Mehmedov Redzhebov

Answer 5

sorted(counter.items(),key = lambda i: i[0])

for example:

arr = [2,3,1,3,2,4,6,7,9,2,19]
c = collections.Counter(arr)
sorted(c.items(),key = lambda i: i[0])

outer:
[(1, 1), (2, 3), (3, 2), (4, 1), (6, 1), (7, 1), (9, 1), (19, 1)]
if you want to get the dictionary format,just

dict(sorted(c.items(),key = lambda i: i[0]))

Answered By: Calab

sorting a counter in python by keys

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