[파이썬] Counter

2021 KAKAO BLIND RECRUITMENT 메뉴 리뉴얼 문제를 푸는데, 다른 사람의 풀이를 보니까 파이썬 모듈을 적절하게 잘 써서 코드를 굉장히 깔끔하게 짠 것을 보고 자극을 받았다.

특히 Counter 모듈을 사용하였는데, most_common() 함수를 사용했다. 알아두면 유용할 것 같아 정리하려고 한다.

Counter

hash 객체를 세기 위한 dictionary 보조 클래스이다. 요소는 dictionary key로 저장되고 개수가 dictionary value로 저장된다.

from collections import Counter

# a new, empty counter
a = Counter()
# a new counter from an iterable
b = Counter('gallahad')

print(b['a']) # 3
print(b['l']) # 2

# a new counter from a mapping
c = Counter({'red': 4, 'blue': 2})

print(c['red']) # 4

# a new counter from keyword args
d = Counter(cats=4, dogs=8)

print(d['cats']) # 4

# a new counter from list objects
e = Counter(['eggs', 'ham'])

print(e['eggs']) # 1
print(e['bacon']) # 0

Counter methods

elements()

c = Counter(a=4, b=2, c=0, d=-2)
print(sorted(c.elements()))
# ['a', 'a', 'a', 'a', 'b', 'b']

most_common([n])

가장 많은 요소 n개와 요소의 개수를 반환. n은 생략가능

c = Counter('abracadabra').most_common(3)
[('a', 5), ('b', 2), ('r', 2)]

subtract([iterable-or-mapping])

c = Counter(a=4, b=2, c=0, d=-2)
d = Counter(a=1, b=2, c=3, d=4)
c.subtract(d)
print(c)
# Counter({'a': 3, 'b': 0, 'c': -3, 'd': -6})

Common patterns for working with Counter objects

# total of all counts
sum(c.values())

# reset all counts
c.clear()

# list unique elements
list(c)

# convert to a set
set(c)

# convert to a regular dictionary
dict(c)

# convert to a list of (elem, cnt) pairs
c.items()

# convert from a list of (elem, cnt) pairs
Counter(dict(list_of_pairs))

# n least common elements
c.most_common()[:-n-1:-1]

# remove zero and negative counts
+c

# 실제 적용

b = Counter('gallahad')

print(set(b))
# {'g', 'h', 'l', 'a', 'd'}

print(dict(b))
# {'g': 1, 'a': 3, 'l': 2, 'h': 1, 'd': 1}

print(b.items())
# dict_items([('g', 1), ('a', 3), ('l', 2), ('h', 1), ('d', 1)])

print(b.most_common())
# [('a', 3), ('l', 2), ('g', 1), ('h', 1), ('d', 1)]

Common patterns for working with Counter objects(mathematical operations)

c = Counter(a=3, b=1)
d = Counter(a=1, b=2, c=1)

print(c+d)
# Counter({'a': 4, 'b': 3, 'c': 1})

print(c-d)
# Counter({'a': 2, 'b': -1, 'c': -1})

print(c&d) # Intersection: min(c[x], d[x])
# Counter({'a': 1, 'b': 1})

print(c|d) # union: max(c[x], d[x])
# Counter({'a': 3, 'b': 2})

# Unary addition and subtraction
e = Counter(a=2, b=-4)

print(+e)
# Counter({'a': 2})

print(-e)
# Counter({'b': 4})