[PG] 메뉴 리뉴얼

문제

me

from itertools import combinations
from collections import defaultdict
def solution(orders, course):
    answer = []
    for o in range(len(orders)):
        orders[o]=''.join(list(sorted(orders[o])))
    for c in course:
        c_set=defaultdict(int)
        for order in orders:
            for cb in combinations(order,c):
                c_set[''.join(cb)]+=1
        mx_cnt = 2
        mx_combs = []
        for key,cnt in c_set.items():
            if mx_cnt < cnt:
                mx_cnt =cnt
                mx_combs=[key]
            elif mx_cnt==cnt:
                mx_combs.append(key)
        answer.extend(mx_combs)
    return sorted(answer)

others

import collections
import itertools

def solution(orders, course):
    result = []

    for course_size in course:
        order_combinations = []
        for order in orders:
            order_combinations += itertools.combinations(sorted(order), course_size)

        most_ordered = collections.Counter(order_combinations).most_common()
        result += [ k for k, v in most_ordered if v > 1 and v == most_ordered[0][1] ]

    return [ ''.join(v) for v in sorted(result) ]

combinations.Counter
dict으로 숫자 세는 거 대신 할 수 있음

def countLetters(word):
    counter = {}
    for letter in word:
        if letter not in counter:
            counter[letter] = 0
        counter[letter] += 1
    return counter
countLetters('hello world'))
# {'h': 1, 'e': 1, 'l': 3, 'o': 2, ' ': 1, 'w': 1, 'r': 1, 'd': 1}

위 코드를 아래와 같이 만들 수 있다.

from collections import Counter
Counter('hello world') # Counter({'l': 3, 'o': 2, 'h': 1, 'e': 1, ' ': 1, 'w': 1, 'r': 1, 'd': 1})

데이터의 개수가 많은 순으로 정렬된 배열을 리턴하는 most_common이라는 메서드를 제공하고 있습니다.

from collections import Counter
Counter('hello world').most_common() # [('l', 3), ('o', 2), ('h', 1), ('e', 1), (' ', 1), ('w', 1), ('r', 1), ('d', 1)]

.most_common(k)하면 k번째까지 반환해준다.