Prepare > Python > Collections > collections.Counter()

collections.Counter() | HackerRank

 

문제

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Task

Mr.R is a shoe shop owner. His shop has S number of shoes.
He has a list containing the size of each shoe he has in his shop.
There are N number of customers who are willing to pay x(i) amount of money only if they get the shoe of their desired size.

Your task is to compute how much money Mr.R earned.

 

Input Format

The first line contains X, the number of shoes.
The second line contains the space separated list of all the shoe sizes in the shop.
The third line contains N, the number of customers.
The next N lines contain the space separated values of the shoe size desired by the customer and x(i), the price of the shoe.

 

Output Format

Print the amount of money earned by Mr.R.

 

=> Mr.R이 가진 신발 수, 신발 사이즈들, 고객들의 희망 사이즈와 지불용의 가격들을 줄거니까 Mr.R의 매출액을 구하라.

 

 

 

 

코드

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from collections import Counter


X = int(input())
x = Counter(list(map(int, input().split())))
N = int(input())


money = 0
for _ in range(N):
    a, b = map(int, input().split())
    
    if x[a] != 0:
       money += b
       x[a] = x[a]-1

    
print(money)

 

 

 

 

노트

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collections.Counter()

from collections import Counter

myList = [1,1,2,3,4,5,3,2,3,4,2,1,2,3]
print(Counter(myList))
# Counter({2: 4, 3: 4, 1: 3, 4: 2, 5: 1})

print(Counter(myList).items()) # items()로 key,value 쌍 튜플로 확인
# [(1, 3), (2, 4), (3, 4), (4, 2), (5, 1)]

print(Counter(myList).keys()) # keys()로 key값들 확인
# [1, 2, 3, 4, 5]

print(Counter(myList).values())# values()로 value값들 확인
# [3, 4, 4, 2, 1]

• 요소가 key, 각 요소의 count()가 value인 dictionary 만듦

 

 

dict에 관하여

• 개별 value 수정

dict = {one:1, three:3}
dict[one] = 3

print(dict)
# {one:3, three: 3}

 

• key, value 쌍 얻기

dict = {one:1, three:3}
print(dict.items())


# (one, 3), (three: 3)

 

• key로 value찾기

dict = {one:1, three:3} 

print(dict[one]) # one이 dict의 key값에 없으면 에러
# 1

print(dict.get(one)) # one이 dict의 key값에 없으면 None 리턴
# 1

 

• key, value 삭제

dict_1 = {'one':1, 'three':3} 

del(dict_1['one']) # key값이 'one'인 쌍 삭제
print(dict_1)
# {'three': 3}


dict_1.clear() # 그냥 사전 비우기. 다 사라짐
print(dict_1)
# {}

 

 

 

두 번째 시도

from collections import Counter 

X = int(input())
shoes = Counter(map(int, input().split()))
N = int(input())

sales = 0
for i in range(N):
    size, price = map(int, input().split())
    
    if shoes[size]:
        sales += price
        shoes[size] -= 1

print(sales)

• if 조건을 직접적인 불린값으로 쓰지 않음!

 

 

 

참조

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02-5 딕셔너리 자료형

[Python] Dictionary 값 수정, 추가, 삭제

HackerRank collections.Counter() solution in Python