[🤷‍♀️ 프로그래머스 Python] 뉴스 클러스터링

📌문제

2018 KAKAO BLIND RECRUITMENT

[1차] 뉴스 클러스터링

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programmers.co.kr/learn/courses/30/lessons/17677

코딩테스트 연습 - [1차] 뉴스 클러스터링

 

 

 

📌

내 풀이

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런타임에러로 장렬하게 실패!!!!😭😭😭😭

# 런타임 에러 - 53점
def solution(str1, str2):
    list1 = [] 
    list2 = []
    
    # 전처리(문자2개씩) list 만들기
    for i in range(len(str1)-1):
        if str1[i:i+2].isalpha():
            list1.add(str1[i:i+2].lower())

    for i in range(len(str2)-1):
        if str2[i:i+2].isalpha():
            list2.add(str1[i:i+2].lower())
    
    # 만약 두개 다 0이면 return 
    if len(list1) == 0 and len(list2) == 0:
        return 65536

    # Count 하기 
    dict1 = {}
    dict2 = {}
    
    for i in list1:
        if i not in dict1:
            dict1[i] = 1
        else:
            dict1[i] += 1
    
    for i in list2:
        if i not in dict2:
            dict2[i] = 1
        else:
            dict2[i] += 1
    
    # 최종 dict1
    for k, v in dict2.items():
        if k not in dict1:
            dict1[k] = v
        else:
            a = dict1[k]
            dict1[k] = [a, v]
    
    hap = []
    kyo = []
    
    #합집합 교집합 만들기
    for k, v in dict1.items():
        if v == 1:
            hap.append(k)
        else:
            for _ in range(min(v)):
                kyo.append(k) 
            for _ in range(max(v)):
                hap.append(k)

    answer = str(len(kyo)/len(hap) * 65536)
    return int(answer[:5])

그저 💩

 

 

 

 

📌

해답

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AND와 OR로 기깔나게 해결,,,,

from collections import Counter as counter
import re
def solution(str1, str2):
    list1 = [
        str1[i:i+2].lower()
        for i in range(len(str1)-1)
        if re.findall('[a-z]{2}', str1[i:i+2].lower())
    ]

    list2 = [
        str2[i:i+2].lower()
        for i in range(len(str2)-1)
        if re.findall('[a-z]{2}', str2[i:i+2].lower())
    ]

            
    # 교집합 계산
    intersection = sum((counter(list1) & counter(list2)).values())
    
    # 합집합 계산
    union = sum((counter(list1) | counter(list2)).values())
    
    # 자카드 유사도
    jaccard_sim = 1 if union == 0 else intersection / union
    return int(jaccard_sim * 65536)

import re
import math

def solution(str1, str2):
    str1 = [str1[i:i+2].lower() for i in range(0, len(str1)-1) if not re.findall('[^a-zA-Z]+', str1[i:i+2])]
    str2 = [str2[i:i+2].lower() for i in range(0, len(str2)-1) if not re.findall('[^a-zA-Z]+', str2[i:i+2])]

    gyo = set(str1) & set(str2)
    hap = set(str1) | set(str2)

    if len(hap) == 0 :
        return 65536

    gyo_sum = sum([min(str1.count(gg), str2.count(gg)) for gg in gyo])
    hap_sum = sum([max(str1.count(hh), str2.count(hh)) for hh in hap])

    return math.floor((gyo_sum/hap_sum)*65536)