[프로그래머스/파이썬] [1차] 뉴스 클러스터링

 

 

나의 풀이

import re
from collections import defaultdict

def solution(str1, str2):
    pool1, pool2 = defaultdict(int), defaultdict(int)
    union, intersection = defaultdict(int), defaultdict(int)

    str1 = re.sub(r'[^a-z]', ' ', str1.lower())
    str2 = re.sub(r'[^a-z]', ' ', str2.lower())

    for i in range(1, len(str1)):
        if str1[i - 1] != ' ' and str1[i] != ' ':
            pool1[str1[i - 1: i + 1]] += 1

    for i in range(1, len(str2)):
        if str2[i - 1] != ' ' and str2[i] != ' ':
            pool2[(str2[i - 1: i + 1])] += 1

    for p in pool1:
        if p in pool2:
            union[p] = max(pool1[p], pool2[p])
        else:
            union[p] = pool1[p]

    for p in pool2:
        if p in pool1:
            union[p] = max(pool1[p], pool2[p])
        else:
            union[p] = pool2[p]

    for p in pool1:
        if p in pool2:
            intersection[p] = min(pool1[p], pool2[p])

    if union:
        return int(sum(intersection.values()) / sum(union.values()) * 65536)
    else:
        return 65536

 

 

집합을 이용한 풀이

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)

 

 

isaplha를 이용한 풀이

def solution(str1, str2):

    list1 = [str1[n:n+2].lower() for n in range(len(str1)-1) if str1[n:n+2].isalpha()]
    list2 = [str2[n:n+2].lower() for n in range(len(str2)-1) if str2[n:n+2].isalpha()]

    tlist = set(list1) | set(list2)
    res1 = [] #합집합
    res2 = [] #교집합

    if tlist:
        for i in tlist:
            res1.extend([i]*max(list1.count(i), list2.count(i)))
            res2.extend([i]*min(list1.count(i), list2.count(i)))

        answer = int(len(res2)/len(res1)*65536)
        return answer

    else:
        return 65536