위클리 챌린지 4주차 직업군 추천하기

https://programmers.co.kr/learn/courses/30/lessons/84325

코딩테스트 연습 - 4주차

파이썬

def solution(table, languages, preference):
    jobs = {}
    for row in table:
        each = row.split()
        jobs[each[0]] = {each[1]: 5, each[2]: 4, each[3]: 3, each[4]: 2, each[5]: 1}
    result = {}
    for each in sorted(list(jobs)):
        result[each] = 0
        for lang, pref in zip(languages, preference):
            if lang in jobs[each]:
                result[each] += pref * jobs[each][lang]
    return max(result.items(), key=lambda x: x[1])[0]

마지막 리턴문을 풀어서 보면 다음과 같다. 

# result
{'CONTENTS': 36, 'GAME': 25, 'HARDWARE': 41, 'PORTAL': 21, 'SI': 29}

# result.items()
dict_items([('CONTENTS', 36), ('GAME', 25), ('HARDWARE', 41), ('PORTAL', 21), ('SI', 29)])

# max(result.items(), key=lambda x: x[1])[0]
HARDWARE

딕셔너리 컴프리헨션을 이용해 보자.
12줄을 4줄로 줄일 수 있다.  

def solution(table, languages, preference):
    jobs = {each[0]: {each[1]: 5, each[2]: 4, each[3]: 3, each[4]: 2, each[5]: 1} for each in
            map(lambda x: x.split(), table)}
    result = {each: sum(pref * jobs[each][lang] for lang, pref in zip(languages, preference) if lang in jobs[each])
              for each in sorted(list(jobs))}
    return max(result.items(), key=lambda x: x[1])[0]

물론 더 줄일 수도 있다. 

def solution(table, languages, preference):
    return max({each[0]: sum(pref * each[1][lang] for lang, pref in zip(languages, preference) if lang in each[1])
                for each in sorted(list({each[0]: {each[1]: 5, each[2]: 4, each[3]: 3, each[4]: 2, each[5]: 1}
for each in map(lambda x: x.split(), table)}.items()))}.items(), key=lambda x: x[1])[0]

 

golang

import "strings"

func solution(table []string, languages []string, preference []int) string {
	tableMap := map[string]map[string]int{}
	for _, v := range table {
		temp := strings.Split(v, " ")
		tableMap[temp[0]] = map[string]int{temp[1]: 5, temp[2]: 4, temp[3]: 3, temp[4]: 2, temp[5]: 1}
	}
	result := map[string]int{}
	maxKey, maxVal := "", 0
	for k, v := range tableMap {
		for i, lang := range languages {
			result[k] += v[lang] * preference[i]
		}
		if maxVal < result[k] {
			maxVal = result[k]
			maxKey = k
		} else if maxVal == result[k] && maxKey > k {
			maxKey = k
		}
	}
	return maxKey
}