使用lodash返回一個唯一值數組，同時刪除非唯一的

Question

我試圖通過survey_id返回訪問者尚未投票的調查。使用下面的示例數組，我期望survey_id：4被返回。

完成此操作的最佳方法是什麼？

這裏是我的示例陣列：

// Survey Object 
 
[{ 
 
    survey_id: 1, 
 
    name: 'Survey 1", 
 
    question: "This is a survey question" 
 
}, { 
 
    survey_id: 2, 
 
    name: 'Survey 2", 
 
    question: "This is a another survey question" 
 
}, { 
 
    survey_id: 3, 
 
    name: 'Survey 3", 
 
    question: "This is a another 3 survey question" 
 
}, { 
 
    survey_id: 4, 
 
    name: 'Survey 4", 
 
    question: "This is a another 4 survey question" 
 
}] 
 

 
// Votes 
 
[{ 
 
    vote_id: 11, 
 
    visitor_token: "aDe4GggAe3", 
 
    survey_id: 1 
 
}, { 
 
    vote_id: 12, 
 
    visitor_token: "aDe4GggAe3", 
 
    survey_id: 2 
 
}, { 
 
    vote_id: 13, 
 
    visitor_token: "aDe4GggAe3", 
 
    survey_id: 3 
 
}]

Answer 1

接受的答案具有爲O（n^2）時間複雜度。 如果性能很重要，應避免嵌套在一起的線性操作 - 如果贊成簡單的代碼，@ S.D.的答案更清晰。這是一個帶有vanilla JavaScript和線性時間O（n）複雜性的版本。

function surveysWithNoVotes(surveyList, voteList) { 
    var surveyIds = surveyList.map(function(survey) { 
    return survey.survey_id 
    }) 
    var votedForIdsMap = voteList.reduce(function(memo, vote) { 
    memo[vote.survey_id] = vote.survey_id 
    }, {}) 
    return surveyIds.filter(function(id) { 
    return !votedForIdsMap[id] 
    }) 
} 

Answer 2

你可以使用的_.filter和_.find的組合來過濾功能應用到您的收藏調查顯示，當匹配survey_id在票收集發現過濾掉的調查。見這個樣本getUnansweredSurveys功能：

function getUnansweredSurveys(surveys, votes) { 
    return _.filter(surveys, function(survey) { 
    return !_.find(votes, {survey_id: survey.survey_id}) 
    }) 
}