Question

我试图通过survey_id返回访问者尚未投票的调查。使用下面的示例数组，我希望返回survey_id：4。

实现这一目标的最佳方法是什么？

以下是我的示例数组：

＆＃13;

// 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
}]

＆＃13;

Answer 1

~~接受的答案有O（n ^ 2）时间复杂度。~~如果性能很重要，应该避免相互嵌套的线性操作 - 如果有利于代码的简洁性，@ SD＆＃39;答案更清晰。这是一个带有vanilla JavaScript和线性时间O（n）复杂度的版本。

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

您可以使用_.filter和_.find的组合将过滤功能应用于您的调查集合，该调查集合会在投票收集中找到匹配survey_id时筛选出调查。请参阅此示例getUnansweredSurveys函数：

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

使用lodash返回唯一值数组，同时删除非唯一值

2 个答案: