我将如何按其中一个子值对该表进行排序，以及如何从中获得最佳结果？ [复制]

Posted 2023-03-14

【中文标题】我将如何按其中一个子值对该表进行排序，以及如何从中获得最佳结果？ [复制]

【英文标题】：How would I sort this table by one of the child values and how would I get the top result from it? [duplicate]

【发布时间】：2021-09-14 18:03:04

【问题描述】：

我目前正在制作一个 discord.js 机器人，它可以帮助使用透视 API 进行自动审核。在对透视图的 API 进行 API 调用后，我对每个“属性”进行排序。问题是我只剩下这种事情了：

但是，我不确定如何按summaryScore.value 对其进行排序，然后从中获得最佳结果。任何帮助表示赞赏！这是我当前的代码：

try 
  google.discoverAPI(DISCOVERY_URL)
    .then(client => 
      const analyzeRequest = 
        comment: 
          text: message.content,
        ,
        requestedAttributes: 
          SEVERE_TOXICITY: 
            scoreThreshold: 0.85,
          ,
          IDENTITY_ATTACK: 
            scoreThreshold: 0.85,
          ,
          THREAT: 
            scoreThreshold: 0.85,
          ,
          INSULT: 
            scoreThreshold: 0.85,
          ,
          PROFANITY: 
            scoreThreshold: 0.85,
          ,
          FLIRTATION: 
            scoreThreshold: 0.85,
          ,
          SEXUALLY_EXPLICIT: 
            scoreThreshold: 0.85,
          ,
          INCOHERENT: 
            scoreThreshold: 0.85,
          ,
          INFLAMMATORY: 
            scoreThreshold: 0.85,
          ,
          SPAM: 
            scoreThreshold: 0.85,
          ,
        ,
        languages: ['en'],
      ;
      client.comments.analyze(
          key: API_KEY,
          resource: analyzeRequest,
        ,
        (err, response) => 
          if (err) console.log(err);
          else 
            complete = false;
            if (response.data.attributeScores) 
              var array = [response.data.attributeScores]
              for (const attributeScore of array) 
               console.log(attributeScore)
                if (attributeScore.name.summaryScore.value > 0.85) 
                  message.reply(`your message has been removed for: "$attributeScore.name".`)
                  message.delete()
                
              
            
          
        );
    )
    .catch(err => 
      console.log(err);
    );

 catch (err) 

非常感谢！任何帮助表示赞赏！我是 javascript 的新手，所以如果这是一个简单的答案，我深表歉意。

【参考方案1】：

您可以将对象转换为数组并按summaryScore.value排序

const obj = 
  SEVERE_TOXICITY: 
    spanScores: [],
    summaryScore:  value: 0.9197861, type: 'PROBABILITY' 
  ,
  PROFANITY: 
    spanScores: [],
    summaryScore:  value: 0.9844065, type: 'PROBABILITY' 
  ,
  INSULT: 
    spanScores: [],
    summaryScore:  value: 0.9806645, type: 'PROBABILITY' 
  ,


const array = Object.entries(obj)
  .map(([key, value]) => (id: key, ...value))

array.sort(
  (a, b) => b.summaryScore.value - a.summaryScore.value
)
console.log(`The top one is $array[0].id with a value of $array[0].summaryScore.value`)

或者，如果您只需要具有最高值的对象，您可以遍历对象条目并检查最高值：

const obj = 
  SEVERE_TOXICITY: 
    spanScores: [],
    summaryScore:  value: 0.9197861, type: 'PROBABILITY' 
  ,
  PROFANITY: 
    spanScores: [],
    summaryScore:  value: 0.9844065, type: 'PROBABILITY' 
  ,
  INSULT: 
    spanScores: [],
    summaryScore:  value: 0.9806645, type: 'PROBABILITY' 
  ,


let highest = null
for (let [id, value] of Object.entries(obj)) 
  if (!highest || value.summaryScore.value > highest.summaryScore.value)
    highest =  id, ...value 


console.log(highest)