计算对象数组上的所有值-javascript

Question

我需要计算对象数组上的每个值，所需的输出应如下所示

[{
    "question": "question1",
    "USA": 2
  }, {
    "question": "question1",
    "AUS": 1
  },

  {
    "question": "question2",
    "item1": 2
  },

  {
    "question": "question2",
    "item1,item2": 1
  }, {
    "question": "question4",
    "3": 1
  }, {
    "question": "question4",
    "2": 1
  }
]

以下是我需要转换为以上输出的输入。 我不知道如何处理无问题，并且当一个问题有2个答案时也有问题。 样本输入

[{"question1":"USA","question2":["item1"],"question4":2}, 
{"question1":"USA","question2":["item1"],"question4":3}, 
{"question1":"AUS","question2":["item1","item2"]}];

 let arr=[{"question1":"USA","question2":["item1"],"question4":2},{"question1":"USA","question2":["item1"],"question4":3},{"question1":"AUS","question2":["item1","item2"]}]; //console.log(arr); function solve(list){ var map = new Map(); var entry = null; for(var item of list){ if(!map.has(item.question1)) map.set(item.question1, {question:'question1'}); entry = map.get(item.question1); if(entry.hasOwnProperty(item.question1)) entry[item.question1] = entry[item.question1] + 1; else entry[item.question1] = 1; if(!map.has(item.question2)) map.set(item.question2, {question: 'question2'}); entry = map.get(item.question2); if(entry.hasOwnProperty(item.question2)) entry[item.question2] = entry[item.question2] + 1; else entry[item.question2] = 1; } return Array.from(map.values()); } console.log(solve(arr)) 

Answer 1

您可以采用对象或任何喜欢的数据结构（以嵌套样式支持键/值结构），然后首先收集所有项，然后重新整理收集的树。

这种方法使用对象，因为键是字符串，这对于作为键的数组很重要。 对于此用例，这用逗号隔开就足够了。

 var data = [{ question1: "USA", question2: ["item1"], question4: 2 }, { question1: "USA", question2: ["item1"], question4: 3 }, { question1: "AUS", question2: ["item1", "item2"] }], hash = data.reduce((hash, o) => { Object.entries(o).forEach(([question, value]) => { var sub = hash[question] = hash[question] || Object.create(null); sub[value] = sub[value] || { question, [value]: 0 }; sub[value][value]++; }); return hash; }, Object.create(null)), result = Object.values(hash).reduce((r, sub) => [...r, ...Object.values(sub)], []); console.log(result); 

 .as-console-wrapper { max-height: 100% !important; top: 0; } 

Answer 2

首先，通过使用reduce获得国家。 然后，对其余部分使用一些嵌套的forEach循环：

 const input = [{"question1":"USA","question2":["item1"],"question4":2}, {"question1":"USA","question2":["item1"],"question4":3}, {"question1":"AUS","question2":["item1","item2"]}]; const countriesOutput = input.reduce((acc, curr) => { if (!acc.some(e => e[curr.question1])) { acc.push({ question: "question1", [curr.question1]: 1 }); } else { acc.find(e => e[curr.question1])[curr.question1]++; } return acc; }, []); let questionsOutput = []; input.forEach(item => { Object.keys(item).forEach(key => { if (key != "question1") { if (Array.isArray(item[key])) { questionsOutput.push({ question: key, [item[key].join(",")]: 1 }); } else { questionsOutput.push({ question: key, [item[key]]: 1 }); } } }); }); const finalOutput = [...countriesOutput, ...questionsOutput]; console.log(finalOutput); 

 .as-console-wrapper { max-height: 100% !important; top: auto; } 

Answer 3

这是使用字典（例如Object）汇总输入并跟踪重复项的问题。 名称/值对的“名称”可以通过将问题和答案与某些定界符组合来唯一标识。

 const input = [{ "question1": "USA", "question2": ["item1"], "question4": 2 }, { "question1": "USA", "question2": ["item1"], "question4": 3 }, { "question1": "AUS", "question2": ["item1", "item2"] } ]; //Sum the input to an array which we can easily search for duplciates var repeatCounter = {}; input.forEach(objItem => { Object.keys(objItem).forEach(propItem => { //Get the counter and the string var s = `${propItem}-${objItem[propItem]}`; var c = repeatCounter[s] || 0; //Modify it or introduce it if absent repeatCounter[s] = c + 1; }) }) var output = Object.keys(repeatCounter).map(element => { var ret = {'question': element.split('-')[0]} ret[element.split('-')[1]] = repeatCounter[element]; return ret; }) console.log(output); 

 .as-console-wrapper { max-height: 100% !important; } 

需要进行一些细微的调整，例如加强定界符，将多个字符串转换为数组项（如问题所示），这需要从实际出发。