Question

我有一个对象数组

[
    {"id":"brand","label":"Brand","value":"value1"},    
    {"id":"waterproof","label":"Waterproof","value":"value1"},
    {"id":"diameter","label":"Diameter","value":""},
    {"id":"brand","label":"Brand","value":"value2"},
    {"id":"waterproof","label":"Waterproof","value":"value2"},
    {"id":"diameter","label":"Diameter","value":""}
]

我需要将其重组为以下结构，

[
    ["Brand", "value1", "value2"], 
    ["Waterproof", "value1","value2"], 
    ["Diameter","",""]
]

关于如何使用reduce方法做到这一点的任何想法。

最佳

Answer 1

必须取Map并收集所有值。

var json = '[{"id":"brand","label":"Brand","value":"value1"},{"id":"waterproof","label":"Waterproof","value":"value1"},{"id":"diameter","label":"Diameter","value":""},{"id":"brand","label":"Brand","value":"value2"},{"id":"waterproof","label":"Waterproof","value":"value2"},{"id":"diameter","label":"Diameter","value":""}]',
    result = Array.from(JSON
        .parse(json)
        .reduce((m, { id, label, value }) => m.set(id, (m.get(id) || [label]).concat(value)), new Map)
        .values()
    );
    
console.log(result);

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Answer 2

请尝试使用Array.reduce和Object.values

let arr = [{"id":"brand","label":"Brand","value":"value1"},{"id":"waterproof","label":"Waterproof","value":"value1"},{"id":"diameter","label":"Diameter","value":""},{"id":"brand","label":"Brand","value":"value2"},{"id":"waterproof","label":"Waterproof","value":"value2"},{"id":"diameter","label":"Diameter","value":""}];

let result = Object.values(arr.reduce((a,c) => {
  if(a[c.id]) a[c.id].push(c.value);
  else a[c.id] = [c.label, c.value];
  return a;
}, {}));
console.log(result);

Answer 3

不是最好的代码，但是我认为这可能对您有用：

let data = [
        {"id":"brand","label":"Brand","value":"value1"},    
        {"id":"waterproof","label":"Waterproof","value":"value1"},
        {"id":"diameter","label":"Diameter","value":""},
        {"id":"brand","label":"Brand","value":"value2"},
        {"id":"waterproof","label":"Waterproof","value":"value2"},
        {"id":"diameter","label":"Diameter","value":""}
    ]

    let reducedData = {}
    data.forEach((row)=>{
        reducedData[row.id] ? reducedData[row.id].push(row.value) : reducedData[row.id] = [row.label, row.value]    
    })
    let newData = Object.keys(reducedData).map((id)=>{return [...reducedData[id]]})

Answer 4

您还可以使用Array.prototype.reduce和Object.keys的简单组合：

const data = [{"id":"brand","label":"Brand","value":"value1"},{"id":"waterproof","label":"Waterproof","value":"value1"},{"id":"diameter","label":"Diameter","value":""},{"id":"brand","label":"Brand","value":"value2"},{"id":"waterproof","label":"Waterproof","value":"value2"},{"id":"diameter","label":"Diameter","value":""}];

const grouped = data.reduce((o, { label, value }) =>
    ({...o, [label]: [...(o[label] || []), value]}), {});

const result = Object.keys(grouped).map(label => [label, ...grouped[label]]);

console.log(result);

Answer 5

在这里，我使用Set来创建profile的唯一列表，然后我id来创建基于这些map的自定义数组。这是通过使用find来获取标签，并使用filter来获取与id相关的所有项目来完成的。我再次在过滤器上映射，以仅返回对象的id。

value

Answer 6

如果JSON数组是某个API中很长的数组，则

var data = [
    {"id":"brand","label":"Brand","value":"value1"},    
    {"id":"waterproof","label":"Waterproof","value":"value1"},
    ...... //some JSON array form some API
];
var json = JSON.stringify(data);

result = Array.from(JSON
    .parse(json)
    .reduce((m, { id, label, value }) => m.set(id, (m.get(id) || [label]).concat(value)), new Map)
    .values()
);
console.log(result);

使用javascript将数组简化为特定结构

6 个答案: