Question

有一系列类似的对象

var a =[
{type: 't1', value: 'x1'},
{type: 't1', value: 'x2'},
{type: 't1', value: 'x3'},
{type: 't1', value: 'x4'},
{type: 't2', value: 'x5'},
{type: 't2', value: 'x1'},
{type: 't2', value: 'x2'},
{type: 't2', value: 'x3'},
{type: 't2', value: 'x4'},
{type: 't2', value: 'x5'}]

如何切片/拼接/过滤此数组以使其返回t1类型的前3个项目和t2类型的前4个项目？

Answer 1

您可以先使用filter()，然后再使用slice()。将0和项目数作为参数传递给slice()

var a =[
{type: 't1', value: 'x1'},
{type: 't1', value: 'x2'},
{type: 't1', value: 'x3'},
{type: 't1', value: 'x4'},
{type: 't2', value: 'x5'},
{type: 't2', value: 'x1'},
{type: 't2', value: 'x2'},
{type: 't2', value: 'x3'},
{type: 't2', value: 'x4'},
{type: 't2', value: 'x5'}]

function items(type,num){
  return a.filter(x => x.type === type).slice(0,num);
}

console.log(items('t1',3));
console.log(items('t2',4));

Answer 2

使用filter，然后使用slice：

const a = [
  {type: 't1', value: 'x1'},
  {type: 't1', value: 'x2'},
  {type: 't1', value: 'x3'},
  {type: 't1', value: 'x4'},
  {type: 't2', value: 'x5'},
  {type: 't2', value: 'x1'},
  {type: 't2', value: 'x2'},
  {type: 't2', value: 'x3'},
  {type: 't2', value: 'x4'},
  {type: 't2', value: 'x5'}
];

const t1 = a.filter(x => x.type === 't1').slice(0, 3);
const t2 = a.filter(x => x.type === 't2').slice(0, 4);

console.log(t1);
console.log(t2);

或使用reduce：

const a = [
  {type: 't1', value: 'x1'},
  {type: 't1', value: 'x2'},
  {type: 't1', value: 'x3'},
  {type: 't1', value: 'x4'},
  {type: 't2', value: 'x5'},
  {type: 't2', value: 'x1'},
  {type: 't2', value: 'x2'},
  {type: 't2', value: 'x3'},
  {type: 't2', value: 'x4'},
  {type: 't2', value: 'x5'}
];

const getByType = (arr, prop, num) =>
  arr.reduce(([out, i], x) => (match = x.type === prop, [
    match && i < num ? [...out, x] : out,
    match ? i + 1 : i
  ]),
  [[], 0])[0];

console.log(getByType(a, 't1', 3));
console.log(getByType(a, 't2', 4));

Answer 3

这很粗糙。但是有效。

var a =[
{type: 't1', value: 'x1'},
{type: 't1', value: 'x2'},
{type: 't1', value: 'x3'},
{type: 't1', value: 'x4'},
{type: 't2', value: 'x5'},
{type: 't2', value: 'x1'},
{type: 't2', value: 'x2'},
{type: 't2', value: 'x3'},
{type: 't2', value: 'x4'},
{type: 't2', value: 'x5'}];

let b = a.filter(x => x.type === 't1').slice(0,3).concat(a.filter(x => x.type === 't2').slice(0,4));

console.log(b);

Answer 4

您可以使用创建一个对象的reduce方法进行自定义功能，然后可以返回Object.values。

var a = [{ type: 't1', value: 'x1' },{ type: 't1', value: 'x2' },{ type: 't1', value: 'x3' },{ type: 't1', value: 'x4' },{ type: 't2', value: 'x5' },{ type: 't2', value: 'x1' },{ type: 't2', value: 'x2' },{ type: 't2', value: 'x3' },{ type: 't2', value: 'x4' },{ type: 't2', value: 'x5' }];

function sliceArr(arr, slice) {
  const obj = arr.reduce((r, { type, ...rest }) => {
    if(slice[type]) {
      if(!r[type]) r[type] = [{ type, ...rest }];
      else if(r[type].length < slice[type]) r[type].push({ type, ...rest })
    }
    return r;
  }, {})

  return [].concat(...Object.values(obj))
}

console.log(sliceArr(a, { t1: 3, t2: 4 }))

通过最大长度拼接对象数组

4 个答案: