Question

[ 
    [ 'TWENTY', 20 ],
    [ 'TWENTY', 20 ],
    [ 'TWENTY', 20 ],
    [ 'TWENTY', 20 ],
    [ 'TEN', 10 ],
    [ 'FIVE', 5 ],
    [ 'ONE', 1 ],
    [ 'QUARTER', 0.25 ],
    [ 'QUARTER', 0.25 ],
    [ 'DIME', 0.1 ],
    [ 'DIME', 0.1 ],
    [ 'PENNY', 0.01 ],
    [ 'PENNY', 0.01 ],
    [ 'PENNY', 0.01 ] 
]

我的数组由相同的数组组成。如何合并相同的内容？

Answer 1

您可以使用reduce对输入数组中唯一索引0的总数求和：

const data = [ [ 'TWENTY', 20 ],
  [ 'TWENTY', 20 ],
  [ 'TWENTY', 20 ],
  [ 'TWENTY', 20 ],
  [ 'TEN', 10 ],
  [ 'FIVE', 5 ],
  [ 'ONE', 1 ],
  [ 'QUARTER', 0.25 ],
  [ 'QUARTER', 0.25 ],
  [ 'DIME', 0.1 ],
  [ 'DIME', 0.1 ],
  [ 'PENNY', 0.01 ],
  [ 'PENNY', 0.01 ],
  [ 'PENNY', 0.01 ] ]
;

const result = Object.entries(data.reduce((a, e) => {
  a[e[0]] = e[1] + (a[e[0]] || 0);
  return a;
}, {}));

console.log(result);

Answer 2

你去了

var a = [
  ["TWENTY", 20],
  ["TWENTY", 20],
  ["TWENTY", 20],
  ["TWENTY", 20],
  ["TEN", 10],
  ["FIVE", 5],
  ["ONE", 1],
  ["QUARTER", 0.25],
  ["QUARTER", 0.25],
  ["DIME", 0.1],
  ["DIME", 0.1],
  ["PENNY", 0.01],
  ["PENNY", 0.01],
  ["PENNY", 0.01]
];
var b = a.reduce((acc, curr) => {
  if (acc[curr[0]]) acc[curr[0]] += curr[1];
  else acc[curr[0]] = curr[1];
  return acc;
}, {});

console.log(b);

Answer 3

数组缩减是您希望用于总计的金额，同时将它们保留为键/值数组，尽管我个人可能将它们映射为{ key: value }，然后缩减以获取更简单的最终结果。

const group = [ 
    [ 'TWENTY', 20 ],
    [ 'TWENTY', 20 ],
    [ 'TWENTY', 20 ],
    [ 'TWENTY', 20 ],
    [ 'TEN', 10 ],
    [ 'FIVE', 5 ],
    [ 'ONE', 1 ],
    [ 'QUARTER', 0.25 ],
    [ 'QUARTER', 0.25 ],
    [ 'DIME', 0.1 ],
    [ 'DIME', 0.1 ],
    [ 'PENNY', 0.01 ],
    [ 'PENNY', 0.01 ],
    [ 'PENNY', 0.01 ] 
];

const totals = group.reduce((accumulator, [key, value]) => {
  const index = accumulator.findIndex(([k,v]) => k === key);
  if (index === -1) {
    return [ ...accumulator, [key, value] ];
  }
  const newTotal = [ key, ( accumulator[index][1] + value ) ];
  return [ ...accumulator.slice(0,index), newTotal, ...accumulator.slice(index + 1) ];
},[]);

const totalsAsObject = group.reduce((accumulator, [key, value]) => {
    const objectKeys = Object.keys(accumulator);
    if (objectKeys.includes(key)) {
      return { ...accumulator, [key]: accumulator[key] + value }
    }
    return { ...accumulator, [key]: value }
  },
  {}
);

console.log(totals);
console.log(totalsAsObject);

Answer 4

您可以使用reduce累加器包装map函数，并使用Array.from将结果转换回数组。

  result = Array.from(arr.reduce((a, cv) => {
    return ([t, amt] = cv,
      (a.has(t)) ? a.set(t, a.get(t) + amt) : a.set(t, amt), a)
  }, new Map()));

let arr = [
    ['TWENTY', 20],
    ['TWENTY', 20],
    ['TWENTY', 20],
    ['TWENTY', 20],
    ['TEN', 10],
    ['FIVE', 5],
    ['ONE', 1],
    ['QUARTER', 0.25],
    ['QUARTER', 0.25],
    ['DIME', 0.1],
    ['DIME', 0.1],
    ['PENNY', 0.01],
    ['PENNY', 0.01],
    ['PENNY', 0.01]
  ],

  result = Array.from(arr.reduce((a, cv) => {
    return ([t, amt] = cv,
      (a.has(t)) ? a.set(t, a.get(t) + amt) : a.set(t, amt), a)
  }, new Map()));

console.log(result);

Answer 5

ES6替代：

const  data = [ [ 'TWENTY', 20 ], 
                [ 'TWENTY', 20 ],
                [ 'TWENTY', 20 ],
                [ 'TWENTY', 20 ],
                [ 'TEN', 10 ],
                [ 'FIVE', 5 ],
                [ 'ONE', 1 ],
                [ 'QUARTER', 0.25 ],
                [ 'QUARTER', 0.25 ],
                [ 'DIME', 0.1 ],
                [ 'DIME', 0.1 ],
                [ 'PENNY', 0.01 ],
                [ 'PENNY', 0.01 ],
                [ 'PENNY', 0.01 ] ]

const result = data.reduce((o, [k, v]) => (o[k] = v + o[k] || v, o), {})

console.log( result )
console.log( Object.entries(result) )

如何合并多个相同的数组？

5 个答案: