Question

我有四个对象（account1、account2、account3、account4）如下，我想获取“所有存款的总和”和“总和”所有提款的”。这里存款和取款显示在移动属性中。这里所有的正值都是存款，负值是取款。但我只想使用没有 flatMap 和 filter 方法的 reduce 方法。我通过执行以下操作获得了所有动作：

const transactions = accounts.reduce(
  (acc, cur) => acc.concat(cur.movements),
[]
);

这将减少到 [200, 450, -400, 3000, -650, -130, 70, 1300, 5000, 3400, -150, -790, -3210, -1000, 8500, -30, 200, -200, 340, -300, -20, 50, 400, -460, 430, 1000, 700, 50, 90]

但是，如何设置运动数组，以便我可以获得正值（存款）和负值（取款）的总和。

const account1 = {
  owner: 'Jonas Schmedtmann',
  movements: [200, 450, -400, 3000, -650, -130, 70, 1300],
  interestRate: 1.2, // %
  pin: 1111,
};

const account2 = {
  owner: 'Jessica Davis',
  movements: [5000, 3400, -150, -790, -3210, -1000, 8500, -30],
  interestRate: 1.5,
  pin: 2222,
};

const account3 = {
  owner: 'Steven Thomas Williams',
  movements: [200, -200, 340, -300, -20, 50, 400, -460],
  interestRate: 0.7,
  pin: 3333,
};

const account4 = {
  owner: 'Sarah Smith',
  movements: [430, 1000, 700, 50, 90],
  interestRate: 1,
  pin: 4444,
};

const accounts = [account1, account2, account3, account4];

Answer 1

您可以使用过滤器运行两次总和减少，或者同时跟踪正/负总和的组合减少：

const amounts = [200, 450, -400, 3000, -650, -130, 70, 1300, 5000, 3400, -150, -790, -3210, -1000, 8500, -30, 200, -200, 340, -300, -20, 50, 400, -460, 430, 1000, 700, 50, 90];
const pos = amounts.filter(am => am > 0).reduce((tot, v) => tot + v);
const neg = amounts.filter(am => am < 0).reduce((tot, v) => tot + v);
console.log('pos:', pos);
console.log('neg:', neg);

// or alternatively
const combo = amounts.reduce(([pos, neg], val) => val > 0 ? [pos + val, neg] : [pos, neg + val], [0, 0]);
console.log('combo:', combo);

Answer 2

可能是这个吗？

const
  account1 = { owner: 'Jonas Schmedtmann',movements: [200, 450, -400, 3000, -650, -130, 70, 1300],interestRate: 1.2, pin: 1111 }
, account2 = { owner: 'Jessica Davis',movements: [5000, 3400, -150, -790, -3210, -1000, 8500, -30],interestRate: 1.5, pin: 2222 }
, account3 = { owner: 'Steven Thomas Williams',movements: [200, -200, 340, -300, -20, 50, 400, -460],interestRate: 0.7, pin: 3333 }
, account4 = { owner: 'Sarah Smith',movements: [430, 1000, 700, 50, 90],interestRate: 1,pin: 4444 }
  ;

const transactions = (...accounts) =>
  accounts.reduce((r,{movements})=>
  { 
  movements.forEach(v =>{ v > 0 ? r.deposits += v : r.withdrawals += v })
  return r
  }, {deposits:0,withdrawals:0})

console.log( transactions( account1, account2, account3, account4 ) )

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

Answer 3

您可以使用 array#reduce 来总结 deposits 和 withdrawals。使用 account 迭代每个 array#reduce，然后使用 movements 迭代每个 array#forEach，在结果累加器中总结 deposits 和 withdrawals。

const result = accounts.reduce((r, {movements}) => {
  movements.forEach(v => {
    v > 0 ? r.deposits += v : r.withdrawals += v;
  });
  return r;
 },{deposits: 0, withdrawals: 0});

const account1 = {
  owner: 'Jonas Schmedtmann',
  movements: [200, 450, -400, 3000, -650, -130, 70, 1300],
  interestRate: 1.2, // %
  pin: 1111,
};

const account2 = {
  owner: 'Jessica Davis',
  movements: [5000, 3400, -150, -790, -3210, -1000, 8500, -30],
  interestRate: 1.5,
  pin: 2222,
};

const account3 = {
  owner: 'Steven Thomas Williams',
  movements: [200, -200, 340, -300, -20, 50, 400, -460],
  interestRate: 0.7,
  pin: 3333,
};

const account4 = {
  owner: 'Sarah Smith',
  movements: [430, 1000, 700, 50, 90],
  interestRate: 1,
  pin: 4444,
};

const accounts = [account1, account2, account3, account4];
const result = accounts.reduce((r, {movements}) => {
  movements.forEach(v => {
    v > 0 ? r.deposits += v : r.withdrawals += v;
  });
  return r;
},{deposits: 0, withdrawals: 0});
console.log(result);

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Answer 4

您可以像这样使用 .reduce 来迭代和累积正负运动。

movements.reduce((acc, movement) => {
  if (movement > 0) acc[0] += movement
  else acc[1] += movement
    
  return acc
}, [0, 0])

这看起来像一个函数。

let allTransactions = [200, 450, -400, 3000, -650, -130, 70, 1300, 5000, 3400, -150, -790, -3210, -1000, 8500, -30, 200, -200, 340, -300, -20, 50, 400, -460, 430, 1000, 700, 50, 90]

function getPosNeg(movements) {
  return movements.reduce((acc, movement) => {
    if (movement > 0) acc[0] += movement
    else acc[1] += movement
    
    return acc
  }, [0, 0])
}

let [positive, negative] = getPosNeg(allTransactions)

console.log({positive: positive, negative: negative})

如何在数组上使用 javascript reduce 方法，该数组是对象属性的值并仅输出正元素的总和？

4 个答案: