Question

我有2个结构类似的数组，并希望使用较小数组中的随机ID遍历较大的数组并为其分配属性

//Actual array length is 140
const users = [
  {
    name: "John Roberts",
    uid: "49ikds_dm3idmssmmi9sz"
  },
  {
    name: "Peter Jones",
    uid: "fmi33_sm39imsz9z9nb"
  }
]

//Actual array lenth is 424
const cars = [
  {
    manufacturer: "BMW",
    model: "320d",
    year: "2010",
    user: null
  },
  {
    manufacturer: "BMW",
    model: "530d",
    year: "2018",
    user: null
  },
  {
    manufacturer: "AUDI",
    model: "RS6",
    year: "2014",
    user: null
  }
]

for(let i = 0; i < cars.length; i++){
  //if index is 2 or greater in this example. users[2] will be undefined
  cars[i].user = users[i].uid;
}

我基本上想重新使用小的users数组。在上面的示例中，一旦变量i为2或更大，则users[2]将为undefined。

任何人都可以推荐一种优雅的解决方案来帮助我解决这个问题。

Answer 1

您可以使用%来始终拥有正确的索引并重复使用用户。

const users = [
  {
    name: 'John Roberts',
    uid: '49ikds_dm3idmssmmi9sz',
  },
  {
    name: 'Peter Jones',
    uid: 'fmi33_sm39imsz9z9nb',
  }
];

const cars = [
  {
    manufacturer: 'BMW',
    model: '320d',
    year: '2010',
    user: null,
  },
  {
    manufacturer: 'BMW',
    model: '530d',
    year: '2018',
    user: null,
  },
  {
    manufacturer: 'AUDI',
    model: 'RS6',
    year: '2014',
    user: null,
  },
];

for (let i = 0; i < cars.length; i += 1){
  cars[i].user = users[i % users.length].uid;
}

console.log(cars);

您不会相信的ES6解决方案！（点击诱饵）

const users = [
  {
    name: 'John Roberts',
    uid: '49ikds_dm3idmssmmi9sz',
  },
  {
    name: 'Peter Jones',
    uid: 'fmi33_sm39imsz9z9nb',
  }
];

const cars = [
  {
    manufacturer: 'BMW',
    model: '320d',
    year: '2010',
    user: null,
  },
  {
    manufacturer: 'BMW',
    model: '530d',
    year: '2018',
    user: null,
  },
  {
    manufacturer: 'AUDI',
    model: 'RS6',
    year: '2014',
    user: null,
  },
];

const ret = cars.map((x, xi) => ({
  ...x,
  
  user: users[xi % users.length].uid,
}));

console.log(ret);

Answer 2

您只需要测试user[i] var并输入默认值即可：

for(let i = 0; i < cars.length; i++){
  //if index is 2 or greater in this example. users[2] will be undefined
  cars[i].user = users[i] && users[i].uid || /*Default value:*/ null;
}

如果满足以下条件，您也可以使用简单的方法实现

：

for(let i = 0; i < cars.length; i++){
  //if index is 2 or greater in this example. users[2] will be undefined 
  if (users[i]) {
    cars[i].user = users[i].uid;
  } else {
    cars[i].user = null; // Or another default
  }
}

------------------更新以考虑评论-----------------

如果要在用户上循环并在完成后重新启动数组，可以使用模%

let user_idx = null
for(let i = 0; i < cars.length; i++){

  user_idx = i % users.length

  //if index is 2 or greater in this example. users[2] will be undefined
  cars[i].user = users[user_idx].uid || null;
}

Answer 3

选项1：

for(let i = 0; i < cars.length; i++){
   cars[i].user = users[i % users.length].uid;
}

选项2：

for(let i = 0; i < cars.length; i++){
   cars[i].user = users[Math.round(Math.random() * (users.length - 1))].uid;
}

Answer 4

这很简单。只需编写以下逻辑即可。

$.grep(users, function(user, i ) {  cars[i].user = user.uid;});

Answer 5

var users = [
	{
		name: "John Roberts",
		uid: "49ikds_dm3idmssmmi9sz"
	},
	{
		name: "Peter Jones",
		uid: "fmi33_sm39imsz9z9nb"
	}
]

//Actual array lenth is 424
var cars = [
	{
		manufacturer: "BMW",
		model: "320d",
		year: "2010",
		user: null
	},
	{
		manufacturer: "BMW",
		model: "530d",
		year: "2018",
		user: null
	},
	{
		manufacturer: "AUDI",
		model: "RS6",
		year: "2014",
		user: null
	}
]

cars.map(function(car){
	var randomIndex = Math.floor(Math.random() * Math.floor(users.length))
	car.user = users[randomIndex].uid;
})
console.log(cars)

2个大小不同的数组，一旦达到最大索引，将在循环中重用一个数组

5 个答案: