Question

考虑到我有一个像下面这样的对象，它可能有很多名字，而其他人可能会有很多名字。可以出现在任何索引处，我如何对具有＆＃39;其他＆＃39;的数组进行排序？始终作为第一个元素，其余的名称按字母顺序排序？

var friends = [
{ id: 1, name: 'Paul' },
{ id: 2, name: 'Mary' },
{ id: 3, name: 'The others' },
{ id: 4, name: 'John' }
];

对于上面的示例数组，所需的结果为：

[
   { id: 3, name: 'The others' }, 
   { id: 4, name: 'John' }, 
   { id: 2, name: 'Mary' }, 
   { id: 1, name: 'Paul' }
]

Answer 1

只需检查排序回调中的任何一个值为The others - 如果-1为a则返回1，如果是b则返回friends.sort(({name: a}, {name:b}) => a == "The others" ? -1 : (b == "The others" ? 1 : a.localeCompare(b))); - 否则返回a和b的localeCompare

friends.sort(function (a, b) {
  if (a.name == "The others") return -1;
  if (b.name == "The others") return 1;
  return a.name.localeCompare(b.name);
});

的“可读”和非ES2015 +版本

var friends = [{
    id: 1,
    name: 'Paul'
  },
  {
    id: 2,
    name: 'Mary'
  },
  {
    id: 3,
    name: 'The others'
  },
  {
    id: 4,
    name: 'John'
  }
];

// https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort
friends.sort(function(a, b) {
  var nameA = a.name.toUpperCase(); // ignore upper and lowercase
  var nameB = b.name.toUpperCase(); // ignore upper and lowercase
  if (nameA < nameB) {
    return -1;
  }
  if (nameA > nameB) {
    return 1;
  }

  // names must be equal
  return 0;
});

var indexOfTheOters = friends.reduce(function(acc, friend, index) {
  if(!acc && friend.name === 'The others') {
    acc = index;
  }
  return acc;
}, null);

// https://stackoverflow.com/questions/5306680/move-an-array-element-from-one-array-position-to-another
Array.prototype.move = function (old_index, new_index) {
    if (new_index >= this.length) {
        var k = new_index - this.length;
        while ((k--) + 1) {
            this.push(undefined);
        }
    }
    this.splice(new_index, 0, this.splice(old_index, 1)[0]);
    return this; // for testing purposes
};
friends.move(indexOfTheOters, 0)
console.log(friends);

Answer 2

以下是您需要遵循的步骤

排序：https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort
移动：Move an array element from one array position to another

请参阅以下工作示例。

original_list.group_by {|x| x[:number]}.values

Answer 3

尝试以下方法：

＆＃13;

The others

＆＃13;

Answer 4

复制您的（filter对象）并使用filter
使用The others
按名称属性对数组进行排序。
使用unshift

compare

this answer

＆＃13;

SELECT c.customer# "C#", o.order# "ORDER#",
       o.orderdate "ORDER DATE",
       NVL2(o.shipdate, 'Not shipped', 'shipped'),
       o.shipdate "Ship date",
       o.shipdate-o.orderdate AS "lead time in days",
TRUNC((o.shipdate-o.orderdate)/30), 2) AS "lead time in months"
FROM customers c, orders o
ORDER BY "lead time in days" DESC, c.customer# ASC, o.order# ASC;

＆＃13;

Answer 5

按这样排序：

＆＃13;

var friends = [
   { id: 1, name: 'Paul' },
   { id: 2, name: 'Mary' },
   { id: 3, name: 'The others' },
   { id: 4, name: 'John' }
];

friends.sort((a, b) => {
     if (a.name == "The others") {
         return -1; // a comes first
     } else if (b.name == "The others") {
         return 1;
     } else {
         return (a.name < b.name ? -1 : 1);
     }
 });

console.log(friends);

＆＃13;

Answer 6

您希望对比较功能进行硬编码，以便“其他人”能够对其进行硬编码。始终具有最低/最高价值。

＆＃13;

var friends = [
{ id: 1, name: 'Paul' },
{ id: 2, name: 'Mary' },
{ id: 3, name: 'The others' },
{ id: 4, name: 'John' }
];

function compare(a, b){
  if(a.name === "The others") {
  	return -1;
  } else if (b.name === "The others") {
  	return 1;
  } else if (a.name > b.name){
  	return -1;
  } else if (a.name < b.name){
  	return 1;
  } else {
  	return 0;
  }
}

console.log(friends.sort(compare))

＆＃13;

Answer 7

这里有一些很棒的想法，但我发现更简单的解决方案是下面两个：

使用sort：

friends.sort((a, b) => a.name !== b.name ? a.name < b.name ? -1 : 1 : 0).sort((a, b) => +(!b.name.localeCompare("The others")));

使用地图和排序：

friends.map(o => o.name).sort((a,b) => a=="The others" ? -1 : b=="The others" ? 1 : a > b);

排序除一个项目之外的名称数组

7 个答案: