从这个最初的问题,我将如何在多个字段应用排序?

使用这种稍作调整的结构,我将如何排序城市(上升)和价格(下降)?

var homes = [
    {"h_id":"3",
     "city":"Dallas",
     "state":"TX",
     "zip":"75201",
     "price":"162500"},
    {"h_id":"4",
     "city":"Bevery Hills",
     "state":"CA",
     "zip":"90210",
     "price":"319250"},
    {"h_id":"6",
     "city":"Dallas",
     "state":"TX",
     "zip":"75000",
     "price":"556699"},
    {"h_id":"5",
     "city":"New York",
     "state":"NY",
     "zip":"00010",
     "price":"962500"}
    ];

我喜欢的事实是,给出的答案提供了一个一般的方法。在我计划使用这段代码的地方,我将不得不对日期以及其他东西进行排序。“启动”对象的能力似乎很方便,如果不是有点麻烦的话。

我试图把这个答案构建成一个很好的通用示例,但我运气不太好。


当前回答

通过添加3个相对简单的帮助程序,可以构建一个非常直观的功能解决方案。在深入研究之前,我们先来了解一下用法:

function usage(homes, { asc, desc, fallback }) { homes.sort(fallback( asc(home => home.city), desc(home => parseInt(home.price, 10)), )); console.log(homes); } var homes = [{ h_id: "3", city: "Dallas", state: "TX", zip: "75201", price: "162500", }, { h_id: "4", city: "Bevery Hills", state: "CA", zip: "90210", price: "319250", }, { h_id: "6", city: "Dallas", state: "TX", zip: "75000", price: "556699", }, { h_id: "5", city: "New York", state: "NY", zip: "00010", price: "962500", }]; const SortHelpers = (function () { const asc = (fn) => (a, b) => (a = fn(a), b = fn(b), -(a < b) || +(a > b)); const desc = (fn) => (a, b) => asc(fn)(b, a); const fallback = (...fns) => (a, b) => fns.reduce((diff, fn) => diff || fn(a, b), 0); return { asc, desc, fallback }; })(); usage(homes, SortHelpers);

如果你向下滚动代码片段,你可能已经看到了helper:

const asc  = (fn) => (a, b) => (a = fn(a), b = fn(b), -(a < b) || +(a > b));
const desc = (fn) => (a, b) => asc(fn)(b, a);
const fallback = (...fns) => (a, b) => fns.reduce((diff, fn) => diff || fn(a, b), 0);

让我快速解释一下这些函数的作用。

asc creates a comparator function. The provided function fn is called for both the comparator arguments a and b. The results of the two function calls are then compared. -1 is returned if resultA < resultB, 1 is returned if resultA > resultB, or 0 otherwise. These return values correspond with an ascending order direction. It could also be written like this: function asc(fn) { return function (a, b) { // apply `fn` to both `a` and `b` a = fn(a); b = fn(b); if (a < b) return -1; if (a > b) return 1; return 0; // or `return -(a < b) || +(a > b)` for short }; } desc is super simple, since it just calls asc but swaps the a and b arguments, resulting in descending order instead of ascending. fallback (there might be a better name for this) allows us to use multiple comparator functions with a single sort. Both asc and desc can be passed to sort by themself. homes.sort(asc(home => home.city)) There is however an issue if you want to combine multiple comparator functions. sort only accepts a single comparator function. fallback combines multiple comparator functions into a single comparator. The first comparator is called with arguments a and b, if the comparator returns the value 0 (meaning that the values are equal) then we fall back to the next comparator. This continues until a non-0 value is found, or until all comparators are called, in which case the return value is 0.

也可以为fallback()提供自定义比较器函数。假设您想使用localeCompare()而不是比较字符串与<和>。在这种情况下,您可以将asc(home => home.city)替换为(a, b) => a.city.localeCompare(b.city)。

homes.sort(fallback(
  (a, b) => a.city.localeCompare(b.city),
  desc(home => parseInt(home.price, 10)),
));

需要注意的一点是,在与<和>比较时,未定义的值总是返回false。因此,如果一个值可能缺失,您可能希望首先根据它的存在进行排序。

homes.sort(fallback(
  // homes with optionalProperty first, true (1) > false (0) so we use desc
  desc(home => home.optionalProperty != null), // checks for both null and undefined
  asc(home => home.optionalProperty),
  // ...
))

因为用localeCompare()比较字符串是一件很常见的事情,所以可以将其作为asc()的一部分。

function hasMethod(item, methodName) {
  return item != null && typeof item[methodName] === "function";
}

function asc(fn) {
  return function (a, b) {
    a = fn(a);
    b = fn(b);

    const areLocaleComparable =
      hasMethod(a, "localeCompare") && hasMethod(b, "localeCompare");

    if (areLocaleComparable) return a.localeCompare(b);

    return -(a < b) || +(a > b);
  };
}

其他回答

这是一个通用的多维排序,允许在每个层次上进行反转和/或映射。

用Typescript编写。对于Javascript,请查看这个JSFiddle

的代码

type itemMap = (n: any) => any;

interface SortConfig<T> {
  key: keyof T;
  reverse?: boolean;
  map?: itemMap;
}

export function byObjectValues<T extends object>(keys: ((keyof T) | SortConfig<T>)[]): (a: T, b: T) => 0 | 1 | -1 {
  return function(a: T, b: T) {
    const firstKey: keyof T | SortConfig<T> = keys[0];
    const isSimple = typeof firstKey === 'string';
    const key: keyof T = isSimple ? (firstKey as keyof T) : (firstKey as SortConfig<T>).key;
    const reverse: boolean = isSimple ? false : !!(firstKey as SortConfig<T>).reverse;
    const map: itemMap | null = isSimple ? null : (firstKey as SortConfig<T>).map || null;

    const valA = map ? map(a[key]) : a[key];
    const valB = map ? map(b[key]) : b[key];
    if (valA === valB) {
      if (keys.length === 1) {
        return 0;
      }
      return byObjectValues<T>(keys.slice(1))(a, b);
    }
    if (reverse) {
      return valA > valB ? -1 : 1;
    }
    return valA > valB ? 1 : -1;
  };
}

用法示例

先按姓排序,再按名排序:

interface Person {
  firstName: string;
  lastName: string;
}

people.sort(byObjectValues<Person>(['lastName','firstName']));

按语言代码的名称排序,而不是按语言代码排序(见地图),然后按降序排序(见反向)。

interface Language {
  code: string;
  version: number;
}

// languageCodeToName(code) is defined elsewhere in code

languageCodes.sort(byObjectValues<Language>([
  {
    key: 'code',
    map(code:string) => languageCodeToName(code),
  },
  {
    key: 'version',
    reverse: true,
  }
]));

这里有一个简单的泛型函数方法。使用数组指定排序顺序。前置减号以指定降序。

var homes = [
    {"h_id":"3", "city":"Dallas", "state":"TX","zip":"75201","price":"162500"},
    {"h_id":"4","city":"Bevery Hills", "state":"CA", "zip":"90210", "price":"319250"},
    {"h_id":"6", "city":"Dallas", "state":"TX", "zip":"75000", "price":"556699"},
    {"h_id":"5", "city":"New York", "state":"NY", "zip":"00010", "price":"962500"}
    ];

homes.sort(fieldSorter(['city', '-price']));
// homes.sort(fieldSorter(['zip', '-state', 'price'])); // alternative

function fieldSorter(fields) {
    return function (a, b) {
        return fields
            .map(function (o) {
                var dir = 1;
                if (o[0] === '-') {
                   dir = -1;
                   o=o.substring(1);
                }
                if (a[o] > b[o]) return dir;
                if (a[o] < b[o]) return -(dir);
                return 0;
            })
            .reduce(function firstNonZeroValue (p,n) {
                return p ? p : n;
            }, 0);
    };
}

编辑:在ES6中它甚至更短!

"use strict"; const fieldSorter = (fields) => (a, b) => fields.map(o => { let dir = 1; if (o[0] === '-') { dir = -1; o=o.substring(1); } return a[o] > b[o] ? dir : a[o] < b[o] ? -(dir) : 0; }).reduce((p, n) => p ? p : n, 0); const homes = [{"h_id":"3", "city":"Dallas", "state":"TX","zip":"75201","price":162500}, {"h_id":"4","city":"Bevery Hills", "state":"CA", "zip":"90210", "price":319250},{"h_id":"6", "city":"Dallas", "state":"TX", "zip":"75000", "price":556699},{"h_id":"5", "city":"New York", "state":"NY", "zip":"00010", "price":962500}]; const sortedHomes = homes.sort(fieldSorter(['state', '-price'])); document.write('<pre>' + JSON.stringify(sortedHomes, null, '\t') + '</pre>')

改编自@chriskelly的回答。


大多数答案都忽略了,如果价值在1万美元以下或超过100万美元,价格将无法正确排序。原因是JS按字母顺序排序。这里回答得很好,为什么JavaScript不能对“5,10,1”排序,这里如何正确地对整数数组排序。

最后,如果我们要排序的字段或节点是一个数字,我们必须做一些计算。我并不是说在这种情况下使用parseInt()是正确的答案,排序结果更重要。

var homes = [{ "h_id": "2", "city": "Dallas", "state": "TX", "zip": "75201", "price": "62500" }, { "h_id": "1", "city": "Dallas", "state": "TX", "zip": "75201", "price": "62510" }, { "h_id": "3", "city": "Dallas", "state": "TX", "zip": "75201", "price": "162500" }, { "h_id": "4", "city": "Bevery Hills", "state": "CA", "zip": "90210", "price": "319250" }, { "h_id": "6", "city": "Dallas", "state": "TX", "zip": "75000", "price": "556699" }, { "h_id": "5", "city": "New York", "state": "NY", "zip": "00010", "price": "962500" }]; homes.sort(fieldSorter(['price'])); // homes.sort(fieldSorter(['zip', '-state', 'price'])); // alternative function fieldSorter(fields) { return function(a, b) { return fields .map(function(o) { var dir = 1; if (o[0] === '-') { dir = -1; o = o.substring(1); } if (!parseInt(a[o]) && !parseInt(b[o])) { if (a[o] > b[o]) return dir; if (a[o] < b[o]) return -(dir); return 0; } else { return dir > 0 ? a[o] - b[o] : b[o] - a[o]; } }) .reduce(function firstNonZeroValue(p, n) { return p ? p : n; }, 0); }; } document.getElementById("output").innerHTML = '<pre>' + JSON.stringify(homes, null, '\t') + '</pre>'; <div id="output"> </div>


用来测试的小提琴

这是一个完全的欺骗,但我认为它为这个问题增加了价值,因为它基本上是一个罐装的库函数,你可以开箱即用。

如果你的代码可以访问lodash或者一个与lodash兼容的库,比如下划线,那么你可以使用_。sortBy方法。下面的代码片段直接复制自lodash文档。

示例中的注释结果看起来像是返回数组的数组,但这只是显示了顺序,而不是实际的结果,它是一个对象数组。

var users = [
  { 'user': 'fred',   'age': 48 },
  { 'user': 'barney', 'age': 36 },
  { 'user': 'fred',   'age': 40 },
  { 'user': 'barney', 'age': 34 }
];

_.sortBy(users, [function(o) { return o.user; }]);
 // => objects for [['barney', 36], ['barney', 34], ['fred', 48], ['fred', 40]]

_.sortBy(users, ['user', 'age']);
// => objects for [['barney', 34], ['barney', 36], ['fred', 40], ['fred', 48]]

我认为这可能是最简单的方法。

https://coderwall.com/p/ebqhca/javascript-sort-by-two-fields

这真的很简单,我尝试了3个不同的键值对,它工作得很好。

这是一个简单的例子,查看链接了解更多细节

testSort(data) {
    return data.sort(
        a['nameOne'] > b['nameOne'] ? 1
        : b['nameOne'] > a['nameOne'] ? -1 : 0 ||
        a['date'] > b['date'] ||
        a['number'] - b['number']
    );
}