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。
下面是一种可扩展的按多个字段排序的方法。
homes.sort(function(left, right) {
var city_order = left.city.localeCompare(right.city);
var price_order = parseInt(left.price) - parseInt(right.price);
return city_order || -price_order;
});
笔记
A function passed to array sort is expected to return negative, zero, positive to indicate less, equal, greater.
a.localeCompare(b) is universally supported for strings, and returns -1,0,1 if a<b,a==b,a>b.
Subtraction works on numeric fields, because a - b gives -,0,+ if a<b,a==b,a>b.
|| in the last line gives city priority over price.
Negate to reverse order in any field, as in -price_order
Add new fields to the or-chain: return city_order || -price_order || date_order;
Date compare with subtraction, because date math converts to milliseconds since 1970.var date_order = new Date(left.date) - new Date(right.date);
Boolean compare with subtraction, which is guaranteed to turn true and false to 1 and 0 (therefore the subtraction produces -1 or 0 or 1). var goodness_order = Boolean(left.is_good) - Boolean(right.is_good)This is unusual enough that I'd suggest drawing attention with the Boolean constructor, even if they're already boolean.
我一直在寻找类似的东西,最后得到了这个:
首先,我们有一个或多个排序函数,总是返回0、1或-1:
const sortByTitle = (a, b): number =>
a.title === b.title ? 0 : a.title > b.title ? 1 : -1;
您可以为想要排序的其他属性创建更多函数。
然后我有一个函数将这些排序函数合并为一个:
const createSorter = (...sorters) => (a, b) =>
sorters.reduce(
(d, fn) => (d === 0 ? fn(a, b) : d),
0
);
这可以用来以一种可读的方式组合上述排序函数:
const sorter = createSorter(sortByTitle, sortByYear)
items.sort(sorter)
当一个排序函数返回0时,将调用下一个排序函数进行进一步排序。
按多个字段排序对象数组的最简单方法:
let 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((a, b) => (a.city > b.city) ? 1 : -1);
输出:
“Bevery山”
“达拉斯”
“达拉斯”
“达拉斯”
“纽约”
一个动态的方法来做到这与多个键:
从排序的每个col/key中过滤唯一的值
按顺序排列或颠倒
根据indexOf(value)键值为每个对象添加weights width zeropad
使用计算的权重进行排序
Object.defineProperty(Array.prototype, 'orderBy', {
value: function(sorts) {
sorts.map(sort => {
sort.uniques = Array.from(
new Set(this.map(obj => obj[sort.key]))
);
sort.uniques = sort.uniques.sort((a, b) => {
if (typeof a == 'string') {
return sort.inverse ? b.localeCompare(a) : a.localeCompare(b);
}
else if (typeof a == 'number') {
return sort.inverse ? b - a : a - b;
}
else if (typeof a == 'boolean') {
let x = sort.inverse ? (a === b) ? 0 : a? -1 : 1 : (a === b) ? 0 : a? 1 : -1;
return x;
}
return 0;
});
});
const weightOfObject = (obj) => {
let weight = "";
sorts.map(sort => {
let zeropad = `${sort.uniques.length}`.length;
weight += sort.uniques.indexOf(obj[sort.key]).toString().padStart(zeropad, '0');
});
//obj.weight = weight; // if you need to see weights
return weight;
}
this.sort((a, b) => {
return weightOfObject(a).localeCompare( weightOfObject(b) );
});
return this;
}
});
Use:
// works with string, number and boolean
let sortered = your_array.orderBy([
{key: "type", inverse: false},
{key: "title", inverse: false},
{key: "spot", inverse: false},
{key: "internal", inverse: true}
]);
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]]