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。
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,
}
]));
另一种方式
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"}
];
function sortBy(ar) {
return ar.sort((a, b) => a.city === b.city ?
b.price.toString().localeCompare(a.price) :
a.city.toString().localeCompare(b.city));
}
console.log(sortBy(homes));
为了简化操作,可以使用这些辅助函数。
您可以根据需要对任意多个字段进行排序。对于每个排序字段,指定属性名,然后可选地指定-1作为排序方向,以降序排序而不是升序排序。
const data = [
{"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"},
{"h_id":"7","city":"New York","state":"NY","zip":"00010","price":"800500"}
]
const sortLexically = (p,d=1)=>(a,b)=>d * a[p].localeCompare(b[p])
const sortNumerically = (p,d=1)=>(a,b)=>d * (a[p]-b[p])
const sortBy = sorts=>(a,b)=>sorts.reduce((r,s)=>r||s(a,b),0)
// sort first by city, then by price descending
data.sort(sortBy([sortLexically('city'), sortNumerically('price', -1)]))
console.log(data)
这是另一个可能更接近您对语法的想法的例子
function sortObjects(objArray, properties /*, primers*/) {
var primers = arguments[2] || {}; // primers are optional
properties = properties.map(function(prop) {
if( !(prop instanceof Array) ) {
prop = [prop, 'asc']
}
if( prop[1].toLowerCase() == 'desc' ) {
prop[1] = -1;
} else {
prop[1] = 1;
}
return prop;
});
function valueCmp(x, y) {
return x > y ? 1 : x < y ? -1 : 0;
}
function arrayCmp(a, b) {
var arr1 = [], arr2 = [];
properties.forEach(function(prop) {
var aValue = a[prop[0]],
bValue = b[prop[0]];
if( typeof primers[prop[0]] != 'undefined' ) {
aValue = primers[prop[0]](aValue);
bValue = primers[prop[0]](bValue);
}
arr1.push( prop[1] * valueCmp(aValue, bValue) );
arr2.push( prop[1] * valueCmp(bValue, aValue) );
});
return arr1 < arr2 ? -1 : 1;
}
objArray.sort(function(a, b) {
return arrayCmp(a, b);
});
}
// just for fun use this to reverse the city name when sorting
function demoPrimer(str) {
return str.split('').reverse().join('');
}
// Example
sortObjects(homes, ['city', ['price', 'desc']], {city: demoPrimer});
演示:http://jsfiddle.net/Nq4dk/2/
编辑:只是为了好玩,这里有一个变化,只需要一个类似sql的字符串,所以你可以做sortObjects(房屋,“城市,价格desc”)
function sortObjects(objArray, properties /*, primers*/) {
var primers = arguments[2] || {};
properties = properties.split(/\s*,\s*/).map(function(prop) {
prop = prop.match(/^([^\s]+)(\s*desc)?/i);
if( prop[2] && prop[2].toLowerCase() === 'desc' ) {
return [prop[1] , -1];
} else {
return [prop[1] , 1];
}
});
function valueCmp(x, y) {
return x > y ? 1 : x < y ? -1 : 0;
}
function arrayCmp(a, b) {
var arr1 = [], arr2 = [];
properties.forEach(function(prop) {
var aValue = a[prop[0]],
bValue = b[prop[0]];
if( typeof primers[prop[0]] != 'undefined' ) {
aValue = primers[prop[0]](aValue);
bValue = primers[prop[0]](bValue);
}
arr1.push( prop[1] * valueCmp(aValue, bValue) );
arr2.push( prop[1] * valueCmp(bValue, aValue) );
});
return arr1 < arr2 ? -1 : 1;
}
objArray.sort(function(a, b) {
return arrayCmp(a, b);
});
}