如何在 Javascript 中根据关联数组的值排序？

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There really isn't any such thing as an "associative array" in JavaScript. What you've got there is just a plain old object. They work kind-of like associative arrays, of course, and the keys are available but there's no semantics around the order of keys.

You could turn your object into an array of objects (key/value pairs) and sort that:

function sortObj(object, sortFunc) {
var rv = [];
for (var k in object) {
if (object.hasOwnProperty(k)) rv.push({key: k, value:  object[k]});
}
rv.sort(function(o1, o2) {
return sortFunc(o1.key, o2.key);
});
return rv;
}

Then you'd call that with a comparator function.

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最佳答案

Javascript doesn't have "associative arrays" the way you're thinking of them. Instead, you simply have the ability to set object properties using array-like syntax (as in your example), plus the ability to iterate over an object's properties.

The upshot of this is that there is no guarantee as to the order in which you iterate over the properties, so there is nothing like a sort for them. Instead, you'll need to convert your object properties into a "true" array (which does guarantee order). Here's a code snippet for converting an object into an array of two-tuples (two-element arrays), sorting it as you describe, then iterating over it:

var tuples = [];


for (var key in obj) tuples.push([key, obj[key]]);


tuples.sort(function(a, b) {
a = a[1];
b = b[1];


return a < b ? -1 : (a > b ? 1 : 0);
});


for (var i = 0; i < tuples.length; i++) {
var key = tuples[i][0];
var value = tuples[i][1];


// do something with key and value
}

You may find it more natural to wrap this in a function which takes a callback:

function bySortedValue(obj, callback, context) {
var tuples = [];


for (var key in obj) tuples.push([key, obj[key]]);


tuples.sort(function(a, b) {
return a[1] < b[1] ? 1 : a[1] > b[1] ? -1 : 0
});


var length = tuples.length;
while (length--) callback.call(context, tuples[length][0], tuples[length][1]);
}


bySortedValue({
foo: 1,
bar: 7,
baz: 3
}, function(key, value) {
document.getElementById('res').innerHTML += `${key}: ${value}<br>`
});

<p id='res'>Result:<br/><br/><p>

小开

i use $.each of jquery but you can make it with a for loop, an improvement is this:

        //.ArraySort(array)
/* Sort an array
*/
ArraySort = function(array, sortFunc){
var tmp = [];
var aSorted=[];
var oSorted={};


for (var k in array) {
if (array.hasOwnProperty(k))
tmp.push({key: k, value:  array[k]});
}


tmp.sort(function(o1, o2) {
return sortFunc(o1.value, o2.value);
});


if(Object.prototype.toString.call(array) === '[object Array]'){
$.each(tmp, function(index, value){
aSorted.push(value.value);
});
return aSorted;
}


if(Object.prototype.toString.call(array) === '[object Object]'){
$.each(tmp, function(index, value){
oSorted[value.key]=value.value;
});
return oSorted;
}
};

So now you can do

    console.log("ArraySort");
var arr1 = [4,3,6,1,2,8,5,9,9];
var arr2 = {'a':4, 'b':3, 'c':6, 'd':1, 'e':2, 'f':8, 'g':5, 'h':9};
var arr3 = {a: 'green', b: 'brown', c: 'blue', d: 'red'};
var result1 = ArraySort(arr1, function(a,b){return a-b});
var result2 = ArraySort(arr2, function(a,b){return a-b});
var result3 = ArraySort(arr3, function(a,b){return a>b});
console.log(result1);
console.log(result2);
console.log(result3);

小开

Instead of correcting you on the semantics of an 'associative array', I think this is what you want:

function getSortedKeys(obj) {
var keys = Object.keys(obj);
return keys.sort(function(a,b){return obj[b]-obj[a]});
}

for really old browsers, use this instead:

function getSortedKeys(obj) {
var keys = []; for(var key in obj) keys.push(key);
return keys.sort(function(a,b){return obj[b]-obj[a]});
}

You dump in an object (like yours) and get an array of the keys - eh properties - back, sorted descending by the (numerical) value of the, eh, values of the, eh, object.

This only works if your values are numerical. Tweek the little function(a,b) in there to change the sorting mechanism to work ascending, or work for string values (for example). Left as an exercise for the reader.

小开

Continued discussion & other solutions covered at How to sort an (associative) array by value? with the best solution (for my case) being by saml (quoted below).

Arrays can only have numeric indexes. You'd need to rewrite this as either an Object, or an Array of Objects.

var status = new Array();
status.push({name: 'BOB', val: 10});
status.push({name: 'TOM', val: 3});
status.push({name: 'ROB', val: 22});
status.push({name: 'JON', val: 7});

If you like the status.push method, you can sort it with:

status.sort(function(a,b) {
return a.val - b.val;
});

小开

Here is a variation of ben blank's answer, if you don't like tuples.

This saves you a few characters.

var keys = [];
for (var key in sortme) {
keys.push(key);
}


keys.sort(function(k0, k1) {
var a = sortme[k0];
var b = sortme[k1];
return a < b ? -1 : (a > b ? 1 : 0);
});


for (var i = 0; i < keys.length; ++i) {
var key = keys[i];
var value = sortme[key];
// Do something with key and value.
}

小开

No unnecessary complication required...

function sortMapByValue(map)
{
var tupleArray = [];
for (var key in map) tupleArray.push([key, map[key]]);
tupleArray.sort(function (a, b) { return a[1] - b[1] });
return tupleArray;
}

小开

@commonpike's answer is "the right one", but as he goes on to comment...

most browsers nowadays just support Object.keys()

Yeah.. Object.keys() is WAY better.

But what's even better? Duh, it's it in coffeescript!

sortedKeys = (x) -> Object.keys(x).sort (a,b) -> x[a] - x[b]


sortedKeys
'a' :  1
'b' :  3
'c' :  4
'd' : -1

[ 'd', 'a', 'b', 'c' ]

小开

Just so it's out there and someone is looking for tuple based sorts. This will compare the first element of the object in array, than the second element and so on. i.e in the example below, it will compare first by "a", then by "b" and so on.

let arr = [
{a:1, b:2, c:3},
{a:3, b:5, c:1},
{a:2, b:3, c:9},
{a:2, b:5, c:9},
{a:2, b:3, c:10}
]


function getSortedScore(obj) {
var keys = [];
for(var key in obj[0]) keys.push(key);
return obj.sort(function(a,b){
for (var i in keys) {
let k = keys[i];
if (a[k]-b[k] > 0) return -1;
else if (a[k]-b[k] < 0) return 1;
else continue;
};
});
}


console.log(getSortedScore(arr))

OUPUTS

 [ { a: 3, b: 5, c: 1 },
{ a: 2, b: 5, c: 9 },
{ a: 2, b: 3, c: 10 },
{ a: 2, b: 3, c: 9 },
{ a: 1, b: 2, c: 3 } ]

小开

The best approach for the specific case here, in my opinion, is the one commonpike suggested. A little improvement I'd suggest that works in modern browsers is:

// aao is the "associative array" you need to "sort"
Object.keys(aao).sort(function(a,b){return aao[b]-aao[a]});

This could apply easily and work great in the specific case here so you can do:

let aoo={};
aao["sub2"]=1;
aao["sub0"]=-1;
aao["sub1"]=0;
aao["sub3"]=1;
aao["sub4"]=0;


let sk=Object.keys(aao).sort(function(a,b){return aao[b]-aao[a]});


// now you can loop using the sorted keys in `sk` to do stuffs
for (let i=sk.length-1;i>=0;--i){
// do something with sk[i] or aoo[sk[i]]
}

Besides of this, I provide here a more "generic" function you can use to sort even in wider range of situations and that mixes the improvement I just suggested with the approaches of the answers by Ben Blank (sorting also string values) and PopeJohnPaulII (sorting by specific object field/property) and lets you decide if you want an ascendant or descendant order, here it is:

// aao := is the "associative array" you need to "sort"
// comp := is the "field" you want to compare or "" if you have no "fields" and simply need to compare values
// intVal := must be false if you need comparing non-integer values
// desc := set to true will sort keys in descendant order (default sort order is ascendant)
function sortedKeys(aao,comp="",intVal=false,desc=false){
let keys=Object.keys(aao);
if (comp!="") {
if (intVal) {
if (desc) return keys.sort(function(a,b){return aao[b][comp]-aao[a][comp]});
else return keys.sort(function(a,b){return aao[a][comp]-aao[a][comp]});
} else {
if (desc) return keys.sort(function(a,b){return aao[b][comp]<aao[a][comp]?1:aao[b][comp]>aao[a][comp]?-1:0});
else return keys.sort(function(a,b){return aao[a][comp]<aao[b][comp]?1:aao[a][comp]>aao[b][comp]?-1:0});
}
} else {
if (intVal) {
if (desc) return keys.sort(function(a,b){return aao[b]-aao[a]});
else return keys.sort(function(a,b){return aao[a]-aao[b]});
} else {
if (desc) return keys.sort(function(a,b){return aao[b]<aao[a]?1:aao[b]>aao[a]?-1:0});
else return keys.sort(function(a,b){return aao[a]<aao[b]?1:aao[a]>aao[b]?-1:0});
}
}
}

You can test the functionalities trying something like the following code:

let items={};
items['Edward']=21;
items['Sharpe']=37;
items['And']=45;
items['The']=-12;
items['Magnetic']=13;
items['Zeros']=37;
//equivalent to:
//let items={"Edward": 21, "Sharpe": 37, "And": 45, "The": -12, ...};


console.log("1: "+sortedKeys(items));
console.log("2: "+sortedKeys(items,"",false,true));
console.log("3: "+sortedKeys(items,"",true,false));
console.log("4: "+sortedKeys(items,"",true,true));
/* OUTPUT
1: And,Sharpe,Zeros,Edward,Magnetic,The
2: The,Magnetic,Edward,Sharpe,Zeros,And
3: The,Magnetic,Edward,Sharpe,Zeros,And
4: And,Sharpe,Zeros,Edward,Magnetic,The
*/


items={};
items['k1']={name:'Edward',value:21};
items['k2']={name:'Sharpe',value:37};
items['k3']={name:'And',value:45};
items['k4']={name:'The',value:-12};
items['k5']={name:'Magnetic',value:13};
items['k6']={name:'Zeros',value:37};


console.log("1: "+sortedKeys(items,"name"));
console.log("2: "+sortedKeys(items,"name",false,true));
/* OUTPUT
1: k6,k4,k2,k5,k1,k3
2: k3,k1,k5,k2,k4,k6
*/

As I already said, you can loop over sorted keys if you need doing stuffs

let sk=sortedKeys(aoo);
// now you can loop using the sorted keys in `sk` to do stuffs
for (let i=sk.length-1;i>=0;--i){
// do something with sk[i] or aoo[sk[i]]
}

Last, but not least, some useful references to Object.keys and Array.sort

小开

A modern approuch to this:

Object.fromEntries(Object.entries(data).sort((a,b)=>b[1]-a[1]).slice(0,5))

P.S: I did an optional slice, you can remove it if you want.