按两个数值字段对 Javascript 数组进行排序

Question

grouperArray.sort(function (a, b) {
    var aSize = a.gsize;
    var bSize = b.gsize;
    var aLow = a.glow;
    var bLow = b.glow;
    console.log(aLow + " | " + bLow);      
    return (aSize < bSize) ? -1 : (aSize > bSize) ? 1 : 0;
});

This code sorts the array by gsize , smallest to largest.

How would I change it to sort first by gsize and then by glow ?

Answer 1

grouperArray.sort(function (a, b) {   
    return a.gsize - b.gsize || a.glow - b.glow;
});

shorter version

Answer 2

grouperArray.sort(function (a, b) {
    var aSize = a.gsize;
    var bSize = b.gsize;
    var aLow = a.glow;
    var bLow = b.glow;
    console.log(aLow + " | " + bLow);

    if(aSize == bSize)
    {
        return (aLow < bLow) ? -1 : (aLow > bLow) ? 1 : 0;
    }
    else
    {
        return (aSize < bSize) ? -1 : 1;
    }
});

Answer 3

grouperArray.sort((a, b) => a.gsize - b.gsize || a.glow - b.glow);

Even shorter version using arrow syntax!

Answer 4

I realize this was asked some time ago, but I thought I would add my solution.

This function generates sort methods dynamically. simply supply each sortable child property name, prepended with +/- to indicate ascending or descending order. Super re-usable, and it doesn't need to know anything about the data structure you've put together. Could be made idiot proof - but doesn't seem necessary.

function getSortMethod(){
    var _args = Array.prototype.slice.call(arguments);
    return function(a, b){
        for(var x in _args){
            var ax = a[_args[x].substring(1)];
            var bx = b[_args[x].substring(1)];
            var cx;

            ax = typeof ax == "string" ? ax.toLowerCase() : ax / 1;
            bx = typeof bx == "string" ? bx.toLowerCase() : bx / 1;

            if(_args[x].substring(0,1) == "-"){cx = ax; ax = bx; bx = cx;}
            if(ax != bx){return ax < bx ? -1 : 1;}
        }
    }
}

example usage:

items.sort(getSortMethod('-price', '+priority', '+name'));

this would sort items with lowest price first, with ties going to the item with the highest priority . further ties are broken by the item name

where items is an array like:

var items = [
    { name: "z - test item", price: "99.99", priority: 0, reviews: 309, rating: 2 },
    { name: "z - test item", price: "1.99", priority: 0, reviews: 11, rating: 0.5 },
    { name: "y - test item", price: "99.99", priority: 1, reviews: 99, rating: 1 },
    { name: "y - test item", price: "0", priority: 1, reviews: 394, rating: 3.5 },
    { name: "x - test item", price: "0", priority: 2, reviews: 249, rating: 0.5 } ...
];

live demo: http://gregtaff.com/misc/multi_field_sort/

EDIT: Fixed issue with Chrome.

Answer 5

I expect the ternary operator ((aSize < bSize)? -1: (aSize > bSize)? 1: 0;) has you confused. You should check out the link to understand it better.

Until then, here's your code blown out into full if/else.

grouperArray.sort(function (a, b) {
    if (a.gsize < b.gsize)
    {
        return -1;
    }
    else if (a.gsize > b.gsize)
    {
        return 1;
    }
    else
    {
        if (a.glow < b.glow)
        {
            return -1;
        }
        else if (a.glow > b.glow)
        {
            return 1;
        }
        return 0;
    }
});

Answer 6

Here's an implementation for those who may want something more generic that would work with any number of fields.

Array.prototype.sortBy = function (propertyName, sortDirection) {

    var sortArguments = arguments;
    this.sort(function (objA, objB) {

        var result = 0;
        for (var argIndex = 0; argIndex < sortArguments.length && result === 0; argIndex += 2) {

            var propertyName = sortArguments[argIndex];
            result = (objA[propertyName] < objB[propertyName]) ? -1 : (objA[propertyName] > objB[propertyName]) ? 1 : 0;

            //Reverse if sort order is false (DESC)
            result *= !sortArguments[argIndex + 1] ? 1 : -1;
        }
        return result;
    });

}

Basically, you may specify any number of property name / sort direction:

var arr = [{
  LastName: "Doe",
  FirstName: "John",
  Age: 28
}, {
  LastName: "Doe",
  FirstName: "Jane",
  Age: 28
}, {
  LastName: "Foo",
  FirstName: "John",
  Age: 30
}];

arr.sortBy("LastName", true, "FirstName", true, "Age", false);
//Will return Jane Doe / John Doe / John Foo

arr.sortBy("Age", false, "LastName", true, "FirstName", false);
//Will return John Foo / John Doe / Jane Doe

Answer 7

Here is an implementation that uses recursion to sort by any number of sort fields from 1 to infinite. You pass it a results array which is an array of result objects to sort, and a sorts array which is an array of sort objects defining the sort. Each sort object must have a "select" key for the key name that it sorts by and an "order" key which is a string indicating "ascending" or "descending".

sortMultiCompare = (a, b, sorts) => {
    let select = sorts[0].select
    let order = sorts[0].order
    if (a[select] < b[select]) {
        return order == 'ascending' ? -1 : 1
    } 
    if (a[select] > b[select]) {
        return order == 'ascending' ? 1 : -1
    }
    if(sorts.length > 1) {
        let remainingSorts = sorts.slice(1)
        return this.sortMultiCompare(a, b, remainingSorts)
    }
    return 0
}

sortResults = (results, sorts) => {
    return results.sort((a, b) => {
        return this.sortMultiCompare(a, b, sorts)
    })
}

// example inputs
const results = [
    {
        "LastName": "Doe",
        "FirstName": "John",
        "MiddleName": "Bill"
    },
    {
        "LastName": "Doe",
        "FirstName": "Jane",
        "MiddleName": "Bill"
    },
    {
        "LastName": "Johnson",
        "FirstName": "Kevin",
        "MiddleName": "Bill"
    }
]

const sorts = [
    {
        "select": "LastName",
        "order": "ascending"
    },
    {
        "select": "FirstName",
        "order": "ascending"
    },
    {
        "select": "MiddleName",
        "order": "ascending"
    }    
]

// call the function like this:
let sortedResults = sortResults(results, sorts)

Answer 8

grouperArray.sort(function (a, b) {
  var aSize = a.gsize;
  var bSize = b.gsize;
  var aLow = a.glow;
  var bLow = b.glow;
  console.log(aLow + " | " + bLow);      
  return (aSize < bSize) ? -1 : (aSize > bSize) ? 1 : ( (aLow < bLow ) ? -1 : (aLow > bLow ) ? 1 : 0 );
});

Answer 9

grouperArray.sort(function (a, b) {
     var aSize = a.gsize;     
     var bSize = b.gsize;     
     var aLow = a.glow;
     var bLow = b.glow;
     console.log(aLow + " | " + bLow);
     return (aSize < bSize) ? -1 : (aSize > bSize) ? 1 : (aLow < bLow) ? -1 : (aLow > bLow) ? 1 : 0); }); 

Answer 10

A dynamic way to do that with MULTIPLE keys:

• filter unique values from each col/key of sort
• put in order or reverse it
• add weights width zeropad for each object based on indexOf(value) keys values
• sort using caclutated weights

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

Answer 11

This is what I use

function sort(a, b) {
    var _a = "".concat(a.size, a.glow);
    var _b = "".concat(b.size, b.glow);
    return _a < _b;
}

concat the two items as a string and they will be sorted by a string value. If you want you could wrap _a and _b with parseInt to compare them as numbers if you know they will be numerical.

Answer 12

Here is the solution for the case, when you have a priority sort key, which might not exist in some particular items, so you have to sort by fallback keys.

An input data example ( id2 is priority sort key):

const arr = [
    {id: 1},
    {id: 2, id2: 3},
    {id: 4},
    {id: 3},
    {id: 10, id2: 2},
    {id: 7},
    {id: 6, id2: 1},
    {id: 5},
    {id: 9, id2: 2},
    {id: 8},
];

And the output should be:

[ { id: 6, id2: 1 },
  { id: 9, id2: 2 },
  { id: 10, id2: 2 },
  { id: 2, id2: 3 },
  { id: 1 },
  { id: 3 },
  { id: 4 },
  { id: 5 },
  { id: 7 },
  { id: 8 } ]

The comparator function will be like:

arr.sort((a,b) => {
  if(a.id2 || b.id2) {
    if(a.id2 && b.id2) {
      if(a.id2 === b.id2) {
        return a.id - b.id;
      }
      return a.id2 - b.id2;
    }
    return a.id2 ? -1 : 1;
  }
  return a.id - b.id
});

PS In case if .id of .id2 can be zeros, consider to use typeof .

Answer 13

Let's simplify.

Say you have an array of arrays:

let tmp = [
    [0, 1],
    [2, 1],
    [1, 1],
    [0, 0],
    [2, 0],
    [1, 0],
    [0, 2],
    [2, 2],
    [1, 2],
]

Executing:

tmp.sort((a, b) => {
    if (a[1] != b[1])
        return a[1] - b[1];
    else
        return a[0] - b[0];
})

Will yield:

[
    [0, 0],
    [1, 0],
    [2, 0],
    [0, 1],
    [1, 1],
    [2, 1],
    [0, 2],
    [1, 2],
    [2, 2]
]

Answer 14

grouperArray.sort(function (a, b) {
    var aSize = a.gsize;
    var bSize = b.gsize;
    if (aSize !== aSize)
        return aSize - bSize;
    return a.glow - b.glow;
});

not tested, but I think that should work.

Answer 15

grouperArray.sort(
  function(a,b){return a.gsize == b.gsize ? a.glow - b.glow : a.gsize - b.gsize}
);

Answer 16

In my case, i sort notification list by param 'important' and by 'date'

• step 1: i filter notifications by 'important' and unImportant

  let importantNotifications = notifications.filter( (notification) => notification.isImportant); let unImportantNotifications = notifications.filter( (notification) =>.notification;isImportant);

• step 2: i sort them by date

   sortByDate = (notifications) => { return notifications.sort((notificationOne, notificationTwo) => { return notificationOne.date - notificationTwo.date; }); };

• step 3: merge them

  [...this.sortByDate(importantNotifications), ...this.sortByDate(unImportantNotifications), ];

Answer 17

If you're happy to use the new tidy.js package you can achieve this with

tidy(input_array,
  arrange(['var1', desc('var2')])
);

Answer 18

Besides the other answers here I got inconsistent data on my arrays where 1 wanted a primary ASC sort on field x and a secondary DESC sort on field y.
The solution is in giving the primary sort more importance by multiplying the number with lets say 1000000000

arrayOfObjects.sort((a, b) => {
    return (
        // Multiply by a high number to the most important sort, that makes them heavier than the second sort

        // First sort ASC (notice the - minus in the end instead of the || in other answers !)
        (a.paramX * 1000000000) -
        (b.paramX * 1000000000) -

        // Second sort DESC (switch them if you want ASC too)
        (a.paramY - b.paramY)
    )
})

for sorting on multiple dates on the object it is this:

// param date1 ASC and param date2 DESC
arrayOfObjects.sort((a, b) => {
    return (
        (a.date1.getTime() * 1000000000) -
        (b.date1.getTime() * 1000000000) -
        (a.date2.getTime() - b.date2.getTime())
    )
})

Answer 19

var items = [
    { name: "z - test item", price: "99.99", priority: 0, reviews: 309, rating: 2 },
    { name: "z - test item", price: "1.99", priority: 0, reviews: 11, rating: 0.5 },
    { name: "y - test item", price: "99.99", priority: 1, reviews: 99, rating: 1 },
    { name: "y - test item", price: "0", priority: 1, reviews: 394, rating: 3.5 },
    { name: "x - test item", price: "0", priority: 2, reviews: 249, rating: 0.5 }];

items.sort(function (a, b) {
    var nameA = a.name.toUpperCase(); 
    var nameB = b.name.toUpperCase();   
    var nameC = a.price.toUpperCase(); 
    var nameD = b.price.toUpperCase(); 

    if (nameA < nameB) {
            return -1;
        }
        if (nameA > nameB || nameC > nameD) {
            return 1;
        }

        // names must be equal
        return 0;
    });`