I have an array of JavaScript objects:
var objs = [
{ first_nom: 'Laszlo', last_nom: 'Jamf' },
{ first_nom: 'Pig', last_nom: 'Bodine' },
{ first_nom: 'Pirate', last_nom: 'Prentice' }
];
How can I sort them by the value of last_nom
in JavaScript?
I know about sort(a,b)
, but that only seems to work on strings and numbers. Do I need to add a toString()
method to my objects?
It's easy enough to write your own comparison function:
function compare( a, b ) {
if ( a.last_nom < b.last_nom ){
return -1;
}
if ( a.last_nom > b.last_nom ){
return 1;
}
return 0;
}
objs.sort( compare );
Or inline (c/o Marco Demaio):
objs.sort((a,b) => (a.last_nom > b.last_nom) ? 1 : ((b.last_nom > a.last_nom) ? -1 : 0))
Or simplified for numeric (c/o Andre Figueiredo):
objs.sort((a,b) => a.last_nom - b.last_nom); // b - a for reverse sort
Answered 2023-09-20 20:13:13
return a.last_nom.localeCompare(b.last_nom)
will work, too. - anyone return a.value - b.value;
(ASC) - anyone charCodeAt
, then use the numeric inline above for a more concise one liner: objs.sort((a,b) => a.last_nom.charCodeAt(0) - b.last_nom.charCodeAt(0));
. This avoids the ugly nested ternary. - anyone You can also create a dynamic sort function that sorts objects by their value that you pass:
function dynamicSort(property) {
var sortOrder = 1;
if(property[0] === "-") {
sortOrder = -1;
property = property.substr(1);
}
return function (a,b) {
/* next line works with strings and numbers,
* and you may want to customize it to your needs
*/
var result = (a[property] < b[property]) ? -1 : (a[property] > b[property]) ? 1 : 0;
return result * sortOrder;
}
}
So you can have an array of objects like this:
var People = [
{Name: "Name", Surname: "Surname"},
{Name:"AAA", Surname:"ZZZ"},
{Name: "Name", Surname: "AAA"}
];
...and it will work when you do:
People.sort(dynamicSort("Name"));
People.sort(dynamicSort("Surname"));
People.sort(dynamicSort("-Surname"));
Actually this already answers the question. Below part is written because many people contacted me, complaining that it doesn't work with multiple parameters.
You can use the function below to generate sort functions with multiple sort parameters.
function dynamicSortMultiple() {
/*
* save the arguments object as it will be overwritten
* note that arguments object is an array-like object
* consisting of the names of the properties to sort by
*/
var props = arguments;
return function (obj1, obj2) {
var i = 0, result = 0, numberOfProperties = props.length;
/* try getting a different result from 0 (equal)
* as long as we have extra properties to compare
*/
while(result === 0 && i < numberOfProperties) {
result = dynamicSort(props[i])(obj1, obj2);
i++;
}
return result;
}
}
Which would enable you to do something like this:
People.sort(dynamicSortMultiple("Name", "-Surname"));
For the lucky among us who can use ES6, which allows extending the native objects:
class MyArray extends Array {
sortBy(...args) {
return this.sort(dynamicSortMultiple(...args));
}
}
That would enable this:
MyArray.from(People).sortBy("Name", "-Surname");
Answered 2023-09-20 20:13:13
In ES6/ES2015 or later you can do it this way:
objs.sort((a, b) => a.last_nom.localeCompare(b.last_nom));
Prior to ES6/ES2015
objs.sort(function(a, b) {
return a.last_nom.localeCompare(b.last_nom)
});
Answered 2023-09-20 20:13:14
localeCompare
. You can use the standard >
operator - like mentioned in the answer by @muasif80 - stackoverflow.com/a/67992215/6908282 - anyone Use Underscore.js]. It’s small and awesome...
sortBy_.sortBy(list, iterator, [context]) Returns a sorted copy of list, ranked in ascending order by the results of running each value through iterator. Iterator may also be the string name of the property to sort by (eg. length).
var objs = [
{ first_nom: 'Lazslo',last_nom: 'Jamf' },
{ first_nom: 'Pig', last_nom: 'Bodine' },
{ first_nom: 'Pirate', last_nom: 'Prentice' }
];
var sortedObjs = _.sortBy(objs, 'first_nom');
Answered 2023-09-20 20:13:14
var sortedObjs = _.sortBy( objs, 'first_nom' );
. objs
will not be sorted itself as a result of this. The function will return a sorted array. That would make it more explicit. - anyone var reverseSortedObjs = _.sortBy( objs, 'first_nom' ).reverse();
- anyone <script src="http://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"> </script>
- anyone Lodash
for the ones who prefer that one - anyone var sortedObjs = _.sortBy( objs, 'first_nom' );
or if you want it in a different order: var sortedObjs = _.orderBy( objs, ['first_nom'],['dsc'] );
- anyone Case sensitive
arr.sort((a, b) => a.name > b.name ? 1 : -1);
Case Insensitive
arr.sort((a, b) => a.name.toLowerCase() > b.name.toLowerCase() ? 1 : -1);
Useful Note
If no change in order (in case of the same strings) then the condition >
will fail and -1
will be returned. But if strings are same then returning 1 or -1 will result in correct output
The other option could be to use >=
operator instead of >
var objs = [
{ first_nom: 'Lazslo', last_nom: 'Jamf' },
{ first_nom: 'Pig', last_nom: 'Bodine' },
{ first_nom: 'Pirate', last_nom: 'Prentice' }
];
// Define a couple of sorting callback functions, one with hardcoded sort key and the other with an argument sort key
const sorter1 = (a, b) => a.last_nom.toLowerCase() > b.last_nom.toLowerCase() ? 1 : -1;
const sorter2 = (sortBy) => (a, b) => a[sortBy].toLowerCase() > b[sortBy].toLowerCase() ? 1 : -1;
objs.sort(sorter1);
console.log("Using sorter1 - Hardcoded sort property last_name", objs);
objs.sort(sorter2('first_nom'));
console.log("Using sorter2 - passed param sortBy='first_nom'", objs);
objs.sort(sorter2('last_nom'));
console.log("Using sorter2 - passed param sortBy='last_nom'", objs);
Answered 2023-09-20 20:13:14
-1
and 1
for eg: from 1 : -1
to -1 : 1
- anyone (a, b) to (b, a)
:) - anyone 1
& -1
more straight forward and logical. - anyone If you have duplicate last names you might sort those by first name-
obj.sort(function(a,b){
if(a.last_nom< b.last_nom) return -1;
if(a.last_nom >b.last_nom) return 1;
if(a.first_nom< b.first_nom) return -1;
if(a.first_nom >b.first_nom) return 1;
return 0;
});
Answered 2023-09-20 20:13:14
b
should come after a
in the array. If a positive number is returned, it means a
should come after b
. If 0
is returned, it means they are considered equal. You can always read the documentation: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… - anyone 1, 0, -1
before I asked this here. I just wasn't finding the info I needed. - anyone As of 2018 there is a much shorter and elegant solution. Just use. Array.prototype.sort().
Example:
var items = [
{ name: 'Edward', value: 21 },
{ name: 'Sharpe', value: 37 },
{ name: 'And', value: 45 },
{ name: 'The', value: -12 },
{ name: 'Magnetic', value: 13 },
{ name: 'Zeros', value: 37 }
];
// sort by value
items.sort(function (a, b) {
return a.value - b.value;
});
Answered 2023-09-20 20:13:14
a.value - b.value
used to compare the object's attributes (numbers in this case) can be adopted for the various times of data. For example, regex can be used to compare each pair of the neighboring strings. - anyone Simple and quick solution to this problem using prototype inheritance:
Array.prototype.sortBy = function(p) {
return this.slice(0).sort(function(a,b) {
return (a[p] > b[p]) ? 1 : (a[p] < b[p]) ? -1 : 0;
});
}
Example / Usage
objs = [{age:44,name:'vinay'},{age:24,name:'deepak'},{age:74,name:'suresh'}];
objs.sortBy('age');
// Returns
// [{"age":24,"name":"deepak"},{"age":44,"name":"vinay"},{"age":74,"name":"suresh"}]
objs.sortBy('name');
// Returns
// [{"age":24,"name":"deepak"},{"age":74,"name":"suresh"},{"age":44,"name":"vinay"}]
Update: No longer modifies original array.
Answered 2023-09-20 20:13:14
.slice(0)
is to make a shallow copy of the array. - anyone Old answer that is not correct:
arr.sort((a, b) => a.name > b.name)
UPDATE
From Beauchamp's comment:
arr.sort((a, b) => a.name < b.name ? -1 : (a.name > b.name ? 1 : 0))
More readable format:
arr.sort((a, b) => {
if (a.name < b.name) return -1
return a.name > b.name ? 1 : 0
})
Without nested ternaries:
arr.sort((a, b) => a.name < b.name ? - 1 : Number(a.name > b.name))
Explanation: Number()
will cast true
to 1
and false
to 0
.
Answered 2023-09-20 20:13:14
arr.sort((a, b) => a.name < b.name ? -1 : (a.name > b.name ? 1 : 0))
- anyone arr.sort((a, b) => a.name > b.name ? 1 : -1
will not work? For Strings i have tested this works great. If you want case insensitive then use a.name.toLowerCase()
and b.name.toLowerCase()
- anyone Lodash (a superset of Underscore.js).
It's good not to add a framework for every simple piece of logic, but relying on well tested utility frameworks can speed up development and reduce the amount of bugs.
Lodash produces very clean code and promotes a more functional programming style. In one glimpse, it becomes clear what the intent of the code is.
The OP's issue can simply be solved as:
const sortedObjs = _.sortBy(objs, 'last_nom');
More information? For example, we have the following nested object:
const users = [
{ 'user': {'name':'fred', 'age': 48}},
{ 'user': {'name':'barney', 'age': 36 }},
{ 'user': {'name':'wilma'}},
{ 'user': {'name':'betty', 'age': 32}}
];
We now can use the _.property shorthand user.age
to specify the path to the property that should be matched. We will sort the user objects by the nested age property. Yes, it allows for nested property matching!
const sortedObjs = _.sortBy(users, ['user.age']);
Want it reversed? No problem. Use _.reverse.
const sortedObjs = _.reverse(_.sortBy(users, ['user.age']));
Want to combine both using chain?
const { chain } = require('lodash');
const sortedObjs = chain(users).sortBy('user.age').reverse().value();
Or when do you prefer flow over chain?
const { flow, reverse, sortBy } = require('lodash/fp');
const sortedObjs = flow([sortBy('user.age'), reverse])(users);
Answered 2023-09-20 20:13:14
(https://lodash.com/docs/4.17.10#orderBy)
This method is like _.sortBy
except that it allows specifying the sort orders of the iteratees to sort by. If orders is unspecified, all values are sorted in ascending order. Otherwise, specify an order of "desc" for descending or "asc" for ascending sort order of corresponding values.
Arguments
collection (Array|Object): The collection to iterate over. [iteratees=[_.identity]] (Array[]|Function[]|Object[]|string[]): The iteratees to sort by. [orders] (string[]): The sort orders of iteratees.
Returns
(Array): Returns the new sorted array.
var _ = require('lodash');
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"}
];
_.orderBy(homes, ['city', 'state', 'zip'], ['asc', 'desc', 'asc']);
Answered 2023-09-20 20:13:14
I haven't seen this particular approach suggested, so here's a terse comparison method I like to use that works for both string
and number
types:
const objs = [
{ first_nom: 'Lazslo', last_nom: 'Jamf' },
{ first_nom: 'Pig', last_nom: 'Bodine' },
{ first_nom: 'Pirate', last_nom: 'Prentice' }
];
const sortBy = fn => {
const cmp = (a, b) => -(a < b) || +(a > b);
return (a, b) => cmp(fn(a), fn(b));
};
const getLastName = o => o.last_nom;
const sortByLastName = sortBy(getLastName);
objs.sort(sortByLastName);
console.log(objs.map(getLastName));
sortBy()
sortBy()
accepts a fn
that selects a value from an object to use in comparison, and returns a function that can be passed to Array.prototype.sort()
. In this example, we're comparing o.last_nom
. Whenever we receive two objects such as
a = { first_nom: 'Lazslo', last_nom: 'Jamf' }
b = { first_nom: 'Pig', last_nom: 'Bodine' }
we compare them with (a, b) => cmp(fn(a), fn(b))
. Given that
fn = o => o.last_nom
we can expand the comparison function to (a, b) => cmp(a.last_nom, b.last_nom)
. Because of the way logical OR (||
) works in JavaScript, cmp(a.last_nom, b.last_nom)
is equivalent to
if (a.last_nom < b.last_nom) return -1;
if (a.last_nom > b.last_nom) return 1;
return 0;
Incidentally, this is called the three-way comparison "spaceship" (<=>
) operator in other languages.
Finally, here's the ES5-compatible syntax without using arrow functions:
var objs = [
{ first_nom: 'Lazslo', last_nom: 'Jamf' },
{ first_nom: 'Pig', last_nom: 'Bodine' },
{ first_nom: 'Pirate', last_nom: 'Prentice' }
];
function sortBy(fn) {
function cmp(a, b) { return -(a < b) || +(a > b); }
return function (a, b) { return cmp(fn(a), fn(b)); };
}
function getLastName(o) { return o.last_nom; }
var sortByLastName = sortBy(getLastName);
objs.sort(sortByLastName);
console.log(objs.map(getLastName));
Answered 2023-09-20 20:13:14
-(fa < fb) || +(fa > fb)
is a mistake here. That's multiple statements being condensed into one line of code. The alternative, written with an if
statement, would be much more readable whilst still being fairly concise. I think it's a mistake to sacrifice readability for prettiness. - anyone fa <=> fb
. - anyone const cmp = (a, b) => -(a < b) || +(a > b);
) Think of ["ä", "a", "c", "b"].sort(cmp)
=> ["a", "b", "c", "ä"]
, where ä
is pushed to the end. Instead you should probably update the comparison function to: const cmp = (a, b) => a.localeCompare(b);
=> ["a", "ä", "b", "c"]
Cheers and thanks for the answer ;-) - anyone localeCompare
removes the ability to sort numbers, and is also significantly slower. - anyone Instead of using a custom comparison function, you could also create an object type with custom toString()
method (which is invoked by the default comparison function):
function Person(firstName, lastName) {
this.firtName = firstName;
this.lastName = lastName;
}
Person.prototype.toString = function() {
return this.lastName + ', ' + this.firstName;
}
var persons = [ new Person('Lazslo', 'Jamf'), ...]
persons.sort();
Answered 2023-09-20 20:13:14
There are many good answers here, but I would like to point out that they can be extended very simply to achieve a lot more complex sorting. The only thing you have to do is to use the OR operator to chain comparison functions like this:
objs.sort((a,b)=> fn1(a,b) || fn2(a,b) || fn3(a,b) )
Where fn1
, fn2
, ... are the sort functions which return [-1,0,1]. This results in "sorting by fn1" and "sorting by fn2" which is pretty much equal to ORDER BY in SQL.
This solution is based on the behaviour of ||
operator which evaluates to the first evaluated expression which can be converted to true.
The simplest form has only one inlined function like this:
// ORDER BY last_nom
objs.sort((a,b)=> a.last_nom.localeCompare(b.last_nom) )
Having two steps with last_nom
,first_nom
sort order would look like this:
// ORDER_BY last_nom, first_nom
objs.sort((a,b)=> a.last_nom.localeCompare(b.last_nom) ||
a.first_nom.localeCompare(b.first_nom) )
A generic comparison function could be something like this:
// ORDER BY <n>
let cmp = (a,b,n)=>a[n].localeCompare(b[n])
This function could be extended to support numeric fields, case-sensitivity, arbitrary data types, etc.
You can use them by chaining them by sort priority:
// ORDER_BY last_nom, first_nom
objs.sort((a,b)=> cmp(a,b, "last_nom") || cmp(a,b, "first_nom") )
// ORDER_BY last_nom, first_nom DESC
objs.sort((a,b)=> cmp(a,b, "last_nom") || -cmp(a,b, "first_nom") )
// ORDER_BY last_nom DESC, first_nom DESC
objs.sort((a,b)=> -cmp(a,b, "last_nom") || -cmp(a,b, "first_nom") )
The point here is that pure JavaScript with functional approach can take you a long way without external libraries or complex code. It is also very effective, since no string parsing have to be done.
Answered 2023-09-20 20:13:14
Try this:
// Ascending sort
items.sort(function (a, b) {
return a.value - b.value;
});
// Descending sort
items.sort(function (a, b) {
return b.value - a.value;
});
// Ascending sort
items.sort((a, b) => a.value - b.value);
// Descending sort
items.sort((a, b) => b.value - a.value);
Answered 2023-09-20 20:13:14
sort
methodThe sort
method can be modified to sort anything like an array of numbers, strings and even objects using a compare function.
A compare function is passed as an optional argument to the sort method.
This compare function accepts 2 arguments generally called a and b. Based on these 2 arguments you can modify the sort method to work as you want.
sort()
method sorts a at a lower index than b. Simply a will come before b.sort()
method leaves the element positions as they are.sort()
method sorts a at greater index than b. Simply a will come after b.Use the above concept to apply on your object where a will be your object property.
var objs = [
{ first_nom: 'Lazslo', last_nom: 'Jamf' },
{ first_nom: 'Pig', last_nom: 'Bodine' },
{ first_nom: 'Pirate', last_nom: 'Prentice' }
];
function compare(a, b) {
if (a.last_nom > b.last_nom) return 1;
if (a.last_nom < b.last_nom) return -1;
return 0;
}
objs.sort(compare);
console.log(objs)
// for better look use console.table(objs)
Answered 2023-09-20 20:13:14
Example Usage:
objs.sort(sortBy('last_nom'));
Script:
/**
* @description
* Returns a function which will sort an
* array of objects by the given key.
*
* @param {String} key
* @param {Boolean} reverse
* @return {Function}
*/
const sortBy = (key, reverse) => {
// Move smaller items towards the front
// or back of the array depending on if
// we want to sort the array in reverse
// order or not.
const moveSmaller = reverse ? 1 : -1;
// Move larger items towards the front
// or back of the array depending on if
// we want to sort the array in reverse
// order or not.
const moveLarger = reverse ? -1 : 1;
/**
* @param {*} a
* @param {*} b
* @return {Number}
*/
return (a, b) => {
if (a[key] < b[key]) {
return moveSmaller;
}
if (a[key] > b[key]) {
return moveLarger;
}
return 0;
};
};
Answered 2023-09-20 20:13:14
1, 0, -1
are used for sort ordering. Even with your explanation above, which looks very good-- I'm still not quite understanding it. I always think of -1
as when using array length property, i.e.: arr.length = -1
means that the item isn't found. I'm probably mixing things up here, but could you help me understand why digits 1, 0, -1
are used to determine order? Thanks. - anyone a
and b
, if a
is greater than b
add 1 to the index of a
and place it behind b
, if a
is less than b
, subtract 1 from a
and place it in front of b
. If a
and b
are the same, add 0 to a
and leave it where it is. - anyone Write short code:
objs.sort((a, b) => a.last_nom > b.last_nom ? 1 : -1)
Answered 2023-09-20 20:13:14
1, -1, 0
- anyone a>b && 1|| -1
is equal to a> b ? 1 : -1
, operator &&
returns first logical false
value, operator ||
returns first logical true
value. - anyone I didn't see any implementation similar to mine. This version is based on the Schwartzian transform idiom.
function sortByAttribute(array, ...attrs) {
// Generate an array of predicate-objects containing
// property getter, and descending indicator
let predicates = attrs.map(pred => {
let descending = pred.charAt(0) === '-' ? -1 : 1;
pred = pred.replace(/^-/, '');
return {
getter: o => o[pred],
descend: descending
};
});
// Schwartzian transform idiom implementation. AKA "decorate-sort-undecorate"
return array.map(item => {
return {
src: item,
compareValues: predicates.map(predicate => predicate.getter(item))
};
})
.sort((o1, o2) => {
let i = -1, result = 0;
while (++i < predicates.length) {
if (o1.compareValues[i] < o2.compareValues[i])
result = -1;
if (o1.compareValues[i] > o2.compareValues[i])
result = 1;
if (result *= predicates[i].descend)
break;
}
return result;
})
.map(item => item.src);
}
Here's an example how to use it:
let games = [
{ name: 'Mashraki', rating: 4.21 },
{ name: 'Hill Climb Racing', rating: 3.88 },
{ name: 'Angry Birds Space', rating: 3.88 },
{ name: 'Badland', rating: 4.33 }
];
// Sort by one attribute
console.log(sortByAttribute(games, 'name'));
// Sort by mupltiple attributes
console.log(sortByAttribute(games, '-rating', 'name'));
Answered 2023-09-20 20:13:14
Since you probably encounter more complex data structures like this array, I would expand the solution.
Are more pluggable version based on @ege-Özcan's very lovely answer.
I encountered the below and couldn't change it. I also did not want to flatten the object temporarily. Nor did I want to use underscore / lodash, mainly for performance reasons and the fun to implement it myself.
var People = [
{Name: {name: "Name", surname: "Surname"}, Middlename: "JJ"},
{Name: {name: "AAA", surname: "ZZZ"}, Middlename:"Abrams"},
{Name: {name: "Name", surname: "AAA"}, Middlename: "Wars"}
];
The goal is to sort it primarily by People.Name.name
and secondarily by People.Name.surname
Now, in the base solution uses bracket notation to compute the properties to sort for dynamically. Here, though, we would have to construct the bracket notation dynamically also, since you would expect some like People['Name.name']
would work - which doesn't.
Simply doing People['Name']['name']
, on the other hand, is static and only allows you to go down the n-th level.
The main addition here will be to walk down the object tree and determine the value of the last leaf, you have to specify, as well as any intermediary leaf.
var People = [
{Name: {name: "Name", surname: "Surname"}, Middlename: "JJ"},
{Name: {name: "AAA", surname: "ZZZ"}, Middlename:"Abrams"},
{Name: {name: "Name", surname: "AAA"}, Middlename: "Wars"}
];
People.sort(dynamicMultiSort(['Name','name'], ['Name', '-surname']));
// Results in...
// [ { Name: { name: 'AAA', surname: 'ZZZ' }, Middlename: 'Abrams' },
// { Name: { name: 'Name', surname: 'Surname' }, Middlename: 'JJ' },
// { Name: { name: 'Name', surname: 'AAA' }, Middlename: 'Wars' } ]
// same logic as above, but strong deviation for dynamic properties
function dynamicSort(properties) {
var sortOrder = 1;
// determine sort order by checking sign of last element of array
if(properties[properties.length - 1][0] === "-") {
sortOrder = -1;
// Chop off sign
properties[properties.length - 1] = properties[properties.length - 1].substr(1);
}
return function (a,b) {
propertyOfA = recurseObjProp(a, properties)
propertyOfB = recurseObjProp(b, properties)
var result = (propertyOfA < propertyOfB) ? -1 : (propertyOfA > propertyOfB) ? 1 : 0;
return result * sortOrder;
};
}
/**
* Takes an object and recurses down the tree to a target leaf and returns it value
* @param {Object} root - Object to be traversed.
* @param {Array} leafs - Array of downwards traversal. To access the value: {parent:{ child: 'value'}} -> ['parent','child']
* @param {Number} index - Must not be set, since it is implicit.
* @return {String|Number} The property, which is to be compared by sort.
*/
function recurseObjProp(root, leafs, index) {
index ? index : index = 0
var upper = root
// walk down one level
lower = upper[leafs[index]]
// Check if last leaf has been hit by having gone one step too far.
// If so, return result from last step.
if (!lower) {
return upper
}
// Else: recurse!
index++
// HINT: Bug was here, for not explicitly returning function
// https://stackoverflow.com/a/17528613/3580261
return recurseObjProp(lower, leafs, index)
}
/**
* Multi-sort your array by a set of properties
* @param {...Array} Arrays to access values in the form of: {parent:{ child: 'value'}} -> ['parent','child']
* @return {Number} Number - number for sort algorithm
*/
function dynamicMultiSort() {
var args = Array.prototype.slice.call(arguments); // slight deviation to base
return function (a, b) {
var i = 0, result = 0, numberOfProperties = args.length;
// REVIEW: slightly verbose; maybe no way around because of `.sort`-'s nature
// Consider: `.forEach()`
while(result === 0 && i < numberOfProperties) {
result = dynamicSort(args[i])(a, b);
i++;
}
return result;
}
}
Working example on JSBin
Answered 2023-09-20 20:13:14
Combining Ege's dynamic solution with Vinay's idea, you get a nice robust solution:
Array.prototype.sortBy = function() {
function _sortByAttr(attr) {
var sortOrder = 1;
if (attr[0] == "-") {
sortOrder = -1;
attr = attr.substr(1);
}
return function(a, b) {
var result = (a[attr] < b[attr]) ? -1 : (a[attr] > b[attr]) ? 1 : 0;
return result * sortOrder;
}
}
function _getSortFunc() {
if (arguments.length == 0) {
throw "Zero length arguments not allowed for Array.sortBy()";
}
var args = arguments;
return function(a, b) {
for (var result = 0, i = 0; result == 0 && i < args.length; i++) {
result = _sortByAttr(args[i])(a, b);
}
return result;
}
}
return this.sort(_getSortFunc.apply(null, arguments));
}
Usage:
// Utility for printing objects
Array.prototype.print = function(title) {
console.log("************************************************************************");
console.log("**** " + title);
console.log("************************************************************************");
for (var i = 0; i < this.length; i++) {
console.log("Name: " + this[i].FirstName, this[i].LastName, "Age: " + this[i].Age);
}
}
// Setup sample data
var arrObj = [{
FirstName: "Zach",
LastName: "Emergency",
Age: 35
},
{
FirstName: "Nancy",
LastName: "Nurse",
Age: 27
},
{
FirstName: "Ethel",
LastName: "Emergency",
Age: 42
},
{
FirstName: "Nina",
LastName: "Nurse",
Age: 48
},
{
FirstName: "Anthony",
LastName: "Emergency",
Age: 44
},
{
FirstName: "Nina",
LastName: "Nurse",
Age: 32
},
{
FirstName: "Ed",
LastName: "Emergency",
Age: 28
},
{
FirstName: "Peter",
LastName: "Physician",
Age: 58
},
{
FirstName: "Al",
LastName: "Emergency",
Age: 51
},
{
FirstName: "Ruth",
LastName: "Registration",
Age: 62
},
{
FirstName: "Ed",
LastName: "Emergency",
Age: 38
},
{
FirstName: "Tammy",
LastName: "Triage",
Age: 29
},
{
FirstName: "Alan",
LastName: "Emergency",
Age: 60
},
{
FirstName: "Nina",
LastName: "Nurse",
Age: 54
}
];
//Unit Tests
arrObj.sortBy("LastName").print("LastName Ascending");
arrObj.sortBy("-LastName").print("LastName Descending");
arrObj.sortBy("LastName", "FirstName", "-Age").print("LastName Ascending, FirstName Ascending, Age Descending");
arrObj.sortBy("-FirstName", "Age").print("FirstName Descending, Age Ascending");
arrObj.sortBy("-Age").print("Age Descending");
Answered 2023-09-20 20:13:14
One more option:
var someArray = [...];
function generateSortFn(prop, reverse) {
return function (a, b) {
if (a[prop] < b[prop]) return reverse ? 1 : -1;
if (a[prop] > b[prop]) return reverse ? -1 : 1;
return 0;
};
}
someArray.sort(generateSortFn('name', true));
It sorts ascending by default.
Answered 2023-09-20 20:13:14
A simple way:
objs.sort(function(a,b) {
return b.last_nom.toLowerCase() < a.last_nom.toLowerCase();
});
See that '.toLowerCase()'
is necessary to prevent erros
in comparing strings.
Answered 2023-09-20 20:13:14
objs.sort( (a,b) => b.last_nom.toLowerCase() < a.last_nom.toLowerCase() );
- anyone Warning!
Using this solution is not recommended as it does not result in a sorted array. It is being left here for future reference, because the idea is not rare.
objs.sort(function(a,b){return b.last_nom>a.last_nom})
Answered 2023-09-20 20:13:14
This is my take on this:
The order
parameter is optional and defaults to "ASC" for ascending order.
It works on accented characters and it's case insensitive.
Note: It sorts and returns the original array.
function sanitizeToSort(str) {
return str
.normalize('NFD') // Remove accented and diacritics
.replace(/[\u0300-\u036f]/g, '') // Remove accented and diacritics
.toLowerCase() // Sort will be case insensitive
;
}
function sortByProperty(arr, property, order="ASC") {
arr.forEach((item) => item.tempProp = sanitizeToSort(item[property]));
arr.sort((a, b) => order === "ASC" ?
a.tempProp > b.tempProp ? 1 : a.tempProp < b.tempProp ? -1 : 0
: a.tempProp > b.tempProp ? -1 : a.tempProp < b.tempProp ? 1 : 0
);
arr.forEach((item) => delete item.tempProp);
return arr;
}
Snippet
function sanitizeToSort(str) {
return str
.normalize('NFD') // Remove accented characters
.replace(/[\u0300-\u036f]/g, '') // Remove diacritics
.toLowerCase()
;
}
function sortByProperty(arr, property, order="ASC") {
arr.forEach((item) => item.tempProp = sanitizeToSort(item[property]));
arr.sort((a, b) => order === "ASC" ?
a.tempProp > b.tempProp ? 1 : a.tempProp < b.tempProp ? -1 : 0
: a.tempProp > b.tempProp ? -1 : a.tempProp < b.tempProp ? 1 : 0
);
arr.forEach((item) => delete item.tempProp);
return arr;
}
const rockStars = [
{ name: "Axl",
lastname: "Rose" },
{ name: "Elthon",
lastname: "John" },
{ name: "Paul",
lastname: "McCartney" },
{ name: "Lou",
lastname: "Reed" },
{ name: "freddie", // Works on lower/upper case
lastname: "mercury" },
{ name: "Ámy", // Works on accented characters too
lastname: "winehouse"}
];
sortByProperty(rockStars, "name");
console.log("Ordered by name A-Z:");
rockStars.forEach((item) => console.log(item.name + " " + item.lastname));
sortByProperty(rockStars, "lastname", "DESC");
console.log("\nOrdered by lastname Z-A:");
rockStars.forEach((item) => console.log(item.lastname + ", " + item.name));
Answered 2023-09-20 20:13:14
Given the original example:
var objs = [
{ first_nom: 'Lazslo', last_nom: 'Jamf' },
{ first_nom: 'Pig', last_nom: 'Bodine' },
{ first_nom: 'Pirate', last_nom: 'Prentice' }
];
Sort by multiple fields:
objs.sort(function(left, right) {
var last_nom_order = left.last_nom.localeCompare(right.last_nom);
var first_nom_order = left.first_nom.localeCompare(right.first_nom);
return last_nom_order || first_nom_order;
});
Notes
a.localeCompare(b)
is universally supported and returns -1,0,1 if a<b
,a==b
,a>b
respectively.||
in the last line gives last_nom
priority over first_nom
.var age_order = left.age - right.age;
return -last_nom_order || -first_nom_order || -age_order;
Answered 2023-09-20 20:13:14
A simple function that sorts an array of object by a property:
function sortArray(array, property, direction) {
direction = direction || 1;
array.sort(function compare(a, b) {
let comparison = 0;
if (a[property] > b[property]) {
comparison = 1 * direction;
} else if (a[property] < b[property]) {
comparison = -1 * direction;
}
return comparison;
});
return array; // Chainable
}
Usage:
var objs = [
{ first_nom: 'Lazslo', last_nom: 'Jamf' },
{ first_nom: 'Pig', last_nom: 'Bodine' },
{ first_nom: 'Pirate', last_nom: 'Prentice' }
];
sortArray(objs, "last_nom"); // Asc
sortArray(objs, "last_nom", -1); // Desc
Answered 2023-09-20 20:13:14
Additional desc parameters for Ege Özcan's code:
function dynamicSort(property, desc) {
if (desc) {
return function (a, b) {
return (a[property] > b[property]) ? -1 : (a[property] < b[property]) ? 1 : 0;
}
}
return function (a, b) {
return (a[property] < b[property]) ? -1 : (a[property] > b[property]) ? 1 : 0;
}
}
Answered 2023-09-20 20:13:14
Sorting objects with Intl.Collator
for the specific case when you want natural sorting (i.e. 1,2,10,11,111
).
const files = [
{name: "1.mp3", size: 123},
{name: "10.mp3", size: 456},
{name: "100.mp3", size: 789},
{name: "11.mp3", size: 123},
{name: "111.mp3", size: 456},
{name: "2.mp3", size: 789},
];
const naturalCollator = new Intl.Collator(undefined, {numeric: true, sensitivity: 'base'});
files.sort((a, b) => naturalCollator.compare(a.name, b.name));
console.log(files);
Note: the undefined
constructor argument for Intl.Collator
represents the locale, which can be an explicit ISO 639-1 language code such as en
, or the system default locale when undefined
.
Answered 2023-09-20 20:13:14
Using Ramda,
npm install ramda
import R from 'ramda'
var objs = [
{ first_nom: 'Lazslo', last_nom: 'Jamf' },
{ first_nom: 'Pig', last_nom: 'Bodine' },
{ first_nom: 'Pirate', last_nom: 'Prentice' }
];
var ascendingSortedObjs = R.sortBy(R.prop('last_nom'), objs)
var descendingSortedObjs = R.reverse(ascendingSortedObjs)
Answered 2023-09-20 20:13:14