Related
my English is not good but i really need your help.
I want to reduce all same values in this 2 Dimensional Array but it doesn't work.
let edge = [[0,1,4],[1,2,4],[0,2,4],[2,3,4],[3,0,4],[0,2,4]];
function removeDup(edge) {
for (let i = 0; i < edge.length; i++) {
if (edge.some((item) => item > 1)) {
edge.splice(i, 1);
}
}
return edge;
}
i want to change array like this.
[[0,1,4],[1,2,4],[0,2,4],[2,3,4],[3,0,4],[0,2,4]]
to
[[0,1,4],[1,2,4],[2,3,4],[3,0,4]]
Use the the filter() array method with the every() array method as shown below:
let edge = [[0,1,4],[1,2,4],[0,2,4],[2,3,4],[3,0,4],[0,2,4]];
function removeDup(edge) {
return edge.filter(
(cur,i) => edge.slice(0,i).every(
prev => !prev.every(elm => cur.includes(elm))
)
);
}
console.log( removeDup(edge) );
//output: [[0,1,4],[1,2,4],[0,2,4],[2,3,4],[3,0,4]]
If your goal is to remove all elements that are repeated, you would need to use the whole array, edge, instead of edge.slice(0,i) and, avoid comparing an element with itself - i !== j - as shown below:
let edge = [[0,1,4],[1,2,4],[0,2,4],[2,3,4],[3,0,4],[0,2,4]];
function removeDup(edge) {
return edge.filter(
(cur,i) => edge.every(
(prev,j) => !prev.every(elm => i !== j && cur.includes(elm))
)
);
}
console.log( removeDup(edge) );
//output: [[0,1,4],[1,2,4],[2,3,4],[3,0,4]]
Another approach would be:
To convert each element array to a single value by first sorting the elements and joining them
Then use filter() to only return the element array that return a frequency of 1
Use an inner filter() to count how many times each converted element appears in the array.
[...cur] ensures original elements are not sorted, since arrays are passed by reference.
let edge = [[0,1,4],[1,2,4],[0,2,4],[2,3,4],[3,0,4],[0,2,4]];
function removeDup(edge) {
const nedge = edge.map(cur => [...cur].sort().join(''));
return edge.filter(
cur => nedge.filter(
n => n === [...cur].sort().join('')
).length === 1
);
}
console.log( removeDup(edge) );
//output: [[0,1,4],[1,2,4],[2,3,4],[3,0,4]]
This may be one possible solution to achieve the desired result:
Copying from the question:
[[0,1,4],[1,2,4],[0,2,4],[2,3,4],[3,0,4],[0,2,4]]
to
[[0,1,4],[1,2,4],[2,3,4],[3,0,4]]
Note: The desired target array has no [0, 2, 4].
Code Sample
const removeDupes = arr => (
Object.entries(
arr.reduce(
(fin, itm) =>
({...fin, [itm]: (fin[itm] || 0) + 1})
,{}
)
).filter(
([k ,v]) => v === 1
).map(x => x[0])
);
Explanation
Iterate over the array using .reduce to generate an object
The key of the resulting object will be each inner array (such as [0, 1, 4]
The value will be a count (number of times the inner array is found in edge)
Once the object is generated, iterate over its entries to filter only those which have a count of exactly 1
Now, use .map to pull only the keys of the object.
Code Snippet
let edge = [[0,1,4],[1,2,4],[0,2,4],[2,3,4],[3,0,4],[0,2,4]];
const removeDupes = arr => (
Object.entries(
arr.reduce(
(fin, itm) =>
({...fin, [itm]: (fin[itm] || 0) + 1})
,{}
)
).filter(
([k ,v]) => v === 1
).map(x => x[0])
);
console.log(removeDupes(edge));
This is the solution by using the reduce() method.
let edge = [
[0, 1, 4],
[1, 2, 4],
[0, 2, 4],
[2, 3, 4],
[3, 0, 4],
[0, 2, 4],
];
function removeDuplicate(edge) {
return edge.reduce((acc, cur) => {
!acc.some((elm) => elm.every((e, i) => cur[i] === e)) && acc.push(cur);
return acc;
}, []);
}
console.log(removeDuplicate(edge));
I have an array of javascript arrays, with each inner array being of equal length. I want to sort one of the inner arrays chronologically (its members are all numbers), and I want the other arrays to re-order in the same way. E.g.
we start with this:
[[2, 3, 1], ["deux", "trois", "un"], ["zwei", "drei", "eins"]]
and the result I want is:
[[1, 2, 3], ["un", "deux", "trois"], ["eins", "zwei", "drei"]]
I've tried different variations on the following, but had no luck:
arr.sort((a, b) => a[0] - b[0])
All I get back is exactly what I put in!
You could get an array of sorted indices and map sorted arrays.
const
arrays = [[2, 3, 1], ["deux", "trois", "un"], ["zwei", "drei", "eins"]],
sorted = arrays.map(
(indices => a => indices.map(i => a[i]))
([...arrays[0].keys()].sort((a, b) => arrays[0][a] - arrays[0][b]))
);
console.log(sorted);
.as-console-wrapper { max-height: 100% !important; top: 0; }
A function for it.
const
sortNumbers = (a, b) => a - b,
sortABC = (a, b) => a.localeCompare(b),
sortArray = (arrays, index, fn) => {
const
source = arrays[index],
indices = [...source.keys()].sort((a, b) => fn(source[a], source[b]));
return arrays.map(a => indices.map(i => a[i]));
},
arrays = [[2, 3, 1], ["deux", "trois", "un"], ["zwei", "drei", "eins"]],
sorted0 = sortArray(arrays, 0, sortNumbers),
sorted1 = sortArray(arrays, 1, sortABC);
sorted2 = sortArray(arrays, 2, sortABC);
console.log(sorted0);
console.log(sorted1);
console.log(sorted2);
.as-console-wrapper { max-height: 100% !important; top: 0; }
This should do the trick, we should get the new indexes for array of numbers then apply the same new indexes to non number arrays:
const arrays = [
[2, 3, 1],
["deux", "trois", "un"],
["zwei", "drei", "eins"],
];
const [arrNbrs] = arrays;
const newIndexes = [...arrNbrs].sort().map((nbr) => arrNbrs.indexOf(nbr));
const sortedArrays = arrays.map((subArray) =>
subArray.map((_, index) => subArray[newIndexes[index]])
);
console.log("sortedArrays", sortedArrays);
const test = [
[3, 2, 1, 3, 2, 1, 2],
["trois", "deux", "un", "trois", "deux", "un", "deux"],
["drei", "zwei", "eins", "drei", "zwei", "eins", "zwei"]
];
function sortListsAccordingToReferenceList(
listOfLists,
listIndex,
referenceComparator
) {
function defaultComparator(a, b) {
return (
((a.localeCompare) && (b.localeCompare))
? a.localeCompare(b)
: (((a < b) && -1) || ((a > b) && 1) || 0)
);
}
const comparator = (typeof referenceComparator === 'function')
? referenceComparator
: defaultComparator;
function compareReferenceItems(a, b) {
return comparator(a.value, b.value);
}
const referenceList = listOfLists[listIndex];
const lastIndexList = referenceList.map((item, idx) => ({
lastIndex: idx,
value: item
})).sort(compareReferenceItems).map(referenceItem =>
referenceItem.lastIndex
);
// console.log('lastIndexList : ', lastIndexList);
// - in case of needing to return the
// mutated original array use this block ...
//
// listOfLists.forEach((list, idx) => {
//
// const duplicates = Array.from(list);
// lastIndexList.forEach((lastIndex, idx) => {
//
// list[idx] = duplicates[lastIndex];
// });
// });
// return listOfLists; // return mutated original array.
// - ... otherwise go with pure mappping.
return listOfLists.map(list =>
list.map((item, idx) => list[lastIndexList[idx]])
);
}
console.log(
'sort every array according to numeric sort order ...',
sortListsAccordingToReferenceList(test, 0)
);
console.log(
'sort everything by French as lexicographic sort reference ...',
sortListsAccordingToReferenceList(test, 1)
);
console.log(
'sort everything by German as lexicographic sort reference ...',
sortListsAccordingToReferenceList(test, 2)
);
console.log(
'sort every array according to numeric custom sort order ...',
sortListsAccordingToReferenceList(test, 0, (a, b) =>
// comparator function for descending sort order
(((a > b) && -1) || ((a < b) && 1) || 0)
)
);
.as-console-wrapper { min-height: 100%!important; top: 0; }
Not a oneliner, but easier to read
let nested = [
[2, 3, 1],
["deux", "trois", "un"],
["zwei", "drei", "eins"]
];
let srcArr;
nested = nested.map((arr, i) => {
if (i === 0) { // the reference
srcArr = arr.slice(0); // take a copy
arr.sort((a, b) => a - b); // sort the nested one
return arr;
}
return arr.map((item, i) => arr[
srcArr.indexOf(nested[0][i]) // return in the order of the reference
]);
})
console.log(nested)
I'm trying to iterate over a simple array using recursion. For this specific case, I'm trying to recreate .map() using recursion (without using .map()!. I currently am only pushing the last element in the original array, but I want to push all into the array.
function recursiveMap (arr, func) {
let newArr = [];
if (arr.length === 1){
newArr.push(func(arr));
}
else {
newArr.push(...recursiveMap(arr.slice(1),func));
}
return newArr;
}
You need to to use func on the current item, and spread the result of calling the function on the rest of the array:
function recursiveMap(arr, func) {
return arr.length ? [func(arr[0]), ...recursiveMap(arr.slice(1), func)] : [];
}
const arr = [1, 2, 3];
const result = recursiveMap(arr, n => n * 2);
console.log(result);
Your base case seems wrong. You will need to check for an empty array:
function recursiveMap (arr, func) {
let newArr = [];
if (arr.length === 0) {
// do nothing
} else {
newArr.push(func(arr[0]));
newArr.push(...recursiveMap(arr.slice(1),func));
}
return newArr;
}
Instead you will need to call func (on the first item) when there is at least one element.
With recursion, I find it is helpful to have the base case be the very first thing you check in your function, and short the execution there. The base case for map is if the array has 0 items, in which case you would return an empty array.
if you haven't seen it before let [a, ...b] is array destructuring and a becomes the first value with b holding the remaining array. You could do the same with slice.
function recursiveMap(arr, func){
if(arr.length == 0) return [];
let [first, ...rest] = arr;
return [func(first)].concat(recursiveMap(rest, func));
}
let test = [1,2,3,4,5,6,7];
console.log(recursiveMap(test, (item) => item * 2));
EDIT
Going back to your sample I see you clearly have seen destructuring before xD, sorry. Leaving it in the answer for future readers of the answer though.
Below are a few alternatives. Each recursiveMap
does not mutate input
produces a new array as output
produces a valid result when an empty input is given, []
uses a single pure, functional expression
Destructuring assignment
const identity = x =>
x
const recursiveMap = (f = identity, [ x, ...xs ]) =>
x === undefined
? []
: [ f (x), ...recursiveMap (f, xs) ]
const square = (x = 0) =>
x * x
console.log (recursiveMap (square, [ 1, 2, 3, 4, 5 ]))
// [ 1, 4, 9, 16, 25 ]
Array slice
const identity = x =>
x
const recursiveMap = (f = identity, xs = []) =>
xs.length === 0
? []
: [ f (xs[0]), ...recursiveMap (f, xs.slice (1)) ]
const square = (x = 0) =>
x * x
console.log (recursiveMap (square, [ 1, 2, 3, 4, 5 ]))
// [ 1, 4, 9, 16, 25 ]
Additional parameter with default assignment – creates fewer intermediate values
const identity = x =>
x
const recursiveMap = (f = identity, xs = [], i = 0) =>
i >= xs.length
? []
: [ f (xs[i]) ] .concat (recursiveMap (f, xs, i + 1))
const square = (x = 0) =>
x * x
console.log (recursiveMap (square, [ 1, 2, 3, 4, 5 ]))
// [ 1, 4, 9, 16, 25 ]
Tail recursive (and cute)
const identity = x =>
x
const prepend = x => xs =>
[ x ] .concat (xs)
const compose = (f, g) =>
x => f (g (x))
const recursiveMap = (f = identity, [ x, ...xs ], then = identity) =>
x === undefined
? then ([])
: recursiveMap
( f
, xs
, compose (then, prepend (f (x)))
)
const square = (x = 0) =>
x * x
console.log (recursiveMap (square, [ 1, 2, 3, 4, 5 ]))
// [ 1, 4, 9, 16, 25 ]
// => undefined
recursiveMap (square, [ 1, 2, 3, 4, 5 ], console.log)
// [ 1, 4, 9, 16, 25 ]
// => undefined
recursiveMap (square, [ 1, 2, 3, 4, 5 ])
// => [ 1, 4, 9, 16, 25 ]
Derived from tail-recursive foldl – Note foldl chooses a similar technique used above: additional parameter with default assignment.
const identity = x =>
x
const foldl = (f = identity, acc = null, xs = [], i = 0) =>
i >= xs.length
? acc
: foldl
( f
, f (acc, xs[i])
, xs
, i + 1
)
const recursiveMap = (f = identity, xs = []) =>
foldl
( (acc, x) => acc .concat ([ f (x) ])
, []
, xs
)
const square = (x = 0) =>
x * x
console.log (recursiveMap (square, [ 1, 2, 3, 4, 5 ]))
// [ 1, 4, 9, 16, 25 ]
You could take another approach by using a third parameter for the collected values.
function recursiveMap(array, fn, result = []) {
if (!array.length) {
return result;
}
result.push(fn(array[0]));
return recursiveMap(array.slice(1), fn, result);
}
console.log(recursiveMap([1, 2, 3, 4, 5], x => x << 1));
console.log(recursiveMap([], x => x << 1));
welcome to Stack Overflow. You could either pass result to itself like in the following example:
function recursiveMap (arr, func,result=[]) {
if (arr.length === 0){
return result;
}
return recursiveMap(
arr.slice(1),
func,
result.concat([func(arr[0])])
);
}
console.log(recursiveMap([1,2,3,4],x=>(x===3)?['hello','world']:x+2));
Or define a recursive function in your function:
function recursiveMap (arr, func) {
const recur = (arr, func,result=[])=>
(arr.length === 0)
? result
: recur(
arr.slice(1),
func,
result.concat([func(arr[0])])
);
return recur(arr,func,[])
}
console.log(recursiveMap([1,2,3,4],x=>(x===3)?['hello','world']:x+2));
Add newArr.push(func(arr[0])); before calling again the function
function recursiveMap (arr, func) {
let newArr = [];
if (arr.length === 1){
newArr.push(func(arr));
}
else {
newArr.push(func(arr[0]));
newArr.push(...recursiveMap(arr.slice(1),func));
}
return newArr;
}
console.log(recursiveMap([1,2,3], function(a){return +a+2}))
Same but modified answer with bugs corrected
function recursiveMap (arr, func) {
let newArr = [];
if(arr.length){
newArr.push(func(arr[0]));
if(arr.length > 1){
newArr.push(...recursiveMap(arr.slice(1),func));
}
}
return newArr;
}
console.log(recursiveMap([1,2,3], function(a){return a+2}))
I'm trying to write a function that implements foldl in JavaScript. I'm trying to use recursion in the function but not being able to implement it.
var foldl = function(f, acc, array) {
if (array.length == 0) {
return acc;
} else {
return f(array[0], foldl(f, acc, array.slice(-1)));
}
}
console.log(foldl(function(x, y) {
return x + y
}, 0, [1, 2, 3]));
console.log(foldl(function(x,y){return x+y}, 0, [1,2,3]));
Error Message..
RangeError: Maximum call stack size exceeded
Your challenge, as mentioned above, is that you're returning an array of the last element. And you're always returning an array of the last element.
What is missing from the answers above is that they're only good for folding to the right.
For the right case, you can just use .slice(1), and that will pull everything after the head.
For the fold left case, you need to specify how far you need to go .slice(0, arr.length - 1).
const foldr = (f, acc, arr) => {
if (!arr.length) {
return acc;
} else {
const head = arr[0];
const tail = arr.slice(1);
return foldr(f, f(acc, head), tail);
}
};
foldr((x, y) => x + y, 0, [1, 2, 3])// 6
const foldl = (f, acc, arr) => {
if (!arr.length) {
return acc;
} else {
const head = arr[arr.length - 1];
const tail = arr.slice(0, arr.length - 1);
return foldl(f, f(acc, head), tail);
}
};
foldl((x, y) => x + y, 0, [3, 2, 1]); // 6
This:
array.slice(-1)
should be:
array.slice(1)
slice(-1) returns the array containing the last element. As you're using the first element of the array, you want the array without that element instead. slice(1) will return the array without the first element.
slice(-1) return only the last element. If you want to recurse over the array, use slice(1) instead, which will return all elements except the first:
var foldl = function(f, acc, array) {
if (array.length == 0) {
return acc;
} else {
return f(array[0], foldl(f, acc, array.slice(1)));
}
}
console.log(foldl(function(x, y) {
return x + y
}, 0, [1, 2, 3]));
Note that foldl and foldr can both be implemented in a way that iterates thru the input list in first-to-rest (left-to-right) order. Awkward negative indexes or calculating precise slice positions are not needed.
const Empty =
Symbol ()
const foldl = (f, acc, [ x = Empty, ...xs ]) =>
x === Empty
? acc
: foldl (f, f (acc, x), xs)
const foldr = (f, acc, [ x = Empty, ...xs ]) =>
x === Empty
? acc
: f (foldr (f, acc, xs), x)
const pair = (a,b) =>
`(${a} ${b})`
const data =
[ 1, 2, 3 ]
console.log (foldl (pair, 0, data))
// (((0 1) 2) 3)
console.log (foldr (pair, 0, data))
// (((0 3) 2) 1)
You can use xs[0] and xs.slice(1) if you don't want to use the destructuring assignment
const foldl = (f, acc, xs) =>
xs.length === 0
? acc
: foldl (f, f (acc, xs[0]), xs.slice (1))
const foldr = (f, acc, xs) =>
xs.length === 0
? acc
: f (foldr (f, acc, xs.slice (1)), xs [0])
const pair = (a,b) =>
`(${a} ${b})`
const data =
[ 1, 2, 3 ]
console.log (foldl (pair, 0, data))
// (((0 1) 2) 3)
console.log (foldr (pair, 0, data))
// (((0 3) 2) 1)
...xs via destructuring assignment used in the first solution and slice used in the second solution create intermediate values and could be a performance hit if xs is of considerable size. Below, a third solution that avoids this
const foldl = (f, acc, xs, i = 0) =>
i >= xs.length
? acc
: foldl (f, f (acc, xs[i]), xs, i + 1)
const foldr = (f, acc, xs, i = 0) =>
i >= xs.length
? acc
: f (foldr (f, acc, xs, i + 1), xs [i])
const pair = (a,b) =>
`(${a} ${b})`
const data =
[ 1, 2, 3 ]
console.log (foldl (pair, 0, data))
// (((0 1) 2) 3)
console.log (foldr (pair, 0, data))
// (((0 3) 2) 1)
Above our foldl and foldr are almost perfect drop-in replacements for natives Array.prototype.reduce and Array.prototype.reduceRight respectively. By passing i and xs to the callback, we get even closer
const foldl = (f, acc, xs, i = 0) =>
i >= xs.length
? acc
: foldl (f, f (acc, xs[i], i, xs), xs, i + 1)
const foldr = (f, acc, xs, i = 0) =>
i >= xs.length
? acc
: f (foldr (f, acc, xs, i + 1), xs[i], i, xs)
const pair = (acc, value, i, self) =>
{
console.log (acc, value, i, self)
return acc + value
}
console.log (foldl (pair, 'a', [ 'b', 'c', 'd' ]))
// a b 0 [ b, c, d ]
// ab c 1 [ b, c, d ]
// abc d 2 [ b, c, d ]
// => abcd
console.log (foldr (pair, 'z', [ 'w', 'x', 'y' ]))
// z y 2 [ x, y, z ]
// zy x 1 [ x, y, z ]
// zyx w 0 [ x, y, z ]
// => zyxw
And finally, reduce and reduceRight accept a context argument. This is important if the folding function f refers to this. If you want to support a configurable context in your own folds, it's easy
const foldl = (f, acc, xs, context = null, i = 0) =>
i >= xs.length
? acc
: foldl ( f
, f.call (context, acc, xs[i], i, xs)
, xs
, context
, i + 1
)
const foldr = (f, acc, xs, context = null, i = 0) =>
i >= xs.length
? acc
: f.call ( context
, foldr (f, acc, xs, context, i + 1)
, xs[i]
, i
, xs
)
const obj =
{ a: 1, b: 2, c: 3, d: 4, e: 5 }
// some function that uses `this`
const picker = function (acc, key) {
return [ ...acc, { [key]: this[key] } ]
}
console.log (foldl (picker, [], [ 'b', 'd', 'e' ], obj))
// [ { b: 2 }, { d: 4 }, { e: 5 } ]
console.log (foldr (picker, [], [ 'b', 'd', 'e' ], obj))
// [ { e: 5 }, { d: 4 }, { b: 2 } ]
Is there a way to return the difference between two arrays in JavaScript?
For example:
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
// need ["c", "d"]
There is a better way using ES7:
Intersection
let intersection = arr1.filter(x => arr2.includes(x));
For [1,2,3] [2,3] it will yield [2,3]. On the other hand, for [1,2,3] [2,3,5] will return the same thing.
Difference
let difference = arr1.filter(x => !arr2.includes(x));
For [1,2,3] [2,3] it will yield [1]. On the other hand, for [1,2,3] [2,3,5] will return the same thing.
For a symmetric difference, you can do:
let difference = arr1
.filter(x => !arr2.includes(x))
.concat(arr2.filter(x => !arr1.includes(x)));
This way, you will get an array containing all the elements of arr1 that are not in arr2 and vice-versa
As #Joshaven Potter pointed out on his answer, you can add this to Array.prototype so it can be used like this:
Array.prototype.diff = function(arr2) { return this.filter(x => !arr2.includes(x)); }
[1, 2, 3].diff([2, 3])
Array.prototype.diff = function(a) {
return this.filter(function(i) {return a.indexOf(i) < 0;});
};
//////////////
// Examples //
//////////////
const dif1 = [1,2,3,4,5,6].diff( [3,4,5] );
console.log(dif1); // => [1, 2, 6]
const dif2 = ["test1", "test2","test3","test4","test5","test6"].diff(["test1","test2","test3","test4"]);
console.log(dif2); // => ["test5", "test6"]
Note .indexOf() and .filter() are not available before IE9.
This answer was written in 2009, so it is a bit outdated, also it's rather educational for understanding the problem. Best solution I'd use today would be
let difference = arr1.filter(x => !arr2.includes(x));
(credits to other author here)
I assume you are comparing a normal array. If not, you need to change the for loop to a for .. in loop.
function arr_diff (a1, a2) {
var a = [], diff = [];
for (var i = 0; i < a1.length; i++) {
a[a1[i]] = true;
}
for (var i = 0; i < a2.length; i++) {
if (a[a2[i]]) {
delete a[a2[i]];
} else {
a[a2[i]] = true;
}
}
for (var k in a) {
diff.push(k);
}
return diff;
}
console.log(arr_diff(['a', 'b'], ['a', 'b', 'c', 'd']));
console.log(arr_diff("abcd", "abcde"));
console.log(arr_diff("zxc", "zxc"));
This is by far the easiest way to get exactly the result you are looking for, using jQuery:
var diff = $(old_array).not(new_array).get();
diff now contains what was in old_array that is not in new_array
The difference method in Underscore (or its drop-in replacement, Lo-Dash) can do this too:
(R)eturns the values from array that are not present in the other arrays
_.difference([1, 2, 3, 4, 5], [5, 2, 10]);
=> [1, 3, 4]
As with any Underscore function, you could also use it in a more object-oriented style:
_([1, 2, 3, 4, 5]).difference([5, 2, 10]);
Plain JavaScript
There are two possible intepretations for "difference". I'll let you choose which one you want. Say you have:
var a1 = ['a', 'b' ];
var a2 = [ 'b', 'c'];
If you want to get ['a'], use this function:
function difference(a1, a2) {
var result = [];
for (var i = 0; i < a1.length; i++) {
if (a2.indexOf(a1[i]) === -1) {
result.push(a1[i]);
}
}
return result;
}
If you want to get ['a', 'c'] (all elements contained in either a1 or a2, but not both -- the so-called symmetric difference), use this function:
function symmetricDifference(a1, a2) {
var result = [];
for (var i = 0; i < a1.length; i++) {
if (a2.indexOf(a1[i]) === -1) {
result.push(a1[i]);
}
}
for (i = 0; i < a2.length; i++) {
if (a1.indexOf(a2[i]) === -1) {
result.push(a2[i]);
}
}
return result;
}
Lodash / Underscore
If you are using lodash, you can use _.difference(a1, a2) (case 1 above) or _.xor(a1, a2) (case 2).
If you are using Underscore.js, you can use the _.difference(a1, a2) function for case 1.
ES6 Set, for very large arrays
The code above works on all browsers. However, for large arrays of more than about 10,000 items, it becomes quite slow, because it has O(n²) complexity. On many modern browsers, we can take advantage of the ES6 Set object to speed things up. Lodash automatically uses Set when it's available. If you are not using lodash, use the following implementation, inspired by Axel Rauschmayer's blog post:
function difference(a1, a2) {
var a2Set = new Set(a2);
return a1.filter(function(x) { return !a2Set.has(x); });
}
function symmetricDifference(a1, a2) {
return difference(a1, a2).concat(difference(a2, a1));
}
Notes
The behavior for all examples may be surprising or non-obvious if you care about -0, +0, NaN or sparse arrays. (For most uses, this doesn't matter.)
A cleaner approach in ES6 is the following solution.
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
Difference
a2.filter(d => !a1.includes(d)) // gives ["c", "d"]
Intersection
a2.filter(d => a1.includes(d)) // gives ["a", "b"]
Disjunctive Union (Symmetric Difference)
[ ...a2.filter(d => !a1.includes(d)),
...a1.filter(d => !a2.includes(d)) ]
To get the symmetric difference you need to compare the arrays in both ways (or in all the ways in case of multiple arrays)
ES7 (ECMAScript 2016)
// diff between just two arrays:
function arrayDiff(a, b) {
return [
...a.filter(x => !b.includes(x)),
...b.filter(x => !a.includes(x))
];
}
// diff between multiple arrays:
function arrayDiff(...arrays) {
return [].concat(...arrays.map( (arr, i) => {
const others = arrays.slice(0);
others.splice(i, 1);
const unique = [...new Set([].concat(...others))];
return arr.filter(x => !unique.includes(x));
}));
}
ES6 (ECMAScript 2015)
// diff between just two arrays:
function arrayDiff(a, b) {
return [
...a.filter(x => b.indexOf(x) === -1),
...b.filter(x => a.indexOf(x) === -1)
];
}
// diff between multiple arrays:
function arrayDiff(...arrays) {
return [].concat(...arrays.map( (arr, i) => {
const others = arrays.slice(0);
others.splice(i, 1);
const unique = [...new Set([].concat(...others))];
return arr.filter(x => unique.indexOf(x) === -1);
}));
}
ES5 (ECMAScript 5.1)
// diff between just two arrays:
function arrayDiff(a, b) {
var arrays = Array.prototype.slice.call(arguments);
var diff = [];
arrays.forEach(function(arr, i) {
var other = i === 1 ? a : b;
arr.forEach(function(x) {
if (other.indexOf(x) === -1) {
diff.push(x);
}
});
})
return diff;
}
// diff between multiple arrays:
function arrayDiff() {
var arrays = Array.prototype.slice.call(arguments);
var diff = [];
arrays.forEach(function(arr, i) {
var others = arrays.slice(0);
others.splice(i, 1);
var otherValues = Array.prototype.concat.apply([], others);
var unique = otherValues.filter(function (x, j) {
return otherValues.indexOf(x) === j;
});
diff = diff.concat(arr.filter(x => unique.indexOf(x) === -1));
});
return diff;
}
Example:
// diff between two arrays:
const a = ['a', 'd', 'e'];
const b = ['a', 'b', 'c', 'd'];
arrayDiff(a, b); // (3) ["e", "b", "c"]
// diff between multiple arrays
const a = ['b', 'c', 'd', 'e', 'g'];
const b = ['a', 'b'];
const c = ['a', 'e', 'f'];
arrayDiff(a, b, c); // (4) ["c", "d", "g", "f"]
Difference between Arrays of Objects
function arrayDiffByKey(key, ...arrays) {
return [].concat(...arrays.map( (arr, i) => {
const others = arrays.slice(0);
others.splice(i, 1);
const unique = [...new Set([].concat(...others))];
return arr.filter( x =>
!unique.some(y => x[key] === y[key])
);
}));
}
Example:
const a = [{k:1}, {k:2}, {k:3}];
const b = [{k:1}, {k:4}, {k:5}, {k:6}];
const c = [{k:3}, {k:5}, {k:7}];
arrayDiffByKey('k', a, b, c); // (4) [{k:2}, {k:4}, {k:6}, {k:7}]
You could use a Set in this case. It is optimized for this kind of operation (union, intersection, difference).
Make sure it applies to your case, once it allows no duplicates.
var a = new JS.Set([1,2,3,4,5,6,7,8,9]);
var b = new JS.Set([2,4,6,8]);
a.difference(b)
// -> Set{1,3,5,7,9}
One Liners
const unique = (a) => [...new Set(a)];
const uniqueBy = (x,f)=>Object.values(x.reduce((a,b)=>((a[f(b)]=b),a),{}));
const intersection = (a, b) => a.filter((v) => b.includes(v));
const diff = (a, b) => a.filter((v) => !b.includes(v));
const symDiff = (a, b) => diff(a, b).concat(diff(b, a));
const union = (a, b) => diff(a, b).concat(b);
const a = unique([1, 2, 3, 4, 5, 5]);
console.log(a);
const b = [4, 5, 6, 7, 8];
console.log(intersection(a, b), diff(a, b), symDiff(a, b), union(a, b));
console.log(uniqueBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
{ id: 1, name: "abc" },
],
(v) => v.id
));
const intersectionBy = (a, b, f) => a.filter((v) => b.some((u) => f(v, u)));
console.log(intersectionBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
],
[
{ id: 1, name: "abc" },
{ id: 3, name: "pqr" },
],
(v, u) => v.id === u.id
));
const diffBy = (a, b, f) => a.filter((v) => !b.some((u) => f(v, u)));
console.log(diffBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
],
[
{ id: 1, name: "abc" },
{ id: 3, name: "pqr" },
],
(v, u) => v.id === u.id
));
TypeScript
playground link
const unique = <T>(array: T[]) => [...new Set(array)];
const intersection = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => array2.includes(v));
const diff = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => !array2.includes(v));
const symDiff = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(diff(array2, array1));
const union = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(array2);
const intersectionBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => array2.some((u) => predicate(v, u)));
const diffBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => !array2.some((u) => predicate(v, u)));
const uniqueBy = <T>(
array: T[],
predicate: (v: T, i: number, a: T[]) => string
) =>
Object.values(
array.reduce((acc, value, index) => {
acc[predicate(value, index, array)] = value;
return acc;
}, {} as { [key: string]: T })
);
function diff(a1, a2) {
return a1.concat(a2).filter(function(val, index, arr){
return arr.indexOf(val) === arr.lastIndexOf(val);
});
}
Merge both the arrays, unique values will appear only once so indexOf() will be the same as lastIndexOf().
With the arrival of ES6 with sets and splat operator (at the time of being works only in Firefox, check compatibility table), you can write the following one liner:
var a = ['a', 'b', 'c', 'd'];
var b = ['a', 'b'];
var b1 = new Set(b);
var difference = [...new Set(a.filter(x => !b1.has(x)))];
which will result in [ "c", "d" ].
to subtract one array from another, simply use the snippet below:
var a1 = ['1','2','3','4','6'];
var a2 = ['3','4','5'];
var items = new Array();
items = jQuery.grep(a1,function (item) {
return jQuery.inArray(item, a2) < 0;
});
It will returns ['1,'2','6'] that are items of first array which don't exist in the second.
Therefore, according to your problem sample, following code is the exact solution:
var array1 = ["test1", "test2","test3", "test4"];
var array2 = ["test1", "test2","test3","test4", "test5", "test6"];
var _array = new Array();
_array = jQuery.grep(array2, function (item) {
return jQuery.inArray(item, array1) < 0;
});
Another way to solve the problem
function diffArray(arr1, arr2) {
return arr1.concat(arr2).filter(function (val) {
if (!(arr1.includes(val) && arr2.includes(val)))
return val;
});
}
diffArray([1, 2, 3, 7], [3, 2, 1, 4, 5]); // return [7, 4, 5]
Also, you can use arrow function syntax:
const diffArray = (arr1, arr2) => arr1.concat(arr2)
.filter(val => !(arr1.includes(val) && arr2.includes(val)));
diffArray([1, 2, 3, 7], [3, 2, 1, 4, 5]); // return [7, 4, 5]
Functional approach with ES2015
Computing the difference between two arrays is one of the Set operations. The term already indicates that the native Set type should be used, in order to increase the lookup speed. Anyway, there are three permutations when you compute the difference between two sets:
[+left difference] [-intersection] [-right difference]
[-left difference] [-intersection] [+right difference]
[+left difference] [-intersection] [+right difference]
Here is a functional solution that reflects these permutations.
Left difference:
// small, reusable auxiliary functions
const apply = f => x => f(x);
const flip = f => y => x => f(x) (y);
const createSet = xs => new Set(xs);
const filter = f => xs => xs.filter(apply(f));
// left difference
const differencel = xs => ys => {
const zs = createSet(ys);
return filter(x => zs.has(x)
? false
: true
) (xs);
};
// mock data
const xs = [1,2,2,3,4,5];
const ys = [0,1,2,3,3,3,6,7,8,9];
// run the computation
console.log( differencel(xs) (ys) );
Right difference:
differencer is trivial. It is just differencel with flipped arguments. You can write a function for convenience: const differencer = flip(differencel). That's all!
Symmetric difference:
Now that we have the left and right one, implementing the symmetric difference gets trivial as well:
// small, reusable auxiliary functions
const apply = f => x => f(x);
const flip = f => y => x => f(x) (y);
const concat = y => xs => xs.concat(y);
const createSet = xs => new Set(xs);
const filter = f => xs => xs.filter(apply(f));
// left difference
const differencel = xs => ys => {
const zs = createSet(ys);
return filter(x => zs.has(x)
? false
: true
) (xs);
};
// symmetric difference
const difference = ys => xs =>
concat(differencel(xs) (ys)) (flip(differencel) (xs) (ys));
// mock data
const xs = [1,2,2,3,4,5];
const ys = [0,1,2,3,3,3,6,7,8,9];
// run the computation
console.log( difference(xs) (ys) );
I guess this example is a good starting point to obtain an impression what functional programming means:
Programming with building blocks that can be plugged together in many different ways.
A solution using indexOf() will be ok for small arrays but as they grow in length the performance of the algorithm approaches O(n^2). Here's a solution that will perform better for very large arrays by using objects as associative arrays to store the array entries as keys; it also eliminates duplicate entries automatically but only works with string values (or values which can be safely stored as strings):
function arrayDiff(a1, a2) {
var o1={}, o2={}, diff=[], i, len, k;
for (i=0, len=a1.length; i<len; i++) { o1[a1[i]] = true; }
for (i=0, len=a2.length; i<len; i++) { o2[a2[i]] = true; }
for (k in o1) { if (!(k in o2)) { diff.push(k); } }
for (k in o2) { if (!(k in o1)) { diff.push(k); } }
return diff;
}
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
arrayDiff(a1, a2); // => ['c', 'd']
arrayDiff(a2, a1); // => ['c', 'd']
The above answer by Joshaven Potter is great. But it returns elements in array B that are not in array C, but not the other way around. For example, if var a=[1,2,3,4,5,6].diff( [3,4,5,7]); then it will output: ==> [1,2,6], but not [1,2,6,7], which is the actual difference between the two. You can still use Potter's code above but simply redo the comparison once backwards too:
Array.prototype.diff = function(a) {
return this.filter(function(i) {return !(a.indexOf(i) > -1);});
};
////////////////////
// Examples
////////////////////
var a=[1,2,3,4,5,6].diff( [3,4,5,7]);
var b=[3,4,5,7].diff([1,2,3,4,5,6]);
var c=a.concat(b);
console.log(c);
This should output: [ 1, 2, 6, 7 ]
Very Simple Solution with the filter function of JavaScript:
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
function diffArray(arr1, arr2) {
var newArr = [];
var myArr = arr1.concat(arr2);
newArr = myArr.filter(function(item){
return arr2.indexOf(item) < 0 || arr1.indexOf(item) < 0;
});
alert(newArr);
}
diffArray(a1, a2);
Array.prototype.difference = function(e) {
return this.filter(function(i) {return e.indexOf(i) < 0;});
};
eg:-
[1,2,3,4,5,6,7].difference( [3,4,5] );
=> [1, 2, 6 , 7]
How about this:
Array.prototype.contains = function(needle){
for (var i=0; i<this.length; i++)
if (this[i] == needle) return true;
return false;
}
Array.prototype.diff = function(compare) {
return this.filter(function(elem) {return !compare.contains(elem);})
}
var a = new Array(1,4,7, 9);
var b = new Array(4, 8, 7);
alert(a.diff(b));
So this way you can do array1.diff(array2) to get their difference (Horrible time complexity for the algorithm though - O(array1.length x array2.length) I believe)
function diffArray(arr1, arr2) {
var newArr = arr1.concat(arr2);
return newArr.filter(function(i){
return newArr.indexOf(i) == newArr.lastIndexOf(i);
});
}
this is works for me
If you have two list of objects
const people = [{name: 'cesar', age: 23}]
const morePeople = [{name: 'cesar', age: 23}, {name: 'kevin', age: 26}, {name: 'pedro', age: 25}]
let result2 = morePeople.filter(person => people.every(person2 => !person2.name.includes(person.name)))
Using http://phrogz.net/JS/ArraySetMath.js you can:
var array1 = ["test1", "test2","test3", "test4"];
var array2 = ["test1", "test2","test3","test4", "test5", "test6"];
var array3 = array2.subtract( array1 );
// ["test5", "test6"]
var array4 = array1.exclusion( array2 );
// ["test5", "test6"]
Pure JavaScript solution (no libraries)
Compatible with older browsers (doesn't use filter)
O(n^2)
Optional fn callback parameter that lets you specify how to compare array items
function diff(a, b, fn){
var max = Math.max(a.length, b.length);
d = [];
fn = typeof fn === 'function' ? fn : false
for(var i=0; i < max; i++){
var ac = i < a.length ? a[i] : undefined
bc = i < b.length ? b[i] : undefined;
for(var k=0; k < max; k++){
ac = ac === undefined || (k < b.length && (fn ? fn(ac, b[k]) : ac == b[k])) ? undefined : ac;
bc = bc === undefined || (k < a.length && (fn ? fn(bc, a[k]) : bc == a[k])) ? undefined : bc;
if(ac == undefined && bc == undefined) break;
}
ac !== undefined && d.push(ac);
bc !== undefined && d.push(bc);
}
return d;
}
alert(
"Test 1: " +
diff(
[1, 2, 3, 4],
[1, 4, 5, 6, 7]
).join(', ') +
"\nTest 2: " +
diff(
[{id:'a',toString:function(){return this.id}},{id:'b',toString:function(){return this.id}},{id:'c',toString:function(){return this.id}},{id:'d',toString:function(){return this.id}}],
[{id:'a',toString:function(){return this.id}},{id:'e',toString:function(){return this.id}},{id:'f',toString:function(){return this.id}},{id:'d',toString:function(){return this.id}}],
function(a, b){ return a.id == b.id; }
).join(', ')
);
To find the difference of 2 arrays without duplicates:
function difference(arr1, arr2){
let setA = new Set(arr1);
let differenceSet = new Set(arr2.filter(ele => !setA.has(ele)));
return [...differenceSet ];
}
1.difference([2,2,3,4],[2,3,3,4]) will return []
2.difference([1,2,3],[4,5,6]) will return [4,5,6]
3.difference([1,2,3,4],[1,2]) will return []
4.difference([1,2],[1,2,3,4]) will return [3,4]
Note: The above solution requires that you always send the larger array as the second parameter. To find the absolute difference, you will need to first find the larger array of the two and then work on them.
To find the absolute difference of 2 arrays without duplicates:
function absDifference(arr1, arr2){
const {larger, smaller} = arr1.length > arr2.length ?
{larger: arr1, smaller: arr2} : {larger: arr2, smaller: arr1}
let setA = new Set(smaller);
let absDifferenceSet = new Set(larger.filter(ele => !setA.has(ele)));
return [...absDifferenceSet ];
}
1.absDifference([2,2,3,4],[2,3,3,4]) will return []
2.absDifference([1,2,3],[4,5,6]) will return [4,5,6]
3.absDifference([1,2,3,4],[1,2]) will return [3,4]
4.absDifference([1,2],[1,2,3,4]) will return [3,4]
Note the example 3 from both the solutions
Here is another solution that can return the differences, just like git diff: (it has been written in typescript, if you're not using typescript version, just remove the types)
/**
* util function to calculate the difference between two arrays (pay attention to 'from' and 'to'),
* it would return the mutations from 'from' to 'to'
* #param { T[] } from
* #param { T[] } to
* #returns { { [x in string]: boolean } } it would return the stringified version of array element, true means added,
* false means removed
*/
export function arrDiff<T>(from: T[], to: T[]): { [x in string]: boolean } {
var diff: { [x in string]: boolean } = {};
var newItems: T[] = []
diff = from.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
for (var i = 0; i < to.length; i++) {
if (diff[JSON.stringify(to[i])]) {
delete diff[JSON.stringify(to[i])]
} else {
newItems.push(to[i])
}
}
return {
...Object.keys(diff).reduce((a, e) => ({ ...a, [e]: false }), {}),
...newItems.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
}
}
Here is a sample of usage:
arrDiff(['a', 'b', 'c'], ['a', 'd', 'c', 'f']) //{"b": false, "d": true, "f": true}
I wanted a similar function which took in an old array and a new array and gave me an array of added items and an array of removed items, and I wanted it to be efficient (so no .contains!).
You can play with my proposed solution here: http://jsbin.com/osewu3/12.
Can anyone see any problems/improvements to that algorithm? Thanks!
Code listing:
function diff(o, n) {
// deal with empty lists
if (o == undefined) o = [];
if (n == undefined) n = [];
// sort both arrays (or this won't work)
o.sort(); n.sort();
// don't compare if either list is empty
if (o.length == 0 || n.length == 0) return {added: n, removed: o};
// declare temporary variables
var op = 0; var np = 0;
var a = []; var r = [];
// compare arrays and add to add or remove lists
while (op < o.length && np < n.length) {
if (o[op] < n[np]) {
// push to diff?
r.push(o[op]);
op++;
}
else if (o[op] > n[np]) {
// push to diff?
a.push(n[np]);
np++;
}
else {
op++;np++;
}
}
// add remaining items
if( np < n.length )
a = a.concat(n.slice(np, n.length));
if( op < o.length )
r = r.concat(o.slice(op, o.length));
return {added: a, removed: r};
}
You can use underscore.js : http://underscorejs.org/#intersection
You have needed methods for array :
_.difference([1, 2, 3, 4, 5], [5, 2, 10]);
=> [1, 3, 4]
_.intersection([1, 2, 3], [101, 2, 1, 10], [2, 1]);
=> [1, 2]
This is working: basically merge the two arrays, look for the duplicates and push what is not duplicated into a new array which is the difference.
function diff(arr1, arr2) {
var newArr = [];
var arr = arr1.concat(arr2);
for (var i in arr){
var f = arr[i];
var t = 0;
for (j=0; j<arr.length; j++){
if(arr[j] === f){
t++;
}
}
if (t === 1){
newArr.push(f);
}
}
return newArr;
}
//es6 approach
function diff(a, b) {
var u = a.slice(); //dup the array
b.map(e => {
if (u.indexOf(e) > -1) delete u[u.indexOf(e)]
else u.push(e) //add non existing item to temp array
})
return u.filter((x) => {return (x != null)}) //flatten result
}