Javascript swap array elements

问题:

Is there any simpler way to swap two elements in an array?

var a = list[x], b = list[y]; list[y] = a; list[x] = b; 

回答1:

You only need one temporary variable.

var b = list[y]; list[y] = list[x]; list[x] = b; 

回答2:

If you want a single expression, using native javascript, remember that the return value from a splice operation contains the element(s) that was removed.

var A = [1, 2, 3, 4, 5, 6, 7, 8, 9], x= 0, y= 1; A[x] = A.splice(y, 1, A[x])[0]; alert(A); // alerts "2,1,3,4,5,6,7,8,9" 

Edit:

The [0] is necessary at the end of the expression as Array.splice() returns an array, and in this situation we require the single element in the returned array.

回答3:

This seems ok....

var b = list[y]; list[y] = list[x]; list[x] = b; 

Howerver using

var b = list[y]; 

means a b variable is going to be to be present for the rest of the scope. This can potentially lead to a memory leak. Unlikely, but still better to avoid.

Maybe a good idea to put this into Array.prototype.swap

Array.prototype.swap = function (x,y) {   var b = this[x];   this[x] = this[y];   this[y] = b;   return this; } 

which can be called like:

list.swap( x, y ) 

This is a clean approach to both avoiding memory leaks and DRY.

回答4:

Well, you don't need to buffer both values - only one:

var tmp = list[x]; list[x] = list[y]; list[y] = tmp; 

回答5:

With numeric values you can avoid a temporary variable by using bitwise xor

list[x] = list[x] ^ list[y]; list[y] = list[y] ^ list[x]; list[x] = list[x] ^ list[y]; 

or an arithmetic sum (noting that this only works if x + y is less than the maximum value for the data type)

list[x] = list[x] + list[y]; list[y] = list[x] - list[y]; list[x] = list[x] - list[y]; 

回答6:

You can swap elements in an array the following way:

list[x] = [list[y],list[y]=list[x]][0] 

See the following example:

list = [1,2,3,4,5] list[1] = [list[3],list[3]=list[1]][0] //list is now [1,4,3,2,5] 

Note: it works the same way for regular variables

var a=1,b=5; a = [b,b=a][0] 

回答7:

According to some random person on Metafilter, "Recent versions of Javascript allow you to do swaps (among other things) much more neatly:"

[ list[x], list[y] ] = [ list[y], list[x] ]; 

My quick tests showed that this Pythonic code works great in the version of JavaScript currently used in "Google Apps Script" (".gs"). Alas, further tests show this code gives a "Uncaught ReferenceError: Invalid left-hand side in assignment." in whatever version of JavaScript (".js") is used by Google Chrome Version 24.0.1312.57 m.

回答8:

Digest from http://www.greywyvern.com/?post=265

var a = 5, b = 9;     b = (a += b -= a) - b;     alert([a, b]); // alerts "9, 5" 

回答9:

To swap two consecutive elements of array

array.splice(IndexToSwap,2,array[IndexToSwap+1],array[IndexToSwap]); 

回答10:

You can swap any number of objects or literals, even of different types, using a simple identity function like this:

var swap = function (x){return x}; b = swap(a, a=b); c = swap(a, a=b, b=c); 

For your problem:

var swap = function (x){return x}; list[y]  = swap(list[x], list[x]=list[y]); 

This works in JavaScript because it accepts additional arguments even if they are not declared or used. The assignments a=b etc, happen after a is passed into the function.

回答11:

var a = [1,2,3,4,5], b=a.length;  for (var i=0; i

回答12:

This didn't exist 7 years ago but no with the new hot sauce of es2015 and up you can use array destructuring to

var a = 1, b = 2; [a, b] = [b, a]; --> a = 2, b = 1 

回答13:

For two or more elements (fixed number)

[list[y], list[x]] = [list[x], list[y]]; 

No temporary variable required!

I was thinking about simply calling list.reverse().
But then I realised it would work as swap only when list.length = x + y + 1.

For variable number of elements

I have looked into various modern Javascript constructions to this effect, including Map and map, but sadly none has resulted in a code that was more compact or faster than this old-fashioned, loop-based construction:

function multiswap(arr,i0,i1) {/* argument immutable if string */     if (arr.split) return multiswap(arr.split(""), i0, i1).join("");     var diff = [];     for (let i in i0) diff[i0[i]] = arr[i1[i]];     return Object.assign(arr,diff); }  Example:     var alphabet = "abcdefghijklmnopqrstuvwxyz";     var [x,y,z] = [14,6,15];     var output = document.getElementsByTagName("code");     output[0].innerHTML = alphabet;     output[1].innerHTML = multiswap(alphabet, [0,25], [25,0]);     output[2].innerHTML = multiswap(alphabet, [0,25,z,1,y,x], [25,0,x,y,z,3]);

Input:
                        
Swap two elements:
            
 

回答14:

There is one interesting way of swapping:

var a = 1; var b = 2; [a,b] = [b,a]; 

(ES6 way)

回答15:

Here's a compact version swaps value at i1 with i2 in arr

arr.slice(0,i1).concat(arr[i2],arr.slice(i1+1,i2),arr[i1],arr.slice(i2+1)) 

回答16:

Here's a one-liner that doesn't mutate list:

let newList = Object.assign([], list, {[x]: list[y], [y]: list[x]})

(Uses language features not available in 2009 when the question was posted!)

回答17:

what about Destructuring_assignment

var arr = [1, 2, 3, 4] [arr[index1], arr[index2]] = [arr[index2], arr[index1]] 

which can also be extended to

[src order elements] => [dest order elements] 

回答18:

Here is a variation that first checks if the index exists in the array:

Array.prototype.swapItems = function(a, b){     if(  !(a in this) || !(b in this) )         return this;     this[a] = this.splice(b, 1, this[a])[0];     return this; } 

It currently will just return this if the index does not exist, but you could easily modify behavior on fail

回答19:

Just for the fun of it, another way without using any extra variable would be:

var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];  // swap index 0 and 2 arr[arr.length] = arr[0];   // copy idx1 to the end of the array arr[0] = arr[2];            // copy idx2 to idx1 arr[2] = arr[arr.length-1]; // copy idx1 to idx2 arr.length--;               // remove idx1 (was added to the end of the array)   console.log( arr ); // -> [3, 2, 1, 4, 5, 6, 7, 8, 9]

回答20:

If need swap first and last elements only:

array.unshift( array.pop() ); 

回答21:

Here is the correct way to do it:

Array.prototype.swap = function(a, b) {   var temp = this[a];   this[a] = this[b];   this[b] = temp; }; 

Usage:

var myArray = [0,1,2,3,4...]; myArray.swap(4,1); 

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