Created
October 21, 2011 21:40
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Easiest way to find duplicate values in a JavaScript array - Native unique function implementation
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//lo sauer, 2011; lsauer.com | |
/** | |
* I saw this thread: http://stackoverflow.com/questions/840781/easiest-way-to-find-duplicate-values-in-a-javascript-array | |
* The solutions above lacked the elegance that can be done a with map-reduce-like operations | |
* Since this implementation works with native functions, the speed is in most circumstances faster | |
* than any solution using scripted-logic | |
* Additionally, I needed to quickly filter duplicate url-entries for: http://lsauer.github.com/chrome-session-restore/ | |
*/ | |
//copy and paste: without error handling | |
Array.prototype.unique = function(){return this.sort().filter( function(v,i,o){if(i>=0 && v!==o[i-1]) return v;});} | |
//copy and paste: with error handling | |
Array.prototype.unique = function(){if(!(this instanceof Array))throw TypeError('Not an Array!'); return this.sort().filter( function(v,i,o){if(i>=0 && v!==o[i-1]) return v;});} | |
/** | |
* Numbers | |
*/ | |
var arr = [324,3,32,5,52,2100,1,20,2,3,3,2,2,2,1,1,1]; | |
//1. sorting / map | |
var a = arr.sort(); | |
>>>[1, 1, 1, 1, 2, 2, 2, 2, 20, 2100, 3, 3, 3, 32, 324, 5, 52] | |
//2. reduce | |
//Note: if you need to copy the array at any point use Array.slice() | |
a.filter( function(v,i,o){if(i>=0 && v!==o[i-1]) return v;}); | |
[2, 20, 2100, 3, 32, 324, 5, 52] | |
/** | |
* Strings | |
*/ | |
var a = 'Magic belongs to Jerry Harry Jerry Harry Potter and Banana Joe'.split(' '); | |
a = a.sort() | |
>>>["Banana", "Harry", "Harry", "Jerry", "Jerry", "Joe", "Magic", "Potter", "and", "belongs", "to"] | |
a.filter( function(v,i,o){if(i>=0 && v!==o[i-1]) return v;}); | |
["Harry", "Jerry", "Joe", "Magic", "Potter", "and", "belongs", "to"] | |
/** | |
* Additional information | |
*/ | |
//you can also use the compact implementation used in is-lib ( github.com/lsauer/is-library ) | |
a.filter( function(v,i,o){ return 1+i&&v!==o[i-1]?v:0;}); | |
//...or a case insensitive function | |
a.filter( function(v,i,o){ return !i||v&&!RegExp(o[i-1],'i').test(v)}); | |
//example | |
var a = 'Magic belongs to Jerry Harry Jerry JERRY AND HARRY Harry Potter and Banana Joe'.split(' '); | |
a.sort() | |
>>>["AND", "Banana", "HARRY", "Harry", "Harry", "JERRY", "Jerry", "Jerry", "Joe", "Magic", "Potter", "and", "belongs", "to"] | |
a.filter( function(v,i,o){ return i&&v&&!RegExp(o[i-1],'i').test(v)?v:0}); | |
>>>["Banana", "HARRY", "JERRY", "Joe", "Magic", "Potter", "and", "belongs", "to"] |
What about strings with numbers such as Zone1, Zone 1 & 2, Zone 1, 2 & 3
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Ok. To regard the index position 0, I could either remove
1+i&&..
or removei&&
altogether. 1+ to mflodin. thanks & cheersI just saw my original filter function correctly stated
1+i&&
all along, but got lost during its evolution in the altered filters underneath... go figure :).