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| function removeDuplicates(arr) { | |
| var clean = [] | |
| var cleanLen = 0 | |
| var arrLen = arr.length | |
| for (var i = 0; i < arrLen; i++) { | |
| var el = arr[i] | |
| var duplicate = false | |
| for (var j = 0; j < cleanLen; j++) { | |
| if (el !== clean[j]) continue | |
| duplicate = true | |
| break | |
| } | |
| if (duplicate) continue | |
| clean[cleanLen++] = el | |
| } | |
| return clean | |
| } |
bendc
commented
Dec 9, 2014
An even faster way would be to add all items to a Set and then pull them all out again.
Tried out @mathiasbynens suggestion in a dumb and dirty JSPerf - http://jsperf.com/dumb-remove-duplicates-from-array
Set can be used to improve the performance of a uniqing function by loading it up with values, use set.add(v) instead of new Set(array) for a wider range of support, and then checking if the value exists in your loop, set.has(v), instead of using an indexOf linear search equiv.
There is a cost to creating and populating the set so I usually don't kick in the set optimization until an array is considered large enough to outweigh the cost, in my case 200, but mileage will vary on your implementation.
Set is great for uniqing objects but not so great at primitives. Since uniqing numbers is a common case I special case numbers using an hash map for better performance.
Quick followup: the following usage of Set seems to be faster than my method for arrays larger than ~130 elements.
var set = new Set(arr)
var clean = []
set.forEach(function(el) { clean.push(el) })
Thank you
Thank you