JavaScript deletes duplicates in an array (uniq) _javascript tips

Source: Internet
Author: User
Code that can be used directly: the cloud-dwelling Community revised edition
<script> function Unique (data) {data = Data | | []; var a = {}; len = data.length; for (var i=0 i<len;i++) {var v = data[i]; if (typeof (A[v]) = = ' undefined ') {a[v] = 1; } }; data.length=0; for (var i in a) {data[data.length] = i; } return data; function test () {var arr = [9,1,3,8,7,7,6,6,5,7,8,8,7,4,3,1]; var arr1 = unique (arr); Alert (Arr1.join (",")); Test (); </script>
[Ctrl + A All SELECT Note: If the need to introduce external JS need to refresh to perform]

Here is the advanced tutorials and instructions, like with in-depth friends can refer to the next.
First of all, let's look at the YUI is how to deal with:
Copy Code code as follows:

var toobject = function (a) {
var o = {};
for (var i = 0; i < a.length i = i+1) {
O[a[i]] = true;
}
return o;
};

var keys = function (o) {
var a=[], I;
For (i in O) {
if (Lang.hasownproperty (o, i)) {//Yui's method
A.push (i);
}
}
return A;
};

var uniq = function (a) {
return keys (Toobject (a));
};

Detailed analysis, see colleagues in the days of sharing "cleverly remove the duplicates in the array."

The way you use is very similar to YUI's way, but only use a loop to complete the deletion of duplicates in the array, as follows:
Copy Code code as follows:

var uniq = function (arr) {
var a = [],
o = {},
I
V
len = arr.length;

if (Len < 2) {
return arr;
}

for (i = 0; i < len; i++) {
v = arr[i];
if (O[v]!== 1) {
A.push (v);
O[V] = 1;
}
}

return A;
}

After a simple test: their use of the way the performance is much higher than the YUI way.

We are welcome to provide a better way to deal with this.

December 28, 2009 Update:

Neither of these function methods can temporarily handle complex arrays of mixed types (thanks to kittens ' questions), such as: [0, "0", 1, "1", 0], ["null", NULL].

We can use the improved function method (thanks to the Closurecache) for the ability to contract the type as a number (note: A number that requires a non-0 start) or an array of strings:
Copy Code code as follows:

var uniq = function (arr) {
var a = [],
o = {},
I
V
CV,//corrected value
len = arr.length;

if (Len < 2) {
return arr;
}

for (i = 0; i < len; i++) {
v = arr[i];

/* Closurecache is using CV = v + 0 in the function provided by
* This will not be able to distinguish a similar [1, 10, "1", "10"] array,
* Because after the Operation => 1, 10, 10, 100, it is obvious that a duplicate identifier appears.
* Is there any problem in front of the plus?
* Some: The array can not appear similar to 01, 001, starting with 0 digits,
* But the applicability is wider than previously.
*/
CV = 0 + V;

if (!O[CV]) {
A.push (v);
O[CV] = true;
}
}

return A;
}

If you want to solve the problem on the basis of ideas, a little more perfect, recommend Dexter.yy method, type judgment, give a unique indicator, see comments on the 20 floor.

There is no best, only the most appropriate way, in fact, using Array.indexof () is also a good idea to choose, for the browser has been supported directly with the native Array.indexof () method, for unsupported, we add Array.indexof () method, as follows:
Copy Code code as follows:

if (! ARRAY.PROTOTYPE.INDEXOF) {
Array.prototype.indexOf = function (obj, fromindex) {
if (Fromindex = = null) {
Fromindex = 0;
else if (Fromindex < 0) {
Fromindex = Math.max (0, this.length + fromindex);
}

for (var i = Fromindex i < this.length; i++) {
if (this[i] = = obj)
return i;
}
return-1;
};
}

Next, the process of implementation is very simple.

For optimization tips for implementing a scenario using the Array.indexof () method: When the same value is found, remove from the array to reduce the amount of the next traversal.
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