Of Stack

Realization of array duplication removal method and efficiency ratio by js

  • 2021-07-21 06:50:36
  • OfStack

The reason why these methods are searched in piles on the reality sheet is mainly because I forgot how other methods are practical after learning 1, so I made a try to see which one is the most efficient. For better display effect, I will first make a try to sum up and sum up the method that is better than others. (Reminder: The following is just a tussle, and it is easy to pollute the overall situation if you don't build a tussle on the prototype under the same situation.)

1. Find the correspondence through the past 


var n = [14,12,2,2,2,5,32,2,59,5,6,33,12,32,6];
Array.prototype.unique1 = function(){
 var obj = {},
  ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(!obj[this[i]]){
   obj[this[i]] = 1;
   ary.push(this[i]);
  }
 }
 return ary.sort(function(a,b){return a - b});
}
console.log(n.unique1());

2. Find a number of positions through the past 


var n = [14,12,2,2,2,5,32,2,59,5,6,33,12,32,6];
Array.prototype.unique2 = function(){
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(ary.indexOf(this[i]) == -1) ary.push(this[i]);
 }
 return ary.sort(function(a,b){return a - b});
}
console.log(n.unique2());

3. It is similar to looking for the position of the number. Is the position of searching for the first appearance of the number word different from the previous position 


var n = [14,12,2,2,2,5,32,2,59,5,6,33,12,32,6];
Array.prototype.unique3 = function(){
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(this.indexOf(this[i]) == i) ary.push(this[i]);
 }
 return ary.sort(function(a,b){return a - b});
}
console.log(n.unique3());

4. Compare with each other, sort first, and then compare with each other. Are the two numbers equal 


var n = [14,12,2,2,2,5,32,2,59,5,6,33,12,32,6];
Array.prototype.unique4 = function(){
 this.sort(function(a,b){return a - b});
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(this[i] !== this[i-1]) ary.push(this[i]);
 }
 return ary;
}
console.log(n.unique4());

Next, there is a less efficiency ratio, which is also the emphasis of this article. We have a number of time functions for 1 time ratio.

First, generate the number of 1 hundred words 


Array.prototype.unique1 = function(){
 var obj = {},
  ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(!obj[this[i]]){
   obj[this[i]] = 1;
   ary.push(this[i]);
  }
 }
 return ary.sort(function(a,b){return a - b});
}
Array.prototype.unique2 = function(){
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(ary.indexOf(this[i]) == -1) ary.push(this[i]);
 }
 return ary.sort(function(a,b){return a - b});
}
Array.prototype.unique3 = function(){
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(this.indexOf(this[i]) == i) ary.push(this[i]);
 }
 return ary.sort(function(a,b){return a - b});
}
Array.prototype.unique4 = function(){
 this.sort(function(a,b){return a - b});
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(this[i] !== this[i-1]) ary.push(this[i]);
 }
 return ary;
}
function randomAry (n) {
 var ary = [],i=0;
 for(; i<n; i++){
  ary.push(Math.ceil(Math.random()*10000));
 }
  console.log(ary)
 return ary;
}
function useTime (fn) {
 var start = new Date();
 fn();
 var end = new Date();
 console.log(' This letter has cost me: ' + (end - start) + ' Milliseconds ');
}
var ary = randomAry(100),
 fn1 = function(){
 ary.unique1()
 },
 fn2 = function(){
 ary.unique2()
 },
 fn3 = function(){
 ary.unique3()
 },
 fn4 = function(){
 ary.unique4()
 };
useTime(fn1);
useTime(fn2);
useTime(fn3);
useTime(fn4);

The knot is made in my Google scout

Method 1: 0 milliseconds

Method 2: 1 millisecond

Method 3: 0 milliseconds

Method 4:00 milliseconds

(Well, sure enough, the current generation of scramblers is incomparable, and they don't spit out the old scramblers.)

Look at 1000 words 


Array.prototype.unique1 = function(){
 var obj = {},
  ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(!obj[this[i]]){
   obj[this[i]] = 1;
   ary.push(this[i]);
  }
 }
 return ary.sort(function(a,b){return a - b});
}
Array.prototype.unique2 = function(){
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(ary.indexOf(this[i]) == -1) ary.push(this[i]);
 }
 return ary.sort(function(a,b){return a - b});
}
Array.prototype.unique3 = function(){
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(this.indexOf(this[i]) == i) ary.push(this[i]);
 }
 return ary.sort(function(a,b){return a - b});
}
Array.prototype.unique4 = function(){
 this.sort(function(a,b){return a - b});
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(this[i] !== this[i-1]) ary.push(this[i]);
 }
 return ary;
}
function randomAry (n) {
 var ary = [],i=0;
 for(; i<n; i++){
  ary.push(Math.ceil(Math.random()*10000));
 }
  console.log(ary)
 return ary;
}
function useTime (fn) {
 var start = new Date();
 fn();
 var end = new Date();
 console.log(' This letter has cost me: ' + (end - start) + ' Milliseconds ');
}
var ary = randomAry(1000),
 fn1 = function(){
 ary.unique1()
 },
 fn2 = function(){
 ary.unique2()
 },
 fn3 = function(){
 ary.unique3()
 },
 fn4 = function(){
 ary.unique4()
 };
useTime(fn1);
useTime(fn2);
useTime(fn3);
useTime(fn4);
 Method 1 : 1-2 Milliseconds 
 Method 2 : 40-50 Milliseconds 
 Method 3 : 40-50 Milliseconds 
 Method 4 : 0-1 Milliseconds 
 Take a look 10000 Counting words (the number is as big as it is, it can't stand it at first, and so on 56 Seconds) 
Array.prototype.unique1 = function(){
 var obj = {},
  ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(!obj[this[i]]){
   obj[this[i]] = 1;
   ary.push(this[i]);
  }
 }
 return ary.sort(function(a,b){return a - b});
}
Array.prototype.unique2 = function(){
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(ary.indexOf(this[i]) == -1) ary.push(this[i]);
 }
 return ary.sort(function(a,b){return a - b});
}
Array.prototype.unique3 = function(){
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(this.indexOf(this[i]) == i) ary.push(this[i]);
 }
 return ary.sort(function(a,b){return a - b});
}
Array.prototype.unique4 = function(){
 this.sort(function(a,b){return a - b});
 var ary = [],
  i= 0,
  len = this.length;
 for(; i<len; i++){
  if(this[i] !== this[i-1]) ary.push(this[i]);
 }
 return ary;
}
function randomAry (n) {
 var ary = [],i=0;
 for(; i<n; i++){
  ary.push(Math.ceil(Math.random()*10000));
 }
  console.log(ary)
 return ary;
}
function useTime (fn) {
 var start = new Date();
 fn();
 var end = new Date();
 console.log(' This letter has cost me: ' + (end - start) + ' Milliseconds ');
}
var ary = randomAry(10000),
 fn1 = function(){
 ary.unique1()
 },
 fn2 = function(){
 ary.unique2()
 },
 fn3 = function(){
 ary.unique3()
 },
 fn4 = function(){
 ary.unique4()
 };
useTime(fn1);
useTime(fn2);
useTime(fn3);
useTime(fn4);

Because the number is so big, I take the number once, and how many times will you refresh the comparison when you are happy

Method 1:10 milliseconds

Method 2: 1258 ms

Method 3: 2972 milliseconds

Method 4: 5 milliseconds

Originally, I wanted to give a bigger reference, but then 100,000 of my pages ran away. . . It's over. . . It's over

After that, I tried 50,000 people and ran away. . . It's over. . . It's over

Then forget it,

Anyway, in 10,000 numbers, the larger the numbers, the increase in milliseconds between Method 1 and Method 4, and the efficiency of Method 2 and Method 3 is touching

Efficiency calculation: 4 > 1 > 2 > 3

Method calculation: Sort first, and then compare whether the first two numbers are equal > Looking for the stability of the image through the past > Go through and find a number of locations > Is the position of the first appearance of the search numeral word 1 different from the previous position

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