Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array

后端未结

关注

 30  2972

I need to check a JavaScript array to see if there are any duplicate values. What\'s the easiest way to do this? I just need to find what the duplicated values are - I don\'

相关标签:

30条回答

别跟我提以往

2020-11-21 05:11

You can add this function, or tweak it and add it to Javascript's Array prototype:

Array.prototype.unique = function () {
    var r = new Array();
    o:for(var i = 0, n = this.length; i < n; i++)
    {
        for(var x = 0, y = r.length; x < y; x++)
        {
            if(r[x]==this[i])
            {
                alert('this is a DUPE!');
                continue o;
            }
        }
        r[r.length] = this[i];
    }
    return r;
}

var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,9];
var unique = arr.unique();
alert(unique);

0 讨论(0)

萌比男神i

2020-11-21 05:12

This should get you what you want, Just the duplicates.

function find_duplicates(arr) {
  var len=arr.length,
      out=[],
      counts={};

  for (var i=0;i<len;i++) {
    var item = arr[i];
    counts[item] = counts[item] >= 1 ? counts[item] + 1 : 1;
    if (counts[item] === 2) {
      out.push(item);
    }
  }

  return out;
}

find_duplicates(['one',2,3,4,4,4,5,6,7,7,7,'pig','one']); // -> ['one',4,7] in no particular order.

0 讨论(0)

礼貌的吻别

2020-11-21 05:13
Shortest vanilla JS:
```
[1,1,2,2,2,3].filter((v,i,a) => a.indexOf(v) !== i) // [1, 2, 2]
```
0 讨论(0)
发布评论:

提交评论
- 加载中...

执念已碎

2020-11-21 05:14

You could sort the array and then run through it and then see if the next (or previous) index is the same as the current. Assuming your sort algorithm is good, this should be less than O(n²):

const findDuplicates = (arr) => {
  let sorted_arr = arr.slice().sort(); // You can define the comparing function here. 
  // JS by default uses a crappy string compare.
  // (we use slice to clone the array so the
  // original array won't be modified)
  let results = [];
  for (let i = 0; i < sorted_arr.length - 1; i++) {
    if (sorted_arr[i + 1] == sorted_arr[i]) {
      results.push(sorted_arr[i]);
    }
  }
  return results;
}

let duplicatedArray = [9, 9, 111, 2, 3, 4, 4, 5, 7];
console.log(`The duplicates in ${duplicatedArray} are ${findDuplicates(duplicatedArray)}`);

In case, if you are to return as a function for duplicates. This is for similar type of case.

Reference: https://stackoverflow.com/a/57532964/8119511

0 讨论(0)

忘了有多久

2020-11-21 05:15

one liner simple way

var arr = [9,1,2,4,3,4,9]
console.log(arr.filter((ele,indx)=>indx!==arr.indexOf(ele))) //get the duplicates
console.log(arr.filter((ele,indx)=>indx===arr.indexOf(ele))) //remove the duplicates

0 讨论(0)

死守一世寂寞

2020-11-21 05:17

Find non-unique values from 3 arrays (or more):

ES2015

//