Checking for duplicate strings in JavaScript array

问题:

I have JS array with strings, for example:

var strArray = [ "q", "w", "w", "e", "i", "u", "r"];

I need to compare for duplicate strings inside array, and if duplicate string exists, there should be alert box pointing to that string.

I was trying to compare it with for loop, but I don't know how to write code so that array checks it`s own strings for duplicates, without already pre-determined string to compare.

回答1:

findDuplicates function compares index of all items in array with index of first occurrence of same item, If indexes are not same returns it as duplicate.

let strArray = [ "q", "w", "w", "w", "e", "i", "u", "r"];
let findDuplicates = arr => arr.filter((item, index) => arr.indexOf(item) != index)

console.log(findDuplicates(strArray)) // All duplicates
console.log([...new Set(findDuplicates(strArray))]) // Unique duplicates

回答2:

Using ES6 features

function checkIfDuplicateExists(w){
    return new Set(w).size !== w.length 
}
console.log(
    checkIfDuplicateExists(["a", "b", "c", "a"])
// true 
);
console.log(
    checkIfDuplicateExists(["a", "b", "c"]))
//false
}

回答3:

var strArray = [ "q", "w", "w", "e", "i", "u", "r", "q"];
var alreadySeen = [];

strArray.forEach(function(str) {
  if (alreadySeen[str])
    alert(str);
  else
    alreadySeen[str] = true;
});

I added another duplicate in there from your original just to show it would find a non-consecutive duplicate.

回答4:

Using some function on arrays: If any item in the array has an index number from the beginning is not equals to index number from the end, then this item exists in the array more than once.

// vanilla js
function hasDuplicates(arr) {
    return arr.some( function(item) {
        return arr.indexOf(item) !== arr.lastIndexOf(item);
    });
}

回答5:

The following code uses a unique-filter (checks if every occurrence of an item is the first occurence) to compare the number of unique items in an array with the total number of items: if both are equal, the array only contains unique elements, otherwise there are some duplicates.

var firstUnique = (value, index, array) => array.indexOf(value) === index;
var numUnique = strArray.filter(firstUnique).length;
var allUnique = strArray.length === numUnique; 

回答6:

Use object keys for good performance when you work with a big array (in that case, loop for each element and loop again to check duplicate will be very slowly).

var strArray = ["q", "w", "w", "e", "i", "u", "r"];

var counting = {};
strArray.forEach(function (str) {
    counting[str] = (counting[str] || 0) + 1;
});

if (Object.keys(counting).length !== strArray.length) {
    console.log("Has duplicates");

    var str;
    for (str in counting) {
        if (counting.hasOwnProperty(str)) {
            if (counting[str] > 1) {
                console.log(str + " appears " + counting[str] + " times");
            }
        }
    }
}

回答7:

   var elems = ['f', 'a','b','f', 'c','d','e','f','c'];

    elems.sort();

    elems.forEach(function (value, index, arr){

        let first_index = arr.indexOf(value);
        let last_index = arr.lastIndexOf(value);

         if(first_index !== last_index){

         console.log('Duplicate item in array ' + value);

         }else{

         console.log('unique items in array ' + value);

         }

    });

回答8:

You could take a Set and filter the values who are alreday seen.

var array = ["q", "w", "w", "e", "i", "u", "r"],
    seen = array.filter((s => v => s.has(v) || !s.add(v))(new Set));

console.log(seen);

回答9:

Simple Javascript (if you don't know ES6)

function hasDuplicates(arr) {
    var counts = [];

    for (var i = 0; i <= arr.length; i++) {
        if (counts[arr[i]] === undefined) {
            counts[arr[i]] = 1;
        } else {
            return true;
        }
    }
    return false;
}

// [...]

var arr = [1, 1, 2, 3, 4];

if (hasDuplicates(arr)) {
  alert('Error: you have duplicates values !')
}