How to completely remove repeated numbers from an array.
For example if: array = [1,1,2,3,1,2,5]
The output would be: [3,5]
CodePudding user response：
You could take an object for keeping track of seen items, by taking an array with the value or set the arrays value to zero.
Finally flat the result set to remove empty arrays.
const
array = [1, 1, 2, 3, 1, 2, 5],
result = array
.reduce((o => (r, v) => {
if (v in o) o[v].length = 0;
else r.push(o[v] = [v]);
return r;
})({}), [])
.flat();
console.log(result);CodePudding user response：
Roko and Ke1vans had answered in functional approaches. Both of them are correct. However, I'd show an answer in imperative approach, which may seems easier for new comer.
Similar to their flow. First, we count the occurrence of each number. Then we select the numbers that has occurred once (hence being non-repeated) into the output array.
let array = [1,1,2,3,1,2,5]
let counts = {}
let output = []
// loop each elements in the array as `item`
for(let item of array) {
// If the item is not set in the counts, `counts[item]` will be `undefined`.
// Using `|| 0` means use zero as fallback value if the items is unseen.
let count = counts[item] || 0
counts[item] = count 1
}
// loop each keys in the object (key-value pairs) as `item`
for(let item in counts) {
let count = counts[item]
if(count == 1) {
// ` item` converts the key from string into number
output.push( item)
}
}
console.log(output) // will print out `[ 3, 5 ]`
CodePudding user response：
You can iterate and create a map of values. later iterate and filter.
const data = [1, 1, 2, 3, 1, 2, 5];
const findUniques = (data = []) => {
const map = data.reduce((m, num) => {
m[num] = (m[num] || 0) 1;
return m;
}, {});
return data.filter((num) => map[num] === 1);
};
console.log(findUniques(data));You can also do the same using 2 set, or 2 array.
const data = [1, 1, 2, 3, 1, 2, 5];
const findUniques2 = (data = []) => {
let unique = new Set();
let seen = new Set();
for (let num of data) {
if (seen.has(num)) unique.delete(num);
else unique.add(num);
seen.add(num);
}
return Array.from(unique);
};
console.log(findUniques2(data));CodePudding user response：
Use an Object where the key is the number, and the value is the number of occurrences. Than reduce it back to the desired array of values:
const arr = [1,1,2,3,1,2,5];
const res = Object.entries(arr.reduce((ob, v) => {
if (!(v in ob)) ob[v] = 0;
ob[v] = 1; // Count occurrences
return ob;
}, {})).reduce((arr, [k, v]) => { // Reduce back to Array
if (v === 1) arr.push( k); // Only keys with 1 occurrence
return arr;
}, []);
console.log(res); // [3, 5]CodePudding user response：
You can use Array.filter() (Array filter article) for this
for Example :
const a = [ 1 , 1 , 2 , 3 , 2 , 4 , 5 , 7];
function filter(value , index , array){
// looping through all the objects in the array
for(let i=0; i<array.length; i ) {
if(i != index && array[i] == value) return false; // return 'false' if the value has a duplicate other than itself
}
// return 'TRUE' if value hasn't been duplicated
return true;
}
const b = a.filter(filter); // [3, 4, 5, 7]
And the short version if this function's going to be used only once:
const a = [ 1 , 1 , 2 , 3 , 2 , 4 , 5 , 7];
const b = a.filter((value , index , array) => {
for(let i=0; i<array.length; i ) if(i != index && array[i] == value) return false;
return true;
});
// [3, 4, 5, 7]
CodePudding user response：
You could find duplicates and then operate a difference.
let a = [1,1,2,3,1,2,5];
const findDuplicates = (nums) => {
nums.sort(); // alters original array
let ans = []
for(let i = 0; i< nums.length; i ){
if(nums[i] === nums[i 1]){
if(ans[ans.length -1] !== nums[i]){
ans.push(nums[i])
}
}
}
return ans;
}
duplicates = new Set(findDuplicates(a))
let difference = new Set([...a].filter(x => !duplicates.has(x)))
console.log(Array.from(difference))
output : [ 3, 5 ]
Note : I grab the findDuplicates function from this link