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How do I remove duplicates from a list from input? (Javascript)

By : Denis Koval
Date : August 01 2020, 11:00 AM
I wish did fix the issue. I need help with detecting duplicate values when entering a link twice or thrice in a row. Whenever I try to add the same string from the input box It add like what it was supposed to but I want to stop adding the same string again. The input should prevent me from adding exactly the same string to the list.
code :
input = ["https://www.youtube.com/watch?v=q07RanslaGM",



output =  Array.from(new Set(input))

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How to remove duplicates from input array and put them in output list?

By : Abdullah Farsid
Date : March 29 2020, 07:55 AM
it helps some times
code :
public static void duplicates(int[] input, int[] output){
    for (int q = 0; q < input.length; q++){
        if (!contains(output, input[q]) || 
                indexOf(output, input[q]) == -1 || 
                count(output, input[q]) == 0){ // # of times the current input appears in 
                // output should be zero
            output[q] = input[q];

python iterate over a list and remove any duplicates found within lists inside a dictionary and remove the duplicates

By : Joe Stubbs
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further You are accessing the dict values in a wrong way. Take a look at the following:
code :
players = {'ray': [1,2,3,6,7,8,], 'al':[1,2,3,4,8,9,]}
lotto =[1,2,3,4,5,6,]

for p, val in players.items():  
    for num in lotto:
        if num in val:

print (players)  # {'al': [8, 9], 'ray': [7, 8]}
players = {'ray': [1,2,3,6,7,8], 'al':[1,2,3,4,8,9]}
lotto = [1,2,3,4,5,6]
lotto = set(lotto)

for p, val in players.items():
    players[p] = list(set(val) - lotto)

print (players)  # {'ray': [8, 7], 'al': [8, 9]}
players = {k: list(set(val) - lotto) for k, v in players.items()}

How to remove duplicates items from the list in javascript

By : user1739103
Date : March 29 2020, 07:55 AM
Hope that helps You could use a Set as closure over the already visited id and remove this object from the result set, if exists.
This approach uses a single loop for the data and for each found duplicate a filtering.
code :
var array = [{ id: 11, type: "sell", quantity: 11, price: 155 }, { id: 11, type: "sell", quantity: 11, price: 155 }, { id: 11, type: "sell", quantity: 11, price: 155 }, { id: 12, type: "buy", quantity: 3, price: 189 }, { id: 13, type: "buy", quantity: 4, price: 189 }, { id: 14, type: "buy", quantity: 2, price: 189 }, { id: 14, type: "buy", quantity: 2, price: 189 }],
    single = array.reduce((s => (r, o) => {
        if (s.has(o.id)) {
            return r.filter(({ id }) => id !== o.id);
        return r;
    })(new Set), []);

.as-console-wrapper { max-height: 100% !important; top: 0; }

How to remove duplicates by key-value in javascript object list?

By : user3640530
Date : March 29 2020, 07:55 AM
This might help you Consider the following array: , You can use reduce to solve the problem:
code :
const list = [
    { url: 'https://url.com/file1', md5: 'fbbbabcc19264ce7b376ce4c726b9b85' },
    { url: 'https://url.com/file2', md5: 'd920d140432b961f07695ec34bd2a8ad' },
    { url: 'https://url.com/file3', md5: 'fbbbabcc19264ce7b376ce4c726b9b85' },
    { url: 'https://url.com/file4', md5: 'bf80655dbe90123324f88a778efa39f7' },
    { url: 'https://url.com/file5', md5: 'fbbbabcc19264ce7b376ce4c726b9b85' },

const removeDup = (arr, key) => {
    return arr.reduce((acc, cur) => {
        if (acc.some(a => a[key] === cur[key])) return acc;
        return acc.concat(cur)
    }, [])

console.log(removeDup(list, 'md5'))

Take a list in input remove all duplicates and count how many they are

By : user3686822
Date : March 29 2020, 07:55 AM
I wish this help you I have tried to make this function but sometimes I get a problem on runtime. I need to take in input a list like 1-2-3-2-4-6-1-2 and the list should be modified as 1-2-3-4-6 so once for every node and I should save how many times is a node repeated so for example 1 is repeated 2 times the 2 is repeated 3 times etc. But sometimes it doesn't work I cannot understand why. If you run it following what I wrote just after you'll see the problem: How many nodes: 5 than write 1, 1, 2, 4, 2 and the output should be 1, 2 -> 2, 2 -> 4 (that means 1 is repeated two times, 2 is repeated two times and 4 is repeated 1 time). If I give in input 1, 2, 3, 2, 1 I get the "right output" plus a garbage value. More are the numbers higher is the probability that it doesn't work :( , I corrected your program :
code :
#include <stdio.h>
#include <stdlib.h>

typedef struct Node{
    int val;
    int rip;
    struct Node *next;
} node;

node *modify(node *head);

void print(node *head2);

int main(){

    int m, i;

    printf("How many nodes: \n");
    scanf("%d", &m);

    node *head = NULL;
    head = (node *)malloc(sizeof(node));
    node *temp = head;
    node *head2 = NULL;

    printf("Write the value in HEAD position : \n");
    scanf("%d", &temp->val);
    temp->next = NULL;

    for(i=0; i < m-1; i++)
        temp->next = (node *)malloc(sizeof(node));
        printf("Write the value in position %d: \n", i);
        temp = temp->next;
        scanf("%d", &temp->val);
        temp->next = NULL;  

    head2 = modify(head);


    return 0;

node *modify(node *head){

    int counter, pass, m;

    node *curr = head;
    node *track = head;
    node *precNode;

    while (track != NULL){
        counter = 0;
        pass = 0;
        m = track->val;
        while( curr != NULL){
            if(m == (curr)->val){
                if(pass > 1){
                    node *removed = curr;

                    precNode->next = (curr)->next;

                    curr = (curr)->next;

                if(pass == 1)
                    precNode = curr;
                    curr = curr->next;
                precNode = curr;
                curr = (curr)->next;
        track->rip = counter;
        track = track->next;
        curr = track;

    return head;

void print(node *head2){

    while(head2 != NULL){
        printf("[%d, %d]  ->   ", head2->val, head2->rip);
        head2 = head2->next;
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