Remove all items that have duplicates in array Javascript - javascript

I have searched on here and have not found a solution. Obviously I will be corrected if I am wrong. What I am trying to do is return values that do not have a duplicates in an array.
Examples:
myArr = [2,1,2,3] // answer [1,3]
myArr = [3,1,2,2,3] // answer [1]
I would post some code but I have not been able to figure this out myself and the only code examples I have found are for removing any duplicate values.
The possible solution above is to return no duplicates... I am trying to return values that are don't have duplicates.

One option is to use the optional second argument to indexOf to find duplicate indexes. Consider that for a given element e and an index i:
if e is the first of two identical elements in the array, indexOf(e) will return i and indexOf(e, i + 1) will return the index of the second element.
if e is the second of two identical elements in the array, indexOf(e) will return the index of the first element, and indexOf(e, i + 1) will return -1
if e is a unique element, indexOf(e) will return i and indexOf(e, i + 1) will return -1.
Therefore:
myArr.filter(function (e, i, a) {
return a.indexOf(e) === i && a.indexOf(e, i + 1) === -1
});

var isUnique = function(v,i,arr){
// return true if the first occurrence is the last occurrence
return ( arr.indexOf(v) === arr.lastIndexOf(v) );
};
var uniqueVals = myArr.filter(isUnique);
console.log( uniqueVals );

If is not an associative array (your case):
var myArr = [1,2,2,3,4,4,1,5];
var myNewArr = [];
if (myArr.length > 0 )
{
myNewArr[0] = myArr[myArr.length-1];
}
var count = 1;
myArr.sort();
for (var i = myArr.length - 2; i >= 0; i--) {
if(myArr[i] != myArr[i-1])
{
myNewArr[count] = myArr[i];
count++;
}
}

var yourArray = [1, 2, 1, 3];
var uniqueValues = [];
$.each(yourArray, function (i, value) { //taking each 'value' from yourArray[]
if ($.inArray(value, uniqueValues) === -1) {
uniqueValues.push(value); // Pushing the non - duplicate value into the uniqueValues[]
}
});
console.log(uniqueValues);
Result: [1,2,3];

Related

How do I find two numbers in an array that are equal to a target number

I'm looking to find two numbers in an array that are equal to a particular target number. I thought this would be a simple task using .filter but for some reason my code only works when I'm looking for a target number of 4 but doesn't work for anything else?
What am I missing here?
var numbers2 = [1,2,3,4];
var target = 3;
var found = numbers2.filter((num) => {
return (num + num) !== target;
});
console returns (4) [1,2,3,4] as opposed to 2[1,2].
var numbers = [1,4,3,2,6,8,12,1,1,1,2,3,4];
var target = 3;
var output = [];
// Use a set to remove duplicate numbers
numbers = [...new Set(numbers)]; // Only do this step if you dont want duplicates ( like 2+2 = 4 so if your target was for 2, would not show up in the list )
// Sort the numbers from lowest to highest
numbers.sort( (a,b) =>a-b);
// Get index of first number that matches the target or is greater than the target
let index;
for( let i =0; i < numbers.length; i++) {
if( numbers[i] >= target ) {
index = i;
break;
}
}
// Remove all numbers from the array starting at the previous index as these are not possible to add up with another number to the target
if( index ) {
numbers.splice(index, numbers.length - index );
}
// Loop through the remianing array to get first number
numbers.forEach( ( num1, index1) => {
// Loop through array again to get second number
numbers.forEach( (num2, index2) => {
// Check if number is same is same index as you dont want to add the same value to itself, then check if the 2 numbers equal the target number
if( index1 !== index2 && num1 + num2 === target ) {
// If number already exists in array dont duplicate otherwise add it to the array
if( output.indexOf( num1 ) == -1 ) {
output.push( num1);
}
// If number already exists in array dont duplicate otherwise add it to the array
if( output.indexOf( num2 ) == -1 ) {
output.push( num2);
}
}
});
});
console.log( output);
You could find the array location of your target number through using a array.forEach, array.indexOf(), array.find(), and array.findIndex():
let numbers2 = [1,2,3,4];
let target = 4;
//Using foreach
numbers2.forEach((item, index)=>{
if (item == target){
console.log("Found the target at array location "+index);
}
});
//Or through using indexOf():
console.log("Found the target at array location "+numbers2.indexOf(target));
//Or through using find():
const found = numbers2.find(element => element == target);
console.log("Found "+target+" in the array.");
//Or through findIndex():
const target1 = (a) => a == target;
console.log("Found the target at array location "+numbers2.findIndex(target1));
Assuming:
you only need one pair
[2,2] does not count when your target is 4 (as '2' only appears once in the array)
One way to go is:
let numbers = [1, 2, 3, 4]
let target = 4;
let output = [];
const N = numbers.length
outer: for (let i = 0; i < N; i++) {
for (let j = i + 1; j < N; j++) {
if (numbers[i] + numbers[j] === target) {
output.push(numbers[i], numbers[j])
break outer;
}
}
}
console.log(output); //[1,3]
Edit: even if you want more than one pair, it's easy to modify to get that effect (now the target is 5):
let numbers = [1, 2, 3, 4]
let target = 5;
let output = [];
const N = numbers.length
for (let i = 0; i < N; i++) {
for (let j = i + 1; j < N; j++) {
if (numbers[i] + numbers[j] === target) {
output.push([numbers[i], numbers[j]])
}
}
}
console.log(output); //[[1,4], [2,3]]
This is an ideal case for the humble for loop. Methods like .forEach() will always try to loop over all the elements in an array, but if we order the data before we start the search we can break early and eliminate a lot of searching.
Ergo...
var numbers = [1,2,3,4];
var target = 5;
var output = [];
// Handling ordered data is much faster than random data so we'll do this first
numbers.sort();
// We want to start the inner search part way up the array, and we also want
// the option to break so use conventional for loops.
for (let i = 0; i<numbers.length; i++) {
for (let j=i+1; j<numbers.length;j++) {
// If the total = target, store the numbers and break the inner loop since later numbers
// will all be too large
if ((numbers[i]+numbers[j])===target) {
output.push([numbers[i], numbers[j]]);
break;
}
}
// Stop searching the first loop once we reach halfway, since any subsequent result
// will already have been found.
if (numbers[i]>(target/2)) {
break;
}
}
console.log( output);
It makes very little sense to get an array of single numbers, because you'll get all the numbers except for the last one unless the array starts at zero or there are numbers skipped. So I've written a function that'll return an array of single numbers or an array of expressions (strings).
First, make a copy of the array:
const array = [1, 2, 3, 4]
const copy = array.slice(0);
Next, use .flatMap() for the first set of iterations:
array.flatMap(num => { // This is the outer loop of numbers
If the third parameter expression is undefined it will default to false. Then .filter() the copy array, the criteria being that the number from the outer loop plus the current number of the inner loop equals the target number AND the numbers cannot be identical.
copy.filter(n => n !== num && target === n + num);
/*
Iterations on the first iteration of outer loop
1 + 1, 1 + 2, 1 + 3,...
*/
If expression is true, then use .flatMap() to return an expression (string) of whatever equals the target number or an empty array (which returns as nothing since .flatMap() flattens it's returns by a level). If both numbers are identical an empty array will be returned.
copy.flatMap(n => n === num ? [] :
target === n + num ? `${n} + ${num}` :
[]
);
If expression is true half of the array is returned so that there isn't any reversed dupes (ex. 6+2 and 2+6)
let half = result.length / 2;
result = result.slice(0, half);
const log = data => console.log(JSON.stringify(data));
// [1, 2, 3,...10]
const array10 = [...new Array(10)].map((_, i) => i + 1);
// [0, 2, 4, 6,...18]
const arrayEven = [...new Array(10)].map((_, i) => i * 2);
function operands(array, target, expression = false) {
const copy = array.slice(0);
let result = array.flatMap(num => {
if (expression) {
return copy.flatMap((n, i) =>
num === n ? [] :
target === n + num ? `${n} + ${num}` :
[]
);
}
return copy.filter(n => n !== num && target === n + num);
});
if (expression) {
let half = result.length / 2;
result = result.slice(0, half);
}
return result;
}
// Return as an array of single numbers
log(array10);
log('3: '+operands(array10, 3));
log('8: '+operands(array10, 8));
log('5: '+operands(array10, 5));
log(arrayEven);
log('2: '+operands(arrayEven, 2));
log('8: '+operands(arrayEven, 8));
log('15: '+operands(arrayEven, 15));
log('=======================');
// Return as an array of expressions (string)
log(array10);
log('3: '+operands(array10, 3, true));
log('8: '+operands(array10, 8, true));
log('5: '+operands(array10, 5, true));
log(arrayEven);
log('2: '+operands(arrayEven, 2, true));
log('8: '+operands(arrayEven, 8, true));
log('15: '+operands(arrayEven, 15, true));

how to print a unique number in a array

The problem is to find the unique number in a array such as [2,2,2,5].
The output should be 5 as it is the 1 unique element in the array.
I have attempted this:
function findUniq(arr) {
var b= arr[0];
var c;
for(var i=0; i<arr.length; i++)
{
if(arr[i]===b )
{
b=arr[i]
}
else
{
c=arr[i];
}
}
return c
console.log(findUniq([3, 5, 3, 3, 3]))
This works fine unless the unique number is the first element in the array. How do I fix this?
You can use indexOf and lastIndexOf to see if a value occurs more than once in the array (if it does, they will be different), and if so, it is not the unique value. Use filter to process the array:
let array = [2,2,2,5];
console.log(array.filter(v => array.indexOf(v) === array.lastIndexOf(v)));
array = [5,3,3,3,3];
console.log(array.filter(v => array.indexOf(v) === array.lastIndexOf(v)));
array = [4,4,5,4];
console.log(array.filter(v => array.indexOf(v) === array.lastIndexOf(v)));
You can create a recursive function that will take the first element of the array and see if it exists in the rest of it, if it does, it will take the next element and do the same, return the element if it doesn't exist in the rest of the array :
const arr = [3, 3, 3, 5, 3];
const find = arr => {
const [f, ...rest] = arr;
if(rest.includes(f))
return find(rest);
else
return f;
}
const result = find(arr);
console.log(result);
Note that this will return the last element if all of them are the same [3,3,3] will return 3
Try something like this using a set, which only stores unique elements:
var set = new Set(arr);
// count instances of each element in set
result = {};
for(var i = 0; i < a.length; ++i) {
if(!result[arr[i]])
result[arr[i]] = 0;
++result[arr[i]];
}
for (var value in result) {
if (value == 1) {
return value;
}
}
// if there isn't any
return false;
This should work, please tell me if it doesn't.
This is another implementation that is surely less efficient than that of #Nick's, but it is a valid algorithm anyway:
function findUniq(arr) {
var elemCount = new Map();
var uniq = [];
// Initialize elements conts
for (var k of arr.values()) {
elemCount.set(k, 0);
}
// Count elements
for (var k of arr.values()) {
elemCount.set(k, elemCount.get(k) + 1);
}
// Add uniq elements to array
for (var [k, v] of elemCount.entries()) {
if (v === 1) uniq.push(k);
}
return uniq;
}
console.log(findUniq([3, 5, 3, 3, 3]))
if you prefer .reduce over .map for your use case (for performance/etc. reasons):
function existance(data) {
return data.reduce((a, c) => (data.indexOf(c) === data.lastIndexOf(c)) ? a.concat(c) : a, []);
}
console.log(existance([1,1,1,2]));
console.log(existance([1,1,2,3,4,5,5,6,6,6]));

Javascript: How to remove only one value from duplicate array values

I have an array ['2530491','2530491','2530491','2530492'] the 2530491 is duplicated thrice, and I want to remove a single value of 2530491 from 3 of them, so the output would be like :
['2530491','2530491','2530492'].
fileArrayAnnounce_size = jQuery.grep(fileArrayAnnounce_size, function(value){
return value != file_size.metas[0].size;
});
I try grip but it removes all value which same. I want to remove only a single value from duplicates.
You can use splice and indexOf to remove the first instance:
fileArrayAnnounce_size.splice(fileArrayAnnounce_size.indexOf('2530491'), 1)
A safer way:
var index = fileArrayAnnounce_size.indexOf('2530491')
if (index > -1) {
fileArrayAnnounce_size.splice(index, 1);
}
Check and remove duplicates:
var mapOfValues = fileArrayAnnounce_size.reduce(function(vals, current) {
if (vals[current]) {
vals[current]++;
} else {
vals[current] = 1;
}
return vals;
}, {});
And now check and remove anything with more than 1 value:
for (var value in mapOfValues) {
if (mapOfValues[value] > 1) {
var idx = fileArrayAnnounce_size.indexOf(value);
fileArrayAnnounce_size.splice(idx, 1);
}
}
Demo: https://jsfiddle.net/1277mxt9/
You could check if given element has dupe elements or not, if so - remove just one duplicate entry of specified element from the original array.
var arr = ['2530491','2530491','2530491','2530492'],
hash = [...new Set(arr)];
hash.forEach((v,i) => arr.indexOf(v) != arr.lastIndexOf(v) ? arr.splice(arr.indexOf(v), 1) : null);
console.log(arr);
You can use filter() and pass one object as thisArg parameter to use it as hash table.
var data = ['2530491','2530491','2530491','2530492', '2530492'];
var result = data.filter(function(e) {
if(!this[e]) this[e] = 1
else if (this[e] == 1) return this[e] = 2, false
return true;
}, {})
console.log(result)

How do I search a string in JavaScript array using jQuery? [duplicate]

This question already has answers here:
How can I access and process nested objects, arrays, or JSON?
(31 answers)
Closed 6 years ago.
I have a JavaScript array:
var j_array = new Array();
j_arry=["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10"];
I need to find how many times the class is coming and its array key, so I use:
found = $.inArray('class', j_array); ` But it returns `-1`;
Then I use:
var search = 'class';
$.each([j_array], function(index, value){
$.each(value, function(key, cell){
if (search.indexOf(cell) !== -1)
console.log('found in array '+index, cell);
});
});
But that is also wrong. How do I solve this?
From this array I want to get the following:
Class coming 4 times, at key 0, 2, 3, and 7
I want to make a separate array of class only, that is,
new_array = ["class:1", "class:2", "class:3", "class:10"];
Currently there are four classes in j_array. How can I get the Nth class value
That is, 1st class value ="class:1", 2nd class value="class:5", etc.
You could filter elements which match in a new array and just return the length of this new array
var j_arry = ["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10"];
var res = j_arry.filter(x => x.includes("class"));
var key = res.map(x => x.split(":")[1]);
console.log("Class coming " + res.length + " , at key " + key.join(","));
console.log("new array = ", res);
Use Array.prototype.filter to filter out the elements of the array that contains the string class - see demo below:
var j_array =["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10"];
var result = j_array.filter(function(e){
return e.indexOf('class')!==-1;
});
console.log(result);
EDIT:
To get the list of indexes too, you can try this:
var j_array =["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10"];
var filteredIndices = []
var filtered = j_array.filter(function(e,i){
if(e.indexOf('class')!==-1) {
filteredIndices.push(i);
return true;
} else {
return false;
}
});
console.log(filtered);
console.log(filteredIndices);
// Nth class value
console.log(filtered[2]); // this prints the 3rd one
.as-console-wrapper{top:0;max-height:100%!important;}
Here is the answer to your questions 1 + 2. It is also 'n' proof so answers your part 3 also. This works by old-fashioned hard graft rather than funky functions. The original array entries are split and filtered then if qualifying we store in an associative array (results) using a pointer array (list) to make it easier to give a sorted result and pull the values from the associative array. The max variable is probably not necessary but included for clarity - could have used list.length instead. Note that the list[] array will be sparse (missing steps) so we test each entry before use in the output steps.
var j_array = new Array();
j_arry=["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10","class:1"];
var a, result = [], list=[], max = -1
for (var i =0; i < j_arry.length; i = i + 1) {
var a = j_arry[i].split(":")
if ( a[0] === "class") {
var key = "c" + a[1]
if ( !result[key] ) { result[key] = {pos:[]}}
result[key].cnt = result[key].cnt ? result[key].cnt + 1 : 1;
result[key].pos.push(i)
list[parseInt(a[1])] = "c" + a[1]
max = parseInt(a[1]) > max ? a[1] : max;
}
}
// say locations
for (var i = 0; i < max; i = i + 1) {
if (list[i]) {
key = "c" + i
console.log("Class " + i + " occurs at " + result[key].pos.toString() )
}
}
// make new array
var newArray=[]
for (var i = 0; i < max; i = i + 1) {
if (list[i]) {
newArray.push("Class:" + i)
}
}
console.log("New array=" + newArray.toString() )
Results are:
Class 1 occurs at 0,8
Class 3 occurs at 3
Class 5 occurs at 2
New array=Class:1,Class:3,Class:5
Single reduce is sufficient here.
var arr = ["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10"],
res = arr.reduce((p,c) => c.includes("class") ? (p.count++, p.keys.push(c.split(":")[1]), p)
: p ,{count:0, keys:[]});
console.log(res);
You can use the filter and map functions to filter your array to have only elements that match the text 'class', and use array index notation to access the nth element in the array. Check the below code snippet I hope it will be of help to you.
The below code snippet uses ES6 arrow syntax.
var arr = ["class:1", "division:a", "class:5", "class:3", "division:b", "division:c", "division:d", "class:10"];
var result = arr.filter(x => x.indexOf('class') !== -1);
var indices = result.map(x => arr.indexOf(x));
console.log(indices);
console.log(result);
var nValue = window.prompt('Enter n value');
console.log(result[nValue]);
If you're using jQuery to support some really old browser that still don't implement the new Array functions, and you don't want to polyfill those because you're already using jQuery, then you can use the jQuery equivalents:
var arr = ["class:1", "division:a", "class:5", "class:3", "division:b", "division:c", "division:d", "class:10"]
var result = $.grep(arr, function (x) { return x.indexOf('class') !== -1 })
var indices = $.map(result, function (x) { return arr.indexOf(x) })
This is the same code as this answer, but using jQuery.
You have to do map first then filter.
var j_array = ["class:1", "division:a", "class:5", "class:3", "division:b", "division:c", "division:d", "class:10"];
var result = j_array.map(function(e, i) {
return e.indexOf('class') > -1 ? '' + i : false;
}).filter(function(e) {
return !!e;
});
console.log(result);

JQuery remove duplicate from array where string contains same text

I have an array with X number of items. Each has variables separated by a pipe character. In a loop I can split on the pipe to get the second item; but how do I splice to remove the duplicate.
"Sometext|22621086|address|333629dc87894a7ea7df5291fa6d1836|PC_E|1803"
"Sometext2|22622138|working|d3e70175ffe942568cd21f1cf96f4d63|PC_E|1803"
"Sometext3|22622138|working|851946e6325445da99c113951590f714|PC_E|1803"
Results should be this.
"Sometext|22621086|address|333629dc87894a7ea7df5291fa6d1836|PC_E|1803"
"Sometext2|22622138|working|d3e70175ffe942568cd21f1cf96f4d63|PC_E|1803"
Note that the duplicate 22622138 is a random number so the solution needs to work for any number in this location (it's always in the arr[1] position).
This is what I tried:
$.each(arr_transcript, function (i, e) {
if (e.length != 0) {
var arr = e.split("|")
var i = arr_transcript.indexOf(arr[1]);
if (i != -1) {
arr_transcript.splice(i, 1);
}
}
});
Here's a generic function:
function uniqBy(a, key) {
let seen = new Set();
return a.filter(item => {
let k = key(item);
return !seen.has(k) && seen.add(k);
});
};
var data = [
"Sometext|22621086|address|333629dc87894a7ea7df5291fa6d1836|PC_E|1803",
"Sometext2|22622138|working|d3e70175ffe942568cd21f1cf96f4d63|PC_E|1803",
"Sometext3|22622138|working|851946e6325445da99c113951590f714|PC_E|1803"
];
var result = uniqBy(data, item => item.split('|')[1]);
console.log(result)
See here for more info.
Create a map of the numbers you want to check against, and then filter based on that
var arr_transcript = [
"Sometext|22621086|address|333629dc87894a7ea7df5291fa6d1836|PC_E|1803",
"Sometext2|22622138|working|d3e70175ffe942568cd21f1cf96f4d63|PC_E|1803",
"Sometext3|22622138|working|851946e6325445da99c113951590f714|PC_E|1803"
];
var map = arr_transcript.map(function(text) {
return text.split('|')[1];
});
var filtered = arr_transcript.filter(function(item, index) {
return index === map.lastIndexOf( map[index] );
});
console.log(filtered)

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