How to create array as a property in javascript? - javascript

I want to create a function that generates random Integers from a certain interval and each time it is called, it should produce a unique one. For that I have created this generateUniqueInt.
function generateUniqueInt() {
var res = Math.floor(Math.random() * 100);
while (generateUniqueInt.used.indexOf(res) !== -1)
res = Math.floor(Math.random() * 100);
generateUniqueInt.used.push(res);
return res;
}
generateUniqueInt.used = new Array;
for (let i = 0; i < 20; i++) {
console.log(generateUniqueInt());
}
generateUniqueInt.used.sort();
console.log(generateUniqueInt.used);
I called this function few times like this and it works.
Then I wanted to check which values were actually generated and for easier inspection I sorted the used property. But as it seems, used is not an array anymore.
I have tried using generateUniqueInt.used = []; as well as Object.defineProperty but the outcome is the same each time. What am I missing here? Is there a way to create used as an array?

Your issue is :The default sort order is according to string Unicode code points.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort
So below code will return following output:
var array1 = [1, 30, 4, 21];
array1.sort();
console.log(array1);
// expected output: Array [1, 21, 30, 4]
In order to solve for numbers you will may follow example from MDN:
var numbers = [4, 2, 5, 1, 3];
numbers.sort(function(a, b) {
return a - b;
});
console.log(numbers);
// [1, 2, 3, 4, 5]
So your example would become following which should give you correct desired output:
generateUniqueInt.used.sort(function(a,b){ return a - b});

By default the sort method sorts elements alphabetically, so your array will be sorted like [1,2,25,3,4 ...]
use this to sort an array of numbers :
generateUniqueInt.used.sort(function(a,b){ return a - b});
let arr = [4,2,65,12,23,5,3]
let sorted1 = arr.sort()
console.log('sorted alphabetically : ', JSON.stringify(sorted1))
let sorted2 = arr.sort(function(a,b){ return a - b})
console.log('sorted numerically : ', JSON.stringify(sorted2))

Related

Runner Up Score - Solution in Javascript

I'm currently still learning the basic JS and wondering how to answer this array type of question.
Question :
Given the participant's score sheet for your University Sports Day, You are required to find the runner-up score. You are given N scores. Store them in a list and find the score of the runner-up.
Input Format :
The first line contains N. The second line contains an array A[] of N integers each separated by a space.
Sample Input: [5,2,3,6,6,5]
Output Format :
Print the runner-up score.
Sample Output: 5
This is my code :
function uniqueScore (value, index, self) {
return self.indexOf(value) === index
}
var score = [5,2,3,6,6,5]
var filter = score.filter(uniqueScore)
var descSort = filter.sort().reverse()
function runnerUpScore(x) {
var runnerUp = descSort
return runnerUp
}
console.log(runnerUpScore(x))
The condition is I'm stuck when already sorting the array and removing duplicate. Hence, I need some guidance how to call the sorted array and only showing the index 1 from array list to show the runner up score.
The problem with your approach is you are sorting the array and then reversing it which costs lots of computation. Here is a simple solution in O(n). It iterates the array only once and finds the second-largest score, which is the runner-up score.
var scores = [1,2,3,4,5];
const runnerUpScore = (scores) => {
scores = [... new Set(scores)] // get unique elements
var largest = -1;
var secondLargest = -1;
scores.forEach((score) => {
if(score >= largest) {
secondLargest = largest;
largest = score;
} else if (score > secondLargest) {
secondLargest = score;
}
});
return secondLargest;
}
console.log(runnerUpScore(scores));
.sort() reverse:
let scores = [5, 2, 3, 6, 6, 5];
let reverseOrder = scores.sort((a, b) => b - a);
// [6, 6, 5, 5, 3, 2]
Filter out duplicates by converting array into a Set() then back to an array:
let set = new Set(reverseOrder)
// {6:6, 5:5, 3:3, 2:2} this is a representation there's much more to a Set.
let unique = [...set]
// [6, 5, 3, 2]
Get the score in the second index position:
unique[1]
// 5
const scores = [5, 2, 3, 6, 6, 5];
const reverseOrderUnique = [...new Set(scores.sort((a, b) => b - a))];
console.log(reverseOrderUnique);
console.log('Runner Up: '+reverseOrderUnique[1]);
You can simply achieve it by sort the array in descending order by using Array.sort() and filtered out the duplicates by using Set() method and then access the runner-up score from 1st index of the array.
Live Demo :
// Input array
const scores = [5,2,3,6,6,5];
// Sort the array in descending order and remove the duplicates by using Set() method.
const sortedScores = [...new Set(scores.sort((a,b) => b-a))];
// Runner Up score by accesing the 1st index value.
console.log(sortedScores[1]);

How can I get the smallest two numbers from an array in js?

Hey I've been trying to return the 2 smallest numbers from an array, regardless of the index. Can you please help me out?
Sort the array in the ascending order.
Use Array#slice to get the first two elements (the smallest ones).
var arr = [5, 4, 7, 2, 10, 1],
res = arr.sort((a,b) => a - b).slice(0, 2);
console.log(res);
While the accepted answer is good and correct, the original array is sorted, which may not be desired
var arr = [5, 4, 7, 2, 10, 1],
res = arr.sort((a,b) => a - b).slice(0, 2);
console.log(arr.join()); // note it has mutated to 1,2,4,5,7,10
console.log(res.join());
You can avoid this by sliceing the original array and sorting on this new copy
I also added code for the lowest two values in descending order, as that may also be useful
const array = [1, 10, 2, 7, 5,3, 4];
const ascending = array.slice().sort((a, b) => a - b).slice(0, 2);
const descending = array.slice().sort((a, b) => b - a).slice(-2);
console.log(array.join()); // to show it isn't changed
console.log(ascending.join());
console.log(descending.join());

javascript push an item into list index by value

Lets say I have a list:
var list = []
Now I want to insert some value into the list in a way that its index should be according to its value
list.push(6)
list.push(2)
This will give me a result of [6, 2]
But what I want is its index should be managed according to its value.
Here 2 is smaller than 6 so 2 should come first and give result like
[2, 6] . And if again I do list.push(1) it should give result like [1,2,6] and so on.
How can I achieve this in javascript ?
You can use Array#sort.
var list = [];
list.push(6);
list.push(2);
list.push(1);
list.sort(function(a, b) {
return a - b;
});
console.log(list);
You simply need to use Array sort function. As it treats array items as string and does not sort integers correctly by default, you need to use custom compare function for this:
var arr = [1, 22, 6, 2 ];
arr.sort(function(a, b){
return a - b;
});
console.log( arr );
You can use the sort() function.
var fruits = ["Banana", "Orange", "Apple", "Mango"];
fruits.sort();
The result of fruits will be:
Apple,Banana,Mango,Orange
Edit:
Sorry, for numbers this solution does not work. You can use this:
Sort numbers in an array in ascending order:
var points = [40, 100, 1, 5, 25, 10];
points.sort(function(a, b){return a-b});
The result of points will be:
1,5,10,25,40,100
We can override push function of list object.
var list = [];
var copyPush = list.push.bind(list);
list.push = function(){
var toret = copyPush.apply(this,arguments);
this.sort(function(a,b){ return a-b});
return toret;
}
list.push(1);
list.push(3);
list.push(2);
list.push(5);
console.log(list); //[1, 2, 3, 5]
These other answers are missing the point, they assume that the list is already populated, and then you push in. What you need to do is compare the number you are going to push into the array, with the array itself, splicing the value in like so:
arr = [1,5,7];
var number = 0;
function findValue(value) {
return function(element, index, array) {
return (element <= value);
}
}
var filtered = arr.filter(findValue(number))
index = arr.indexOf(filtered[filtered.length-1]) +1;
arr.splice(index, 0, number);
alert(arr);
Here's a working fiddle:
https://jsfiddle.net/GerardSimpson/h3t6vcka/

Merging of two arrays, store unique elements, and sorting in jQuery

var Arr1 = [1,3,4,5,6];
var Arr2 = [4,5,6,8,9,10];
I am trying to do merge these two arrays and output coming is [1,3,4,5,6,4,5,6]
I have used $.merge(Arr1, Arr2); this piece to merge them. Using alert I can see the merged array like above.
Now my question is how can I get the following output:
[1,3,4,5,6,8,9,10]
i.e. the elements should be unique as well as sorted in the same manner I have mentioned.
Please help.
You can use Array.prototype.sort() to do a real numeric sort and use Array.prototype.filter() to only return the unique elements.
You can wrap it into a helper similar to this:
var concatArraysUniqueWithSort = function (thisArray, otherArray) {
var newArray = thisArray.concat(otherArray).sort(function (a, b) {
return a > b ? 1 : a < b ? -1 : 0;
});
return newArray.filter(function (item, index) {
return newArray.indexOf(item) === index;
});
};
Note that the custom sort function works with numeric elements only, so if you want to use it for strings or mix strings with numbers you have to update it off course to take those scenarios into account, though the rest should not change much.
Use it like this:
var arr1 = [1, 3, 4, 5, 6];
var arr2 = [4, 5, 6, 8, 9, 10];
var arrAll = concatArraysUniqueWithSort(arr1, arr2);
arrAll will now be [1, 3, 4, 5, 6, 8, 9, 10]
DEMO - concatenate 2 arrays, sort and remove duplicates
There is many ways of doing this I'm sure. This was just the most concise I could think off.
merge two or more arrays + remove duplicities + sort()
jQuery.unique([].concat.apply([],[[1,2,3,4],[1,2,3,4,5,6],[3,4,5,6,7,8]])).sort();
One line solution using just javascript.
var Arr1 = [1,3,4,5,6];
var Arr2 = [4,5,6,8,9,10];
const sortedUnion = [... new Set([...Arr1,... Arr2].sort((a,b)=> a-b))]
console.log(sortedUnion)
This looks like a job for Array.prototype.indexOf
var arr3 = arr1.slice(), // clone arr1 so no side-effects
i; // var i so it 's not global
for (i = 0; i < arr2.length; ++i) // loop over arr2
if (arr1.indexOf(arr2[i]) === -1) // see if item from arr2 is in arr1 or not
arr3.push(arr2[i]); // it's not, add it to arr3
arr3.sort(function (a, b) {return a - b;});
arr3; // [1, 3, 4, 5, 6, 8, 9, 10]
a = [1, 2, 3]
b = [2, 3, 4]
$.unique($.merge(a, b)).sort(function(a,b){return a-b}); -> [1, 2, 3, 4]
Update:
This is a bad idea, since the 'unique' function is not meant for use on numbers or strings.
However, if you must then the sort function needs to be told to use a new comparator since by default it sorts lexicographically.
Using underscore.js:
_.union([1, 2, 3], [101, 2, 1, 10], [2, 1]).sort(function(a,b){return a-b});
=> [1, 2, 3, 10, 101]
This example is taken directly from underscore.js, a popular JS library which complements jQuery
I did that as follows, where t1 and t2 are my two tables.
The first command put the values of the table t2 to the t1. The second command removes the duplicate values from the table.
$.merge(t1, t2);
$.unique(t1);
function sortUnique(matrix) {
if(matrix.length < 1 || matrix[0].length < 1) return [];
const result = [];
let temp, ele;
while(matrix.length > 0) {
temp = 0;
for(let j=0; j<matrix.length; j++) {
if(matrix[j][0] < matrix[temp][0]) temp = j;
}
if(result.length === 0 || matrix[temp][0] > result[result.length-1]) {
result.push(matrix[temp].splice(0,1)[0]);
} else {
matrix[temp].splice(0,1);
}
if(matrix[temp].length===0) matrix.splice(temp, 1);
}
return result;
}
console.log(sortUnique([[1,4,8], [2,4,9], [1,2,7]]))
Using JavaScript ES6 makes it easier and cleaner. Try this:
return [...Arr1, ...Arr2].filter((v,i,s) => s.indexOf(v) === i).sort((a,b)=> a - b);
and there you have it. You could build it in a function like:
function mergeUniqueSort(Arr1, Arr2){
return [...Arr1, ...Arr2].filter((v,i,s) => s.indexOf(v) === i).sort((a,b)=> a - b);
}
and that settles it. You can also break it down using ES6. Use a Spread Operator to combine arrays:
let combinedArrays = [...Arr1, ...Arr2]
then get the unique elements using the filter function:
let uniqueValues = combinedArrays.filter((value, index, self ) => self.indexOf(value) === index)
Lastly you now sort the uniqueValue object:
let sortAscending = uniqueValues.sort((a-b) => a - b) // 1, 2, 3, ....10
let sortDescending = uniqueValues.sort((b-a) => b - a) // 10, 9, 8, ....1
So you could use any part, just in case.

How to count certain elements in array?

I have an array:
[1, 2, 3, 5, 2, 8, 9, 2]
I would like to know how many 2s are in the array.
What is the most elegant way to do it in JavaScript without looping with for loop?
[this answer is a bit dated: read the edits, in the notion of 'equal' in javascript is ambiguous]
Say hello to your friends: map and filter and reduce and forEach and every etc.
(I only occasionally write for-loops in javascript, because of block-level scoping is missing, so you have to use a function as the body of the loop anyway if you need to capture or clone your iteration index or value. For-loops are more efficient generally, but sometimes you need a closure.)
The most readable way:
[....].filter(x => x==2).length
(We could have written .filter(function(x){return x==2}).length instead)
The following is more space-efficient (O(1) rather than O(N)), but I'm not sure how much of a benefit/penalty you might pay in terms of time (not more than a constant factor since you visit each element exactly once):
[....].reduce((total,x) => (x==2 ? total+1 : total), 0)
or as a commenter kindly pointed out:
[....].reduce((total,x) => total+(x==2), 0)
(If you need to optimize this particular piece of code, a for loop might be faster on some browsers... you can test things on jsperf.com.)
You can then be elegant and turn it into a prototype function:
[1, 2, 3, 5, 2, 8, 9, 2].count(2)
Like this:
Object.defineProperties(Array.prototype, {
count: {
value: function(value) {
return this.filter(x => x==value).length;
}
}
});
You can also stick the regular old for-loop technique (see other answers) inside the above property definition (again, that would likely be much faster).
2017 edit:
Whoops, this answer has gotten more popular than the correct answer. Actually, just use the accepted answer. While this answer may be cute, the js compilers probably don't (or can't due to spec) optimize such cases. So you should really write a simple for loop:
Object.defineProperties(Array.prototype, {
count: {
value: function(query) {
/*
Counts number of occurrences of query in array, an integer >= 0
Uses the javascript == notion of equality.
*/
var count = 0;
for(let i=0; i<this.length; i++)
if (this[i]==query)
count++;
return count;
}
}
});
You could define a version .countStrictEq(...) which used the === notion of equality. The notion of equality may be important to what you're doing! (for example [1,10,3,'10'].count(10)==2, because numbers like '4'==4 in javascript... hence calling it .countEq or .countNonstrict stresses it uses the == operator.)
Caveat:
Defining a common name on the prototype should be done with care. It is fine if you control your code, but bad if everyone wants to declare their own [].count function, especially if they behave differently. You may ask yourself "but .count(query) surely sounds quite perfect and canonical"... but consider perhaps you could do something like [].count(x=> someExpr of x). In that case you define functions like countIn(query, container) (under myModuleName.countIn), or something, or [].myModuleName_count().
Also consider using your own multiset data structure (e.g. like python's 'collections.Counter') to avoid having to do the counting in the first place. This works for exact matches of the form [].filter(x=> x==???).length (worst case O(N) down to O(1)), and modified will speed up queries of the form [].filter(filterFunction).length (roughly by a factor of #total/#duplicates).
class Multiset extends Map {
constructor(...args) {
super(...args);
}
add(elem) {
if (!this.has(elem))
this.set(elem, 1);
else
this.set(elem, this.get(elem)+1);
}
remove(elem) {
var count = this.has(elem) ? this.get(elem) : 0;
if (count>1) {
this.set(elem, count-1);
} else if (count==1) {
this.delete(elem);
} else if (count==0)
throw `tried to remove element ${elem} of type ${typeof elem} from Multiset, but does not exist in Multiset (count is 0 and cannot go negative)`;
// alternatively do nothing {}
}
}
Demo:
> counts = new Multiset([['a',1],['b',3]])
Map(2) {"a" => 1, "b" => 3}
> counts.add('c')
> counts
Map(3) {"a" => 1, "b" => 3, "c" => 1}
> counts.remove('a')
> counts
Map(2) {"b" => 3, "c" => 1}
> counts.remove('a')
Uncaught tried to remove element a of type string from Multiset, but does not exist in Multiset (count is 0 and cannot go negative)
sidenote: Though, if you still wanted the functional-programming way (or a throwaway one-liner without overriding Array.prototype), you could write it more tersely nowadays as [...].filter(x => x==2).length. If you care about performance, note that while this is asymptotically the same performance as the for-loop (O(N) time), it may require O(N) extra memory (instead of O(1) memory) because it will almost certainly generate an intermediate array and then count the elements of that intermediate array.
Modern JavaScript:
Note that you should always use triple equals === when doing comparison in JavaScript (JS). The triple equals make sure, that JS comparison behaves like double equals == in other languages (there is one exception, see below). The following solution shows how to solve this the functional way, which will ensure that you will never have out of bounds error:
// Let has local scope
let array = [1, 2, 3, 5, 2, 8, 9, 2]
// Functional filter with an Arrow function
// Filter all elements equal to 2 and return the length (count)
array.filter(x => x === 2).length // -> 3
The following anonymous Arrow function (lambda function) in JavaScript:
(x) => {
const k = 2
return k * x
}
may be simplified to this concise form for a single input:
x => 2 * x
where the return is implied.
Always use triple equals: === for comparison in JS, with the exception of when checking for nullability: if (something == null) {} as it includes a check for undefined, if you only use double equals as in this case.
Very simple:
var count = 0;
for(var i = 0; i < array.length; ++i){
if(array[i] == 2)
count++;
}
2017:
If someone is still interested in the question, my solution is the following:
const arrayToCount = [1, 2, 3, 5, 2, 8, 9, 2];
const result = arrayToCount.filter(i => i === 2).length;
console.log('number of the found elements: ' + result);
Here is an ES2017+ way to get the counts for all array items in O(N):
const arr = [1, 2, 3, 5, 2, 8, 9, 2];
const counts = {};
arr.forEach((el) => {
counts[el] = counts[el] ? (counts[el] + 1) : 1;
});
You can also optionally sort the output:
const countsSorted = Object.entries(counts).sort(([_, a], [__, b]) => a - b);
console.log(countsSorted) for your example array:
[
[ '2', 3 ],
[ '1', 1 ],
[ '3', 1 ],
[ '5', 1 ],
[ '8', 1 ],
[ '9', 1 ]
]
If you are using lodash or underscore the _.countBy method will provide an object of aggregate totals keyed by each value in the array. You can turn this into a one-liner if you only need to count one value:
_.countBy(['foo', 'foo', 'bar'])['foo']; // 2
This also works fine on arrays of numbers. The one-liner for your example would be:
_.countBy([1, 2, 3, 5, 2, 8, 9, 2])[2]; // 3
Weirdest way I can think of doing this is:
(a.length-(' '+a.join(' ')+' ').split(' '+n+' ').join(' ').match(/ /g).length)+1
Where:
a is the array
n is the number to count in the array
My suggestion, use a while or for loop ;-)
Not using a loop usually means handing the process over to some method that does use a loop.
Here is a way our loop hating coder can satisfy his loathing, at a price:
var a=[1, 2, 3, 5, 2, 8, 9, 2];
alert(String(a).replace(/[^2]+/g,'').length);
/* returned value: (Number)
3
*/
You can also repeatedly call indexOf, if it is available as an array method, and move the search pointer each time.
This does not create a new array, and the loop is faster than a forEach or filter.
It could make a difference if you have a million members to look at.
function countItems(arr, what){
var count= 0, i;
while((i= arr.indexOf(what, i))!= -1){
++count;
++i;
}
return count
}
countItems(a,2)
/* returned value: (Number)
3
*/
I'm a begin fan of js array's reduce function.
const myArray =[1, 2, 3, 5, 2, 8, 9, 2];
const count = myArray.reduce((count, num) => num === 2 ? count + 1 : count, 0)
In fact if you really want to get fancy you can create a count function on the Array prototype. Then you can reuse it.
Array.prototype.count = function(filterMethod) {
return this.reduce((count, item) => filterMethod(item)? count + 1 : count, 0);
}
Then do
const myArray =[1, 2, 3, 5, 2, 8, 9, 2]
const count = myArray.count(x => x==2)
Most of the posted solutions using array functions such as filter are incomplete because they aren't parameterized.
Here goes a solution with which the element to count can be set at run time.
function elementsCount(elementToFind, total, number){
return total += number==elementToFind;
}
var ar = [1, 2, 3, 5, 2, 8, 9, 2];
var elementToFind=2;
var result = ar.reduce(elementsCount.bind(this, elementToFind), 0);
The advantage of this approach is that could easily change the function to count for instance the number of elements greater than X.
You may also declare the reduce function inline
var ar = [1, 2, 3, 5, 2, 8, 9, 2];
var elementToFind=2;
var result = ar.reduce(function (elementToFind, total, number){
return total += number==elementToFind;
}.bind(this, elementToFind), 0);
Really, why would you need map or filter for this?
reduce was "born" for these kind of operations:
[1, 2, 3, 5, 2, 8, 9, 2].reduce( (count,2)=>count+(item==val), 0);
that's it! (if item==val in each iteration, then 1 will be added to the accumulator count, as true will resolve to 1).
As a function:
function countInArray(arr, val) {
return arr.reduce((count,item)=>count+(item==val),0)
}
Or, go ahead and extend your arrays:
Array.prototype.count = function(val) {
return this.reduce((count,item)=>count+(item==val),0)
}
It is better to wrap it into function:
let countNumber = (array,specificNumber) => {
return array.filter(n => n == specificNumber).length
}
countNumber([1,2,3,4,5],3) // returns 1
I use this:
function countElement(array, element) {
let tot = 0;
for(var el of array) {
if(el == element) {
tot++;
}
}
return tot;
}
var arr = ["a", "b", "a", "c", "d", "a", "e", "f", "a"];
console.log(countElement(arr, "a")); // 4
var arrayCount = [1,2,3,2,5,6,2,8];
var co = 0;
function findElement(){
arrayCount.find(function(value, index) {
if(value == 2)
co++;
});
console.log( 'found' + ' ' + co + ' element with value 2');
}
I would do something like that:
var arrayCount = [1,2,3,4,5,6,7,8];
function countarr(){
var dd = 0;
arrayCount.forEach( function(s){
dd++;
});
console.log(dd);
}
I believe what you are looking for is functional approach
const arr = ['a', 'a', 'b', 'g', 'a', 'e'];
const count = arr.filter(elem => elem === 'a').length;
console.log(count); // Prints 3
elem === 'a' is the condition, replace it with your own.
Array.prototype.count = function (v) {
var c = 0;
for (let i = 0; i < this.length; i++) {
if(this[i] === v){
c++;
}
}
return c;
}
var arr = [1, 2, 3, 5, 2, 8, 9, 2];
console.log(arr.count(2)); //3
Solution by recursion
function count(arr, value) {
if (arr.length === 1) {
return arr[0] === value ? 1 : 0;
} else {
return (arr.shift() === value ? 1 : 0) + count(arr, value);
}
}
count([1,2,2,3,4,5,2], 2); // 3
Create a new method for Array class in core level file and use it all over your project.
// say in app.js
Array.prototype.occurrence = function(val) {
return this.filter(e => e === val).length;
}
Use this anywhere in your project -
[1, 2, 4, 5, 2, 7, 2, 9].occurrence(2);
// above line returns 3
Here is a one liner in javascript.
Use map. Find the matching values (v === 2) in the array, returning an array of ones and zeros.
Use Reduce. Add all the values of the array for the total number found.
[1, 2, 3, 5, 2, 8, 9, 2]
.map(function(v) {
return v === 2 ? 1 : 0;
})
.reduce((a, b) => a + b, 0);
The result is 3.
Depending on how you want to run it:
const reduced = (array, val) => { // self explanatory
return array.filter((element) => element === val).length;
}
console.log(reduced([1, 2, 3, 5, 2, 8, 9, 2], 2));
// 3
const reducer = (array) => { // array to set > set.forEach > map.set
const count = new Map();
const values = new Set(array);
values.forEach((element)=> {
count.set(element, array.filter((arrayElement) => arrayElement === element).length);
});
return count;
}
console.log(reducer([1, 2, 3, 5, 2, 8, 9, 2]));
// Map(6) {1 => 1, 2 => 3, 3 => 1, 5 => 1, 8 => 1, …}
You can use built-in function Array.filter()
array.filter(x => x === element).length;
var arr = [1, 2, 3, 5, 2, 8, 9, 2];
// Count how many 2 there are in arr
var count = arr.filter(x => x === 2).length;
console.log(count);
One-liner function
const countBy = (a,f)=>a.reduce((p,v,i,x)=>p+!!f(v,i,x), 0)
countBy([1,2,3,4,5], v=>v%2===0) // 2
There are many ways to find out. I think the easiest way is to use the array filter method which is introduced in es6.
function itemCount(array, item) {
return array.filter(element => element === item).length
}
const myArray = [1,3,5,7,1,2,3,4,5,1,9,0,1]
const items = itemCount(myArray, 1)
console.log(items)
Something a little more generic and modern (in 2022):
import {pipe, count} from 'iter-ops';
const arr = [1, 2, 3, 5, 2, 8, 9, 2];
const n = pipe(arr, count(a => a === 2)).first; //=> 3
What's good about this:
It counts without creating a new array, so it is memory-efficient
It works the same for any Iterable and AsyncIterable
Another approach using RegExp
const list = [1, 2, 3, 5, 2, 8, 9, 2]
const d = 2;
const counter = (`${list.join()},`.match(new RegExp(`${d}\\,`, 'g')) || []).length
console.log(counter)
The Steps follows as below
Join the string using a comma Remember to append ',' after joining so as not to have incorrect values when value to be matched is at the end of the array
Match the number of occurrence of a combination between the digit and comma
Get length of matched items
I believe you can use the new Set array method of JavaScript to have unique values.
Example:
var arr = [1, 2, 3, 5, 2, 8, 9, 2]
var set = new Set(arr);
console.log(set);
// 1,2,3,5,8,9 . We get unique values as output.
You can use length property in JavaScript array:
var myarray = [];
var count = myarray.length;//return 0
myarray = [1,2];
count = myarray.length;//return 2

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