for learning purposes I want to get every second element of an array. I succeeded with a for loop:
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
function filterEverySecond(arr) {
let everySecondEl = [];
for (let i = 0; i < arr.length; i += 2) {
everySecondEl.push(arr[i]);
}
return everySecondEl;
}
console.log({
numbers,
result: filterEverySecond(numbers)
});
Now I want to achieve the same without a for loop, but by using array methods (forEach, filter, map or reduce). Can anyone recommend which method would be best here?
you can do it easily with filter
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const filtered = numbers.filter((_, i) => i % 2 === 0)
console.log(filtered)
you just filter out the elements that have a odd index
You can use filter and check if the index of the current array element is divisible by 2:
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
console.log({
numbers,
result: numbers.filter((_, index) => index % 2 === 0)
});
You can use filter for index.
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
console.log({
numbers,
result: numbers.filter((n,i)=>i%2==0)
});
You could use a for each loop like so to get the same desired output:
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const result = [];
numbers.forEach(number => {
if (number % 2 != 0) {
result.push(number);
}
});
console.log(numbers);
console.log(result);
The modulus operator returns the remainder of the two numbers when divided. For example: 1 % 2 will return 1 as the remainder. So in the if statement we are checking if the number is not divisible by 2.
Related
var numbers = [1, 2, 3, 4, 5, 6, 7, 8];
need to check if 2, 3, 4 exist in the array. Only has to be 1 of these numbers to return true...not all. What's the best approach. I was thinking lodash includes, but I believe I can only pass in a single value.
Using Array#some and Array#includes:
const hasAny = (arr = [], nums = []) =>
nums.some(n => arr.includes(n));
console.log( hasAny([1, 2, 3, 4, 5, 6, 7, 8], [2, 3, 4]) );
If you want to get the intersection, you can filter the first array by the second array, or rather, whether or not the second array contains each element of the first. You can see which members were included, or check the size of the new array > 0 if you need a boolean (or use Mr Badawi's .some method).
var numbers = [1, 2, 3, 4, 5, 6, 7, 8];
var needs = [2, 3, 4 ];
var check = numbers.filter(x=>needs.includes(x));
console.log(check);
var numbers = [3, 5, 6, 7, 8, 12, 20];
var needs = [2, 3, 4 ];
var check = numbers.filter(x=>needs.includes(x));
console.log(check);
I have two arrays (X and Y) and I need to create array Z that contains all elements from array X except those, that are present in array Y p times where p is a prime number.
I am trying to write this in JS.
For Example:
Array X:
[2, 3, 9, 2, 5, 1, 3, 7, 10]
Array Y:
[2, 1, 3, 4, 3, 10, 6, 6, 1, 7, 10, 10, 10]
Array Z:
[2, 9, 2, 5, 7, 10]
So far I have this:
const arrX = [2, 3, 9, 2, 5, 1, 3, 7, 10]
const arrY = [2, 1, 3, 4, 3, 10, 6, 6, 1, 7, 10, 10, 10]
const arrZ = []
const counts = [];
// count number occurrences in arrY
for (const num of arrY) {
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
// check if number is prime
const checkPrime = num => {
for (let i = 2; i < num; i++) if (num % i === 0) return false
return true
}
console.log(counts[10]);
// returns 4
Any hint or help appreciated. Thanks!
You're on the right track. counts should be an object mapping elements in arrY to their number of occurrences. It's easily gotten with reduce.
The prime check needs a minor edit, and the last step is to filter arrX. The filter predicate is just a prime check on the count for that element.
// produce an object who's keys are elements in the array
// and whose values are the number of times each value appears
const count = arr => {
return arr.reduce((acc, n) => {
acc[n] = acc[n] ? acc[n]+1 : 1;
return acc;
}, {})
}
// OP prime check is fine, but should handle the 0,1 and negative cases:
const checkPrime = num => {
for (let i = 2; i < num; i++) if (num % i === 0) return false
return num > 1;
}
// Now just filter with the tools you built...
const arrX = [2, 3, 9, 2, 5, 1, 3, 7, 10]
const arrY = [2, 1, 3, 4, 3, 10, 6, 6, 1, 7, 10, 10, 10]
const counts = count(arrY);
const arrZ = arrX.filter(n => checkPrime(counts[n]));
console.log(arrZ)
With an array of: [1, 2, 3, 4, 5, 6]
I would like to delete between 2 indices such as 2 and 4 to produce [1, 2, null, null, 5, 6]. What's the easiest way to do this?
Hopefully better than this:
const array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
let i = 2;
const rangeEnd = 9;
while (i < rangeEnd) {
delete array[i];
i++;
}
console.log(array)
If you want to use some native API you can actually do this with splice(). Otherwise, you should iterate a for loop through your array and change the value in each iteration.
Here is an example of how it would be done:
const array = [1, 2, 3, 4, 5, 6]
array.splice(3, 2, null, null) // the First element is beginning index and the second is count one will indicate how many indexes you need to traverse from the first one, then you should provide replace element for each of them.
console.log(array)
Note: For more info about it you can read more here.
There is a possible workaround for large scale replacement, so I will give it a touch here:
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
var anotherArr = Array(2).fill(null); // or you can simply define [null, null, ...]
Array.prototype.splice.apply(arr, [3, anotherArr.length].concat(anotherArr));
console.log(arr);
As you mean the range (2, 4] so you can follow this:
The range is: lower limit exclusive and the upper limit inclusive.
const arr = [1, 2, 3, 4, 5, 6];
const deleteRange = (arr, f, t) => {
return arr.map((item, i) => {
if (i + 1 > f && i + 1 <= t) {
return null;
}
return item;
})
}
console.log(deleteRange(arr, 2, 4));
I'm trying to have a forEach loop over an array, but only the last few entries.
I'd know how to do this in a for loop, that'd look a bit like this:
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
/* This will loop over the last 3 entries */
for(var x = arr.length; x >= 7; x--){
console.log(arr[x]);
}
Would there be any way of achieving the same results in a forEach loop?
You can use slice() and reverse() methods and then forEach() loop on that new array.
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
arr.slice(-3).reverse().forEach(e => console.log(e))
This is how you do it with forEach loop:
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
arr.forEach((element, index) => {
if(index>7) console.log(arr[index]);
})
You could take a classic approach by taking the count of the last elements and use it as counter and an offset for the index.
Then loop with while by decrementing and checking the counter.
var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
last = 3,
offset = array.length - last;
while (last--) {
console.log(array[last + offset]);
}
I'm beginner and have one question.
I have one um array and need show just 2 in 2 elements of array elements, my code is:
var datas = [];
followTotalsData.forEach(function(dataByDay) {
datas.push(dataByDay[1]);
});
datas = [1, 2 , 3 , 4, 5, 6, 7, 8, 9, 10];
But I need the array formated as
var newDatas = [2, 4, 6, 8, 10];
I'm interpreting your question as
How can I only get Numbers that are multiples of 2 from my Array of Numbers
Use the remainder operator %.
If x % y is 0 then y divides x.
var datas = [1, 2 , 3 , 4, 5, 6, 7, 8, 9, 10];
var newDatas = datas.filter(function (e) {return e % 2 === 0;});
// [2, 4, 6, 8, 10]
Another solution:
var followTotalsData = [1, 2, 3, 4];
var result=[];
followTotalsData.forEach(function(data) {
if (data % 2 == 0 ) result.push(data);
});
console.log(result);
If you want every other element of the array, use the modulus function on the index of the array:
var datas = [1, 2 , 3 , 4, 5, 6, 7, 8, 9, 10];
var newDatas = datas.filter(function (e, index) {return index % 2 === 1;});
// [2, 4, 6, 8, 10]
var newDatas = datas.filter(function (e, index) {return index % 2 === 0;});
// [1, 3, 5, 7, 9]