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How to generate an array of incremental values from a given array
the idea is to create a kind of diamond shape where the arrays start decreasing in size once they reach the middle of the array. In other words the longest array is going to be the one that is containing the middle value of the array or (array.length/2 + 1)
and in cases where the elements are short to complete the array on the second half just replace it with 'E' to indicate empty space just like on the second example.
example 1
var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p']
//the longest array in length is containing 'i' which is the value at
array.length/2 + 1
var output = [
['a'],
['b','c'],
['d','e','f'],
['g','h','i','j'],
['k','l','m'],
['n','o'],
['p']
]
example 2
var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t']
enter code here
//the longest array in length is containing 'k' which is the value at array.length/2 + 1
var output = [
['a'],
['b','c'],
['d','e','f'],
['g','h','i','j'],
['k','l','m','n','o'],
['p','q','r','s'],
['t','E','E'],
['E','E'],
['E]
]
Here is the code i have tried:
const values = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16];
const halfLen = values.length/2 + 1;
var topArr = [];
for(let i = 0; i < values.length; i ++) {
if(i <= halfLen) {
topArr.push(values[i])
}
}
console.log(topArr)
var filTopArr = [];
for(let i = 0; i <= topArr.length; i ++) {
let prevIndex = i - 1;
if(i === 0) {
filTopArr.push(topArr[i])
} else if(i === 1) {
filTopArr.push(topArr.slice(i, i + i + 1))
} else {
filTopArr.push(topArr.slice(i, i + i ))
}
}
console.log(filTopArr)
my idea here was to separate the array into two different arrays which are going to be the top part that is incrementing in size and the second/bottom part that is going to be decreasing in size.
The above code had this output
[1, [2, 3], [3, 4], [4, 5, 6], [5, 6, 7, 8], [6, 7, 8, 9], [7, 8, 9], [8, 9], [9], []]
Some observations:
The number of strings in the output (including the padding "E" strings) is always a perfect square (1, 4, 9, 16, 25, ...etc)
In order to know how many "E" strings need to be added, we thus need to know which is the least perfect square that is not less than the input size.
The longest (middle) subarray in the output has a size that is the square root of that perfect square.
The number of subarrays is the double of that number minus 1.
This leads to the following implementation:
function diamond(array) {
// Get least perfect square that is not less than the array length
const sqrt = Math.ceil(Math.sqrt(array.length));
const size = sqrt ** 2;
// Pad the array with "E" strings so to reach that perfect square size
const all = [...array, ..."E".repeat(size - array.length)];
const length = 2 * sqrt;
return Array.from({length}, (_, width) => {
return all.splice(0, Math.min(width, length - width));
}).slice(1); // Skip the first subarray that was produced (empty array)
}
// Demo using the two provided examples:
var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p'];
console.log(diamond(array));
var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t'];
console.log(diamond(array));
Here's a recursive version. Note that display is just for presentation purposes.
The only real work is in diamond:
const diamond = (xs, [len = xs.length, up = true, n = 1] = []) => n == 0 ? [] : [
Object .assign (Array (n) .fill ('E'), xs .slice (0, n)),
...diamond (xs .slice (n), up && n * n < len ? [len, true, n + 1] : [len, false, n - 1])
]
const display = (xss) => console .log (`${xss .map (
(xs, i) => `${' '.repeat (Math .abs ((xss .length - 1) / 2 - i) + 1)}${xs .join (' ')
}`) .join ('\n')}`)
const demos = [
['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p'],
['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u'],
[1, 2, 3, 4, 5, 6, 7, 8]
]
demos .forEach (array => display (diamond (array)))
.as-console-wrapper {max-height: 100% !important; top: 0}
We track the length of the current string (n, defaulting to 1), the length of the original array (len) , and a boolean flag to tell whether our length is moving up or down (up). We increase n on initial iterations, adding the next n characters from our input as the next subarray. When n hits zero we return an empty array. When n ** n is greater than or equal to len, we switch up to false and start subtracting one from n from then on. The only other necessity is to fill our remaining array with 'E's. We do this with an Object .assign call.
If you want formatted output more like an array literal form, you could use this version of display:
const display = (xss) => console .log (`[\n${xss .map (
(xs, i) => `${' '.repeat (Math .abs ((xss .length - 1) / 2 - i) + 1)}['${xs .join (`','`)
}']`) .join ('\n')}\n]`)
to get output like this:
[
['a']
['b','c']
['d','e','f']
['g','h','i','j']
['k','l','m','n','o']
['p','q','r','s']
['t','u','E']
['E','E']
['E']
]
Note though that this recursion is a little overbearing, with three separate defaulted recursive variables. I would be as likely to go with trincot's solution as this. But it's good to have alternatives.
What is the fastest way to split an array into small chunks, then apply math function in the small array?
The math function could be simple such as the total of the chunk mod 26.
What I mean, I have array [1,2,3,4,5,6], I need to create chunks of it to have every 3 elements in 1 chunk, so I will get from the main array:
[1,2,3]
[4,5,6]
The apply total of [1,2,3] 6mod26, and [4,5,6] 15mod26.
So, the final array will be [6,15].
Also, ignore the remaining of the main array if the remaining elements less the required chunk size, or add it in array alone.
simple way ..
[1,2,3,4,5,6,7] => chunk1[1,2,3] , chunk2[4,5,6] , chunk3[7]
result = [ (total([1,2,3]) mod 26 ) , (total([4,5,6]) mod 26 ) ]
I hope this is clear.
You could slice the array with the wanted size and reduce the array by adding all items and push the result of the remainder value.
var add = (a, b) => a + b,
array = [1, 2, 3, 4, 5, 6],
size = 3,
result = [],
i = 0;
while (i < array.length) {
result.push(array.slice(i, i += size).reduce(add) % 26);
}
console.log(result);
Sorry, but I don't know how to make more clear than this, simple way:
1. chunk the array
2. get the total sum of each chunk
3. mod the total sum of each chunk
4. create a new array from the mod.
I reached to away, but I think there is a better way:
splitArrayIntoChunksOfLen(arr, len) {
let chunks = [], i = 0, n = arr.length;
while (i < n) {
chunks.push(arr.slice(i, i += len));
}
return chunks;
}
let add = this.splitArrayIntoChunksOfLen(this.stringsArrays, 5)
let newAdd = add.map(function (item) {
return item.reduce(function (acc, curr) {
acc += curr;
return acc - 92 * Math.floor(acc / 92);
}, 0)
});
I want to sort only odd numbers without moving even numbers. For example, when I write :
sortArray([5, 3, 2, 8, 1, 4])
The expected result is :
[1, 3, 2, 8, 5, 4]
I am new to JavaScript and I came across a challenge on the Internet that has me perplexed. I normally wouldn't post asking for a solution on the Internet, BUT I have tried for hours and I would like to learn this concept in JavaScript.
The challenge states :
You have an array of numbers.
Your task is to sort ascending odd numbers but even numbers must be on their places.
Zero isn't an odd number and you don't need to move it. If you have an empty array, you need to return it.
Here is my code so far, please take it easy on me I am in the beginning stages of programming.
function sortArray(array) {
let oddNums = [];
for(let i = 0; i < array.length; i++) {
if(array[i] % 2 !== 0) {
oddNums.push(array[i]);
}
}
oddNums = oddNums.sort((a,b)=> a-b);
array.concat(oddNums);
array = array.sort((a,b) => a-b);
return array;
}
You could take a helper array for the odd indices and another for the odd numbers, sort them and apply them back on the previously stored indices of the original array.
var array = [5, 3, 2, 8, 1, 4],
indices = [];
array
.filter((v, i) => v % 2 && indices.push(i))
.sort((a, b) => a - b)
.forEach((v, i) => array[indices[i]] = v);
console.log(array);
Here's a solution using mostly the built-in array methods. Get a list of just the odds, sort it, then map through the original, replacing each item with the first sorted odd if the item is odd, or itself if even:
const array = [5, 3, 2, 8, 1, 4] // to: [1, 3, 2, 8, 5, 4]
function sortOddsOnly(arr) {
const odds = arr
.filter(x => x%2)
.sort((a, b) => a - b);
return arr
.map(x => x%2 ? odds.shift() : x);
}
console.log(sortOddsOnly(array));
I have a solution like this.
Build a sorted odd number array 1st, and then fill the rest of even numbers in order:
const arr = [5, 3, 2, 8, 1, 4];
const odd = arr.filter(i => i%2 !== 0).sort();
let i = 0,
result = [];
arr.forEach(e => {
if (e%2 === 0) {
result.push(e)
} else {
result.push(odd[i]);
i++;
}
});
console.log(result);
just do:
arr.sort((a, b) => a%2 && b%2 ? a - b : 0)
If that works depends on the sort algorithm your browser uses.
A browserindependent version:
for(const [i1, v1] of arr.entries())
for(const [i2, v2] of arr.entries())
if( v1%2 && v2%2 && (i1 < i2) === (v1 > v2))
([arr[i1], arr[i2]] = [v2, v1]);
One of the possible solutions is this. What I have done is created new array odd(array with odd position in original array using Array.prototype.filter) and then sort that array using Array.prototype.sort. Then using Array.prototype.map change value of all odd element of original array with odd array.
x1=[5, 3, 2, 8, 1, 4];
function sortArray(array) {
var odd = array.filter((x,i) => (i+1) % 2 ).sort((a,b) => a > b); //sort odd position and store that in new array
return array.map((x,i) => (i+1) % 2 ? odd.shift() : x ); //if i is odd then replace it with element from
//odd array otherwise keep the element as it is
}
console.log(sortArray(x1));
Here is a possible solution using a slightly customized selection sort :
var xs = [5, 3, 2, 8, 1, 4];
console.log(sortOddsOnly(xs));
function sortOddsOnly (xs) {
var n = xs.length;
for (var i = 0; i < n - 1; i++) {
if (xs[i] % 2 === 1) {
for (var j = i + 1; j < n; j++) {
if (xs[j] % 2 === 1) {
if (xs[i] > xs[j]) {
var min = xs[j];
xs[j] = xs[i];
xs[i] = min;
}
}
}
}
}
return xs;
}
The first two if guarantee that we swap only odd numbers (x % 2 === 1 means "x is odd").
def sort_array(source_array):
b = sorted([n for n in source_array if n % 2 != 0])
c = -1
d = []
for i in source_array:
c = c+1
if i % 2 != 0 :
d.append(c)
for x in range (len(d)):
z = d[x]
source_array[z] = b[x]
return source_array
Important edit : I can't use filter - the purpose is pedagogic.
I have an array in which I would want to count the number of its elements that verify a boolean, using only map and reduce.
Count of the array's size
I already wrote something that counts the array's size (ie. : the number of all of its elements), using reduce :
const array_numbers = [12, 15, 1, 1]; // Size : 4
console.log(
array_numbers.reduce((acc) => {
return acc + 1;
}, 0)
);
Count of the array's elements checking a boolean condition
Now I would want to count only the elements that verify a boolean condition. Thus, I must use map before reduce and the elements of the map's returned array will be only the good elements.
So I wrote this code but it doesn't work... Indeed, I put null when I encounter a not-good element (and null is counted as en element unfortunately).
NB : here, the boolean condition is "is the element even ? (%2 == 0)".
const array_numbers = [12, 15, 1, 1]; // Size : 4
console.log(
array_numbers.map((current_value) => {
if(current_value % 2 == 0) {
return current_value;
}
return null;
}).reduce((acc) => {
return acc + 1;
}, 0)
);
Array#filter the array and check the length:
const array_numbers = [12, 15, 1, 1];
const result = array_numbers.filter((n) => n % 2 === 0).length;
console.log(result);
Or count using Array#reduce:
const array_numbers = [12, 15, 1, 1, 4];
const result = array_numbers.reduce((r, n) => n % 2 ? r : r + 1, 0);
console.log(result);
Or if you must, you can use Array#map with Array#reduce:
const array_numbers = [12, 15, 1, 1, 4];
const result = array_numbers
.map((n) => n % 2 === 0 ? 1 : 0) // map to 1 or 0 according to the condition
.reduce((r, n) => r + n); // sum everything
console.log(result);
You can use Array.prototype.filter to filter the even numbers - and you don't need the reduce() function - you can use length of the array returned by the filter() function.
Or you can use reduce() method alone like below:
See demos below:
const array_numbers = [12, 15, 1, 1]; // Size : 4
// using filter
console.log(
array_numbers.filter((current_value) => {
return current_value % 2 == 0;
}).length
);
// using reduce
console.log(
array_numbers.reduce((prev, curr) => {
return curr % 2 == 0 ? prev + 1 : prev;
}, 0)
);
Since per your comment you must use reduce for some reason:
arr.reduce((acc, n) => { n % 2 ? acc + n : acc }, 0);
The map is unecessary.
As Jared Smith mentioned, you don't need to use map for this task.
Array.reduce() gets the current element as a second argument in the callback function which you can use to check if it satisfies your given condition.
So, again assuming that you must use either map and/or reduce:
const myArray = [1,2,3,4,5,6];
const condition = function(a) {
// let's say
return a %2 == 0;
}
let result = myArray.reduce((acc, val) => {
return condition(val) ? acc + 1 : acc
}, 0);
console.log(result);
I am trying to solve a math problem where I take a number e.g. 45, or 111 and then split the number into separate digits e.g. 4 5 or 1 1 1. I will then save each number to a var to run a method on. Does anyone know how to split a number into individual digitals?
For example I have a loop that runs on an array :
for (var i = 0; i < range.length; i++) {
var n = range[i];
}
For each number, I would like to split its digits and add them together?
var num = 123456;
var digits = num.toString().split('');
var realDigits = digits.map(Number)
console.log(realDigits);
var number = 12354987,
output = [],
sNumber = number.toString();
for (var i = 0, len = sNumber.length; i < len; i += 1) {
output.push(+sNumber.charAt(i));
}
console.log(output);
/* Outputs:
*
* [1, 2, 3, 5, 4, 9, 8, 7]
*/
UPDATE: Calculating a sum
for (var i = 0, sum = 0; i < output.length; sum += output[i++]);
console.log(sum);
/*
* Outputs: 39
*/
You can also do it in the "mathematical" way without treating the number as a string:
var num = 278;
var digits = [];
while (num != 0) {
digits.push(num % 10);
num = Math.trunc(num / 10);
}
digits.reverse();
console.log(digits);
One upside I can see is that you won't have to run parseInt() on every digit, you're dealing with the actual digits as numeric values.
This is the shortest I've found, though it does return the digits as strings:
let num = 12345;
[...num+''] //["1", "2", "3", "4", "5"]
Or use this to get back integers:
[...num+''].map(n=>+n) //[1, 2, 3, 4, 5]
I will provide a variation on an answer already given so you can see a different approach that preserves the numeric type all along:
var number = 12354987,
output = [];
while (number) {
output.push(number % 10);
number = Math.floor(number/10);
}
console.log(output.reverse().join(',')); // 1,2,3,5,4,9,8,7
I've used a technique such as the above to good effect when converting a number to Roman numerals, which is one of my favorite ways to begin to learn a programming language I'm not familiar with. For instance here is how I devised a way to convert numbers to Roman numerals with Tcl slightly after the turn of the century: http://code.activestate.com/recipes/68379-conversion-to-roman-numerals/
The comparable lines in my Tcl script being:
while {$arabic} {
set digit [expr {$arabic%10}]
set arabic [expr {$arabic/10}]
// Split positive integer n < 1e21 into digits:
function digits(n) {
return Array.from(String(n), Number);
}
// Example:
console.log(digits(1234)); // [1, 2, 3, 4]
You can work on strings instead of numbers to achieve this. You can do it like this
(111 + '').split('')
This will return an array of strings ['1','1','1'] on which you can iterate upon and call parseInt method.
parseInt('1') === 1
If you want the sum of individual digits, you can use the reduce function (implemented from Javascript 1.8) like this
(111 + '').split('').reduce(function(previousValue, currentValue){
return parseInt(previousValue,10) + parseInt(currentValue,10);
})
Use String, split and map :
String(number).split("").map(Number);
function splitNum(num) {
return String(num).split("").map(Number);
}
console.log(splitNum(1523)); // [1, 5, 2, 3]
console.log(splitNum(2341)); // [2, 3, 4, 1]
console.log(splitNum(325)); // [3, 2, 5]
Without converting to string:
function toDigits(number) {
var left;
var results = [];
while (true) {
left = number % 10;
results.unshift(left);
number = (number - left) / 10;
if (number === 0) {
break;
}
}
return results;
}
Using String, ... and map
const num = 7890;
const digits = [...String(num)].map(Number);
console.log(digits)
Alternatively, using ... and reduce to get digits and their sum.
const sumOfDigits = num => [...""+num].reduce((acc, dig) => acc + +dig, 0);
console.log('Sum of digits: ', sumOfDigits(7890));
Separate each 2 parametr.
function separator(str,sep) {
var output = '';
for (var i = str.length; i > 0; i-=2) {
var ii = i-1;
if(output) {
output = str.charAt(ii-1)+str.charAt(ii)+sep+output;
} else {
output = str.charAt(ii-1)+str.charAt(ii);
}
}
return output;
}
console.log(separator('123456',':')); //Will return 12:34:56
With ES6, you could use Array.from with a stringed number as iterables and Number as mapping function.
const getDigits = n => Array.from(n.toString(), Number);
console.log(getDigits(12345));
A fun introduction to recursion. This answer takes a Number and returns an array of Number digits. It does not convert the number to a string as an intermediate step.
Given n = 1234,
n % 10 will return first (right-moist) digit, 4
n / 10 will return 123 with some remainder
Using Math.floor we can chop the remainder off
Repeating these steps, we can form the entire result
Now we just have to build the recursion condition,
If the number is already a single digit (n < 10), return an array singleton of the digit
otherwise (inductive) the number is 10 or greater; recur and prepend to the first digit
const digits = (n = 0) =>
n < 10
? [ n ]
: [ ... digits (Math.floor (n / 10)), n % 10 ]
console.log (digits ()) // [ 0 ]
console.log (digits (1)) // [ 1 ]
console.log (digits (12)) // [ 1, 2 ]
console.log (digits (123)) // [ 1, 2, 3 ]
console.log (digits (11234)) // [ 1, 2, 3, 4 ]
console.log (digits (123456789012))
// [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2 ]
This also works:
var number = 12354987;
console.log(String(number).split('').map(Number));
Shadow Wizard , extended version by Orien
var num:Number = 1523;
var digits:Array = [];
var cnt:int = 0;
while (num > 0) {
var mod:int = num % 10;
digits.push(mod * Math.pow(10, cnt))
num = Math.floor(num / 10);
cnt++;
}
digits.reverse();
trace(digits);
output:1000,500,20,3
A functional approach in order to get digits from a number would be to get a string from your number, split it into an array (of characters) and map each element back into a number.
For example:
var number = 123456;
var array = number.toString()
.split('')
.map(function(item, index) {
return parseInt(item);
});
console.log(array); // returns [1, 2, 3, 4, 5, 6]
If you also need to sum all digits, you can append the reduce() method to the previous code:
var num = 123456;
var array = num.toString()
.split('')
.map(function(item, index) {
return parseInt(item);
})
.reduce(function(previousValue, currentValue, index, array) {
return previousValue + currentValue;
}, 0);
console.log(array); // returns 21
As an alternative, with ECMAScript 2015 (6th Edition), you can use arrow functions:
var number = 123456;
var array = number.toString().split('').map((item, index) => parseInt(item));
console.log(array); // returns [1, 2, 3, 4, 5, 6]
If you need to sum all digits, you can append the reduce() method to the previous code:
var num = 123456;
var result = num.toString()
.split('')
.map((item, index) => parseInt(item))
.reduce((previousValue, currentValue) => previousValue + currentValue, 0);
console.log(result); // returns 21
I used this simple way of doing it.
To split digits
var N = 69;
var arr = N.toString().split('').map(Number)
// outputs [6,9]
console.log( arr );
To add them together
console.log(arr.reduce( (a,b) => a+b )); // 15
And the easiest.... num_string.split('').map(Number)
Try below:
console.log((''+123).split('').map(Number))
To just split an integer into its individual digits in the same order, Regular Expression is what I used and prefer since it prevents the chance of loosing the identity of the numbers even after they have been converted into string.
The following line of code convert the integer into a string, uses regex to match any individual digit inside the string and return an array of those, after which that array is mapped to be converted back to numbers.
const digitize = n => String(n).match(/\d/g).map(Number);
I might be wrong, but a solution picking up bits and pieces. Perhaps, as I still learning, is that the functions does many things in the same one. Do not hesitate to correct me, please.
const totalSum = (num) => [...num + ' '].map(Number).reduce((a, b) => a + b);
So we take the parameter and convert it to and arr, adding empty spaces. We do such operation in every single element and push it into a new array with the map method. Once splited, we use reduce to sum all the elements and get the total.
As I said, don't hesitate to correct me or improve the function if you see something that I don't.
Almost forgot, just in case:
const totalSum = (num) => ( num === 0 || num < 0) ? 'I need a positive number' : [...num + ' '].map(Number).reduce((a, b) => a + b);
If negatives numbers or just plain zero go down as parameters. Happy coding to us all.
I am posting this answer to introduce the use of unshift which is a modern solution. With push, you add to the end of an array while unshift adds to the beginning. This makes the mathematical approach more powerful as you won't need to reverse anymore.
let num = 278;
let digits = [];
while (num > 0) {
digits.unshift(num % 10);
num = parseInt(num / 10);
}
console.log(digits);
var num = 111,
separateDigits = num.toString().split(""), i, l = separateDigits.length;
for( i = 0; i < l; ++i ) {
someObject.someMethod( +separateDigits[i] );
}
You can try this.
var num = 99;
num=num.toString().split("").map(value=>parseInt(value,10)); //output [9,9]
Hope this helped!
function iterateNumber(N, f) {
let n = N;
var length = Math.log(n) * Math.LOG10E + 1 | 0;
for (let i = 0; i < length; i++) {
const pow = Math.pow(10, length - i - 1)
let c = (n - (n % pow)) / pow
f(c, i)
n %= pow
}
}
('' + 123456789).split('').map( x => +x ).reduce( (a,b) => a+b ) === 45
true
or without map
('' + 123456789).split('').reduce( (a,b) => (+a)+(+b) ) === 45
true
You can do it in single line, seperate each digits than add them together :
var may = 12987;
var sep = (""+may).split("").map(n=>+n).reduce((a,b)=>a+b);
This is my short solution.. with sum of number
function sum (num) {
let sNumber = num
.toString()
.split('')
.reduce((el1, el2) => {
return Number(el1) + Number(el2)
}, 0)
return sNumber
}
console.log(sum(123))
console.log(sum(456))
javascript has a function for it and you can use it easily.
console.log(new Intl.NumberFormat().format(number));
for example :
console.log(new Intl.NumberFormat().format(2334325443534));
==> 2,334,325,443,534
Iterate through each number with for...of statement.
By adding a + sign before a String, it will be converted into a number.
const num = 143,
digits = [];
for (const digit of `${num}`) {
digits.push(+digit)
}
console.log(digits);
Inspired by #iampopov You can write it with spread syntax.
const num = 143;
const digits = [...`${num}`].map(Number);
console.log(digits);
And as a one liner.
console.log(Number.MAX_SAFE_INTEGER.toString().split('').reduce((pv, v) => Number(v) + pv, 0));