So I have an array of ids something like this:
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
I need a function that will be called like mergeArray(arr, 3), and it should return comma separated values with maximum of 3 elements like this:
const newArr = ['1,2,3', '4,5,6', '7,8,9', '10,11'];
How can I do this? If possible with ES6 functions for simpler code.
slice your array into 3 lengths arrays and directly join them
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
const mergeArray = (arr, size) => {
let res = [];
for (i = 0; i < arr.length; i += size) {
res.push(arr.slice(i, i + size).join(','));
}
return res;
}
console.log(mergeArray(arr, 3));
You can split() the array into the specific size and join() them before pushing into the resulting array:
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
var i, j, newArr=[], size = 3;
for (i=0,j=arr.length; i<j; i+=size) {
newArr.push(arr.slice(i, i+size).join());
}
console.log(newArr);
One of the ways to do it is with Array.prototype.reduce and Array.prototype.map:
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
function mergeArray(arr, n) {
return arr
.reduce((all, el, i) => {
const ind = Math.floor(i/n);
all[ind] = [...all[ind] || [], el]
return all;
},[])
.map(a => a.join(','))
}
console.log(mergeArray(arr, 3));
You could join the array and match the wanted parts with a regular expression.
var data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
result = data.join(',').match(/\d+(,\d+(,\d+)?)?/g)
console.log(result);
Related
function distinctUnion(arr, arr2) {
let merged = [...arr, ...arr2];
var result = [];
var map = {}
for (let i = 0; i < merged.length; i++) {
if (!map.hasOwnProperty(merged[i])) {
map[merged[i]] = true; // Line 3 --> if I remove this line, it prints duplicates
console.log('map', JSON.stringify(map, 2, null));
result.push(merged[i]);
}
}
return result;
}
let arr = [3, 4, 5, 6, 6, 4, 5, 8, 9, 10, 10];
let arr2 = [11, 11, 11, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6];
console.log('unique ', JSON.stringify(distinctUnion(arr, arr2), 2, null));
All we are setting here is map[merged[i]] = true; for all keys in object
map {"3":true,"4":true,"5":true,"6":true,"8":true,"9":true,"10":true,"11":true}
then how result.push(merged[i]) has only unique values?
I mean to say merged[i] inside loop should still have all array values including duplicates right?
I am not able to understand the link between map[merged[i]] = true; and result.push(merged[i])
If you do not set the property to anything, map.hasOwnProperty(...) will spuriously return false for the next time that value is encountered, thus allowing duplicates. You don't need to set it to true, as it is just used to indicate the presence of a key; any value is fine—even undefined!
function distinctUnion(arr, arr2) {
let merged = [...arr, ...arr2];
var result = [];
var map = {}
for (let i = 0; i < merged.length; i++) {
if (!map.hasOwnProperty(merged[i])) {
map[merged[i]] = undefined;
result.push(merged[i]);
}
}
return result;
}
let arr = [3, 4, 5, 6, 6, 4, 5, 8, 9, 10, 10];
let arr2 = [11, 11, 11, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6];
console.log('unique ', JSON.stringify(distinctUnion(arr, arr2), 2, null));
To make your code works, you just need to replace map[merged[i]] = true; with map[merged[i]] = undefined;.
However, you can make your function more simplified as follows:
function distinctUnion(arr, arr2) {
let map = {};
[...arr, ...arr2].forEach((x)=>{map[x] = x});
return Object.values(map);;
}
let arr = [3, 4, 5, 6, 6, 4, 5, 8, 9, 10, 10];
let arr2 = [11, 11, 11, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6];
console.log('Unique ', distinctUnion(arr, arr2));
I'm wanting to create a function which accepts 2 arguments, first argument is an array, second argument is a number of index positions to move all the array items.
So for example if I passed exampleFunc([1,2,3,4,5], 2) it should move all items 2 places to the right, so returns [4,5,1,2,3]. I've done the following, however is there a more eloquent / efficient way of doing this? Also if I wanted to reverse the direction and condense into 1 function and not two as done below, any suggestions how to do this other than putting conditionals around the different part of each function? Tried using .splice() method but didn't really got anywhere. Any help would really be appreciated!
const moveArrayPositionRight = (array, movePositions) => {
let newArray = new Array(array.length);
for (i = 0; i < array.length; i++) {
let newIndex = i - movePositions;
if (newIndex < 0) {
newIndex += array.length;
}
newArray[i] = array[newIndex];
}
return newArray;
};
console.log(moveArrayPositionRight([2, 4, 6, 8, 10], 2)); // output: [8, 10, 2, 4, 6]
const moveArrayPositionLeft = (array, movePositions) => {
let newArray = new Array(array.length);
for (i = 0; i < array.length; i++) {
let newIndex = i - movePositions;
if (newIndex < 0) {
newIndex += array.length - 1;
}
newArray[i] = array[newIndex];
}
return newArray;
};
console.log(moveArrayPositionLeft([3, 6, 9, 12, 15], 2)); // output: [9,12,15,3,6]
You have the index of the position where you want to slice the array up and rearrange it, so you can use .slice to do exactly that - extract the sub-arrays that need to be rearranged, and put into a new array:
const moveArrayPositionRight = (array, movePositions) => [
...array.slice(array.length - movePositions),
...array.slice(0, array.length - movePositions)
];
console.log(moveArrayPositionRight([2, 4, 6, 8, 10], 2)); // output: [8, 10, 2, 4, 6]
console.log(moveArrayPositionRight([2, 4, 6, 8, 10], 3)); // expected [6, 8, 10, 2, 4]
.slice can also take negative indicies to slice an amount from the end instead of from the beginning:
const moveArrayPositionRight = (array, movePositions) => [
...array.slice(-movePositions),
...array.slice(0, -movePositions)
];
console.log(moveArrayPositionRight([2, 4, 6, 8, 10], 2)); // output: [8, 10, 2, 4, 6]
console.log(moveArrayPositionRight([2, 4, 6, 8, 10], 3)); // expected [6, 8, 10, 2, 4]
Can also use .concat instead of spread
const moveArrayPositionRight = (array, movePositions) => array
.slice(array.length - movePositions)
.concat(array.slice(0, array.length - movePositions));
console.log(moveArrayPositionRight([2, 4, 6, 8, 10], 2)); // output: [8, 10, 2, 4, 6]
console.log(moveArrayPositionRight([2, 4, 6, 8, 10], 3)); // expected [6, 8, 10, 2, 4]
Same sort of thing for moveArrayPositionLeft:
const moveArrayPositionLeft = (array, movePositions) => [
...array.slice(movePositions),
...array.slice(0, movePositions)
];
console.log(moveArrayPositionLeft([3, 6, 9, 12, 15], 2)); // output: [9,12,15,3,6]
I have 2 arrays, one has 10 elements and the other one 3, I need to create a new array with the same size of the biggest vector, with a boolean checking true in the position where exist some element from the array of 3 elements
I have the following arrays
array1 = [1,2,3,4,5,6,7,8,9,10]
array2 = [4,6,10]
I tried making 2 for loops
for(var i=0; i<array1.lenght; i++){
for(var j=0; i<array2.lenght; i++){
if(array1[i]==array2[j]){
array3.push(true)
}else{
array3.push(false)
}
}
}
the vector that I need would be
array3 = [false, false, false, true, false, true, false, false, false, true]
You can forEach first array and use include method to check if item existed in array as
let array1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
let array2 = [4, 6, 10];
let array3 = [];
array1.forEach(function (c) {
if (array2.includes(c)) {
array3.push(true)
} else {
array3.push(false);
}
})
console.log(array3)
Use map like so with shift like so:
const array1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const array2 = [4, 6, 10];
const array3 = array1.map(e => {
if (array2[0] == e) {
array2.shift();
return true;
}
return false;
});
console.log(array3);
.as-console-wrapper { max-height: 100% !important; top: auto; }
If you just want a basic check as for whether the element is in the array, not the order, then use includes.
const array1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const array2 = [4, 6, 10];
const array3 = array1.map(e => array2.includes(e));
console.log(array3);
.as-console-wrapper { max-height: 100% !important; top: auto; }
You can also instead of another array use a Set and then Array.map the first away checking if the value is in the Set:
let array1 = [1,2,3,4,5,6,7,8,9,10],
set = new Set([4,6,10])
let result = array1.map(x => set.has(x))
console.log(result)
I would suggest to keep things simple and to use Array#indexOf method to determine if array contains another element.
const array1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const array2 = [4, 6, 10];
const b = array1.map(el => {
return array2.indexOf(el) !== -1;
});
console.log(b);
const array1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const array2 = [4, 6, 10];
const finalArray = [];
for (let data of array1) {
finalArray.push(array2.includes(data));
}
I couldn't find an answer to this specific question on S.O.
Let's say I have an array of strings, or in this case, numbers:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
I'd like the output to be:
var output = [[1,1,1], [2], [3,3,3,3,3], [4], [5, 5, 5]];
I was hoping to use Lodash but most of that stuff tends to remove duplicates rather chunk them together into their own array. Maybe some kind of .map iterator?
The order of the output doesn't really matter so much. It just needs to chunk the duplicates into separate arrays that I'd like to keep.
You can use reduce to group the array elements into an object. Use Object.values to convert the object into an array.
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
var result = Object.values(x.reduce((c, v) => {
(c[v] = c[v] || []).push(v);
return c;
}, {}));
console.log(result);
Shorter version:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
var result = Object.values(x.reduce((c, v) => ((c[v] = c[v] || []).push(v), c), {}));
console.log(result);
You can do this with Array.reduce in a concise way like this:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5]
let result = x.reduce((r,c) => (r[c] = [...(r[c] || []), c],r), {})
console.log(Object.values(result))
The exact same with lodash would be:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5]
let result = _.values(_.groupBy(x))
console.log(result)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
Using _.values to extract the values of the grouping object and _.groupBy to get the actual groupings
Use Array#prototype#reduce to group them:
const x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
let helperObj = {};
const res = x.reduce((acc, curr) => {
// If key is not present then add it
if (!helperObj[curr]) {
helperObj[curr] = curr;
acc.push([curr]);
}
// Else find the index and push it
else {
let index = acc.findIndex(x => x[0] === curr);
if (index !== -1) {
acc[index].push(curr);
}
}
return acc;
}, []);
console.log(res);
Since you're hoping to use Lodash, you might be interested in groupBy. It returns on object, but the _.values will give you the nested array:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
let groups = _.values(_.groupBy(x))
console.log(groups)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
Here's an imperative solution:
var x = [1, 1, 1, 2, 3, 3, 5, 3, 3, 5, 4, 5];
x.sort();
var res = [];
for (const [i, n] of x.entries()) {
if (n !== x[i-1]) res.push([n]);
else res[res.length-1].push(n);
}
console.log(res);
How can I convert the elements of the array into another array?
let a = [2, 4, 0, 8, 9, 15]
and the result will be :
a = [[2], [4], [0], [8], [9], [15]]
Use Array#map to iterate the array, and wrap each item in an array:
let a = [2, 4, 0, 8, 9, 15]
const result = a.map((n) => [n])
console.log(JSON.stringify(result))
let a = [2, 4, 0, 8, 9, 15];
let b = a.map((item) => [item]);
a = [2, 4, 0, 8, 9, 15]
console.log(result = a.map((n) => [n]))
b=[]
//you can try below also
for (i in a){
b.push([i]);
}