A problem with nested Arrays and accessing them

Question

So, i'm like really new to coding in general and i'm going to tryout a coding bootcamp. So right now i'm trying to prepare myself to the "entry test" that will be focused on Javascript basics. I have this problem:

const charactersArray = [
    ['☆', 'D', 'a', '☆', '☆', 'r'],
    ['t', '☆', 'h', '☆', 'H', 'a'],
    ['c', 'k', '☆', 'e', '☆', 'r'],
    ['L', '☆', 'o', 'o', '☆', 'p'],
    ['S', '☆', 'k', 'y', '☆', 'w'],
    ['a', 'l', '☆', 'k', '☆', '☆'],
    ['☆', 'e', '☆', '☆', 'r', '☆']
];
let newArr = [];
for (i = 0; i < charactersArray.length; i++) {
    for (j = 0; i < charactersArray[i]; j++) {
        }
    }

so I've got this far but i get stuck after this. I need to take off that star special letter and return a new array with only the characters. Any helping hand? I want the array to be multidimensional, yes.

Answer 1

Here you go

let resultAr = charactersArray.map(arr => arr.filter(item => item != "☆"));

Adding one more solution with RegEx for finer control on inclusion/exclusion

let resultAr = charactersArray.map(arr => arr.filter(item => /[^\W]/i.test(item)));

Answer 2

You can also simply use Array.reduce and Array.filter to get the desired result:

If you need one dimensional array then you can just spread the filtered arrays into a new one:

 const arr = [ ['☆', 'D', 'a', '☆', '☆', 'r'], ['t', '☆', 'h', '☆', 'H', 'a'], ['c', 'k', '☆', 'e', '☆', 'r'], ['L', '☆', 'o', 'o', '☆', 'p'], ['S', '☆', 'k', 'y', '☆', 'w'], ['a', 'l', '☆', 'k', '☆', '☆'], ['☆', 'e', '☆', '☆', 'r', '☆'] ]; const result = arr.reduce((r,c) => (r.push(...c.filter(x => x !== '☆')), r), []) console.log(result) 

If you need the same dimensions but filtered is the same without the spread :

 const arr = [ ['☆', 'D', 'a', '☆', '☆', 'r'], ['t', '☆', 'h', '☆', 'H', 'a'], ['c', 'k', '☆', 'e', '☆', 'r'], ['L', '☆', 'o', 'o', '☆', 'p'], ['S', '☆', 'k', 'y', '☆', 'w'], ['a', 'l', '☆', 'k', '☆', '☆'], ['☆', 'e', '☆', '☆', 'r', '☆'] ]; const result = arr.reduce((r,c) => (r.push(c.filter(x => x !== '☆')), r), []) console.log(result) 

Same can be achieved in more concise manner via Array.from :

 const arr = [ ['☆', 'D', 'a', '☆', '☆', 'r'], ['t', '☆', 'h', '☆', 'H', 'a'], ['c', 'k', '☆', 'e', '☆', 'r'], ['L', '☆', 'o', 'o', '☆', 'p'], ['S', '☆', 'k', 'y', '☆', 'w'], ['a', 'l', '☆', 'k', '☆', '☆'], ['☆', 'e', '☆', '☆', 'r', '☆'] ]; const result = Array.from(arr, x => x.filter(y => y != '☆')) console.log(result) 

If you need a string:

 const arr = [ ['☆', 'D', 'a', '☆', '☆', 'r'], ['t', '☆', 'h', '☆', 'H', 'a'], ['c', 'k', '☆', 'e', '☆', 'r'], ['L', '☆', 'o', 'o', '☆', 'p'], ['S', '☆', 'k', 'y', '☆', 'w'], ['a', 'l', '☆', 'k', '☆', '☆'], ['☆', 'e', '☆', '☆', 'r', '☆'] ]; const result = arr.reduce((a,c) => a.concat(c.filter(x => x !== '☆').join('')), "") console.log(result) 

Answer 3

Given OP's purpose, I don't think he/she is interested in reduce/map/filter/...(at least right now). Just a simple for loop to manipulate two-dimension array.

for (i = 0; i < charactersArray.length; i++) {
    for (j = 0; j < charactersArray[i].length; j++) {
        if(charactersArray[i][j] != "☆") {
            newArr.push(charactersArray[i][j]);
        }
    }
}

Answer 4

You can use Array.prototype.concat() with the spread operator to flatten the two-dimensional array, and then use Array.prototype.filter() to filter out the stars:

const result = [].concat(...charactersArray).filter(c => c !== '☆');

---

Complete snippet:

 const charactersArray = [ ['☆', 'D', 'a', '☆', '☆', 'r'], ['t', '☆', 'h', '☆', 'H', 'a'], ['c', 'k', '☆', 'e', '☆', 'r'], ['L', '☆', 'o', 'o', '☆', 'p'], ['S', '☆', 'k', 'y', '☆', 'w'], ['a', 'l', '☆', 'k', '☆', '☆'], ['☆', 'e', '☆', '☆', 'r', '☆'] ]; const result = [].concat(...charactersArray).filter(c => c !== '☆'); console.log(result.join('')); // DarthHackerLoopSkywalker 

Answer 5

charactersArray.map((a)=>{
    return a.filter((c)=>{
        return c != '☆'});
});

->

[ [ 'D', 'a', 'r' ],
  [ 't', 'h', 'H', 'a' ],
  [ 'c', 'k', 'e', 'r' ],
  [ 'L', 'o', 'o', 'p' ],
  [ 'S', 'k', 'y', 'w' ],
  [ 'a', 'l', 'k' ],
  [ 'e', 'r' ] ]

It is what you want?

Answer 6

Here is what you might want from this requirement

I need to take off that star special letter and return a new array with only the characters.

 const charactersArray = [ ['☆', 'D', 'a', '☆', '☆', 'r'], ['t', '☆', 'h', '☆', 'H', 'a'], ['c', 'k', '☆', 'e', '☆', 'r'], ['L', '☆', 'o', 'o', '☆', 'p'], ['S', '☆', 'k', 'y', '☆', 'w'], ['a', 'l', '☆', 'k', '☆', '☆'], ['☆', 'e', '☆', '☆', 'r', '☆'] ]; let newArr = []; charactersArray.forEach(function(subArrays){ let subArray = subArrays.filter(subArray => subArray !== '☆'); subArray.forEach(function(character){ newArr.push(character); }); }); console.log(newArr);