I want to sort string inside an array, not a normal sorting because if I do so the result will be:
var a = ["dcb","acb","gfe"];
console.log(a.sort());
the answer would be:
["acb","dcb","gfe"]
which I don't want, the sorted array I want is sorting of string inside the array, For eg:
var a = ["dcb","acb","gfe"];
expected answer :
["bcd","abc","efg"]
hope my question is clear to you :)
Use Array.map() and sort each string seperatly:
const arr = ["dcb","acb","gfe"];
const result = arr.map(str => str
.split('')
.sort()
.join('')
)
console.log(result);
a.map(s=>[...s].sort().join(''))
.split('') admittedly has more symmetry with .join('') that makes the implementation easier to understand, even if it is slightly more characters to type... I don't think "being succinct" is necessarily the most important quality here when comparing these two approaches.To perform an operation on each element of an array (in this case, sort a string), use a loop. The Array.prototype.map function is an appropriate "loop-function" when you want to map an array of length n to an array of the same length (i.e. when you want to transform each element of an array).
let a = ["dcb", "acb", "gfe"];
let sorted = a.map(a => [...a].sort().join(''));
console.log(sorted);
a.map(s=>[...s].sort().join(''))The .map() method iterates through each element of the array.
Use the ES6 arrow function and Destructuring assignment operator to simplify the code, if you want to use ES5 the use regular function and use the .split('') method.
The .sort() method to sort the individual characters after splitting it.
The .join('') method to joining them back.
Example (ES6):
let a = ["dcb","acb","gfe"];
let sorted = a.map( s => [...s].sort().join('') );
consloe.log(sorted);
Example (ES5):
let a = ["dcb","acb","gfe"];
let sorted = a.map(
function(s) {
return s.split('').sort().join('');
});
consloe.log(sorted);
a.map(s => s.split('').sort().join(''))a.map(s=>[...s].sort().join('')).map()method iteratates through each element of the array, and code within map callback sorts that individual element. This will work.