♦ pattern:
e,b,c,d,i,f,g,h,o,j,k,l,m,n,u,p,q,r,s,t,a,v,w,x,y,z
I want to sort my words from arr alphabetically by 1st letter and then by 2nd letter of a similar word matched by 1st letter according to my given pattern.
[‘aobcdh’, ‘aibcdh’, ‘aabcdh’, ‘aacbdh’, ‘cfghjd’, ‘cighjd’]
♦ output should be:
[‘aibcdh’, ‘aobcdh’, ‘aabcdh’, ‘aacbdh’, ‘cighjd’, ‘cfghjd’ ]
♦ or:
aibcdh
aobcdh
aabcdh
aacbdh
cighjd
cfghjd
My Code here:
let pattern = ['e', 'b', 'c', 'd', 'i', 'f', 'g', 'h', 'o', 'j', 'k', 'l', 'm', 'n', 'u', 'p', 'q', 'r', 's', 't', 'a', 'v', 'w', 'x', 'y', 'z']
let arr = ['aobcdh', 'aibcdh', 'aabcdh', 'aacbdh', 'cfghjd', 'cighjd']
let arrSorted = arr.sort() //Natural sorting
console.log(arrSorted)
// output in array
const newArr = arrSorted.sort((a, b) => pattern.indexOf(a[1]) - pattern.indexOf(b[1]))
console.log(newArr) //Sorted by its 2nd character with given pattern
// single output without array
for (let i = 0; i < pattern.length; i++) {
for (let j = 0; j < arrSorted.length; j++) {
if (pattern[i] === arrSorted[j][1]) {
console.log(arrSorted[j]) //Sorted by its 2nd character with given pattern
}
}
}
>Solution :
You could sort the first by string and the second by custom value.
const
pattern = 'ebcdifghojklmnupqrstavwxyz',
array = ['aobcdh', 'aibcdh', 'aabcdh', 'aacbdh', 'cfghjd', 'cighjd'],
order = Object.fromEntries(Array.from(pattern, (l, i) => [l, i + 1]));
array.sort((a, b) =>
a[0].localeCompare(b[0]) ||
order[a[1]] - order[b[1]]
);
console.log(array);