Reputation: 1381
So I have an array of strings to be sorted out.
Let's say I have this array:
let array = ["8AD", "8AB", "8A6", "8BC", "86F", "835", "81D"];
Now, there two kinds of sorting that needs to be implemented:
Now, I needed to sort using either of these two for every character.
So, a Numerical-Numerical-Alphabetical order would give me:
"81D","835","86F","8AB","8AD","8A6","8BC"
While a Numerical-Alphabetical-Numerical order would give me:
"8A6","8AB","8AD","8BC","81D","835","86F"
I'm thinking of assigning every single digit number and all characters to a double-digit integer:
let alpha = {
A= 11, B= 12 , C= 13 , D= 14 , E= 15 , F= 16 , G= 17 , H= 18 , I= 19 , J= 20 , K= 21,
L= 22 , M= 23 , N= 24 , O= 25 , P= 26 , Q= 27 , R= 28 , S= 29 , T= 30 , U= 31 , V= 32,
W= 33 , X= 34 , Y= 35 , Z= 36 , 0=37, 1= 38 , 2= 39 , 3= 40 , 4= 41 , 5= 42 , 6= 43 , 7= 44,
8= 45 , 9= 46 };
let numeral = {
0=11, 1=12, 2=13, 3=14, 4=15, 5=16, 6=17, 7=18, 8=19, 9=20, A=21,
B=22, C=23, D=24, E=25, F=26, G=27, H=28, I=29, J=30, K=31, L=32,
M=33, N=34, O=35, P=36, Q=37, R=38, S=39, T=40, U=41, V=42, W=43,
X=44, Y=45, Z=46 }
And then replacing every characters to whichever order is needed. Does anyone have a simpler or more efficient way to achieve what is needed to do?
Upvotes: 1
Views: 75
Reputation: 214959
Here's some generic/abstract stuff
// given a sequence and an array of "order" strings, create a comparable "key" string
let multiSortKey = (subject, orders) =>
[...subject].map(
(c, i) =>
String(orders[i].indexOf(c)).padStart(16, '0')
).join();
// generic comparison function
let cmp = (a, b) => (a > b) - (a < b);
// generic sort-by-map, aka Schwartzian, function
let sortBy = (xs, key) => xs
.map(x => [x, key(x)])
.sort((x, y) => cmp(x[1], y[1]))
.map(x => x[0]);
// applied to the task at hand:
data = ["8AD", "8AB", "8A6", "8BC", "86F", "835", "81D"]
N = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'
A = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'
console.log(sortBy(data, x => multiSortKey(x, [N, N, A])))
console.log(sortBy(data, x => multiSortKey(x, [N, A, N])))
Upvotes: 1
Reputation: 386604
An approach with looking at each letter and a pattern for sorting with map.
function sort(array, pattern) {
return array
.map((value, index) => ({ index, value: Array.from(value, c => pattern.map(fn => fn(c) ? c : ' ').join('')).join('') }))
.sort(({ value: a }, { value: b }) => a.localeCompare(b))
.map(({ index }) => array[index]);
}
const
isDigit = c => /^\d$/.test(c),
isNonDigit = c => /^\D$/.test(c),
digitNonDigit = [isNonDigit, isDigit], // revers sorted
nondigitDigit = [isDigit, isNonDigit], // revers sorted
array = ["8AD", "8AB", "8A6", "8BC", "86F", "835", "81D"];
console.log(sort(array, digitNonDigit));
console.log(sort(array, nondigitDigit));
.as-console-wrapper { max-height: 100% !important; top: 0; }
Upvotes: 0
Reputation: 258
Try to use parseInt()
function with radix
argument for transfer your string to numeric value.
Also .sort()
method to sorting your array.
Example:
let array = ["8AB", "8A6", "8BC", "86F", "835", "81D", "22", "33", "0"];
array.sort((x,y) => parseInt(x, 32) - parseInt(y, 32));
console.log(array);
Upvotes: 0