按字母或数字的字符对字符串数组进行排序

Question

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:

1. Alphabetical characters are prioritized over numerical characters
2. Numerical characters are prioritized over alphabetical characters

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?

Answer 1

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])))

Answer 2

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));

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Answer 3

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);