JavaScript：根据数组项中的数字对数组重新排序

Question

I am trying to unscramble an array. The array has string items out of order with a number attached to specify what the order should be. I want to take the number within the array item and reassign the array item index to reflect that number. For example var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"] . I have a function that searches the array for the number attached and returns the value. Now I want to take the value returned and turn it into the new item's index in a new array, like var unscrambled = [] . This is what I have so far:

function unscramblePhrase() {
    var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"];   
    var unscrambled = [];
    for (var counter = 0; counter < scrambled.length; counter++) {
        numPosition = scrambled[counter].search('[0-9]');
        arrayIndex = scrambled[counter].substring(numPosition);
        console.log(arrayIndex);
        unscrambled.push(scrambled[arrayIndex]);
    }
 console.log(unscrambled)
}

I see that my arrayIndex is pulling the numbers from the end of the scrambled array items but my attempt at assigning index position based off this variable is producing a newly scrambled array: ["want1", "I0", "pizza4", "eat3", "to2"] .

Answer 1

This is not the best way to sort an array, but your issue is that your using Array.prototype.push , when you should just be assigning the value to an index in unscrambled .

unscrambled[arrayIndex] = scrambled[counter];

But the real way to do this is with Array.prototype.sort

function getNum(str){
    return Number(str.substring(str.length -1));
}


unscrambled.sort((a, b) => getNum(a) - getNum(b));

Note: This method sorts your array in-place . This may be good, or not good depending on your requirements

But you can always use it on a clone:

 [...unscrambled].sort((a, b) => getNum(a) - getNum(b)); 

Answer 2

You could simplify this by using RegEx to split the array into value and index, sort them, and then remove the additional information using a secondary .map to return the string array.

scrambled.map(i => 
[i.replace(/\d/g,""), +i.replace(/\D/g,"")])
.sort((a, b) => a[1] - b[1])).map(i => i[0]);

 var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"]; var unscrambled = scrambled.map(i => [i.replace(/\\d/g,""), +i.replace(/\\D/g,"")]) .sort((a, b) => a[1] - b[1]) .map( i => i[0]); console.log(unscrambled); 

Answer 3

试试这个（如果可行，不知道）：

function unscramblePhrase() { var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"];
var unscrambled = []; for (var counter = 0; counter < scrambled.length; counter++) { numPosition = scrambled[counter].search('[0-9]'); arrayIndex = scrambled[counter].substring(numPosition); console.log(arrayIndex); unscrambled[arrayIndex] = scrambled[counter]; } console.log(unscrambled) }

Answer 4

You could take a single loop and take the parts of the string and the index, assign the value to the index and return the array.

 var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"], result = scrambled.reduce((r, string) => { var [s, i] = string.match(/\\D+|\\d+/g); r[i] = s; return r; }, []); console.log(result); 

More fun with objects.

 var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"], result = Object.assign( [], ...scrambled.map(s => (([v, k]) => ({ [k]: v }))(s.match(/\\D+|\\d+/g))) ); console.log(result); 

Answer 5

You can attempt with "Array.sort" as well like below

 var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"] let getNumberIndex = (d) => [...d].findIndex(v => Number.isInteger(+v)) let getNumber = (d) => d.slice(getNumberIndex(d)) let unscrambled = scrambled.slice(0).sort((a,b) => getNumber(a) - getNumber(b)) console.log(unscrambled) 

Answer 6

const words = ["I0", "want1", "to2", "eat3", "pizza4"]

words
    .map (w => parseInt (w [--w.length]))
    .sort ()
    .map (i => words.filter (w => w [--w.length] == i) [0].replace (i, ''))

// ["I", "want", "to", "eat", "pizza"]