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Sort two arrays the same way

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爱一瞬间的悲伤
爱一瞬间的悲伤 2020-11-27 15:48

For example, if I have these arrays:

var name = [\"Bob\",\"Tom\",\"Larry\"];
var age =  [\"10\", \"20\", \"30\"];

And I use name.sort

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12条回答
  • 伪装坚强ぢ
    2020-11-27 16:37

    This solution (my work) sorts multiple arrays, without transforming the data to an intermediary structure, and works on large arrays efficiently. It allows passing arrays as a list, or object, and supports a custom compareFunction.

    Usage:

    let people = ["john", "benny", "sally", "george"];
let peopleIds = [10, 20, 30, 40];

sortArrays([people, peopleIds]);
[["benny", "george", "john", "sally"], [20, 40, 10, 30]] // output

sortArrays({people, peopleIds});
{"people": ["benny", "george", "john", "sally"], "peopleIds": [20, 40, 10, 30]} // output

    Algorithm:

    • Create a list of indexes of the main array (sortableArray)
    • Sort the indexes with a custom compareFunction that compares the values, looked up with the index
    • For each input array, map each index, in order, to its value

    Implementation:

    /**
 *  Sorts all arrays together with the first. Pass either a list of arrays, or a map. Any key is accepted.
 *     Array|Object arrays               [sortableArray, ...otherArrays]; {sortableArray: [], secondaryArray: [], ...}
 *     Function comparator(?,?) -> int   optional compareFunction, compatible with Array.sort(compareFunction)
 */
function sortArrays(arrays, comparator = (a, b) => (a < b) ? -1 : (a > b) ? 1 : 0) {
    let arrayKeys = Object.keys(arrays);
    let sortableArray = Object.values(arrays)[0];
    let indexes = Object.keys(sortableArray);
    let sortedIndexes = indexes.sort((a, b) => comparator(sortableArray[a], sortableArray[b]));

    let sortByIndexes = (array, sortedIndexes) => sortedIndexes.map(sortedIndex => array[sortedIndex]);

    if (Array.isArray(arrays)) {
        return arrayKeys.map(arrayIndex => sortByIndexes(arrays[arrayIndex], sortedIndexes));
    } else {
        let sortedArrays = {};
        arrayKeys.forEach((arrayKey) => {
            sortedArrays[arrayKey] = sortByIndexes(arrays[arrayKey], sortedIndexes);
        });
        return sortedArrays;
    }
}

    See also https://gist.github.com/boukeversteegh/3219ffb912ac6ef7282b1f5ce7a379ad

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  • 醉酒成梦
    2020-11-27 16:41

    I was having the same issue and came up with this incredibly simple solution. First combine the associated ellements into strings in a seperate array then use parseInt in your sort comparison function like this:

    <html>
<body>
<div id="outPut"></div>
<script>
var theNums = [13,12,14];
var theStrs = ["a","b","c"];
var theCombine = [];

for (var x in theNums)
{
    theCombine[x] = theNums[x] + "," + theStrs;
}

var theSorted = theAr.sort(function(a,b)
{
    var c = parseInt(a,10);
    var d = parseInt(b,10);
    return c-d;
});
document.getElementById("outPut").innerHTML = theS;
</script>
</body>
</html>
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  • 小蘑菇
    2020-11-27 16:43

    You could get the indices of name array using Array.from(name.keys()) or [...name.keys()]. Sort the indices based on their value. Then use map to get the value for the corresponding indices in any number of related arrays

    const indices = Array.from(name.keys())
indices.sort( (a,b) => name[a].localeCompare(name[b]) )

const sortedName = indices.map(i => name[i]),
const sortedAge = indices.map(i => age[i])

    Here's a snippet:

    const name = ["Bob","Tom","Larry"],
      age =  ["10", "20", "30"],
      
      indices = Array.from(name.keys())
                     .sort( (a,b) => name[a].localeCompare(name[b]) ),
                     
      sortedName = indices.map(i => name[i]),
      sortedAge = indices.map(i => age[i])

console.log(indices)
console.log(sortedName)
console.log(sortedAge)

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  • 花落未央
    2020-11-27 16:46

    If performance matters, there is sort-ids package for that purpose:

    var sortIds = require('sort-ids')
var reorder = require('array-rearrange')

var name = ["Bob","Larry","Tom"];
var age =  [30, 20, 10];

var ids = sortIds(age)
reorder(age, ids)
reorder(name, ids)

    That is ~5 times faster than the comparator function.

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  • 悲哀的现实
    2020-11-27 16:46

    inspired from @jwatts1980's answer, and @Alexander's answer here I merged both answer's into a quick and dirty solution; The main array is the one to be sorted, the rest just follows its indexes

    NOTE: Not very efficient for very very large arrays

     /* @sort argument is the array that has the values to sort
   @followers argument is an array of arrays which are all same length of 'sort'
   all will be sorted accordingly
   example:

   sortMutipleArrays(
         [0, 6, 7, 8, 3, 4, 9], 
         [ ["zr", "sx", "sv", "et", "th", "fr", "nn"], 
           ["zero", "six", "seven", "eight", "three", "four", "nine"] 
         ]
   );

  // Will return

  {  
     sorted: [0, 3, 4, 6, 7, 8, 9], 
     followed: [
      ["zr", th, "fr", "sx", "sv", "et", "nn"], 
      ["zero", "three", "four", "six", "seven", "eight", "nine"]
     ]
   }
 */

    You probably want to change the method signature/return structure, but that should be easy though. I did it this way because I needed it

    var sortMultipleArrays = function (sort, followers) {
  var index = this.getSortedIndex(sort)
    , followed = [];
  followers.unshift(sort);
  followers.forEach(function(arr){
    var _arr = [];
    for(var i = 0; i < arr.length; i++)
      _arr[i] = arr[index[i]];
    followed.push(_arr);
  });
  var result =  {sorted: followed[0]};
  followed.shift();
  result.followed = followed;
  return result;
};

var getSortedIndex = function (arr) {
  var index = [];
  for (var i = 0; i < arr.length; i++) {
    index.push(i);
  }
  index = index.sort((function(arr){
  /* this will sort ints in descending order, change it based on your needs */
    return function (a, b) {return ((arr[a] > arr[b]) ? -1 : ((arr[a] < arr[b]) ? 1 : 0));
    };
  })(arr));
  return index;
};
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  • 借酒劲吻你
    2020-11-27 16:47

    You can sort the existing arrays, or reorganize the data.

    Method 1: To use the existing arrays, you can combine, sort, and separate them: (Assuming equal length arrays)

    var names = ["Bob","Tom","Larry"];
var ages =  ["10", "20", "30"];

//1) combine the arrays:
var list = [];
for (var j = 0; j < names.length; j++) 
    list.push({'name': names[j], 'age': ages[j]});

//2) sort:
list.sort(function(a, b) {
    return ((a.name < b.name) ? -1 : ((a.name == b.name) ? 0 : 1));
    //Sort could be modified to, for example, sort on the age 
    // if the name is the same.
});

//3) separate them back out:
for (var k = 0; k < list.length; k++) {
    names[k] = list[k].name;
    ages[k] = list[k].age;
}

    This has the advantage of not relying on string parsing techniques, and could be used on any number of arrays that need to be sorted together.

    Method 2: Or you can reorganize the data a bit, and just sort a collection of objects:

    var list = [
    {name: "Bob", age: 10}, 
    {name: "Tom", age: 20},
    {name: "Larry", age: 30}
    ];

list.sort(function(a, b) {
    return ((a.name < b.name) ? -1 : ((a.name == b.name) ? 0 : 1));
});

for (var i = 0; i<list.length; i++) {
    alert(list[i].name + ", " + list[i].age);
}
​

    For the comparisons,-1 means lower index, 0 means equal, and 1 means higher index. And it is worth noting that sort() actually changes the underlying array.

    Also worth noting, method 2 is more efficient as you do not have to loop through the entire list twice in addition to the sort.

    http://jsfiddle.net/ghBn7/38/

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