Tips for generating a list of sorted indexes by value within an object

Question

Tips for generating a list of sorted indexes by value within an object

In my project, I have multiple objects named stations, each containing a property called money.

stations[0].money = 2000
stations[1].money = 500
stations[2].money = 1200
stations[3].money = 2200

My goal is to create an array of station indexes (0, 1, 2, and 3 in this case) but arranged based on the amount of money each station has, sorted in ascending order. So the desired output should be:

var moneyArray = [1, 2, 0, 3]

I'm looking for the most elegant solution to achieve this result. Any suggestions?

javascript arrays sorting

Answer 1

Answer №1

To begin, create a regular array of indexes (ideally generated within a loop) and then organize it based on the values of each object in your stations array:

[0, 1, 2, 3].sort(function(ai, bi) {
    return stations[ai].money - stations[bi].money;
})

Answer 2

To begin, create a regular array of indexes (ideally generated within a loop) and then organize it based on the values of each object in your stations array:

[0, 1, 2, 3].sort(function(ai, bi) {
    return stations[ai].money - stations[bi].money;
})

Answer 3

Answer №2

For more information, feel free to check out this resource: Sorting with map

// sample array for sorting
var items = [
        { price: 2000 },
        { price: 500 },
        { price: 1200 },
        { price: 2200 },
    ];

// create a new array with index and value pairs
var mapped = items.map(function (el, i) {
    return { index: i, value: el.price };
})

// sort the mapped array based on values
mapped.sort(function (a, b) {
    return a.value - b.value;
});

// reorder the original array based on sorted values
var result = mapped.map(function (el) {
    return items[el.index];
});

// extract the indices of sorted elements
var keys = mapped.map(function (el) {
    return el.index;
});

document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(keys, 0, 4) + '</pre>');

Answer 4

For more information, feel free to check out this resource: Sorting with map

// sample array for sorting
var items = [
        { price: 2000 },
        { price: 500 },
        { price: 1200 },
        { price: 2200 },
    ];

// create a new array with index and value pairs
var mapped = items.map(function (el, i) {
    return { index: i, value: el.price };
})

// sort the mapped array based on values
mapped.sort(function (a, b) {
    return a.value - b.value;
});

// reorder the original array based on sorted values
var result = mapped.map(function (el) {
    return items[el.index];
});

// extract the indices of sorted elements
var keys = mapped.map(function (el) {
    return el.index;
});

document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(keys, 0, 4) + '</pre>');

Answer 5

Answer №3

To achieve this, one method is to utilize a Temp variable. Alternatively, if making modifications to the main data is feasible, then eliminating the temp variable is an option.

var stations = [{money :2000},{money :500},{money :1200},{money :2200}]

var tempArray = stations.slice();

tempArray.forEach(function (value, i) {
    value.index = i;
});

tempArray.sort(function(a, b){return a.money-b.money});

var finalResult = tempArray.map(function(a) {return a.index;});

document.write(JSON.stringify(finalResult));

Answer 6

To achieve this, one method is to utilize a Temp variable. Alternatively, if making modifications to the main data is feasible, then eliminating the temp variable is an option.

var stations = [{money :2000},{money :500},{money :1200},{money :2200}]

var tempArray = stations.slice();

tempArray.forEach(function (value, i) {
    value.index = i;
});

tempArray.sort(function(a, b){return a.money-b.money});

var finalResult = tempArray.map(function(a) {return a.index;});

document.write(JSON.stringify(finalResult));

Tips for generating a list of sorted indexes by value within an object

Answer №1

Answer №2

Answer №3

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