Merge two arrays together and arrange them in ascending order based on the dateTime field

Question

Merge two arrays together and arrange them in ascending order based on the dateTime field

I have combined two sorted arrays into a new listC.

listA = [
  {id:"1234435", name:"apple", dateTime_ISO:"2019-01-15 17:27:30"},
  {id:"1234435", name:"orange", dateTime_ISO:"2019-01-15 10:25:30"},
  {id:"1234435", name:"banana", dateTime_ISO:"2019-01-15 10:25:02"},
  {id:"1234435", name:"pear", dateTime_ISO:"2019-01-15 07:21:52"},
  {id:"1234435", name:"lemon", dateTime_ISO:"2019-01-15 07:22:24"},
]

listB = [
  {id:"1234435", name:"bread", dateTime:"2019-01-15 17:27:34"},
  {id:"1234435", name:"rice", dateTime:"2019-01-15 09:25:30"},
  {id:"1234435", name:"noodle", dateTime:"2019-01-15 07:25:02"},
  {id:"1234435", name:"pie", dateTime:"2019-01-15 07:06:52"},
  {id:"1234435", name:"cake", dateTime:"2019-01-15 06:22:24"},
]

listC = this.listA.concat(this.listB)

What's the best way to sort listC based on the dateTime?

One approach could involve creating a new list of only the dateTimes, sorting that list, and then sorting listC accordingly.

dateTimeList = this.listA
                .map(x => x.dateTime_ISO)
                .concat(this.listB.map(x => x.dateTime));

Is there a more robust method for sorting this dateTimeList?

If the issue is due to the different field names in the two lists (dateTime_ISO and dateTime), we can consider treating them as equivalent. The database can be updated accordingly.

Any help would be greatly appreciated.

javascript arrays

Answer 1

Answer №1

If you want to efficiently sort the combined list with different properties using short circuiting, here's an example:

let firstList = [{id:"1234435", name:"apple", dateTime_ISO:"2019-01-15 17:27:30"},{id:"1234435", name:"orange", dateTime_ISO:"2019-01-15 10:25:30"},{id:"1234435", name:"banana", dateTime_ISO:"2019-01-15 10:25:02"},{id:"1234435", name:"pear", dateTime_ISO:"2019-01-15 07:21:52"},{id:"1234435", name:"lemon", dateTime_ISO:"2019-01-15 07:22:24"},]
let secondList = [{id:"1234435", name:"bread", dateTime:"2019-01-15 17:27:34"},{id:"1234435", name:"rice", dateTime:"2019-01-15 09:25:30"},{id:"1234435", name:"noodle", dateTime:"2019-01-15 07:25:02"},{id:"1234435", name:"pie", dateTime:"2019-01-15 07:06:52"},{id:"1234435", name:"cake", dateTime:"2019-01-15 06:22:24"},]
  
let combinedList = firstList.concat(secondList)

combinedList.sort((a, b) => {
    return (a.dateTime_ISO || a.dateTime).localeCompare((b.dateTime_ISO || b.dateTime))
})
console.log(combinedList)

Answer 2

If you want to efficiently sort the combined list with different properties using short circuiting, here's an example:

let firstList = [{id:"1234435", name:"apple", dateTime_ISO:"2019-01-15 17:27:30"},{id:"1234435", name:"orange", dateTime_ISO:"2019-01-15 10:25:30"},{id:"1234435", name:"banana", dateTime_ISO:"2019-01-15 10:25:02"},{id:"1234435", name:"pear", dateTime_ISO:"2019-01-15 07:21:52"},{id:"1234435", name:"lemon", dateTime_ISO:"2019-01-15 07:22:24"},]
let secondList = [{id:"1234435", name:"bread", dateTime:"2019-01-15 17:27:34"},{id:"1234435", name:"rice", dateTime:"2019-01-15 09:25:30"},{id:"1234435", name:"noodle", dateTime:"2019-01-15 07:25:02"},{id:"1234435", name:"pie", dateTime:"2019-01-15 07:06:52"},{id:"1234435", name:"cake", dateTime:"2019-01-15 06:22:24"},]
  
let combinedList = firstList.concat(secondList)

combinedList.sort((a, b) => {
    return (a.dateTime_ISO || a.dateTime).localeCompare((b.dateTime_ISO || b.dateTime))
})
console.log(combinedList)

Answer 3

Answer №2

To sort the list, you can utilize the sort() method and include a condition within the sorting logic to check for the existence of a specific property (e.g., dateTime_ISO). If the property does not exist, you can then use an alternative property name (e.g., dateTime):

const listA = [
  {id:"1234435", name:"apple", dateTime_ISO:"2019-01-15 17:27:30"},
  {id:"1234435", name:"orange", dateTime_ISO:"2019-01-15 10:25:30"},
  {id:"1234435", name:"banana", dateTime_ISO:"2019-01-15 10:25:02"},
  {id:"1234435", name:"pear", dateTime_ISO:"2019-01-15 07:21:52"},
  {id:"1234435", name:"lemon", dateTime_ISO:"2019-01-15 07:22:24"},
]

const listB = [
  {id:"1234435", name:"bread", dateTime:"2019-01-15 17:27:34"},
  {id:"1234435", name:"rice", dateTime:"2019-01-15 09:25:30"},
  {id:"1234435", name:"noodle", dateTime:"2019-01-15 07:25:02"},
  {id:"1234435", name:"pie", dateTime:"2019-01-15 07:06:52"},
  {id:"1234435", name:"cake", dateTime:"2019-01-15 06:22:24"},
]

let listC = listA.concat(listB);

// Now, let's proceed with the sorting.

let sortedListC = listC.sort((a, b) =>
{
    let x = (a.dateTime_ISO || a.dateTime);
    let y = (b.dateTime_ISO || b.dateTime);

    return ((x > y) && 1) || ((x < y) && -1) || 0;
});

console.log(sortedListC);

Answer 4

To sort the list, you can utilize the sort() method and include a condition within the sorting logic to check for the existence of a specific property (e.g., dateTime_ISO). If the property does not exist, you can then use an alternative property name (e.g., dateTime):

const listA = [
  {id:"1234435", name:"apple", dateTime_ISO:"2019-01-15 17:27:30"},
  {id:"1234435", name:"orange", dateTime_ISO:"2019-01-15 10:25:30"},
  {id:"1234435", name:"banana", dateTime_ISO:"2019-01-15 10:25:02"},
  {id:"1234435", name:"pear", dateTime_ISO:"2019-01-15 07:21:52"},
  {id:"1234435", name:"lemon", dateTime_ISO:"2019-01-15 07:22:24"},
]

const listB = [
  {id:"1234435", name:"bread", dateTime:"2019-01-15 17:27:34"},
  {id:"1234435", name:"rice", dateTime:"2019-01-15 09:25:30"},
  {id:"1234435", name:"noodle", dateTime:"2019-01-15 07:25:02"},
  {id:"1234435", name:"pie", dateTime:"2019-01-15 07:06:52"},
  {id:"1234435", name:"cake", dateTime:"2019-01-15 06:22:24"},
]

let listC = listA.concat(listB);

// Now, let's proceed with the sorting.

let sortedListC = listC.sort((a, b) =>
{
    let x = (a.dateTime_ISO || a.dateTime);
    let y = (b.dateTime_ISO || b.dateTime);

    return ((x > y) && 1) || ((x < y) && -1) || 0;
});

console.log(sortedListC);

Answer 5

Answer №3

Considering listA and listB are already sorted in descending order, leveraging this fact to merge them can be more efficient. This approach should have a time complexity of O(n) instead of (nlog(n)) if you were to sort the result post concatenation.

You can try implementing something similar to the following:

var listA = [
  {id:"1234435", name:"apple", dateTime_ISO:"2019-01-15 17:27:30"},
  {id:"1234435", name:"orange", dateTime_ISO:"2019-01-15 10:25:30"},
  {id:"1234435", name:"banana", dateTime_ISO:"2019-01-15 10:25:02"},
  {id:"1234435", name:"pear", dateTime_ISO:"2019-01-15 07:21:52"},
  {id:"1234435", name:"lemon", dateTime_ISO:"2019-01-15 07:22:24"},
];

var listB = [
  {id:"1234435", name:"bread", dateTime:"2019-01-15 17:27:34"},
  {id:"1234435", name:"rice", dateTime:"2019-01-15 09:25:30"},
  {id:"1234435", name:"noodle", dateTime:"2019-01-15 07:25:02"},
  {id:"1234435", name:"pie", dateTime:"2019-01-15 07:06:52"},
  {id:"1234435", name:"cake", dateTime:"2019-01-15 06:22:24"},
];

function merge(a, b) {
        var ret = [];
        while (a.length && b.length) {
                var smallest = a[0].dateTime_ISO > b[0].dateTime ? a.shift() : b.shift();
                ret.push(smallest);
        }
        return ret.concat(a.length ? a : b);
};

console.log(JSON.stringify(merge(listA, listB), null, 2));

Answer 6

Considering listA and listB are already sorted in descending order, leveraging this fact to merge them can be more efficient. This approach should have a time complexity of O(n) instead of (nlog(n)) if you were to sort the result post concatenation.

You can try implementing something similar to the following:

var listA = [
  {id:"1234435", name:"apple", dateTime_ISO:"2019-01-15 17:27:30"},
  {id:"1234435", name:"orange", dateTime_ISO:"2019-01-15 10:25:30"},
  {id:"1234435", name:"banana", dateTime_ISO:"2019-01-15 10:25:02"},
  {id:"1234435", name:"pear", dateTime_ISO:"2019-01-15 07:21:52"},
  {id:"1234435", name:"lemon", dateTime_ISO:"2019-01-15 07:22:24"},
];

var listB = [
  {id:"1234435", name:"bread", dateTime:"2019-01-15 17:27:34"},
  {id:"1234435", name:"rice", dateTime:"2019-01-15 09:25:30"},
  {id:"1234435", name:"noodle", dateTime:"2019-01-15 07:25:02"},
  {id:"1234435", name:"pie", dateTime:"2019-01-15 07:06:52"},
  {id:"1234435", name:"cake", dateTime:"2019-01-15 06:22:24"},
];

function merge(a, b) {
        var ret = [];
        while (a.length && b.length) {
                var smallest = a[0].dateTime_ISO > b[0].dateTime ? a.shift() : b.shift();
                ret.push(smallest);
        }
        return ret.concat(a.length ? a : b);
};

console.log(JSON.stringify(merge(listA, listB), null, 2));

Answer 7

Answer №4

Sorting a list of dates can be achieved elegantly using JavaScript. Here is a snippet that demonstrates how to sort a list of dates in descending order:

This code snippet effectively sorts the "dateTimeList" array based on the date attribute.

Answer 8

Sorting a list of dates can be achieved elegantly using JavaScript. Here is a snippet that demonstrates how to sort a list of dates in descending order:

This code snippet effectively sorts the "dateTimeList" array based on the date attribute.

Merge two arrays together and arrange them in ascending order based on the dateTime field

Answer №1

Answer №2

Answer №3

Answer №4

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