How to Sort an Array of Objects with Null Properties in TypeScript ? - GeeksforGeeks (2024)

Sorting an array of objects in TypeScript involves arranging objects based on specified properties. When dealing with null properties, ensure consistency by placing them either at the start or end of the sorted array to maintain predictability in sorting outcomes.

Below are the approaches used to sort an array of objects with null properties in TypeScript:

Approach 1: Using Array.prototype.sort with a ternary operator

In this approach, we Sort an array of objects with null properties using Array.prototype.sort and a ternary operator for comparison. Prioritizes sorting based on existing properties or default sorting criteria if properties are null.

Example: Here, we sort arr based on the name. Objects with non-null name properties are sorted alphabetically, and null properties are sorted by id.

interface MyObject {id: number;name?: string;}const arr: MyObject[] = [{ id: 3, name: "Rohit" },{ id: 1, name: "Amit" },{ id: 2, name: "Rahul" }];const result = arr.sort((a, b) => (a.name && b.name) ?  a.name.localeCompare(b.name) :  (a.name ? -1 : b.name ? 1 : a.id - b.id));console.log(result);

Output:

[
 { id: 1, name: 'Amit' },
 { id: 2, name: 'Rahul' },
 { id: 3, name: 'Rohit' }
]

Approach 2: Using Array.prototype.reduce

In this approach we uses Array.prototype.reduce with a custom sorting function. It iterates over elements, inserting them into a sorted array based on their properties. Handles cases where properties are null.

Example: Here, we Sorts arr based on name. Uses Array.prototype.reduce with functional programming to insert elements into a new array at the correct position. Prints the result.

interface MyObject {id: number;name?: string;}const arr: MyObject[] = [{ id: 3, name: "Rohit" },{ id: 1, name: "Amit" },{ id: 2, name: "Rahul" }];const result =  arr.reduce((acc: MyObject[], curr: MyObject) => {const index = acc.findIndex((obj: MyObject) => { if (curr.name && obj.name) { return curr.name.localeCompare(obj.name) < 0; } else { return !obj.name; }});return index !== -1 ? [...acc.slice(0, index), curr, ...acc.slice(index)] : [...acc, curr];}, []);console.log(result);

Output:

[
 { id: 1, name: 'Amit' },
 { id: 2, name: 'Rahul' },
 { id: 3, name: 'Rohit' }
]

Approach 3: Using Array.prototype.filter() and concat()

Filter out objects with null properties, sort the remaining objects based on their non-null property, then concatenate them with objects having null properties to maintain original order.

Example: In this example we filter out objects with defined names, sorts them by name, then concatenates them with objects having undefined names, resulting in the sorted array.

interface MyObject { id: number; name?: string;}const arr: MyObject[] = [ { id: 3, name: "Rohit" }, { id: 1, name: "Amit" }, { id: 2, name: "Rahul" }];const sortedArray = arr // Filter out objects with undefined names .filter(obj => obj.name !== undefined)  // Sort by name .sort((a, b) => a.name!.localeCompare(b.name!))  // Concatenate with objects having undefined names .concat(arr.filter(obj => obj.name === undefined));  console.log(sortedArray);

Output:

[
 { id: 1, name: 'Amit' },
 { id: 2, name: 'Rahul' },
 { id: 3, name: 'Rohit' }
]

Approach 4: Using Array.prototype.sort with a custom comparison function

In this approach, we’ll use Array.prototype.sort with a custom comparison function that handles cases where properties are null or undefined. This approach provides flexibility in defining sorting criteria and can handle various scenarios efficiently.

Example: Here, we sort arr based on the id property. Objects with non-null id properties are sorted numerically, and objects with null or undefined id properties are moved to the end of the array while preserving their original order.

interface MyObject { id?: number; name?: string;}const arr: MyObject[] = [ { id: 3, name: "Rohit" }, { name: "Amit" }, { id: 1, name: "Nikunj" }, { id: 2, name: "John" }, { name: "Smith" }];const result = arr.sort((a, b) => { if (a.id !== undefined && b.id !== undefined) { return a.id - b.id; } else if (a.id !== undefined) { return -1; } else if (b.id !== undefined) { return 1; } else { // If both objects have undefined id properties, // compare based on name property return (a.name || '').localeCompare(b.name || ''); }});console.log(result);

Output:

[{
 "id": 1,
 "name": "Nikunj"
}, {
 "id": 2,
 "name": "John"
}, {
 "id": 3,
 "name": "Rohit"
}, {
 "name": "Amit"
}, {
 "name": "Smith"
}] 

Approach 5: Using Type Assertions

Using type assertions to sort an array with null properties involves asserting the property as a defined type and providing a default value if it is null. This ensures TypeScript treats the property as defined for comparison.

Example

interface Item { name?: string | null; value?: number | null;}const items: Item[] = [ { name: 'item1', value: 10 }, { name: 'item2', value: null }, { name: null, value: 5 }, { name: 'item3', value: 15 },];items.sort((a, b) => { return (a.value ?? 0) - (b.value ?? 0);});console.log(items);

Output:

[
 { name: 'item2', value: null },
 { name: null, value: 5 },
 { name: 'item1', value: 10 },
 { name: 'item3', value: 15 }
]