Discussion on: 8 Best JavaScript One-Liners

View post

3) can be shorter:

const sortBy = (arr, key) => arr.sort((a, b) => a[key] > b[key] || -1)

5) You have not shown how to use the function, and it will only delay the execution of code chained with the Promise. Statements immediately after will proceed as normal with no delay.

6) Your input is missing the key.

7) Your input shown is wrong, and this can also be shorter - using the built-in method designed for the purpose:

const insert = (arr, index, newItem, a=[...arr]) => (a.splice(index, 0, newItem), a)

8) Will not function correctly as it will never return #ffffff, and also many other colours as you are not padding the string with leading zeroes. A correct version:

const randomColor = "#" + (~~(Math.random() * 8**8)).toString(16).padStart(6,0)

Shivam Singh • Sep 15 '23

1. The `chill` Function

You're correct; my initial example did not show how to actually use the function in a meaningful way. It's a Promise-based function, so you'd typically use it with async/await or .then() to actually cause a delay in the code's execution. Here's how you could use it:

(async () => {
  console.log("Before waiting");
  await chill(2000);
  console.log("After waiting for 2 seconds");
})();

2. The `sortBy` Function

You bring up a good point. The function could indeed be simplified to:

const sortBy = (arr, key) => arr.sort((a, b) => a[key] > b[key] ? 1 : -1);

This approach assumes that a[key] and b[key] will never be equal if sorting is based on unique identifiers, which might be okay for some cases but not all.

3. The `pluck` Function

You're right; the input was incomplete without specifying the key. A more full example could be:

const pluck = (arr, key) => arr.map(obj => obj[key]);
console.log(pluck([{x: 1}, {x: 2}], 'x')); // Output: [1, 2]

4. The `insert` Function

Your point is valid; the input example could have been more complete. Also, your shorter version using splice is more concise. However, one benefit of the original slice approach is that it does not mutate the original array.

const insert = (arr, index, newItem, a=[...arr]) => (a.splice(index, 0, newItem),a);

5. The `randomColor` Function

You are correct again; the original version can indeed skip some colors and never reach #FFFFFF. Your version with padding zeros is more accurate in generating the full spectrum of colors:

const randomColor = "#" + (~~(Math.random() * 8**8)).toString(16).padStart(6,0);

Thank you for pointing out these details; your observations are insightful and enhance the accuracy and efficiency of the examples.
I will update it according to it!

Jon Randy 🎖️ • Sep 15 '23 • Edited

Try the shorter sortBy with repeated identifiers. Works just fine. If the two identifiers are equal it makes no difference what order they are sorted in.

My splice example doesn't mutate the original array.

Jon Randy 🎖️ • Sep 15 '23

Might want to fix the glaring typo in the header image too! 😉

Shivam Singh • Sep 15 '23

@jonrandy which one?

Jon Randy 🎖️ • Sep 15 '23

"8 One-Linear You Will Love"

Shivam Singh • Sep 15 '23 • Edited

ohh yes you are correct! 😊
I am open for any other suggestions

Shivam Singh • Sep 15 '23

1. The `sortBy` Function with Repeated Identifiers

You're right that if two elements have the same identifier, the order doesn't matter for those particular elements. In many sorting algorithms, including JavaScript's native .sort(), elements that compare as equal remain in their original order (stable sort). So, your simplified version is equally effective:

const sortBy = (arr, key) => arr.sort((a, b) => a[key] > b[key] ? 1 : -1);

This will work just fine even with repeated identifiers.

2. The `splice` Example and Array Mutation

I apologize for the misunderstanding. You're correct; your example using splice actually doesn't mutate the original array because you made a shallow copy (a=[...arr]) before applying the splice.

Here's the function for clarity:

const insert = (arr, index, newItem, a=[...arr]) => (a.splice(index, 0, newItem),a);

In this function, a=[...arr] creates a new array that is a copy of the original arr. Then, splice is called on this new array (a), so the original arr remains unchanged. Therefore, it is a non-mutating operation on the original array.

Thank you for your keen observations; they serve to improve the quality and accuracy of the discussion.