Your inquiry is focused on the process of "removing odds" rather than retaining evens. While the end result remains unchanged, there are various approaches to achieving this outcome. Not every scenario will have a clear-cut opposite function that can be directly applied with Array.prototype.filter. As a result, the solutions provided here will concentrate on eliminating elements based on the identification of odd values, as opposed to preserving even elements. In the following section, diverse strategies will be outlined to address your issue before delving into an evaluation of your existing code.
Insight from Your Friend:
I've included a console.log statement within your inner loop to facilitate visualization of each element being assessed. It's apparent that the current solution involves more computational effort than necessary - utilizing multiple loops to iterate through the array of numbers. One loop should suffice for traversing the array efficiently.
Additionally, it's imperative to exercise caution when altering the length of an array while iterating over it. For instance, employing arr.splice(i,1) results in certain consequences during iteration:
- All elements to the right of "i" shift left by one position.
- The overall length of the array diminishes by one.
Considering these outcomes, how should the loop adapt accordingly?
- If all elements shift left after a splice operation, we must reevaluate the same index
i once more since it now corresponds to a new value.
- Given the reduction in length by one, adjustments in the loop’s exit condition are necessary to conclude one iteration earlier.
The response titled mutRejectOdds subsequently tackles these specific concerns.
Recursive Approach:
Boasting high readability and simplicity, yet not immune to stack overflow risks.
const isOdd = x => x % 2 !== 0;
const removeOdds = ([x,...xs]) => {
if (x === undefined)
return [];
else if (isOdd(x))
return removeOdds(xs);
else
return [x, ...removeOdds(xs)];
}
let data = [1,2,3,4,5,6,7,8];
console.log(removeOdds(data)); // [2,4,6,8]
console.log(data); // [1,2,3,4,5,6,7,8]
Linear Iterative Method using Accumulator:
An approach that ensures stack safety and practical utility.
const isOdd = x => x % 2 !== 0;
const removeOdds = xs => {
let acc = [];
for (let x of xs) {
if (!isOdd(x)) {
acc.push(x);
}
}
return acc;
}
let data = [1,2,3,4,5,6,7,8];
console.log(removeOdds(data)); // [2,4,6,8]
console.log(data); // [1,2,3,4,5,6,7,8]
Continuation Passing Style Technique:
An intricate recursive solution that could benefit from trampolining for enhanced stack safety.
const isOdd = x => x % 2 !== 0;
const identity = x => x;
const removeOdds = xs => {
const aux = ([x,...xs], k) => {
if (x === undefined)
return k([]);
else if (isOdd(x))
return aux(xs, k);
else
return aux(xs, acc => k([x, ...acc]));
};
return aux(xs, identity);
}
let data = [1,2,3,4,5,6,7,8];
console.log(removeOdds(data)); // [2,4,6,8]
console.log(data); // [1,2,3,4,5,6,7,8]
Higher-Order Function Implementation:
Similar to the recursive methodology, but incorporating an extra argument representing a function identifying elements to skip - amenable to linear iterative or continuation passing style implementation.
const isOdd = x => x % 2 !== 0;
const reject = (f, [x,...xs]) => {
if (x === undefined)
return [];
else if (f(x))
return reject(f, xs);
else
return [x, ...reject(f, xs)];
}
let data = [1,2,3,4,5,6,7,8];
console.log(reject(isOdd, data)); // [2,4,6,8]
console.log(data); // [1,2,3,4,5,6,7,8]
Function Composition with Array.prototype.filter:
A pragmatic utilization of the built-in Array.prototype.filter, reversing the output via function composition involving not.
const isOdd = x => x % 2 !== 0;
const comp = f => g => x => f(g(x));
const not = x => !x;
const reject = (f, xs) =>
xs.filter(comp(not)(f));
let data = [1,2,3,4,5,6,7,8];
console.log(reject(isOdd, data)); // [2,4,6,8]
console.log(data); // [1,2,3,4,5,6,7,8]
In-Place Mutation via Linear Iteration:
All foregoing methods avoid mutating the original data. However, under specific circumstances where creating a copy devoid of odd values is unnecessary, direct modification to data may be preferred.
This proposed solution rectifies the issues inherent in your initial code snippet.
const isOdd = x => x % 2 !== 0;
const mutRejectOdds = xs => {
for (let i = 0, len = xs.length; i < len; i++) {
if (isOdd(xs[i])) {
xs.splice(i, 1);
i--;
len--;
}
}
}
let data = [1,2,3,4,5,6,7,8];
console.log(mutRejectOdds(data)); // undefined
console.log(data); // [2,4,6,8]