Have an array
const arr = [1, 'abc', [], ['John'], {}, {name: 'Smith'}, null, 0];
How to get new array without empty values? (empty array and empty object are also reckoned as empty values).
My variant seems to be a bit of hardcoded:
const newArr = arr.filter(elem =>
(elem && Object.keys(elem).length !== 0)
|| (typeof(elem) == 'number' && elem !== 0));
If is it possible to write less or simplier conditions?
If you have specific variants you can try
const emptyVariants = ['', 0, null,] // all expected variants
const newArr = arr.filter(elem => !emptyVariants.includes(elem);
Another approach that you can try is using != but it takes all falsy values
const newArr = arr.filter(elem => (elem != null));
For more information see != and == operator https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comparison_Operators
emptyVariants[1] contains that specific empty object, most likely that same object doesn't occur anywhere else. Also detecting null seems to fail ...For the falsy values, empty object and array you have to use this part:
elem && Object.keys(elem).length !== 0
But you can minimise it like:
elem && Object.keys(elem).length
as any value for length > 0 will return as truthy.
Also, you can minimise the number logic like:
arr.filter(elem => Number(elem));
So, the final version will be like:
const arr = [1, 'abc', [], ['John'], {}, {name: 'Smith'}, null, 0];
const newArr = arr.filter(a => (a && Object.keys(a).length) || Number(a));
console.log(newArr)How about this?
arr.filter(i => (typeof i === 'number') || (i && i.length>0) || (( i && Object.keys(i).length)>0) || (i===0));
!= null, but it takes all falsy values, like false and {}