Is it possible to sort an array but override precedence for certain words by placing them at the end using a place_last_lookup array?
input_place_last_lookup = ["not","in"];
input_array = [ "good", "in", "all", "are", "not", "programmers", "the", "world"]
input_array.sort(function(){/*override*/})
expected_output = ["all", "are", "good", "programmers", "the", "world", "not", "in"]
The words "not" and "in" should be placed at the end based on the lookup array.
If you want to prioritize certain strings in a list, you can compare them with a lookup table first before resorting to normal comparison methods:
let always_last = ["not","in"];
let array = [ "good", "in", "all", "are", "not", "programmers", "the", "world"]
array.sort((a, b) => always_last.includes(a) - always_last.includes(b)
|| a.localeCompare(b));
console.log(array);Key Points:
The strings matching the lookup list will be sorted among themselves before other strings.
If the lookup array is large, consider using a Set for faster lookups (constant time complexity).
To maintain the order of matches in the lookup array, use indexOf instead of includes:
let always_last = ["not","in"];
let array = [ "good", "in", "all", "are", "not", "programmers", "the", "world"]
array.sort((a, b) => always_last.indexOf(a) - always_last.indexOf(b)
|| a.localeCompare(b));
console.log(array);For sorting items that should come first or last, adjust your comparison logic accordingly:
let always_first = ["programmers"];
let always_last = ["not","in"];
let array = [ "good", "in", "all", "are", "not", "programmers", "the", "world"]
array.sort((a, b) => always_last.includes(a) - always_last.includes(b)
|| always_first.includes(b) - always_first.includes(a)
|| a.localeCompare(b));
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