Imagine we have array like this:
const words = ['luigi', 'Bar', 'Test', 'zac1', 'Alf0'];Now I would like to sort this array alphabetically:
words.sort((a, b) => a.localeCompare(b));Imagine we have array of super words. This words have high priority and should be on the top of 'words list'.
const superWords = ["zac1", "Test"]So the desired result of words array will be:
const desiredResult = ["Test", "zac1","Alf0", "Bar", "luigi"]How can I do it?
You could convert superWords into a Set for O(1) lookup time using has rather than O(n) using includes when superWords is an Array (where n is the number of words in superWords):
const words = ['luigi', 'Bar', 'Test', 'zac1', 'Alf0'];
const superWords = ['zac1', 'Test'];
const superWordsSet = new Set(superWords); // This is O(n) but you only need to do this once.
words.sort((a, b) => {
if (superWordsSet.has(a) && !superWordsSet.has(b)) {
return -1;
} else if (!superWordsSet.has(a) && superWordsSet.has(b)) {
return 1;
}
return a.localeCompare(b);
});
console.log(words);Performance
The
Sethasmethod checks if a value is in aSetobject, using an approach that is, on average, quicker than testing most of the elements that have previously been added to theSetobject. In particular, it is, on average, faster than theArray.prototype.includesmethod when anArrayobject has alengthequal to aSetobject'ssize.
you can do this
const words = ['luigi', 'Bar', 'Test', 'zac1', 'Alf0'];
words.sort((a, b) => {
const superWords = ["zac1", "Test"]
if(superWords.includes(a) && !superWords.includes(b)){
return -1
}else if(superWords.includes(b) && !superWords.includes(a)){
return 1
}
return a.localeCompare(b)
})
console.log(words)First step find doubles and the your can sort both arrays and afterwards you can concat the arrays.
let words = ['luigi', 'Bar', 'Test', 'zac1', 'Alf0'];
const superWords = ["zac1", "Test"]
words = words.filter(w => ! superWords.includes(w));
words.sort((a, b) => a.localeCompare(b));
superWords.sort((a, b) => a.localeCompare(b));
console.log(superWords.concat(words));