How would one go about looping through this array to get every second "note" starting at the note of choice eg start at F and get f,a,c. Any help would be greatly appricated :)
let notes = ['c','d','e','f','g','a','b'];
CodePudding user response:
You can use findIndex and filter using remainder operator
I assume you want to wrap around
If I wrap I get f,a,c,e If not I get f,a, so I assume your expected output was missing the e
?
let notes = ['c', 'd', 'e', 'f', 'g', 'a', 'b'];
const findNotes = (startNote,gap) => {
const start = notes.findIndex(note => note === startNote)
if (start !=-1) return notes.slice(start).concat(notes.slice(0,start)).filter((note,i) => i%gap===0)
return "not found"
};
console.log(findNotes("f",2))
CodePudding user response:
You can use a for loop and start the loop at the index of the element you want to start at. For example this will output: 'f', 'a'
.
let notes = ['c', 'd', 'e', 'f', 'g', 'a', 'b'];
let startIndex = 3; // start at 'f'
for (let i = startIndex; i < notes.length; i = 2) {
let note = notes[i];
console.log(note);
}
CodePudding user response:
If this is an operation that you repeat often, you can map each note with its index. Then create an array with 2 copies of notes to easily get next values in a circular manner.
let notes = ['c', 'd', 'e', 'f', 'g', 'a', 'b'],
indexMap = notes.reduce((map, n, i) => map.set(n, i), new Map),
twoNotes = [...notes, ...notes]
Then, create a function which gets the initial index from the mapper. And then items at the next 2 indices
function findNext(note) {
const index = indexMap.get(note)
return [ twoNotes[index], twoNotes[index 2], twoNotes[index 4] ]
}
or, if you want more indices in the future, you can make it generic one and get the indices array as an argument.
function findNext(note) {
const index = notesIndex.get(note)
return [0, 2, 4].map(i => twoNotes[i index])
}
let notes = ['c', 'd', 'e', 'f', 'g', 'a', 'b'],
indexMap = notes.reduce((map, n, i) => map.set(n, i), new Map),
twoNotes = [...notes, ...notes]
function findNext(note) {
const index = indexMap.get(note)
return [0, 2, 4].map(i => twoNotes[i index])
}
console.log(...findNext('f'))
console.log(...findNext('c'))
CodePudding user response:
You can use the modulo operator so that your index variable wraps around to 0 once it reaches the length of the notes array:
let notes = ['c','d','e','f','g','a','b'];
let startIndex = notes.indexOf('f');
for (let i = startIndex; i !== startIndex-1; i = (i 2) % notes.length)) {
console.log(notes[i]);
}
There is another super performant way using the for loop
in some cases...
let notes = ['c','d','e','f','g','a','b'];
let startIndex = notes.indexOf('f')-2;
for (
let i = startIndex;
i !== startIndex-1;
(i = (i 2) % notes.length)==console.log(notes[i])
);
Note: the ;
is very important after the for loop without {}
.
Another method would be to slice
the two parts of the array into the proper order first:
let notes = ['c','d','e','f','g','a','b'];
let startIndex = notes.indexOf('f');
notes = [
...notes.slice(startIndex),
...notes.slice(0, startIndex)
];
for (let i = 0; i < notes.length - 2; i = 2) {
let note = notes[i];
console.log(note);
}