JS中如何从索引开始遍历数组

Question

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'];

Answer 1

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))

Answer 2

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'))

Answer 3

You can use the modulo operator so that your index variable wraps around to 0 once it reaches the length of the notes array(outputs f,a,c, ):

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)) {
     document.write(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)==document.write(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];
      document.write(note+",");
}

Answer 4

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); }