英文:
How to loop through an array starting at an index in JS
问题
如何循环遍历该数组以获取从选择的音符开始的每第二个音符，例如从 F 开始获取 f、a、c。任何帮助将不胜感激 :)
让音符 = ['c', 'd', 'e', 'f', 'g', 'a', 'b'];
<details><summary>英文:</summary>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'];
</details># 答案1**得分**: 1你可以使用 findIndex 和 filter 以及余数运算符。我假设你想要循环处理。如果我循环，我得到 f, a, c, e，如果不循环，我只得到 f, a，所以我猜你期望的输出可能缺少 `e`？```jslet 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));
英文:
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 ?
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
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))
<!-- end snippet -->
答案2
得分: 1
以下是您要翻译的内容：
如果这是您经常重复的操作，您可以将每个音符与其索引进行映射。然后创建一个包含音符的两个副本的数组，以便轻松地以循环方式获取下一个值。
let notes = ['c', 'd', 'e', 'f', 'g', 'a', 'b'],indexMap = notes.reduce((map, n, i) => map.set(n, i), new Map),twoNotes = [...notes, ...notes]
然后，创建一个函数，该函数从映射器获取初始索引。然后获取下一个2个索引处的项。
function findNext(note) {const index = indexMap.get(note)return [ twoNotes[index], twoNotes[index+2], twoNotes[index+4] ]}
或者，如果您将来需要更多的索引，您可以将其制作为通用函数，并将索引数组作为参数获取。
function findNext(note) {const index = notesIndex.get(note)return [0, 2, 4].map(i => twoNotes[i + index])}
请注意，这是一段JavaScript代码，用于处理音符和获取下一个音符的值。如果您需要进一步的解释或帮助，请告诉我。
英文:
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])}
<!-- begin snippet: js hide: true console: true babel: false -->
<!-- language: lang-js -->
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'))
<!-- end snippet -->
答案3
得分: 1
你可以使用模运算符，使你的索引变量在达到音符数组的长度后重新从0开始（输出 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]+",");}
还有另一种在某些情况下非常高效的方法，使用for循环...
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]+","));
注意：在没有{}的情况下，for循环之后非常重要要加上 ;。
另一种方法是先将数组的两部分按正确的顺序“切片”：
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+",");}
英文:
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+",");}
答案4
得分: 0
你可以使用一个for循环，并从你想要开始的元素的索引开始循环。例如，这将输出：'f', 'a'。
let notes = ['c', 'd', 'e', 'f', 'g', 'a', 'b'];let startIndex = 3; // 从 'f' 开始for (let i = startIndex; i < notes.length; i += 2) {let note = notes[i];console.log(note);}
英文:
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'.
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
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);}
<!-- end snippet -->
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