Question

我有两个数组：array1和array2。因此，array1将根据内部元素array2进行拆分。例如：

array1["1","2","3","4","5","6"]
array2["2","5"]

我的代码：

var prev = 0;
newArray = []; 
for (var ii in array2) {
    var index = array1.indexOf(array2[ii]);

    if (index != prev) {
        newArray.push(array1.slice(prev, index));
        prev = index;
    }
 }
 newArray.push(array1.slice(prev));

结果将是：

["1"],["2","3","4"],["5","6"]

但是现在我面临的问题是array1的元素可能没有按顺序排列。例如：["1","5","3","4","2","6"]。所以根据我的代码，它将错误地拆分array1，因为array2中的第一个元素是＆＃34; 2＆＃34;，所以它已经将array1拆分为两个["1","5","3","4"],["2","6"]。接下来当来到＆＃34; 5＆＃34;时，它找不到它。

预期结果为：["1"],["5","3","4"],["2","6"]

那么如何基于array2拆分array1，无论array1是升序，降序还是随机顺序。对不起，我的英语不好。希望你们能理解。

Answer 1

有关详细信息，请参阅Set和Array.prototype.reduce()。

＆＃13;

// Split Up.
const splitup = (array, keys) => (set => array.reduce((output, value) => {
  if (set.has(value)) output.push([value]) // Split.
  else output[output.length-1].push(value) // Append.
  return output
}, [[]]))(new Set(keys))

// Output.
const output1 = splitup(["1","2","3","4","5","6"], ["2","5"])
console.log(...output1) // ["1"],["2","3","4"],["5","6"]
const output2 = splitup(["1","5","3","4","2","6"], ["2","5"])
console.log(...output2) // ["1"],["5","3","4"],["2","6"]

＆＃13;

Answer 2

试试这个。它循环遍历array1并将每个项目推送到临时数组中。当在array2中找到该项时，临时数组将被推入最终数组，然后重置。

var array1 = ["1", "5", "3", "4", "2", "6"];
var array2 = ["2", "5"];

var newArray = [];
var currArray = [];

for (let i = 0; i < array1.length; i++) {

  // Item exists in array2. Add to newArray and reset currArray
  if (i > 0 && array2.includes(array1[i])) {
    newArray.push(currArray);
    currArray = [];
  }

  currArray.push(array1[i]);
}

newArray.push(currArray); // Add final currArray to newArray

console.log(newArray); // print result

另一方面，如果索引顺序很重要（如在您的情况下那样），最好不要使用for..in来遍历数组，因为它不会以任何特定顺序返回索引。更多信息here。

如何根据另一个数组中的元素拆分数组

2 个答案: