Question

我有n个数组。每行中的行数相同：Array0，Array1，...，Arrayn。

现在我想遍历所有这些人并最终加入他们。将每个数组转置到新创建的ResultArray的列中

for (var ID = 0; ID < n+1; ID++) {
    for (var i = 0; i < ("Array" + ID).length; i++) {
      ResultArray[i][ID] = ("Array" + ID)[i]
        }
  }

以下是我想要的示例：

Array0 = [Apple, Banana, Chicken]
Array1 = [5, 7, 8]
Array2 = [High, Low, Low]
ResultArray = [[Apple, 5, High], [Banana, 7, Low], [Chicken, 8, Low]]

Answer 1

您具有以下数组。

    Array0 = ["Apple", "Banana", "Chicken"];
    Array1 = [5, 7, 8];
    Array2 = ["High", "Low", "Low"];

您要将上面的数组转换为下面的数组。

    [["Apple",5,"High"],["Banana",7,"Low"],["Chicken",8,"Low"]]

在您的示例数组中，每个数组的数组长度都是相同的。

如果我的理解是正确的，那么该修改如何？请认为这只是几个答案之一。

修改后的脚本1：

var Array0 = ["Apple", "Banana", "Chicken"];
var Array1 = [5, 7, 8];
var Array2 = ["High", "Low", "Low"];

var arrays = [Array0, Array1, Array2]; // Please prepare this.

var res = [];
for (var i = 0; i < arrays[0].length; i++) {
  var temp = [];
  for (var j = 0; j < arrays.length; j++) {
    var ar = arrays[j];
    temp.push(ar[i]);
  }
  res.push(temp);
}
Logger.log(res)

修改后的脚本2：

var Array0 = ["Apple", "Banana", "Chicken"];
var Array1 = [5, 7, 8];
var Array2 = ["High", "Low", "Low"];

var numberOfArrays = 3; // Please prepare this. In your case, it's 3.

var res = [];
for (var i = 0; i < Array0.length; i++) {
  var temp = [];
  for (var j = 0; j < numberOfArrays; j++) {
    var ar = eval("Array" + j);
    temp.push(ar[i]);
  }
  res.push(temp);
}
Logger.log(res)

注意：

请根据您的情况选择上述脚本之一。

如果我误解了您的问题，而这不是您想要的结果，我深表歉意。

Answer 2

您还可以使用Array.reduce以更简洁的方式解决此问题：

const arr1 = ["Apple", "Banana", "Chicken"], arr2 = [5, 7, 8], arr3 = ["High", "Low", "Low"],
      data = [arr1, arr2, arr3]

let result = data.reduce((acc, c) => 
 (c.forEach((x,k) => acc[k].push(x)), acc), Array.from({length: data.length}, () => []))

console.log(result)

或更易读的形式：

const arr1 = ["Apple", "Banana", "Chicken"], arr2 = [5, 7, 8], arr3 = ["High", "Low", "Low"],
      data = [arr1, arr2, arr3]

let result = data.reduce((acc, c) => {
  c.forEach((x,i) => acc[i].push(x))
  return acc
}, Array.from({length: data.length}, () => []))

console.log(result)

想法是merge这些数组，然后在Array.reduce已经是accumulator形式的情况下执行[[],[],[]]，因此我们只填充每个子数组。

Answer 3

使用Array.reduce()查找最长的数组。使用Array.map()迭代最长的数组，并获取行索引。然后使用第二个Array.map()从数组中获取行项。

const zip = (...arrs) => arrs
  .reduce((r, c) => c.length > r.length ? c : r, []) // get the longest array
  .map((_, i) => arrs.map(a => a[i])) // map the longest array to get the current row number, then create the row

const arr1 = ["Apple", "Banana", "Chicken"], arr2 = [5, 7, 8], arr3 = ["High", "Low", "Low"]

const result = zip(arr1, arr2, arr3)

console.log(result)

如何遍历几个数组

3 个答案:

修改后的脚本1：

修改后的脚本2：

注意：