从多维数组中提取列的子数组并改变原始数组以删除这些列JavaScript

Question

我想从多维数组中提取

Columns

的子数组，并改变原始数组以删除 JavaScript 中的这些列

例如：如果我有一个数组

originalArray =
[
 [A, B, C, D, E, F],
 [a1,b1,c1,d1,e1,f1],
 [a2,b2,c2,d2,e2,f2],
 [a3,b3,c3,d3,e3,f3]
[  

其中

A,B,C,D,E,F

是标题 我想按这个顺序提取列 [4,0,3]

subArray =
[
 [E, A, D],
 [e1,a1,d1],
 [e2,a2,d2],
 [e3,a3,d3]
[


originalArray = 
[
 [B, C, F],
 [b1,c1,f1],
 [b2,c2,f2],
 [b3,c3,f3]
[  

我发现了这个： 从多维数组中提取子数组并对原始数组进行变异

对行执行此操作

let subArray = originalArray.reduce((accumulator, current, index) => {
    if (idxList.includes(index)) {
      //if the current index is included in the idxList array, 
      //push the current value to the accumulator
      accumulator.push(current);
    } else {
      //push the current value to the `tempArray`.
      tempArray.push(current)
    }
    return accumulator;
  }, []);

然后我将连接这两个数组

reorderedArray = [...subArray, ...originalArray]

Reduce 似乎是一种有趣的方法，看起来应该是一个简单的修复，但我不得不承认我很难理解

Reduce

预先感谢您的协助

Answer 1

您可以将缺失的索引添加到

order

并通过将嵌套数组与顺序索引映射来映射数组。

let
    data = [['A', 'B', 'C', 'D', 'E', 'F'], ['a1', 'b1', 'c1', 'd1', 'e1', 'f1'], ['a2', 'b2', 'c2', 'd2', 'e2', 'f2'], ['a3', 'b3', 'c3', 'd3', 'e3', 'f3']],
    order = [4, 0, 3],
    seen = new Set(order),
    i = 0;

while (order.length < data[0].length) {
    while (seen.has(i)) ++i;
    order.push(i++);
}

data = data.map(a => order.map(i => a[i]));

data.forEach(a => console.log(...a));

Answer 2

常规映射似乎可行：

function myFunction() {
  var originalArray = SpreadsheetApp.getActive().getRange('Sheet1!A:F').getValues();
  var subarray = originalArray.map(e=>[e[4],e[0],e[3]]);
  originalArray = originalArray.map(e=>[e[1],e[2],e[5]]);

}

Answer 3

可能

function rearrangeCols(data, orderFront, orderBack) {
  orderFront = orderFront.map( x => data[0].includes(x) ? data[0].indexOf(x) : null).filter(x => x != null);
  orderBack  = orderBack.map( x => data[0].includes(x) ? data[0].indexOf(x) : null).filter(x => x != null);

  let
    dataIdx  = data[0].map( x => data[0].indexOf(x)),
    remove   = [...orderFront,...orderBack],           
    newOrder = [...orderFront,
                ... dataIdx.filter(item => !remove.includes(item)),
                ...orderBack
               ];

  //Rearange data array
  data = data.map(a => newOrder.map(i => a[i]));
  
  return data;
}