有没有更好的方法来根据条件（Google Apps 脚本）合并行数据？

Question

好的...首先我要说我是自学的 Google Apps 脚本...说得够多了，对吧！？下面的脚本正在运行，但我想优化它或想出另一种方法来实现相同的结果。该脚本采用 18000 行和 86 列数据，并根据 id 列表将它们组合成单行。 id 列表长约 13000 行。简短的版本是这样的......它通过 id 过滤数据，然后检查每一列是否有提交数据的最后一行并返回该单元格。例如：

//sample data
[[311112, 1, 2, 4, 5,"","","","","","", 2, 3],
[323223,"","","","","", 2, 4, 4,"","","",""],
[321321, 1, 2, 4, 5,"","","","","","", 2, 3],
[311112, 4, 1, 6, 7,"", 3,"", 3,"","", 5, 3],
[321233,"","","","","","", 4, 3, 1, 5,"",""],
[321321,"","","","","","","","", 1 ,4,"",""],
[323223,"","","","","", 2, 3,"","","","",""],
[323153,"", 2, 3, 6,"","","","","","","",""],
[321321,"","","","","", 2, 3,"","","","",""],
[321321,"", 5, 3,"", 1,"","","","","","",""]]

//filtered Data by id 321321
[[321321, 1, 2, 4, 5,"","","","","","", 2, 3],
[321321,"","","","","","","","", 1, 4,"",""],
[321321,"","","","","", 2, 3,"","","","",""],
[321321,"", 5, 3,"", 1,"","","","","","",""]]

// returned row is getting the last nonempty value for each column from the filtered data.

[[321321, 1, 5, 3, 5, 1, 2, 3,"", 1, 4, 2, 3]]

脚本完成大约需要 16-18 分钟。有没有更好的方法来完成这个或任何优化建议？

function combineR(startRow, startRange) {
  var ss = SpreadsheetApp.getActiveSpreadsheet();
  var sheets = ss.getSheets();
  var testSheet = ss.getSheetByName('Raw Scores');
  var cSheet = ss.getSheetByName('Combined Scores');
  var gradingResults = testSheet.getRange(1, 1, testSheet.getLastRow(), testSheet.getLastColumn()).getValues();

  if (startRow > cSheet.getLastRow()) {
    return;
  }

  if (startRow + startRange > cSheet.getLastRow()) {
    startRange = cSheet.getLastRow() - startRow;
  }

  var sID = cSheet.getRange(startRow, 2, startRange).getValues();
  var maxScores = [];
  for (var x = 0; x < sID.length; x++) {
    var filtered = gradingResults.filter(function (dataRow) {
      return dataRow[0] === sID[x][0];
    });

    if (isFinite(filtered)) {
      maxScores.push(['', '', '', '', '', '', '', '', '', '',
        '', '', '', '', '', '', '', '', '', '',
        '', '', '', '', '', '', '', '', '', '',
        '', '', '', '', '', '', '', '', '', '',
        '', '', '', '', '']);
      continue;
    } else {
      maxScores.push(['', getMaxLetter(filtered, 3), lastGraded(filtered, 4), lastGraded(filtered, 5), lastGraded(filtered, 6), lastGraded(filtered, 7), lastGraded(filtered, 8), lastGraded(filtered, 9), lastGraded(filtered, 10), lastGraded(filtered, 11),
        lastGraded(filtered, 12), lastGraded(filtered, 13), lastGraded(filtered, 14), lastGraded(filtered, 15), lastGraded(filtered, 16), lastGraded(filtered, 17), lastGraded(filtered, 18), lastGraded(filtered, 19), lastGraded(filtered, 20), lastGraded(filtered, 21),
        lastGraded(filtered, 22), lastGraded(filtered, 23), lastGraded(filtered, 24), lastGraded(filtered, 25), lastGraded(filtered, 26), lastGraded(filtered, 27), lastGraded(filtered, 28), lastGraded(filtered, 29), lastGraded(filtered, 30), lastGraded(filtered, 31),
        lastGraded(filtered, 32), lastGraded(filtered, 33), lastGraded(filtered, 34), lastGraded(filtered, 35), lastGraded(filtered, 36), lastGraded(filtered, 37), lastGraded(filtered, 38), lastGraded(filtered, 39), lastGraded(filtered, 40), lastGraded(filtered, 41),
        lastGraded(filtered, 42), lastGraded(filtered, 43), lastGraded(filtered, 44), lastGraded(filtered, 45), lastGraded(filtered, 46)]);
    }
  }
  cSheet.getRange(startRow, 11, maxScores.length, maxScores[0].length).setValues(maxScores)
}

function getMaxLetter(arr, idx) {
  var letter = arr.map(function (e) { return e[idx] }).sort().pop();
  return letter;
}

function lastGraded(arr, idx) {
  var newArray = arr.map(function (e) { return e[idx] });
  newArray.reverse();
  for (var x = 0; x < newArray.length; x++) {
    if (typeof newArray[x] == 'number') {
      return newArray[x];
    }
  }
  return '';
}

A 列有重复的 Id 需要合并

B 列具有作为最终合并产品的唯一值

Answer 1

问题：

脚本似乎有各种问题，但主要问题似乎是多次使用各种索引调用lastGraded 函数。这对map、reverse 以及每个索引的其他所有操作都有效，而且会耗费时间。

解决办法：

鉴于您的样本数据，我建议采用以下方法：

• 获取1个二维数组中的所有输入数据

• Reduce 输入数据到Map。该地图将每个id 与key 匹配，并将与key 匹配的所有行作为每个key 的二维array。这将以内存为代价大大提高性能/速度。这比按每个 id 过滤数组要好，因为，

  • 您只循环输入数组一次
  • 而arr.filter 必须为每个 id 循环数组

• 一旦缩减为map，则反向循环遍历map 中的每个array，以查找最后一行中的每个元素以查找非空元素。

示例 sn-p：

const arrMain =
  //sample data
  [
    [311112, 1, 2, 4, 5, '', '', '', '', '', '', 2, 3],
    [323223, '', '', '', '', '', 2, 4, 4, '', '', '', ''],
    [321321, 1, 2, 4, 5, '', '', '', '', '', '', 2, 3],
    [311112, 4, 1, 6, 7, '', 3, '', 3, '', '', 5, 3],
    [321233, '', '', '', '', '', '', 4, 3, 1, 5, '', ''],
    [321321, '', '', '', '', '', '', '', '', 1, 4, '', ''],
    [323223, '', '', '', '', '', 2, 3, '', '', '', '', ''],
    [323153, '', 2, 3, 6, '', '', '', '', '', '', '', ''],
    [321321, '', '', '', '', '', 2, 3, '', '', '', '', ''],
    [321321, '', 5, 3, '', 1, '', '', '', '', '', '', ''],
  ];

//reduce input array to a map of id=>rows
const map = arrMain.reduce((map, row) => {
  if (!map.has(row[0])) map.set(row[0], [row]);
  else map.get(row[0]).push(row);
  return map;
}, new Map());

const out = [];
map.forEach(arr2d => {
  const l = arr2d.length - 1,
    lastRow = arr2d[l].slice(0);
  //iterate lastrow of this id's column elements
  for (let j = 0; j < lastRow.length; ++j) {
    if (lastRow[j] === '') {
      //iterate each row of this id
      for (let i = l; i >= 0; --i) {
        if (arr2d[i][j] !== '') {
          lastRow[j] = arr2d[i][j];
          break;
        }
      }
    }
  }
  out.push(lastRow);
});
console.log(out);