刪除 Google App Script FAST 中的特定列

Question

下面是使用for循環刪除除“G”和“M”列之外的整個列的代碼，但是這個過程太慢了，有沒有辦法更快地執行它？

  var lastCol = newSheet.getLastColumn();
  var keep = [7,13]; 

  for (var col=lastCol; col > 0; col--) {
  if (keep.indexOf(col) == -1) {
  newSheet.deleteColumn(col);
                }
     }

Answer 1

以下是使用 SpreadsheetApp API 執行此操作的方法：

function test() {
  const keep = [7, 13]; // columns G and M
  const newSheet = SpreadsheetApp.getActiveSheet();
  deleteColumns_(newSheet, keep);
}


/**
* Deletes all columns in sheet except the ones whose column numbers
* are listed in columnsToKeep.
*
* The columnsToKeep array [1, 2, 7, 13] means that columns A, B, G and M
* will remain while other columns are deleted.
*
* @param {Sheet} sheet The sheet where to delete columns.
* @param {Number[]} columnsToKeep Array of column numbers to keep.
*/
function deleteColumns_(sheet, columnsToKeep) {
  // version 1.0, written by --Hyde, 13 June 2022
  //   - see https://stackoverflow.com/q/72600890/13045193
  const columnsToDelete = [];
  for (let i = 1, maxColumns = sheet.getMaxColumns(); i <= maxColumns; i++) {
    if (!columnsToKeep.some(columnNumber => i === columnNumber)) {
      columnsToDelete.push(i);
    }
  }
  const tuples = getRunLengths_(columnsToDelete).reverse();
  tuples.forEach(([columnStart, numColumns]) => sheet.deleteColumns(columnStart, numColumns));
}


/**
* Counts consecutive numbers in an array and returns a 2D array that
* lists the first number of each run and the number of items in each run.
*
* The numbers array [1, 2, 3, 5, 8, 9, 11, 12, 13, 5, 4] will get
* the result [[1, 3], [5, 1], [8, 2], [11, 3], [5, 1], [4, 1]].
*
* For best results, sort the numbers array like this:
* const runLengths = getRunLengths_(numbers.sort((a, b) => a - b));
* Note that duplicate values in numbers will give duplicates in result.
*
* @param {Number[]} numbers The numbers to group into runs.
* @return {Number[][]} The numbers grouped into runs, or [] if the array is empty.
*/
function getRunLengths_(numbers) {
  // version 1.1, written by --Hyde, 31 May 2021
  if (!numbers.length) {
    return [];
  }
  return numbers.reduce((accumulator, value, index) => {
    if (!index || value !== 1 + numbers[index - 1]) {
      accumulator.push([value]);
    }
    const lastIndex = accumulator.length - 1;
    accumulator[lastIndex][1] = (accumulator[lastIndex][1] || 0) + 1;
    return accumulator;
  }, []);
}

Answer 2

我相信你的目標如下。

• 您想降低腳本的處理成本。
• 您想刪除除“G”和“M”列之外的所有列。

在這種情況下，使用 Sheets API 怎么樣？ 使用 Sheets API 時，您的腳本如下所示。 我認為當使用 Sheets API 時，處理成本可能會降低一點。

修改后的腳本：

在使用此腳本之前， 請在 Advanced Google services 中啟用 Sheets API 。

function myFunction() {
  var sheetName = "Sheet1";  // Please set the sheet name.
  var keep = [7, 13];

  var ss = SpreadsheetApp.getActiveSpreadsheet();
  var newSheet = ss.getSheetByName(sheetName);
  var lastCol = newSheet.getLastColumn(); // or newSheet.getMaxColumns()
  var sheetId = newSheet.getSheetId();
  var requests = [...Array(lastCol)].reduce((ar, _, i) => {
    if (!keep.includes(i + 1)) {
      ar.push({ deleteDimension: { range: { sheetId, startIndex: i, endIndex: i + 1, dimension: "COLUMNS" } } });
    }
    return ar;
  }, []).reverse();
  if (requests.length == 0) return;
  Sheets.Spreadsheets.batchUpdate({ requests }, ss.getId());
}

• 運行此腳本時，除“G”和“M”列之外的所有列都將被刪除。
• 從您的腳本中， var lastCol = newSheet.getLastColumn(); 用來。 在這種情況下，使用數據范圍。 如果要檢查所有列，請使用var lastCol = newSheet.getMaxColumns(); 而不是它。

參考：

• 方法：電子表格.batchUpdate
• 刪除維度請求