谷歌应用程序脚本数组从顶部删除重复的值 [英] google app script array delete duplicate value from top

问题描述

我有这个工作代码，它通过比较B列中的值来删除重复项.如果第2行和第3行在B列中具有相同的值，则删除第3行.

I've got this working code that removes duplicates by comparing values in column B. If row 2 and 3 have the same values in column B then row 3 is deleted.

function removeDuplicates() {
  var sheet = SpreadsheetApp.getActiveSheet();
  var data = sheet.getDataRange().getValues();
  var newData = new Array();
  for(i in data){
    var row = data[i];
    var duplicate = false;
    for(j in newData){
      if(row[1] == newData[j][1]){
        duplicate = true;
}
    }
    if(!duplicate){
      newData.push(row);
    }
  }
  sheet.clearContents();
  sheet.getRange(1, 1, newData.length, newData[0].length).setValues(newData);
}

如何更改此代码以通过编辑此代码来构建阵列，以便删除第2行而不是删除第3行?

How can I change this code to build the array from edit this so that instead of row 3 being deleted, row 2 is deleted?

推荐答案

我认为这可以做到.

  function removeDuplicates() {
  var ss=SpreadsheetApp.getActive();
  var sh=ss.getActiveSheet();
  var rg=sh.getDataRange();
  var row=rg.getRow();
  var col=rg.getColumn();
  var vA=rg.getValues();
  var nA=[];
  var duplicate=true;
  for(var i=0;i<vA.length;i++)
  {
    duplicate=false;
    for(var j=0;j<nA.length;j++)
    {
      if(vA[i][1]==nA[j][1])
      {
        duplicate=true;
        nA[j]=vA[i];
      }
    }
    if(!duplicate)
    {
      nA.push(vA[i]);
    }
  }
  rg.clearContent();
  sh.getRange(row, col, nA.length, nA[0].length).setValues(nA);
}

外部循环遍历活动工作表的所有行，并且每次循环都将重复项设置为false.内循环搜索nA []来查找columnB匹配项，如果找到匹配的列，则将该副本设置为true.如果重复为true，则不会将其添加到nA []中.第一次通过nA.length为0，因此内部循环不执行任何重复操作都为false，因此该元素被添加到nA []中.它会一直这样做，直到没有更多的行并且nA中的行成为唯一行为止.那就是我第一次做的时候的运行方式.但是由于您想保留最后一个而不是第一个重复，因此我添加了nA[j]=vA[i];，它将当前元素替换为当前匹配项.

The outer loop is iterating through all of the rows of the active sheet and each time through it sets duplicate to false. The inner loop searches through nA[] looking for columnB matches if it finds one it sets duplicate to true. If duplicate is true then it doesn't get added to nA[]. The first time through nA.length is 0 so the inner loop doesn't do anything duplicate is false and so that element gets added to nA[]. It keeps doing this until there are no more rows and the rows that are in nA become the unique row. That's the way it use to run when I first did it. But since you wanted to keep the last duplicate instead of the first then I added nA[j]=vA[i]; which replaces the current element with the current match.

只需设置一些虚假数据并进行处理，您将开始了解其工作原理.

Just setup some fake data and play with it and you'll start to see how it works.

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