Get count of duplicates based on two columns in a multidivisional array and add count to the array Google Script

[Timothy Hanson]

Get the count of duplicates based on two columns in a multidivisional array and add the count to the array Google Script.

Have

Need

I need to count the duplicates in an array based on the conditions.

values[i][2]==values[i+1][2] && values[i][1]==values[i+1][1])

 and add the duplicate value to the column `Total Gifts`

And sum the totals for `ig_Flow` and output the results to the column `Total Flow`

I have tried modifying this but with no success

function count_duplicate() {
  let sheet = SpreadsheetApp.getActiveSpreadsheet().getSheetByName("Need");
  let values = sheet.getDataRange().getValues();

  let counts = new Map();

  for (let i = 0; i < values.length; i++) {
    const data = counts.get(values[i].toString());
    if (data) data.count++;
    else counts.set(values[i].toString(), { value: values[i], count: 1 });
  }

  const result = [];
  for (let [, { value, count }] of counts) {
    result.push([...value, count]);
  }
  console.log(result);
  return result;
}

Here is a Google sheet with the data

Thanks for assistance

 

[Tanaike]

In your situation, how about the following sample script?

let sheet = SpreadsheetApp.getActiveSpreadsheet().getSheetByName("Need");

let [, ...values] = sheet.getDataRange().getValues();
const obj = values.reduce((o, [, b, c, , e]) => (o[b + c] = o[b + c] ? [++o[b + c][0], o[b + c][1] + e] : [1, e], o), {});
const res = values.map(([, b, c]) => obj[b + c] || null);
console.log(res);

When your sample Spreadsheet is used, `res` is `[[1,1],[2,5],[2,5],[1,15],[3,9],[3,9],[3,9],[1,1],[1,7],[4,12],[4,12],[4,12],[4,12]]`. I thought that in this case, the order of search objects is not related to the result. So, I used a JSON object for searching the values instead of the Map object.

If I misunderstood your goal, I apologize.

[Jason Cobble]

You do realize.... Youre helping these assholes hack and harass ppl right? Or do you even care?

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[Timothy Hanson]

Tanaike, thank you!  That works great.  I will have to look up reduce and see what I can make of your code.

Thanks again

[Timothy Hanson]

Tanaike, wow, do I not understand this part :)

const obj = values.reduce((o, [, b, c, , e]) => (o[b + c] = o[b + c] ? [++o[b + c][0], o[b + c][1] + e] : [1, e], o), {});

Can you explain it some?

Thanks

[Timothy Hanson]

I have been studying and what exactly is o[b + c]

[Tanaike]

About "const obj = values.reduce((o, [, b, c, , e]) => (o[b + c] = o[b + c] ? [++o[b + c][0], o[b + c][1] + e] : [1, e], o), {}); Can you explain it some?",

In this part, an object for searching the values of column "E" and the count. And, the values of columns "B" and "C" are used as a key. From your sample Spreadsheet, I thought that this key can be used as the unique value.

About "I have been studying and what exactly is o[b + c]",

In this part, the unique key is used using the values of columns "B" and "C".