Split array in array of arrays based on sum of attribute with google app script-CodePudding

I would like to split the following sample array into arrays of array, where the sum of each sub array should not exceed 3 (here var sumMax).

Below is my current codes:

function splitArrByDuration(arr, sumMax) {
  var sumMax = 3;
  var arr = [["a", 2], ["b", 1], ["c", 2], ["d", 1]];
  var splittedArr = [];
  var temp_subArr = [];
  if (arr.length) {
    for (var i = 0; n = arr.length, i < n; i  ) {
      temp_subArr.push(arr[i]);
      if (sumAtIndex(temp_subArr, index = 1) > sumMax && arr.length) {
        var extraRow_temp_subArr = temp_subArr.splice(temp_subArr.length - 1, 1); //remove last element of "temp_subArr" because with it the sum will be greater than "sumMax"
        //Logger.log(extraRow_temp_subArr); 
        arr.unshift(extraRow_temp_subArr[0]); //add the extra from temp_subArr to first index of "arr"
        //i = i - 1;
        splittedArr.push(temp_subArr);
        temp_subArr = [];
      }
    }
  }
  Logger.log(splittedArr);
}

function sumAtIndex(arr, index = 1) {
  var sumIndex = 0;
  if (arr.length) {
    for (var i = 0; n = arr.length, i < n; i  ) {
      sumIndex = sumIndex   arr[i][index];
    }
  }
  //Logger.log(sumIndex);
  return sumIndex;
}

The result from Logger.log(splittedArr) is currently : [[[a, 2.0], [b, 1.0]]]

What is expected should be [[[a, 2.0], [b, 1.0]], [[c, 2.0], [d, 1.0]]]

Can you please help me to get the right result? Thanks in advance,

CodePudding user response：

I thought that when I saw your script, it might be required to process the case that i is the last index. So, when your showing script is modified, how about the following modification?

Modified script:

function splitArrByDuration(arr, sumMax) {
  var sumMax = 3;
  var arr = [["a", 2], ["b", 1], ["c", 2], ["d", 1]];
  var splittedArr = [];
  var temp_subArr = [];
  if (arr.length) {
    for (var i = 0; n = arr.length, i < n; i  ) {
      temp_subArr.push(arr[i]);
      if (sumAtIndex(temp_subArr, index = 1) > sumMax && arr.length) {
        var extraRow_temp_subArr = temp_subArr.splice(temp_subArr.length - 1, 1); //remove last element of "temp_subArr" because with it the sum will be greater than "sumMax"
        //Logger.log(extraRow_temp_subArr); 
        arr.unshift(extraRow_temp_subArr[0]); //add the extra from temp_subArr to first index of "arr"
        //i = i - 1;
        splittedArr.push(temp_subArr);
        temp_subArr = [];

      // Added
      } else if (i == arr.length - 1) {
        splittedArr.push(temp_subArr);
      }

    }
  }
  Logger.log(splittedArr);
}

Added:

From your following comment,

if I change var arr into var arr = [["a", 2], ["b", 1], ["c", 2], ["d", 1], ["e", 4]]; the script never stop to run.

In your script, when the value is ["e", 4], by arr.unshift(extraRow_temp_subArr[0]), arr is continued to be increased. By this, the infinite loop occurs. In order to achieve your new condition by modifying your script, I thought that it is required to add the process for checking the value of each element. The sample modified script is as follows.

Modified script:

function splitArrByDuration(arr, sumMax) {
  var sumMax = 3;
  var arr = [["a", 2], ["b", 1], ["c", 2], ["d", 1], ["e", 4]];
  var splittedArr = [];
  var temp_subArr = [];
  if (arr.length) {
    for (var i = 0; n = arr.length, i < n; i  ) {
      temp_subArr.push(arr[i]);

      // Added
      if (arr[i][1] >= sumMax) {
        var t = temp_subArr.pop();
        splittedArr.push(temp_subArr);
        splittedArr.push([t]);
        temp_subArr = [];

      } else if (sumAtIndex(temp_subArr, index = 1) > sumMax && arr.length) {
        var extraRow_temp_subArr = temp_subArr.splice(temp_subArr.length - 1, 1);
        arr.unshift(extraRow_temp_subArr[0]);
        splittedArr.push(temp_subArr);
        temp_subArr = [];

      // Added
      } else if (i == arr.length - 1) {
        splittedArr.push(temp_subArr);
      }

    }
  }
  Logger.log(splittedArr);
}

In this case, [[["a",2],["b",1]],[["c",2],["d",1]],[["e",4]]] is obtained.