Complete array with nulls in javascript

Question

This is a similar question as this one: Complete missing sequence in array with zeros in javascript

However, I can't seem to go around this problem. This is my array:

const array = [
  [5, 'a', 2.3],
  [6, 'a', 1.7],
  [7, 'a', 5.4],
  [8, 'a', 2.8],
  [9, 'a', 8.5],
  [10, 'a', 9.2],
  [2, 'b', 1.6],
  [5, 'b', 5.7],
  [6, 'b', 8.9],
  [7, 'b', 3.5],
  [8, 'b', 6.1],
  [9, 'b', 1.8],
  [10, 'b', 7.4],
];

console.log(array);

First element: this is my reference value, it ranges from 1 to 10.

Second element: this is a category value.

Third element: this is a value that belongs to the second element, which happened at a timestamp that belongs to the first element.

My issue: I need to make sure that all the unique categories in the second element of the array (e.g., a and b) have the following sequence in the first element: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]. If they do not have one of these numbers, then I need to create it, and then assign null to the third element.

Therefore, this is my expected output:

[
  [1, 'a', null],
  [2, 'a', null],
  [3, 'a', null],
  [4, 'a', null],
  [5, 'a', 2.3],
  [6, 'a', 1.7],
  [7, 'a', 5.4],
  [8, 'a', 2.8],
  [9, 'a', 8.5],
  [10, 'a', 9.2],
  [1, 'b', null],
  [2, 'b', 1.6],
  [3, 'b', null],
  [4, 'b', null],
  [5, 'b', 5.7],
  [6, 'b', 8.9],
  [7, 'b', 3.5],
  [8, 'b', 6.1],
  [9, 'b', 1.8],
  [10, 'b', 7.4],
];

Any ideas?

boehm_s · Accepted Answer · 2021-04-26 14:28:13Z

2

You can create a range from 1 to 10, loop over it and when you can't find an association in your array, create a new element and push it. Do that for every category and you're good.

const range = new Array(0).fill().map((_, i) => i + 1); // from 1 to 10

const categories = array
                     .map(x => x[1]) // get categories
                     .filter((val, i, self) => self.indexOf(val) === i) // uniq

categories.forEach(categ => {
  range.forEach(n => {
    const alreadyInArray = array.some(x => x[0] === n && x[1] === categ);
    
    if (!alreadyInArray) {
      const newEntry = [n, categ, null];

      array.push(newEntry);
    }
  });
})

You can of course replace the forEach with classic for loops

answered Apr 26, 2021 at 14:28

boehm_s

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2 Comments

Ajith Gopi Over a year ago

Thanks. I have learnt to get unique elements with filter. I had the logic, but didn't know that I could use the filter's 3rd argument xD

Johannes H. Over a year ago

This requires looping over all elements in the array for every one of the elements that need to be present. It works, but is inefficient for large amounts of data. In this case, an indexed data structure should be used (an implementation of a HashMap for example).

Nicoowr · Accepted Answer · 2021-04-26 14:41:28Z

A functional solution, first get the categories, then for each category fill the corresponding array.

const array = [
  [5, 'a', 2.3],
  [6, 'a', 1.7],
  [7, 'a', 5.4],
  [8, 'a', 2.8],
  [9, 'a', 8.5],
  [10, 'a', 9.2],
  [2, 'b', 1.6],
  [5, 'b', 5.7],
  [6, 'b', 8.9],
  [7, 'b', 3.5],
  [8, 'b', 6.1],
  [9, 'b', 1.8],
  [10, 'b', 7.4],
];

const getMissingIndicesFromCategory = (tuples) => {
  const indices = tuples.map(tuple => tuple[0])
  const fullIndices = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
  return fullIndices.filter(index => !indices.includes(index));
}

const createMissingTuples = (missingIndices, category) => {
  return missingIndices.map(index => [index, category, null])
}


const completeCategoryTuples = (array, category) => {
  const categoryTuples = array.filter(tuple => tuple[1] === category)
  const missingIndices = getMissingIndicesFromCategory(categoryTuples)
  const missingTuples = createMissingTuples(missingIndices, category)
  return [...categoryTuples, ...missingTuples].sort((tuple1, tuple2) => tuple1[0] > tuple2[0] ? 1 : -1)
}

const getAllUniqueCategories = (array) => Array.from(new Set(array.map(tuple => tuple[1])))

const fillArray = (array) => {
  const categories = getAllUniqueCategories(array)
  return categories.flatMap(category => completeCategoryTuples(array, category))
}


const fullArray = fillArray(array)
console.log(fullArray)

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