Most readable and efficient way to check if an item exists inside of array in an if statement even if item is 0

Question

Given the following array of arrays of numbers

numbers = [[7, 4, 0, 1], // see 0 at position [0][2]
           [5, 6, 2, 2], 
           [6, 9, 7, 8], 
           [1, 4, 2, 0]]

I'd like to check if a number exists in inner array array[j] and a array[j+2]. array[j] and array[j+2] could be 0 at times. In an if statement like the below, I'm having problems due to 0 being a falsy value.

for (let i = 0; i < numbers.length; i++) {
  for (let j = 0; j < numbers[0].length; j++) {
    if (numbers[i][j] && numbers[i][j+2]) {
      // how to evaluate this to true even if numbers[i][j] or numbers[i][j+2] is the number 0?
     }
  }
}

I tried adding an extra || check to see if that number was 0, e.g. (numbers[i][j] || numbers[i][j] === 0) && (numbers[i][j+2] || numbers[i][j+2] === 0) which I think looks terrible.

`numbers.flat().includes(n)`? – Yevhen Horbunkov May 12 '21 at 10:48 — Yevhen Horbunkov, May 12 '21 at 10:48

score 1 · Answer 1 · answered May 12 '21 at 10:43

If the array entry is not present, JavaScript will return undefined. Therefore, the following code should work:

for (let i = 0; i < numbers.length; i++) {
  for (let j = 0; j < numbers[0].length; j++) {
    if (numbers[i][j] !== undefined && numbers[i][j+2] !== undefined) {
      // how to evaluate this to true even if numbers[i][j] or numbers[i][j+2] is the number 0?
     }
  }
}

Most readable and efficient way to check if an item exists inside of array in an if statement even if item is 0

1 Answers1