For循环中的嵌套循环

Question

I have a long line of code which runs fine, but I was wondering if a nested If loop within the outer For loop would shorten the code, and if so, what would be the appropriate way to write this nested loop.

This is for a JavaScript array which has nested objects called by counter i. Within each object are key:value pairs. Myopia, Maple, and truckdriver are the keys, and the returned values are numbers 0 to infinity.

for (let i = 0; i < z; i++) {
  for (let i = 0; i < 49; i++) {
      y[i].truckdriver = 0;
  }
  for (let i = 49; i < (z - 1); i++) {
      if (y[i].Myopia > 0 && y[i].Maple < y[i].Myopia) {
           y[i].truckdriver = 0;
      } else if ((i + 1) >= z) {
           y[i].truckdriver = 0;
      } else if (y[i].Myopia > 0 && y[i + 1].Maple < y[i].Myopia) {
           y[i].truckdriver = 1;
      } else if ((i + 2) >= z) {
           y[i].truckdriver = 0;
      } else if (y[i].Myopia > 0 && y[i + 2].Maple < y[i].Myopia) {
           y[i].truckdriver = 2;
      } else if ((i + 3) >= z) {
           y[i].truckdriver = 0;
      } else if (y[i].Myopia > 0 && y[i + 3].Maple < y[i].Myopia) {
           y[i].truckdriver = 3;
      } else {
           y[i].truckdriver = 0;
      }
  }
  for (let i = (z - 1); (i < z); i++) {
      y[i].truckdriver = 0;
  }

}

I hope to shorten this code while retaining its functionality.

Answer 1

Some minor room for shortening.

• You can dedupe the 1st and 3rd for loops with a single for loop setting all truckdriver values to 0 initially.
• This also allows removing the last else case, as truckdriver is already 0
• You can remove the branch (i+1) >= z , as always false since you iterate i while < (z - 1)
• Extracting the check y[i].Myopia > 0 makes your code longer in # of lines, but shorter in character count and much easier to read
• Lastly, as you yourself mentioned, the if statements and their then clauses are very similar and can be reused inside a for loop.

for (let i = 0; i < z; i++)
  y[i].truckdriver = 0;

for (let i = 49; i < (z - 1); i++) {
  if (y[i].Myopia > 0)
    continue;

  for (let j = 0; j <= 3; j++)
    if (i + j < z && y[i + j].MAPLE < y[i].Myopia) {
      y[i].truckdriver = j;
      break;
    }
}