0

My problem is that I am trying to iterate over an array several times, however, my for loop will only iterate over the array once (Or so it seems to me) which causes the outcome to be wrong.

What I am trying to do is loop on this array: let arr = ["e5", "b2", "a1", "c3","d4"]; and grabbing the string based on its number to move it to the first, second, third... position in the array depending on that number.

My code does not throw any error, however, seems like it iterates only once, grabs "a1", moves it to position 0 in the array and that's it. It doesn't iterate again to grab b2 and move it to position 1 in the array.

Next I would like to show you my code:

function order(words) {
  let sp = words.split(" ");   // Array to split string
  let pos = 1;   //variable to store number 1 to find in string
  let arr = []; //New array to push into, since trying to move values in same array wasn' tworking.
  for (let i = 0; i < sp.length; i++) {

    if (sp[i].includes(pos)) {
      arr.push(sp[i]);
      pos = pos + 1;

    }
    console.log(i);
  }

  return arr;
}

order("is2 Thi1s T4est 3a");

When I execute the above code, it returns:

0
1
2
3
[ 'Thi1s' ]

So as you can see, the result tells me that it iterates once, grabs "thi1s" and pushes it to the new array, however, it stops at that point and doesn´t do the other iterations or the rest of the pushing.

I am logging i to the console to see how many times it's iterating.

Link to the REPL: https://repl.it/@Otho01/Variables#index.js

Nimantha
  • 6,405
  • 6
  • 28
  • 69
Luis Gonzalez
  • 45
  • 1
  • 7

3 Answers3

1

The algorithm provided only loops the array of words once. For what you stated your goal is to find the position of each word based on a number inside each word. There are several ways of solving this but I will explain an easy algorithm.

You can accomplish the desired result by creating a loop over the positions {1..N} and inside this loop another loop on the words {word1...wordN}. So for each position you will try to find what word belongs to that position. The following algorithm is one of many solution for this problem:

function order(words){
  let sp = words.split(" ");
  let arr = [];

  for (let pos = 1; pos <= sp.length; pos++) {
    for (let i = 0; i < sp.length; i++) {
      if (sp[i].includes(pos)) {
        arr.push(sp[i])
      }
    }
  }
  
  return arr;
}

order("is2 Thi1s T4est 3a");

The result of this code is [ 'Thi1s', 'is2', '3a', 'T4est' ]

Francisco de Castro
  • 749
  • 6
  • 9
  • Thanks a lot for your answer, it works perfectly. Would you please tell me why my code was iterating only once? This is not the first time that I fail to make a for loop iterate more than once. – Luis Gonzalez Jan 27 '21 at 00:29
  • Note that in the algorithm I provided there are two loops. One `for` loop over the positions and another `for` loop over the words. So in the execution what happens is that for each `pos` value it loops all the words. In this case for `pos` = 1 there is a loop for the words and after the last word `pos` will become 2 and the loop on the words will happen again until pos is at the last possible position defined in its loop. The algorithm you provided you added one `for` loop over the words so once it gets to the last word the loop ends. – Francisco de Castro Jan 27 '21 at 00:38
0

I would also recommend looking into the array.sort() method as that might simplify your code Combine that with looking for the number within the string, and that should solve your issue

Array.sort() Find number in string

Joel Durham
  • 36
  • 3
0

The entire approach has flaws. Let's try order("is2 Thi123s T24est do9 3a"). I'd expect the result to be

["is2", "3a", "do9", "Te4est", "Thi123s"]

but instead I get

["Thi123s", "is2", "Thi123s", "T24est", "Thi123s", "3a", "T24est"]

and do9 ain't even in the result.

  • Thi123s matches for pos 1, 2 and 3 not pos 123, that'S why you get duplicates, and you get them at the wrong places in the result.
  • you don't remove a processed element from sp so it can be matched again by a different pos, again that's why you get duplicates. And it means that you check the first elements over and over and over for every pos
  • You don't even count up to 123, but if you did, you'd check the entire array for every number from 25 to 122 without pushing a single value to the returned result.

Summary:

This entire approach is incredibly expensive (for what it does) and would become even more so trying to fix it. So let's try a different approach; let's parse the number in the word and sort the array by that number.

const logCalls = [];
function getNumber(word){
  logCalls.push(word);
  return Number(word.match(/\d+/));
}

function order(words){
  return words.split(" ").sort((a,b) => getNumber(a) - getNumber(b));
}

console.log(order("is2 Thi123s T24est do9 3a"));

console.log("getNumber calls:");
console.log(logCalls.join("\n"));
.as-console-wrapper{top:0;max-height:100%!important}

But as you can see, getNumber is called quite a few times. And this list grows exponentially with the number of words you sort.

So let's improve on that and use String#matchAll to match all numbers only once and then sort:

function order(words){
  // matchAll() returns an iterator. To sort that we first have to make this an array
  const matches = [...words.matchAll(/\b[a-z]*(\d+)\w*\b/gi)];
  console.log("matches", matches);

  const sorted = matches.sort((a,b) => a[1] - b[1]);
  console.log("sorted", sorted);

  return sorted.map(match => match[0]);
    
// summary:   
// return [...words.matchAll(/\b[a-z]*(\d+)\w*\b/gi)]
//   .sort((a,b) => a[1] - b[1])
//   .map(match => match[0])
}

console.log(order("is2 Thi123s T24est do9 3a"));
.as-console-wrapper{top:0;max-height:100%!important}

Luis, you seem quite new to programming in general, so here's a little

FAQ:

Thomas
  • 11,958
  • 1
  • 14
  • 23