Shift certain array elements to front of array

Question

I want the function to return all 3-digit integers to the front of the array. When I console log this, I get 8 arrays, none of which is the output I'm looking for. I suspect the error starts when I begin using for ().

Also, any tips on how to further clean up the code is appreciated as I am learning best practices as I go.

    let numList = [1, 324,34, 3434, 304, 2929, 23, 444]
    			
    function checkLength (num){
    	num.forEach(function (n){ 
    		var stringLength = n.toString().length; //n = 324, 204, 444 
    		for (i=0; i<num.length; i++){
    			if (stringLength == 3){ 
    			let a = num.splice (i,1);
    			num.unshift (a[0]); 
    			} 
    		}   	
    	})
      console.log(num);
   	}
    
    checkLength(numList);

Did you look here: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — Jonathan
– Jonathan, Commented Apr 17, 2018 at 20:11

Ori Drori · Accepted Answer · 2018-04-17 20:20:41Z

1

Reduce the array into an array of 3 digits, and array of the rest, then flatten them to a single array by spreading into Array.concat():

function checkLength(array) {
  return [].concat(...array.reduce((r, n) => {   
    r[n > 99 && n < 1000 ? 0 : 1].push(n);
    
    return r;
  }, [[], []]));
}

console.log(checkLength([1, 324, 34, 3434, 304, 2929, 23, 444]));

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Apr 17, 2018 at 20:20

Ori Drori

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Comments

Zahid Saeed · Accepted Answer · 2018-04-17 20:16:04Z

Please have a look at the code. I hope that it is self-explanatory to you. Basically, you have 2 options here:

Mutate the original array.
Create a new array from the function and return it.

I'm using the first approach in this code below:

function shiftThreeDigitValuesInFrontOfArray(array)
{
    var threeDigitValue; // Will hold the value later,

    array.forEach(function(element, index)
    {
        // We are only interested in 3 digit values
    
        if(element.toString().length === 3)
        {
            // Remove the value from the array
            threeDigitValue = array.splice(index, 1);
            
            // And put it in front of the array
            array.unshift(threeDigitValue[0]);
            
        }
    });
}

var array = [1, 324,34, 3434, 304, 2929, 23, 444];

// Let the magic begin!
shiftThreeDigitValuesInFrontOfArray(array);

// View the results
console.log(array);

Please let me know if this works for you :)

Luis felipe De jesus Munoz · Accepted Answer · 2018-04-17 20:10:57Z

0

I think you are confused with how Shift works...

The shift() method removes the first element from an array and returns that removed element

It wont bring to the front all 3-digit integers, It will only remove the first element from the array and that is whats gonna be returned.

I think you are looking either sorting or filtering the array.

If you want to sort the array by the length of its value this is a way to do it: .sort(a=>a.toString().length!=3?1:-1)

let numList = [1, 324,34, 3434, 304, 2929, 23, 444]

console.log(numList.sort(a=>a.toString().length!=3?1:-1))

If you want to Filter is almost the same:

let numList = [1, 324,34, 3434, 304, 2929, 23, 444]

console.log(numList.filter(a=>a.toString().length==3?1:0))

answered Apr 17, 2018 at 20:10

Luis felipe De jesus Munoz

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2 Comments

Nina Scholz Over a year ago

sort is not necessarily stable.

Chi Lee Over a year ago

Thanks for sharing alternate string methods!

Nina Scholz · Accepted Answer · 2018-04-17 20:12:01Z

Basically you need an index of the first items of length with three and another index for iterating the array.

If an item has the string lenght of three splice this item and put it back to the stored index. Then increment that index.

Proceed with the loop.

This proposal keeps the order of the items. It just mooves the items with a string length to to while maintaining the order of these items as well.

function checkLength(array) {
    var i = 0,
        j;

    for (j = 0; j < array.length; j++) {
        if (array[j].toString().length === 3) {
            array.splice(i++, 0, array.splice(j, 1)[0]);
        }
    }
    return array;
}

console.log(checkLength([1, 324, 34, 3434, 304, 2929, 23, 444]));

.as-console-wrapper { max-height: 100% !important; top: 0; }

Liviath · Accepted Answer · 2018-04-17 20:16:03Z

0

Why do you use a nested loop?

You can do it like this:

var arr = [1, 324,34, 3434, 304, 2929, 23, 444];

function checkLength (arr){
    arr.forEach(function (n, index){ 
        if (n >= 100 && n < 1000) { 
            arr.unshift(arr.splice(index, 1)[0]);
        }
    });
    return arr;
}

checkLength(arr);

Please note that modifying an array you are looping over can be dangerous and might result in something you would not expect.

answered Apr 17, 2018 at 20:16

Liviath

411 bronze badge

Comments

Mohammad Usman · Accepted Answer · 2018-04-17 20:26:45Z

0

Instead of converting number to string you can just check the range. Once a 3 digit number is found, we will remove it from the array using splice() and add it on front of the array using .unshift().

let numList = [1, 324, 34, 3434, 304, 2929, 23, 444];

function checkLength(num) {
  num.forEach((v, i, a) => {
    if(v > 99 && v < 1000) {
      a.unshift(...a.splice(i, 1));
    }
  });
    
  return num;
}

console.log(checkLength(numList));

edited Apr 17, 2018 at 20:26

answered Apr 17, 2018 at 20:14

Mohammad Usman

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2 Comments

Chi Lee Over a year ago

What is the use of multiple periods for in this line -> a.unshift(...a.splice(i, 1));

Mohammad Usman Over a year ago

@ChiLee This is spread syntax

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