How can i get loop and filter at once in Javascript

Question

    const count = [
        {number: 1},
        {number: 2},
        {number: 3},
        {number: 4},
        {number: 5},
        {number: 6},
        {number: 7},
        {number: 8},
        {number: 9},
        {number: 10},
        {number: 11},
        {number: 12},
        {number: 13},
        {number: 14},
        {number: 15}
        ];
 
 
    for(let i = 0;i < count.length; i++) {

    document.write(count[i].number);
    }

If the main number what i want to get out the loop is 5. But i want to write out one above 5 and one under 5. So 456. How can i achieve that?

@JacobSchneider but watch out for an out of bounds exception! — Our_Benefactors
– Our_Benefactors, Commented Jan 22, 2019 at 5:20

Nick Parsons · Accepted Answer · 2019-01-22 07:51:53Z

2

In order to filter out the number which you want you can use an if statement to check if the current object's number count[i].number is the value you want, and then if it is, print out its associated values.

You can access the object before and after by changing your value of i. Using i-1 you will get the element before your current value. Similarly, using i+1 you will go to the object after your current value.

Lastly, you should check the value of i to make sure that it falls within the array. For instance, if you chose to find the number 1 it has no before object, and so you need to check for this case!

See working example below:

const count = [
        {number: 1},
        {number: 2},
        {number: 3},
        {number: 4},
        {number: 5},
        {number: 6},
        {number: 7},
        {number: 8},
        {number: 9},
        {number: 10},
        {number: 11},
        {number: 12},
        {number: 13},
        {number: 14},
        {number: 15}
        ];
 
const toFind = 5;
for(let i = 0; i < count.length; i++) {
  if(count[i].number == toFind) {
    if(i-1 >= 0) console.log(count[i-1].number);
    console.log(count[i].number);
    if(i+1 < count.length) console.log(count[i+1].number);
    break; // complete loop cycle
  }
}

edited Jan 22, 2019 at 7:51

answered Jan 22, 2019 at 5:20

Nick Parsons

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1 Comment

mhodges Over a year ago

This is the most algorithmically efficient solution. One small optimization to make is to break; at the end of the if() statement, as you have already found your number and iterating the rest of the array is unnecessary.

nicholaswmin · Accepted Answer · 2019-01-22 05:42:32Z

2

Assuming the needle is unique, you can .filter the numbers you want, then .forEach to print them.

const needle = 5
const haystack = [
  {number: 1},
  {number: 2},
  {number: 3},
  {number: 4},
  {number: 5},
  {number: 6},
  {number: 7},
  {number: 8},
  {number: 9},
  {number: 10},
  {number: 11},
  {number: 12},
  {number: 13},
  {number: 14},
  {number: 15}
]

haystack
  .sort((a, b) => a.number - b.number) // skip this if haystack is already sorted.
  .filter(el => [needle - 1, needle, needle + 1].includes(el.number))
  .forEach(el => {
    document.write(el.number)
  })

Note that you have to guard against needle = 0 which has no -1 and needle = 15 which has no +1.

edited Jan 22, 2019 at 5:42

answered Jan 22, 2019 at 5:18

nicholaswmin

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6 Comments

Our_Benefactors Over a year ago

More generally I believe their query is "I need to get X number, and show the values at x-1 and x+1"

nicholaswmin Over a year ago

@Our_Benefactors Ah, true. Edited.

Code Maniac Over a year ago

No need of filter here. why not do it simply with one forEach ?

nicholaswmin Over a year ago

@CodeManiac You mean using an if within forEach?

Code Maniac Over a year ago

@NikKyriakides yes. there's no need of one extra loop

|

Kuncheria · Accepted Answer · 2019-01-22 05:51:48Z

1

        const count = [
      { number: 1 },
      { number: 2 },
      { number: 3 },
      { number: 4 },
      { number: 5 },
      { number: 6 },
      { number: 7 },
      { number: 8 },
      { number: 9 },
      { number: 10 },
      { number: 11 },
      { number: 12 },
      { number: 13 },
      { number: 14 },
      { number: 15 }
    ];
    var index = count.findIndex(item => item.number === 5);
    document.write(count[index - 1].number);
    document.write(count[index].number);
    document.write(count[index + 1].number);

edited Jan 22, 2019 at 5:51

answered Jan 22, 2019 at 5:48

Kuncheria

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1 Comment

mhodges Over a year ago

Make sure to bounds check your index-1 and index+1 or you'll get an exception for undefined.number

Code Maniac · Accepted Answer · 2019-01-22 05:24:52Z

0

You can directly use the forEach and based on range you can write to document.

const count = [{number: 1},{number: 2},{number: 3},{number: 4},{number: 5},{number: 6},{number: 7},{number: 8},{number: 9},{number: 10},{number: 11},{number: 12},{number: 13},{number: 14},{number: 15}];
        
count.forEach(({number})=> {
  if(number>3 && number<7){
    document.write(number)
  }
})

answered Jan 22, 2019 at 5:24

Code Maniac

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