43

I get an Array with an unknown Number of data. But I only have an predefined amount of data to be shown/store. How can I take every nth Element of the initial Array and reduce it in JavaScript?

Eg.: I get an Array with size=10000, but are only able to show n=2k Elements.

I tried it like that: delta= Math.round(10*n/size)/10 = 0.2 -> take every 5th Element of the initial Array.

for (i = 0; i < oldArr.length; i++) {
  arr[i] = oldArr[i].filter(function (value, index, ar) {
    if (index % delta != 0) return false;
    return true;
  });
}

With 0.2 it´s always 0, but with some other deltas (0.3) it is working. Same for delta=0.4, i works, but every second Element is taken with that. What can I do to get this to work?

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6
  • What is n? What is k? What is delta? What is oldArr? Commented Nov 2, 2015 at 16:57
  • How about delta = size / n? Commented Nov 2, 2015 at 16:58
  • 1
    0.2 evenly divides all the integers, so someInt % 0.2 == 0 always. I think you want someInt % (1 / 0.2), ie someInt % 5 Commented Nov 2, 2015 at 17:01
  • 1
    @James 1 % 0.2 produces 0.19999999999999996 for me, because floating point math is broken Commented Nov 2, 2015 at 17:06
  • @Oriol you're right! 1 % 0.25 works (because 0.25 can be exactly represented in binary, I suppose), I guess using modulus operator on non-integer operands is a bad idea in JS. Commented Nov 2, 2015 at 17:10

7 Answers 7

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64

Maybe one solution :

avoid filter because you don't want to loop over 10 000 elements ! just access them directly with a for loop !

 
var log = function(val){document.body.innerHTML+='<div></pre>'+val+'</pre></div>'} 

var oldArr = [0,1,2,3,4,5,6,7,8,9,10]
var arr = [];

var maxVal = 5;

var delta = Math.floor( oldArr.length / maxVal );

// avoid filter because you don't want
// to loop over 10000 elements !
// just access them directly with a for loop !
//                                 |
//                                 V
for (i = 0; i < oldArr.length; i=i+delta) {
  arr.push(oldArr[i]);
}


log('delta : ' + delta + ' length = ' + oldArr.length) ;
log(arr);

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1 Comment

Thanks, I guess that´s it. Especially since that should be faster than with the modulo-check. And I don´´t have 10k Elements. It´s an 2D-Array, so in the other Version there would even be 100k checks.
33

Filter itself returns an array. If I'm understanding you correctly, you don't need that surrounding loop. So:

newArr = oldArr.filter(function(value, index, Arr) {
    return index % 3 == 0;
});

will set newArr to every third value in oldArr.

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Comments

12

Try

arr = oldArr.filter(function (value, index, ar) {
    return (index % ratio == 0);
} );

where ratio is 2 if you want arr to be 1/2 of oldArr, 3 if you want it to be 1/3 of oldArr and so on.

ratio = Math.ceil(oldArr.length / size); // size in the new `arr` size

You were calling filter() on each element of oldAdd inside a loop and you're supposed to call filter() on the whole array to get a new filtered array back.

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1 Comment

Shouldn't it be Math.ceil() to prevent the new array from being larger than the desired size?
1

this also works by using map to create the new array without iterating over all elements in the old array..

// create array with 10k entries
const oldArr = [ ...Array( 10000 ) ].map( ( _el, i ) => i );
const max = 10;
const delta = Math.floor( oldArr.length / max );

const newArr = [ ...Array( max ) ].map( ( _el, i ) => (
  oldArr[ i * delta ]
) );

console.log( newArr );

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0

Borrowing from @anonomyous0day's solution, generate a new Array with the desired indices from the given array:

(Take every 3 items)

Array.prototype.take = function(n) {
  if (!Number(n) && n !== 0) {
    throw new TypeError(`Array.take requires passing in a number.  Passed in ${typeof n}`);
  } else if (n <= 0) {
    throw new RangeError(`Array.take requires a number greater than 0.  Passed in ${n}`);
  }

  const selectedIndicesLength = Math.floor(this.length / n);
  return [...Array(selectedIndicesLength)].map((item, index) => this[index * n + 1]);
};

[1, 2, 3, 4, 5, 6, 7, 8].take(2); // => 2, 4, 6, 8
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0

may help!

 const myFunction = (a, n) => {

   let array = []


    for(i = n; i <= a.length; i += n){
      array.push(a[i-1]);

   }

  return array;

}
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Comments

0

To separate array every 5 element need to make another 2d array to store every 5 element separately it's my solution if can help to your problem

const arrayParts = [];
var separateEach = 5;
var startLen = 0;

var partCount = Math.ceil(infos.length / separateEach);
console.log('partCount', partCount);

for (var i = 1; i <= partCount; i++) {
  arrayParts.push(infos.slice(startLen, separateEach * i));
  startLen = separateEach * i;
}

console.log('Parts ', arrayParts);
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