looping through javascript object without sorting based on keys

Question

this is the problem description: Given an array of integers, calculate the fractions of its elements that are positive, negative, and are zeros. Print the decimal value of each fraction on a new line. for example given the array arr=[1,1,0,-1,-1] output should be:

0.400000
0.400000
0.200000

I know there is more more simple solution for it ,and i am sorry for my silly simple question but i wanna make my code work, my code sorts the output based on the key and removes duplicates. for this arr, my code output is:

0.200000
0.400000

thank you so much in advance for any help.

function plusMinus(arr) {
    var freq = {};
    for (var i = 0; i < arr.length; i++){
        if (freq[arr[i]]) {
            freq[arr[i]]++;
        } else {
            freq[arr[i]] = 1;
        }
    } for(var key in freq){

        console.log((freq[key]/arr.length).toFixed(6));
    }
    }

Nina Scholz · Accepted Answer · 2019-03-24 09:04:07Z

2

You could take an object with predifined properties, this prevents each loop for checking the existence and take an array of keys for getting the result in a wanted order.

function plusMinus(arr) {
    var freq = { 1: 0, '-1': 0, 0: 0 },
        i, key;

    for (i = 0; i < arr.length; i++) {
        freq[arr[i]]++;
    }

    for (key of [1, -1, 0]) {
        console.log((freq[key] / arr.length).toFixed(6));
    }
}

plusMinus([1, 1, 0, -1, -1]);

answered Mar 24, 2019 at 9:04

Nina Scholz

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4 Comments

adiga Over a year ago

The number can be any integer according the question. You might have to do: freq[Math.sign(arr[i])]++

Nina Scholz Over a year ago

@adiga, right, for any integer, the sign sould be sufficient.

Nasim Over a year ago

@NinaScholz the order is right now but still deletes duplicates, the output should be 0.4,0.4,0.2 but its giving 0.4,0.2 , why is that?

Nina Scholz Over a year ago

i don't understand the last output. what do you get?

Dipak · Accepted Answer · 2019-03-24 09:11:51Z

0

Let's make sure the order of key in the map by defining it first.

function plusMinus(arr) {
  var freq = {
    posetive: 0,
    negative: 0,
    zero: 0
  };
  for (var i = 0; i < arr.length; i++){
    if( arr[i] < 0) {
      freq.negative++;
    } else if(arr[i] > 0) {
      freq.posetive++;
    } else {
      freq.zero++;
    }
  } 
  for(var key in freq){
    console.log((freq[key]/arr.length).toFixed(6));
  }
}
plusMinus([1,1,0,-1,-1]);

answered Mar 24, 2019 at 9:11

Dipak

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Comments

Code Maniac · Accepted Answer · 2019-03-24 09:03:45Z

0

You can use reduce.

Here idea is

First loop through original array and check for the value.
If value is zero we increment count of zero key.
If value is positive we increment count of pos key.
If value is negative we increment count of neg key.
Finally we divide each count by length of array.

let arr = [1,1,0,-1,-1]

let op = arr.reduce((op,inp)=>{
  if(inp === 0){
    op.zero.count++
  } else if (inp > 0){
    op.pos.count++;
  } else {
    op.neg.count++;
  }
  return op
},{zero:{count:0},pos:{count:0},neg:{count:0}})


let final = Object.entries(op).map(([key,value])=>({
  [key] : value.count / arr.length
}))

console.log(final)

answered Mar 24, 2019 at 9:03

Code Maniac

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Comments

Jack Bashford · Accepted Answer · 2019-03-24 09:11:13Z

0

Use reduce, map and filter:

const arr = [1, 1, 0, -1, -1];

const counts = arr.reduce((acc, curr) => {
  if (!curr) acc[0]++;
  else if (curr > 0) acc[1]++;
  else acc[2]++;
  return acc
}, [0, 0, 0]);

const result = counts.map(e => e / arr.length).filter((e, i, a) => a.indexOf(e) == i);

console.log(result);

answered Mar 24, 2019 at 9:11

Jack Bashford

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Comments

Amardeep Bhowmick · Accepted Answer · 2019-03-24 09:20:19Z

0

You can try using Array.reduce and the resulting array will have the fraction of positive number at the '0'th index, negative at '1'st and zero at the '2'nd index.

Now if you want to control the count of the number of elements after decimal point, use Array.map at the end to transform it.

const array = [1,1,0,-1,-1];
function plusMinus(arr){
  const output = arr.reduce((acc, ele) => {
  if(ele > 0){
    acc[0] = ((acc[0] || 0 ) + 1 / arr.length);
  }
  if(ele < 0){
    acc[1] = ((acc[1] || 0 ) + 1 / arr.length);
  }
  if(ele === 0) {
    acc[2] = ((acc[2] || 0 ) + 1 / arr.length);
  }
  return acc;
  }, []).map(ele => ele.toFixed(6));
 console.log(...output);
}
plusMinus(array);

edited Mar 24, 2019 at 9:20

answered Mar 24, 2019 at 9:07

Amardeep Bhowmick

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Comments

Doppio · Accepted Answer · 2019-03-24 09:28:09Z

0

Math.sign is your friend here. Math.sign

Also lodash would really help this snippet to be cleaner, I highly recommend _.countBy. Lodash .countBy

Here's the code.

const plusMinus = (numbers) => {
  // Count by Sign (-1, 0 1)
  const countSign = _.countBy(numbers, Math.sign);
  // _.countBy return object, of counted { '1': 2, '0': 1, '-1': 2 } 
  
  // Print them in orders
  const printOrder = [1, -1, 0];
  printOrder.forEach(sign => {
    console.log((countSign[sign] / numbers.length).toFixed(6));
  });
}

const testArr = [1,1,0,-1,-1];
plusMinus(testArr);

<script src="https://cdn.jsdelivr.net/npm/[email protected]/lodash.min.js"></script>

answered Mar 24, 2019 at 9:28

Doppio

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Comments

Andriy · Accepted Answer · 2019-03-24 09:35:57Z

0

Here is another one-line solution using Array.reduce() and Array.forEach() functions:

const plusMinus = arr => arr
  .reduce((res, curr) => ++res[!curr ? 2 : curr < 0 ? 1 : 0] && res, [0, 0, 0])
  .forEach(i => console.log((i / arr.length).toFixed(6)));

plusMinus([1, 1, 0, -1, -1]);

answered Mar 24, 2019 at 9:35

Andriy

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