返回奇数和空数组的函数

Question

I've a task to write function that returns the average value of even numbers from an array. If there are no even numbers in the array, return null. Here are arrays and expected result: [1,2,3,4,5,6,7] - should return "4" and [1,1,1,1] - should return "null". Where shall I put condition for null?I've tried different options but nothing works but maybe my code is wrong.

 function getEvenAverage(array) { const even = array.filter(function (element) { return element % 2 === 0 }); const sum = even.reduce(function (sum, element) { return (sum + element); }); return sum / even.length; } const result1 = getEvenAverage([1,2,3,4,5,6,7]) console.log(result1); const result2 = getEvenAverage([1,1,1,1]) console.log(result2);

Answer 1

You can achieve this in 1 loop as well:

Logic:

• Create an object(say map ) to hold values
• Loop on array.
• Check if current item is even, compute total and length. Set it to map
• At the end, check for length and return value

 function getEvenAverage(array) { const map = array.reduce((acc, item) => { if (item % 2 === 0) { return { total: acc.total + item, length: acc.length + 1 } } return acc }, { total: 0, length: 0 }); return map.length ? map.total / map.length : null } const result1 = getEvenAverage([1, 2, 3, 4, 5, 6, 7]) console.log(result1); const result2 = getEvenAverage([1, 1, 1, 1]) console.log(result2);

Answer 2

check this code

  function getEvenAverage (arr) {
  let temp = [];
  let sum = 0;
  let count = 0;
  arr.map(item => {
    if(item % 2 === 0) {
      temp = [...temp, item]
    }
  })
  temp.map(item => {
    sum += item;
    count +=1;
  })
  if(temp.length ===0) {
    return null
  }
  return sum/count;
}

Answer 3

You need to do the following changes:

1. Add null as an initial value for Array.prototype.reduce
2. Check if sum === null return sum , otherwise calculate an average value

You can find the code with changed applied below

 function getEvenAverage(array) { const even = array.filter(function (element) { return element % 2 === 0 }); const sum = even.reduce(function (sum, element) { return (sum + element); }, null); return sum === null ? sum : sum / even.length; } const result1 = getEvenAverage([1,2,3,4,5,6,7]) console.log(result1); const result2 = getEvenAverage([1,1,1,1]) console.log(result2);

Answer 4

Originally posted as a simple edit to this answer . Reposting as a separate answer below (as the other answer-er, Rajesh rolled back & noted that this be posted separately).

Added comments as this is now posted as a separate answer.

 // method to get average of even numbers in array const getEvenAverage = (arr) => ( arr // use optional-chainging "?." to handle bad "arr" values ?.reduce( // iterate over the array (acc, itm, idx) => { // access "acc" (accumulator), "itm" and "idx" if (itm % 2 === 0) { // if itm is even number acc.tot += itm; // add it to "acc" total (ie, "tot" prop) acc.len++; // increment the "len" prop of "acc" }; if ( // if processing last "itm" idx === arr.length - 1 && acc.len > 0 // and have previously processed even number "itm" ) { // update "result" prop of "acc" acc.result = acc.tot / acc.len; }; return acc; // return the updated "acc" for each iteration }, // initialize "acc" to have three props { tot: 0, len: 0, result: null } ) // simply send the "result" back to the caller ?.result ); console.log( 'call fn with array: 1,2...7\naverage-of-even-numbers: ', getEvenAverage([1, 2, 3, 4, 5, 6, 7]) ); console.log( 'call fn with array: 1,1,1,1\naverage-of-even-numbers: ', getEvenAverage([1, 1, 1, 1]) ); console.log( 'call fn with empty-array: []\naverage-of-even-numbers: ', getEvenAverage([]) ); console.log( 'call fn with undefined: \naverage-of-even-numbers: ', getEvenAverage() );

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