Javascript代码将偶数和奇数分开，然后获得各自的总和和平均值

Question

So I have this code that requires a user to specify the length of an array,store some values in the array then separate the odd and even numbers. I have done everything so far but i am getting some incorrect output. I have attached my code.

 //Variable and array declaration var arrayNum = []; //Create an array with no size var arrayEven = []; //Array to hold even numbers var arrayOdd = []; //Array to hold odd numbers var i; //Variable to store the index of arrayNum() above var NUM_INPUTS; //Variable to store array size which is the number of elements var sumOdd = 0.0; //variable to store the sum if the odd numbers var sumEven = 0.0; //variable to store the sum if the odd numbers var avgO = 0.0; var avgE = 0.0; NUM_INPUTS = parseInt(prompt("Enter the number of inputs you need: ")); //Ask the user to specify size of the array //loop for entering values into the array for (i = 0; i < NUM_INPUTS; i++) { arrayNum.push(parseFloat(prompt("Enter the numbers: " + (i + 1)))); //Check whether a number is odd and add it to arrayOdd[] if ((arrayNum[i] % 2) === 1) { arrayOdd.push(arrayNum[i]); for (var x = 0; x < arrayOdd.length; x++) { sumOdd += arrayOdd[x]; } //calculate the sum of odd numers avgO = sumOdd / arrayOdd.length; } //Check whether a number is eve and add it to arrayEven[] else { arrayEven.push(arrayNum[i]); for (var y = 0; y < arrayOdd.length; y++) { sumEven += arrayEven[y]; } //sumEven+=arrayEven[i]; //calculate sum of even numbers avgE = sumEven / arrayEven.length; } } //Output results document.write("All numbers in the array are: " + arrayNum); document.write("<br/>All even numbers in the array are: " + arrayEven); document.write("<br/> The sum of all even numbers is: " + sumEven + " and average of the even numbers is: " + avgE); document.write("<br/>"); document.write("<br/>All odd numbers in the array are: " + arrayOdd); document.write("<br/> The sum of all odd numbers is: " + sumOdd + " and the average of the odd numbers is " + avgO); 

Answer 1

In cases like this, explain your code to a duck! (No I'm not kidding, this is known as rubber duck debugging )

The user enters 2 as the count of numbers he wants to insert.

The loop will iterate 2 times and ask the user for a number.

The user enters 1 as the first number.

1 is odd, therefore it gets pushed to arrayOdd , it also gets added to sumOdd which is 1 then.

The user enters 3 as the second number. As it is odd, it will also get pushed into arrayOdd . The inner loop runs again, iterates over arrayOdd , takes the first element ( 1 ) and adds it to sumOdd (!), which is 2 now, then it takes the second element ( 3 ) and adds it to sumOdd , sumOdd is 5.

But wait ... the sum of 1 and 3 is 4, not 5. Summing up all numbers of sumOdd and sumEven in the loop seems to be a mistake.

 for (var x = 0; x < arrayOdd.length; x++) {
  sumOdd += arrayOdd[x];
}

That has to be:

 sumOdd += /* left as an exercise */;

Answer 2

You are doing following things wrong

• You are adding all the previous values of arrayOdd and arrayEven again and again. You should add the new value to the sumOdd and sumEven .
• You are calculating average before all inputs are completed. You should calculate avgO and avgE at end of loop.

 var arrayNum = []; var arrayEven = []; var arrayOdd = []; var i; var NUM_INPUTS; var sumOdd = 0.0; var sumEven = 0.0; var avgO = 0.0; var avgE = 0.0; NUM_INPUTS = parseInt(prompt("Enter the number of inputs you need: ")); for (i = 0; i < NUM_INPUTS; i++) { arrayNum.push(parseFloat(prompt("Enter the numbers: " + (i + 1)))); if ((arrayNum[i] % 2) === 1) { arrayOdd.push(arrayNum[i]); sumOdd += arrayNum[i]; } else { arrayEven.push(arrayNum[i]); sumEven += arrayNum[i]; } } avgO = sumOdd / arrayOdd.length; avgE = sumEven / arrayEven.length; //Output results document.write("All numbers in the array are: " + arrayNum); document.write("<br/>All even numbers in the array are: " + arrayEven); document.write("<br/> The sum of all even numbers is: " + sumEven + " and average of the even numbers is: " + avgE); document.write("<br/>"); document.write("<br/>All odd numbers in the array are: " + arrayOdd); document.write("<br/> The sum of all odd numbers is: " + sumOdd + " and the average of the odd numbers is " + avgO); 

Answer 3

The mistake I found is that you use oddnum arrays for calculating even numbers.

for (var y = 0; y < arrayOdd.length; y++) {
      sumEven += arrayEven[y];
    }

 //Variable and array declaration var arrayNum = []; //Create an array with no size var arrayEven = []; //Array to hold even numbers var arrayOdd = []; //Array to hold odd numbers var i; //Variable to store the index of arrayNum() above var NUM_INPUTS; //Variable to store array size which is the number of elements var sumOdd = 0.0; //variable to store the sum if the odd numbers var sumEven = 0.0; //variable to store the sum if the odd numbers var avgO = 0.0; var avgE = 0.0; NUM_INPUTS = parseInt(prompt("Enter the number of inputs you need: ")); //Ask the user to specify size of the array //loop for entering values into the array for (i = 0; i < NUM_INPUTS; i++) { arrayNum.push(parseFloat(prompt("Enter the numbers: " + (i + 1)))); //Check whether a number is odd and add it to arrayOdd[] if ((arrayNum[i] % 2) === 1) { arrayOdd.push(arrayNum[i]); for (var x = 0; x < arrayOdd.length; x++) { sumOdd += arrayOdd[x]; } //calculate the sum of odd numers avgO = sumOdd / arrayOdd.length; } //Check whether a number is eve and add it to arrayEven[] else { arrayEven.push(arrayNum[i]); for (var y = 0; y < arrayEven.length; y++) { sumEven += arrayEven[y]; } //sumEven+=arrayEven[i]; //calculate sum of even numbers avgE = sumEven / arrayEven.length; } } //Output results document.write("All numbers in the array are: " + arrayNum); document.write("<br/>All even numbers in the array are: " + arrayEven); document.write("<br/> The sum of all even numbers is: " + sumEven + " and average of the even numbers is: " + avgE); document.write("<br/>"); document.write("<br/>All odd numbers in the array are: " + arrayOdd); document.write("<br/> The sum of all odd numbers is: " + sumOdd + " and the average of the odd numbers is " + avgO)