检查数组中可以成功转换为数字的字符串数

Question

Here's a description of the task I'm working on:

Write a function called countNumbers, which accepts an array of strings. The function should return a count of the number of strings in the array that can be successfully converted into a number. For example, the string "1" can be successfully converted to the number 1, but the string "hello" cannot be converted into a number.

countNumbers(['a','b','3','awesome','4']); // 2
countNumbers(['32', '55', 'awesome', 'test', '100']); // 3
countNumbers([]); // 0
countNumbers(['4','1','0','NaN']); // 3
countNumbers(['7', '12', 'a', '', '6', '8', ' ']); // 4

My code:

function countNumbers(arr) {
    var count = 0;
    for (num in arr) {
        if (Number(arr[num]) !== NaN) {
            count++;
        }
    }
    return count;
}

When I console log (Number(arr[num]), I do see NaN in the console, however when I'm comparing in the loop, the count variable does not increase and I end up with a return value equal to the length of the array.

Any tips on what I could be doing wrong or what I overlooked are much appreciated.

Answer 1

You could filter with isFinite and take the length.

 +'' -> 0 +' ' -> 0

 function countNumbers(array) { return array.filter(isFinite).length; } console.log(countNumbers(['a','b','3','awesome','4'])); // 2 console.log(countNumbers(['32', '55', 'awesome', 'test', '100'])); // 3 console.log(countNumbers([])); // 0 console.log(countNumbers(['4','1','0','NaN'])); // 3 console.log(countNumbers(['7', '12', 'a', '', '6', '8', ' '])); // 4, now 6

If you really like only digits, then you could check with a regular expression.

 function countNumbers(array) { return array.filter(RegExp.prototype.test, /^\d+$/).length; } console.log(countNumbers(['a','b','3','awesome','4'])); // 2 console.log(countNumbers(['32', '55', 'awesome', 'test', '100'])); // 3 console.log(countNumbers([])); // 0 console.log(countNumbers(['4','1','0','NaN'])); // 3 console.log(countNumbers(['7', '12', 'a', '', '6', '8', ' '])); // 4

Answer 2

Filter your arrays and get their lengths.

 console.log(countNumbers(['a', 'b', '3', 'awesome', '4'])); // 2 console.log(countNumbers(['32', '55', 'awesome', 'test', '100'])); // 3 console.log(countNumbers([])); // 0 console.log(countNumbers(['4', '1', '0', 'NaN'])); // 3 console.log(countNumbers(['7', '12', 'a', '', '6', '8', ' '])); // 4 function countNumbers(arr) { return arr.filter(function(el) { return parseFloat(el) == el; }).length; }

Answer 3

You must pass n through parseInt() before passing to isNaN() , otherwise the space will evaluate as a 0 and count towards the total.

 const isNum = (n) =>;isNaN(parseInt(n)). const countNumbers = (arr) => arr.filter(isNum);length. console,log( countNumbers(['a','b','3','awesome','4']), // 2 countNumbers(['32', '55', 'awesome', 'test', '100']), // 3 countNumbers([]), // 0 countNumbers(['4','1','0','NaN']), // 3 countNumbers(['7', '12', 'a', '', '6', '8'; ' ']) // 4 );

Answer 4

Two things straight up:

1. Loops.
2. Variables.

When you solve a problem using loops and variables, you now have two problems.

Remove the inline console.log to return the actual count as a number.

There may be more exhaustive or smarter checks than that, but you can easily modify the isNum function, without touching any of the other code.

No loops, no variables, and a separation of concerns - the isNum logic is not mixed with control flow.

Update: the parseFloat check in Alex's answer is the smarter check you are looking for.

 const isNum = maybeNum => (typeof maybeNum === 'number' || typeof +maybeNum === 'number') &&.isNaN(+maybeNum) && maybeNum.= '' && maybeNum.= ' ' const countNumbers = arr => console,log(arr,filter(isNum),length) countNumbers(['a','b';'3','awesome','4']), // 2 countNumbers(['32', '55'; 'awesome'; 'test', '100']), // 3 countNumbers([]), // 0 countNumbers(['4';'1','0','NaN']), // 3 countNumbers(['7', '12', 'a', ''; '6', '8', ' ']); // 4

Here is the version with the smarter check from Alex's answer. The editor doesn't support nullish coalescing, so instead of parseFloat(maybeNum)?? false parseFloat(maybeNum)?? false , you have to do it like it's still the 90's:

parseFloat(maybeNum) || parseFloat(maybeNum) === 0

 const isNum = maybeNum => parseFloat(maybeNum) || parseFloat(maybeNum) === 0 const countNumbers = arr => arr.filter(isNum).length const test = arr => console.log(countNumbers(arr)) test(['a','b','3','awesome','4']); // 2 test(['32', '55', 'awesome', 'test', '100']); // 3 test([]); // 0 test(['4','1','0','NaN']); // 3 test(['7', '12', 'a', '', '6', '8', ' ']); // 4

Answer 5

Reduce over it and evaluate edge cases

 console.log(countNumbers(['a','b','3','awesome','4'])); // 2 console.log(countNumbers(['32', '55', 'awesome', 'test', '100'])); // 3 console.log(countNumbers([])); // 0 console.log(countNumbers(['4','1','0','NaN'])); // 3 console.log(countNumbers(['7', '12', 'a', '', '6', '8', ' '])); // 4 function countNumbers(arrayStrings) { return arrayStrings.reduce((count, string) => isNaN(string) || string.trim() === ''? count: count + 1, 0) }

Answer 6

I think with reduce it can be done

 const countNumbers = arr => { return arr.reduce((acc, value) => isNaN(Number(value))? acc: acc + 1, 0) } const num = countNumbers(['a','b','3','awesome','4']); console.log(num)

Answer 7

You can make use of in-built 'isNaN()' function.
Simply replace your if statement with below:

if (!isNaN(arr[num])){

Answer 8

The other solutions testing isNaN(Number(character)) are better. I forgot about isNaN and wrote this. It performs slower and is much longer.

function countNumbers(array) {
    let count = 0
    for (number of array) {
      if(typeof number === 'number') {
        count++
      } else if(typeof number === 'string') {
        let isNumber
        let numberCharacters = number.split('')
        const numberRegularExpression = new RegExp(/[0-9]/)
        for(character of numberCharacters) {
          if(numberRegularExpression.test(character) !== true) {
            isNumber = false
            break
          } else {
            isNumber = true
          }
        }
        if(isNumber === true) count++
      }
    }
    return count;
}