使用递归在数组中添加数字，而忽略任何其他数据类型

Question

Forgive me if this is to basic or have been asked already, but I'm stuck at this problem and I feel like it must be something simple I'm not seeing.

I want to add all numbers in the array using recursion(done!) and I'm missing a statement to ignore all other types of values. For example:

var arr = [1,'a','b',2,[1],'c']
sumValues(arr) // => 4 . 

function sumValues(arr){
  if(arr.length === 0){
    return 0;
  } // if the array is empty
  if(arr.length > 0){
    if(Array.isArray(arr[0])){
      return sumValues(arr[0]);
    } // if the next element is an array
    return arr.shift() + sumValues(arr);
  }
}

Answer 1

You can use Number.isFinite(value) to determine whether a variable is a number other than NaN or Infinity .

Based on this test, check and conditionally add values to the summation.

 function sumValues(arr){ if (arr.length === 0){ return 0; } // if the array is empty if (arr.length > 0) { if (Array.isArray(arr[0])){ return sumValues(arr[0]); } // if the next element is an array // pop the first element off the array var value = arr.shift(); // check its type and conditionally add it to the summation return (Number.isFinite(value) ? value : 0) + sumValues(arr); } } var arr = [1,'a','b',2,[1],'c'] console.log(sumValues(arr)); // 4 arr = [1,'3','b',2,[1,[4,3]],'c']; console.log(sumValues(arr)); // 11 (because '3' is ignored) 

Answer 2

You can use isNaN() to test if something is Not a Number. By using the boolean inversion operator ! you can test if it's a number: !isNaN(variable)

Basically it says: if variable is Not Not A Number
A double negative is a positive, so it becomes: if variable Is a Number

 function sumValues(arr){ if(Array.isArray(arr)) { if(arr.length > 0) { // Get the first value and remove it from the array. var first = arr.shift(); // Test if it's numeric. if(!isNaN(first)) { // If it is, parse the value, add the rest resulting array values. return parseInt(first) + parseInt(sumValues(arr)); } // if the first item is an array, we need to iterate that one too. if(Array.isArray(first)) { return parseInt(sumValues(first)) + parseInt(sumValues(arr)); } // It isn't a number, just continue with what's left of the array. return parseInt(sumValues(arr)); } // The array is empty, return 0. return 0; } // It isn't an array else { // Is it an number? if(!isNaN(arr)) { // return the number return parseInt(arr); } // return 0, it's a dead end. return 0; } } var arr = [1,'a','b',2,[1],'c']; console.log(sumValues(arr)) // => 4 . arr = [1,'3','b',2,[1,[4,3]],'c']; console.log(sumValues(arr)) // => 14 . arr = 'foobar'; console.log(sumValues(arr)) // => 0 . arr = '10'; console.log(sumValues(arr)) // => 10 . arr = ['foobar',10,'20',[10,20,'foobar']]; console.log(sumValues(arr)) // => 60 . 

Answer 3

You're squashing an array, there is no need to create your own reduction function. Use Array#reduce to squash the array. Upon each iteration, check if the current element is a number; if it is, add it to the accumulator; if not, move on.

To make it recursive (as shown in the example below) check if the current element is an array; if it is, add the result of calling the function on the element to the accumulator.

 const input = [1,'a','b',2,[1],'c']; const sumValues = i => i.reduce((m, e) => m + (e instanceof Array ? sumValues(e) : (isNaN(e) ? 0 : e)), 0); console.log(sumValues(input));