Question

我在Codewars上做了一个基本的Javascript Kata，如果它们不是Number原语，那么挑战就是过滤数组中的所有项目。

我写了以下内容：

const list = [ '123', 123, 2, 'aasf', '1', 1 ];

function filter_list(l) {

  const filteredArray = l.filter((item) => {
   if ( Number(item) !== NaN ) return item;
  });

  return filteredArray;

}

console.log(filter_list(list));

我是编程新手，无法理解为什么这不起作用。你的解决方案是什么？

非常感谢

Answer 1

This answer永远不等于isNaN本身！ NaN

使用[Ref]测试值。 NaN函数确定值是否为const list = ['123', 123, 2, 'aasf', '1', 1]; function filter_list(l) { const filteredArray = l.filter((item) => { return isNaN(item); }); //could be written as l.filter(isNaN); return filteredArray; } console.log(filter_list(list));。

package mergesort;

import java.util.concurrent.ArrayBlockingQueue;
import java.util.concurrent.Callable;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;
import java.util.concurrent.LinkedBlockingQueue;
import java.util.concurrent.ThreadPoolExecutor;
import java.util.concurrent.TimeUnit;

// 18s
public class Main {
  public static final ExecutorService ex = new ThreadPoolExecutor(100, 100, 5, TimeUnit.SECONDS,
      new ArrayBlockingQueue<>(10000), new ThreadPoolExecutor.CallerRunsPolicy());

  public static void main(String[] args) throws InterruptedException, ExecutionException {
    int n = 1_000_000;
    Future<int[]> T1 = ex.submit(new Callable<int[]>() {

      @Override
      public int[] call() throws Exception {
        // TODO Auto-generated method stub
        return mergesort(generate(n));
      }
    });

    int[] ret = T1.get();
    for (int i : ret) {
      System.out.println(i);
    }
    System.out.println("done");
    ex.shutdownNow();
  }

  public static int[] generate(int n) {
    int[] nums = new int[n];
    for (int i = 0; i < n; i++) {
      nums[i] = (int) Math.floor(Math.random() * n);
    }
    return nums;
  }

  public static int[] mergesort(int[] nums) throws InterruptedException, ExecutionException {
    final int[] B;
    if (nums.length < 2) {
      return nums;
    }
    final int[] A = new int[nums.length / 2];
    if (nums.length % 2 == 0) {
      B = new int[nums.length / 2];
    } else {
      B = new int[nums.length / 2 + 1];
    }
    for (int i = 0; i < nums.length; i++) {
      if (i < nums.length / 2) {
        A[i] = nums[i];
      } else {
        B[i - nums.length / 2] = nums[i];
      }
    }

    Future<int[]> T2 = ex.submit(new Callable<int[]>() {

      @Override
      public int[] call() throws Exception {
        // TODO Auto-generated method stub
        return mergesort(B);
      }
    });
    Future<int[]> T1 = ex.submit(new Callable<int[]>() {

      @Override
      public int[] call() throws Exception {
        // TODO Auto-generated method stub
        return mergesort(A);
      }
    });
    Future<int[]> T3 = ex.submit(new Callable<int[]>() {
      @Override
      public int[] call() throws Exception {

        return merge(T1.get(), T2.get());
      }
    });
    return T3.get();
  }

  public static int[] merge(int[] A, int[] B) {
    int[] ret = new int[A.length + B.length];
    int i = 0;
    int j = 0;
    int k = 0;
    while (i < A.length && j < B.length) {
      if (A[i] < B[j]) {
        ret[k] = A[i];
        i++;
      } else {
        ret[k] = B[j];
        j++;
      }
      k++;
    }

    while (j < B.length) {
      ret[k] = B[j];
      j++;
      k++;
    }
    while (i < A.length) {
      ret[k] = A[i];
      i++;
      k++;
    }
    return ret;
  }
}

Answer 2

您可以使用const list = [ '123', 123, 2, 'aasf', '1', 1 ]; function filter_list(l) { const filteredArray = l.filter((item) => { if ( isNaN(Number(item)) ) return item; }); return filteredArray; } console.log(filter_list(list));

isNaN

NaN功能的必要性

与JavaScript中的所有其他可能值不同，不可能依赖等于运算符（==和===）来确定值是否为NaN == NaN，因为NaN === NaN都是isNaN并且varstocases /make DX from DX1 to DX25. freq DX.评估为假。因此，dataset name OrigData. dataset copy ForRestr. dataset activate ForRestr. varstocases ..... freq .... dataset activate OrigData.函数的必要性。

来源：isNaN

与NaN的比较并不像我期望的那样发挥作用？

2 个答案: