javascript - 检查一个数组中的至少一个值是否存在于另一个数组中失败

Question

以下内容应返回 true，但报告为 false:

var array1 = ['fred'];
var array2 = ['sue', 'fred'];

var modified = [];

//create a new modified array that converts initials to names and vice versa 
    array1.forEach(function(element) {
            var index = array2.findIndex(el => el.startsWith(element[0]));
            if (index > -1) {
                modified.push(array2[index]);
                array2.splice(index, 1);
            } else {
                modified.push(element);
            }
    });

console.log(modified); // should be the same as array1
console.log(modified.some(v => array2.includes(v))); //false should be true

我正在尝试检查 array2 中是否至少存在一个modified值。

相反也是错误的:

console.log(array2.some(v => modified.includes(v))); //false should be true

JSFiddle

Answer 1

问题出在这一行:

array2.splice(index, 1);

您实际上是从 array2 中删除找到的项目，所以当然，如果您稍后在 array2 中查找该项目，它就会获胜不会被发现。观察:

var array1 = ['fred'];
var array2 = ['sue', 'fred'];
var modified = [];

//create a new modified array that converts initials to names and vice versa 
array1.forEach(function(element) {
  var index = array2.findIndex(el => el.startsWith(element[0]));
  if (index > -1) {
    modified.push(array2[index]);
    array2.splice(index, 1);
  } else {
    modified.push(element);
  }
});

console.log("modified: ", ...modified); // should be the same as array1
console.log("array2: ", ...array2);     // array2 has been modified

一个快速解决方案是在开始修改数组 array2 之前对其进行克隆，然后在克隆上进行工作:

var array1 = ['fred'];
var array2 = ['sue', 'fred'];
var modified = [];
var filtered = [...array2];

//create a new modified array that converts initials to names and vice versa 
array1.forEach(function(element) {
  var index = filtered.findIndex(el => el.startsWith(element[0]));
  if (index > -1) {
    modified.push(array2[index]);
    filtered.splice(index, 1);
  } else {
    modified.push(element);
  }
});

console.log(modified.some(v => array2.includes(v))); // true