使用javascript在对象中搜索数组而不使用循环

Question

我想找到一个更好、更优雅的解决方案，而不是使用 2 次迭代来查找对象中的值。我查看了在数组中搜索的函数，例如 find、indexOf，但想知道我可以在不使用 for 或 forEach 的情况下搜索对象。

假设我想找到所有游泳者。到目前为止我得到了什么。

members = [
  { name: "Sue", hobby: ["Swimming", "Running", "Dancing"] },
  { name: "Sam", hobby: ["Cars", "Travelling"] },
  { name: "John", hobby: ["Reading", "Swimming"] },
  { name: "Rob", hobby: ["Running", "Coding"] },      
];
function findSwimmers(members, hobby) {      
  let swimmers = [];      
  members.forEach(function (e){        
    e.hobby.forEach(function (element){          
      if (element.toLowerCase() === hobby.toLowerCase()) {
        swimmers.push(e);
      }
    });
  });
  return swimmers;
}

Answer 1

您可以将filter 与some 结合使用，方法是将回调 函数作为参数传递。

let members = [
  { name: "Sue", hobby: ["Swimming", "Running", "Dancing"] },
  { name: "Sam", hobby: ["Cars", "Travelling"] },
  { name: "John", hobby: ["Reading", "Swimming"] },
  { name: "Rob", hobby: ["Running", "Coding"] },      
];

function findSwimmers(members, hobby) {      
  return members_swimming = members.filter(({hobby}) => hobby.some(item => item == "Swimming"));
                                
}

console.log(findSwimmers(members, "Swimming"));

Answer 2

你可以用一行来实现：

let swimmers = members.filter(person => person.hobby.includes('Swimming'));

Answer 3

您可以将Array.filter() 和Array.map() 用于解构：

var members = [
  { name: "Sue", hobby: ["Swimming", "Running", "Dancing"] },
  { name: "Sam", hobby: ["Cars", "Travelling"] },
  { name: "John", hobby: ["Reading", "Swimming"] },
  { name: "Rob", hobby: ["Running", "Coding"] },      
];
var swimmers = members.filter(({hobby}) => hobby.includes('Swimming')).map(({name}) => name);
console.log(swimmers)

Answer 4

任何解决方案都将涉及循环。唯一的问题是循环是在你的代码还是你调用的函数中。

在您的情况下，filter 和 some 似乎是可以使用的工具：

function findSwimmers(members, hobby) {
  hobby = hobby.toLowerCase();
  return members.filter(member =>
    member.hobby.some(h => h.toLowerCase() === hobby)
  );
}

您可能需要不同的函数名称。 :-)

现场示例：

const members = [
  { name: "Sue", hobby: ["Swimming", "Running", "Dancing"] },
  { name: "Sam", hobby: ["Cars", "Travelling"] },
  { name: "John", hobby: ["Reading", "Swimming"] },
  { name: "Rob", hobby: ["Running", "Coding"] },      
];
function findSwimmers(members, hobby) {
  hobby = hobby.toLowerCase();
  return members.filter(member =>
    member.hobby.some(h => h.toLowerCase() === hobby)
  );
}

console.log(findSwimmers(members, "swimming"));

.as-console-wrapper {
  max-height: 100% !important;
}

请注意，您的原始代码在找到爱好匹配时并没有停止循环（如果爱好数组中有重复的条目，则会多次添加相同的成员 - 但可能并不意味着在该数组中重复）。

您也可以将String#localeCompare 与sensitivity: "base" 选项一起使用，而不是比较小写字符串。

Answer 5

第一个解决方案：有点破解但是，嘿，没有循环

第二种解决方案：使用Array.findIndex 可能会减少内部循环中的迭代次数（嗯，至少给你一些概率更少的迭代）因为（MDN）：

...如果找到这样的元素，findIndex 立即返回索引 对于那个迭代......

const findMembersWithValueForKey = (members, key, value) => 
  members.filter(member => member[key] instanceof Array
    && `#${member[key].join("#")}#`.toLowerCase()
         .indexOf(`#${value.toLowerCase()}#`) > -1
    || false);
    
const findMembersWithValueForKey2 = (members, key, value) => {
    value = value.toLowerCase();
    return members.filter(member => 
      member[key] instanceof Array
      && member[key].findIndex(v => v.toLowerCase() === value) > -1
      || false);
};
    
const members = [
  { name: "Sue", hobby: ["Swimming", "Running", "Dancing"] },
  { name: "Sam", hobby: ["Cars", "Travelling"] },
  { name: "John", hobby: ["Reading", "Swimming"] },
  { name: "Rob", hobby: ["Running", "Coding"] },      
];

console.log(findMembersWithValueForKey(members, "hobby", "swimming"));

console.log(findMembersWithValueForKey2(members, "hobby", "swimming"));