在比较两个数组javascript时找到第一个匹配值

I'm trying to find a match between the breeds array and the dogs array. How can I return the first match that is found. This is what i have that doesn't seem to work and I cant quite find the error. Thanks

我正在尝试找到品种阵列和狗阵列之间的匹配。如何返回找到的第一个匹配项。这是我所拥有的似乎不起作用，我不能找到错误。谢谢

var dogs = ["pug", "husky", "hound", "poodle"];

function findTheBreed(dogs) {
    var breeds = ["terrier", "mix", "lab", "hound"];
 for(let i = 0; i < dogs.length; i++) {
   for (let b = 0; b < breeds.length; b++) {
     if(dogs[i] === breeds[b]) {
       return breeds[b]
      } else {
       return "no match"
      }
    }
   }
 }

5 个解决方案

#1

You're returning false on the first non-match, whereas you should continue iterating.

你在第一次不匹配时返回false，而你应该继续迭代。

move the return "no match" to after everything has been iterated through.

一切都经过迭代后，将返回“不匹配”移动到。

jsfiddle

的jsfiddle

You can alternatively also use the indexOf here:

您也可以在这里使用indexOf：

function findTheBreed(dogs) {
    var breeds = ["terrier", "mix", "lab", "hound"];
    for(let i = 0; i < dogs.length; i++) {
      if(breeds.indexOf(dogs[i]) != -1){
      return dogs[i];
    }
  }   
  return "no match";
}

jsfiddle

的jsfiddle

One other (in my opinion the best) route would be to use the Array.filter functionality and retrieve the entire array of matches.

另一个（在我看来最好的）路线是使用Array.filter功能并检索整个匹配数组。

  var breeds = ["terrier", "mix", "lab", "hound", "snickerdoodle"];
    var filtered = breeds.filter(
    function (elem) {
      return dogs.indexOf(elem) != -1
    }
  );
  return filtered;

jsfiddle

的jsfiddle

#2

It would be simpler to turn one array into a dictionary and do a direct lookup.

将一个数组转换为字典并进行直接查找会更简单。

For example, if dog is turned into a dictionary

例如，如果将狗变成字典

var dictionary = {};
for(var i=0;i<dogs.length;i++){
  dictionary[dogs[i]] = true;
}

for(var i=0;i<breeds.length;i++){
  if(dictionary[breeds[i]] === true) return true; //found match
}

#3

Or shorter:

或更短：

var findtheBreed=dogs=>["terrier","mix","lab","hound"].find(breed=>dogs.find(dog=>dog===breed))||"no match";

alert(findtheBreed(["pug","husky","hound","poodle"]));

http://jsbin.com/mubatiyebo/edit?console

http://jsbin.com/mubatiyebo/edit?console

using Array.prototype.find , Arrow functions and the brilliant OR operator...

使用Array.prototype.find，Arrow函数和出色的OR运算符......

#4

To fix your implementation, move the return "no match" out of the loop body to the end of your findTheBreed function.

要修复您的实现，请将循环体中的返回“不匹配”移动到findTheBreed函数的末尾。

However, you can find the first match faster - in constant time - by leveraging constant time Set.has lookup:

但是，通过利用常量时间Set.has查找，您可以更快地找到第一个匹配 - 在恒定时间内：

function match(dogs, breeds) {
  return dogs.find(Set.prototype.has, new Set(breeds));
}

let dogs = ["pug", "husky", "hound", "poodle"];
let breeds = ["terrier", "mix", "lab", "hound"];

console.log(match(dogs, breeds));

Also, I think you should consider renaming your arrays as there is currently no distinction between a dog and a breed. Also, you might consider declaring the breeds as a Set from the beginning on to avoid redundant conversion.

另外，我认为您应该考虑重命名数组，因为目前狗和品种之间没有区别。此外，您可以考虑从一开始就将品种声明为Set，以避免冗余转换。

#5

You're returning out of the function too early. If you want the same data structure, maybe try this:

你太早退出了这个功能。如果你想要相同的数据结构，也许试试这个：

var dogs = ["pug", "husky", "hound", "poodle"];
function findBreed(dogs) {
    var breeds = ["terrier", "mix", "lab", "hound"];
    for (let i = 0; i < dogs.length; i++){
        var dog = dogs[i];
        var successIdx = breeds.indexOf(dog)
        if (successIdx !== -1){
            return dog
        }

    }
    return "no match"
 }