当另一个数组有一个数字（值）时如何匹配2个数组的字符串值

Question

我是 Javascript 新手。我试图获取数组（arr1）的Math.max（...arr1 [i] [1]），只有当它与第二个数组（arr2）匹配时。然后，将其推送到数组（arr3）上，结果应该没有重复值。

我尝试迭代两个数组（arr1 和 arr2），然后使用“if 语句”来匹配它们。

var arr1 = [ [ 'abandon', -2 ],
[ 'abandon', 1 ],
[ 'abandon', -2 ],
[ 'abduct', 1 ],
[ 'abduct', -2 ],
[ 'abduct', -2 ],
[ 'abhor', -3 ],
[ 'abhor', 1 ],
[ 'abhor', -1 ],
[ 'abil', 2 ],
[ 'abil', 4 ] ];


var arr2 = [ [ 'abandon' ],
[ 'abil' ],
[ 'abhor' ],
[ 'abduct' ],
['test'],
['hey'],
['testAgain'],
['array']];

var arr3 = [];

const mapping = arr2.map(word => {
    return word
})

for(var i = 0; i < arr1.length; i++){

    if(arr1[i][0] === mapping){
        arr3.push(Math.max(...arr1[i][1]))
    }

}
let arr4 = [...new Set(arr3)]

//example result:

var arr4 = [[abandon, -2],
[abduct, -2],
[abhor, -3],
[abil, 4]... and so on]

我知道我做错了什么，而且我别无选择。需要帮助。

Answer 1

您可以更好地直接使用Set 而不是arr2 之类的数组来查找匹配项，因为在Set 的情况下它将是恒定时间查找。

然后使用Array.prototype.filter 过滤数组arr1 并获取arr2 中的那些数组。

最后Array.prototype.reduce 将帮助您创建一个对象，其中键是单词，值是arr1 中该单词的最大值，您可以使用从reduce 返回的对象中的Object.entries 来获取二维数组形式的数据：

var arr1 = [ [ 'abandon', -2 ],
[ 'abandon', 1 ],
[ 'abandon', -2 ],
[ 'abduct', 1 ],
[ 'abduct', -2 ],
[ 'abduct', -2 ],
[ 'abhor', -3 ],
[ 'abhor', 1 ],
[ 'abhor', -1 ],
[ 'abil', 2 ],
[ 'abil', 4 ] ];


var arr2 = [ [ 'abandon' ],
[ 'abil' ],
[ 'abhor' ],
[ 'abduct' ],
['test'],
['hey'],
['testAgain'],
['array']];

var lookup = new Set(arr2.flat());
var mapping = arr1.filter(([word, val]) => lookup.has(word));
var data = Object.entries(mapping.reduce((acc, o, i) => {
    if(acc[o[0]]){
        acc[o[0]] = Math.max(o[1], acc[o[0]]);
    }else{
        acc[o[0]] = o[1];
    }
    return acc;
},{}));

console.log(data);

编辑

形成您的评论我假设您使用的是旧版本的节点运行时，其中flat() 不存在于Array.prototype 中。所以你可以使用下面编辑的sn-p：

var arr1 = [ [ 'abandon', -2 ],
[ 'abandon', 1 ],
[ 'abandon', -2 ],
[ 'abduct', 1 ],
[ 'abduct', -2 ],
[ 'abduct', -2 ],
[ 'abhor', -3 ],
[ 'abhor', 1 ],
[ 'abhor', -1 ],
[ 'abil', 2 ],
[ 'abil', 4 ] ];


var arr2 = [ [ 'abandon' ],
[ 'abil' ],
[ 'abhor' ],
[ 'abduct' ],
['test'],
['hey'],
['testAgain'],
['array']];

//flatten using Array.prototype.concat
var lookup = new Set([].concat.apply([], arr2)); 
//If Set doesn't work use the array, but this will not be a constant time lookup
//var lookup = [].concat.apply([], arr2);  

var mapping = arr1.filter(([word, val]) => lookup.has(word));
//If you are not using Set and going with an array, use Array.prototype.includes, so search won't be O(1)
//var mapping = arr1.filter(([word, val]) => lookup.includes(word));

var data = Object.entries(mapping.reduce((acc, o, i) => {
    if(acc[o[0]]){
        acc[o[0]] = Math.max(o[1], acc[o[0]]);
    }else{
        acc[o[0]] = o[1];
    }
    return acc;
},{}));

console.log(data);

Answer 2

我会尽可能少地使用循环。

var arr1 = [ [ 'abandon', -2 ], [ 'abandon', 1 ], [ 'abandon', -2 ], [ 'abduct', 1 ], [ 'abduct', -2 ], [ 'abduct', -2 ], [ 'abhor', -3 ], [ 'abhor', 1 ], [ 'abhor', -1 ], [ 'abil', 2 ], [ 'abil', 4 ] ];

var arr2 = [ [ 'abandon' ], [ 'abil' ], [ 'abhor' ], [ 'abduct' ], ['test'], ['hey'], ['testAgain'], ['array']];


function getHighestResults(dataArr, filterArr) {
  var name = '';    // higher readability in code and created outside 
  var number = -1;  // for loops to avoid creating variables inside of them
  
  function existsInObject(obj, name) {
    return obj.hasOwnProperty(name);
  }
  
  function filterBasedOn(arr, obj) {
    var filteredArr = [];
    
    for (let i = 0; i < arr.length; i++) {
      name = arr[i][0];
      
      if (existsInObject(obj, name)) {
        filteredArr.push([name, obj[name]]);
      }
    }
    
    return filteredArr;
  }
  
  function getHighestValuesAsObj(arr) {
    var dataObj = {};

    for (let i = 0; i < arr.length; i++) {
      name = arr[i][0];
      number = arr[i][1];
  
      if (!existsInObject(dataObj, name) || dataObj[name] < number) {
        dataObj[name] = number;
      }
    }
    
    return dataObj;
  }
    
  return filterBasedOn(filterArr, getHighestValuesAsObj(dataArr));
}


console.log(getHighestResults(arr1, arr2));