javascript - 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 的新手。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 & 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，我别无选择。需要帮忙。

标签： javascriptarraysloopsiterable

Answer 1

您可以更好地使用 aSet而不是直接使用数组arr2来查找匹配项，因为在 a 的情况下它将是恒定时间查找Set。

然后使用 aArray.prototype.filter过滤数组arr1并获取arr2.

最后Array.prototype.reduce将帮助您创建一个对象，其中键是单词，值是该单词中的最大值，arr1您可以使用Object.entries从返回的对象reduce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，flat()而Array.prototype. 所以你可以使用下面的编辑片段：

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);