使用其他数组的元素.filter（）数组 [英] .filter() array using another array's elements

问题描述

我的人的名字与他们的语文知识沿着阵列。我想要做的就是通过一个过滤器上的语言列，并过滤掉不符合任何结果。

这是样品阵列

  myarray的无功= [[Steppen，西班牙波兰]，
                  [狼，西班牙波兰菲律宾语]，
                  [阿曼达，西班牙]
                  [阿达，波兰]，
                  [朗达，西班牙他加禄语]];
 

至于传递过滤器，它可以是一种语言或多个。即使从过滤器一种语言相匹配 - 结果应该返回。因此，例如，他加禄语的过滤器应该返回 - 狼和朗达。 西班牙波兰的过滤器应该回报大家 - 有无论是在西班牙或波兰的比赛

我写的过滤功能，但由于某种原因，它被卡住，当我通过过滤器菲律宾语只迭代到第二单元阵列（西班牙波兰菲律宾语），并重演多次，而不是前进。

我在做什么错了，我应该是迭代不同？

  VAR userPassedFilter =新的Array（）;
 userPassedFilter [0] =他加禄语; newArray = consolidatedFilters（myarray的，userPassedFilter）;
 的console.log（newArray）; 功能consolidatedFilters（passedArray，passedFilter）
 {
 VAR filteredArray = passedArray.filter（
    功能（EL）
    {
        对于（VAR I = 0; I＆LT; passedArray.length;我++）
         {
            的console.log（我是+ I）;
             对（在passedFilter变种j）条
            {
                （顺利通过过滤器J+ passedFilter [J]）的console.log;
                的console.log（传递的数组我+ passedArray [I] [1]）;
                的console.log（字符串搜索+ passedArray [I] [1] .search（passedFilter [J]））;                如果（passedArray [I] [1] .search（passedFilter [j]的）！=  -  1）
                {
                    返回true;
                }
            }
        }
         返回false;
    }
 ）;
 返回filteredArray;
 }
 

解决方案

对我来说，好像你正在做的有点太复杂了。

1. 遍历三次（过滤器，为循环，在循环）。


2. 使用了数组的在循环。


3. 同时使用新阵列和 [...] 。

我更新了这一点，它看起来这是你想要什么： http://jsfiddle.net/pimvdb/RQ6an/

  myarray的无功= [[Steppen，西班牙波兰]，
              [狼，西班牙波兰菲律宾语]，
              [阿曼达，西班牙]
              [阿达，波兰]，
              [朗达，西班牙他加禄语]];VAR userPassedFilter = [菲律宾语];newArray = consolidatedFilters（myarray的，userPassedFilter）;
的console.log（newArray）;功能consolidatedFilters（passedArray，passedFilter）{
    VAR filteredArray = passedArray.filter（
    功能（EL）{//对每个人执行
        对于（VAR I = 0; I＆LT; passedFilter.length;我++）{//对滤波器迭代
            如果（EL [1] .indexOf（passedFilter [I]）！=  -  1）{
                返回true; //如果这个人知道这种语言
            }
        }
        返回false;
    }
    ）;
    返回filteredArray;
}
 

I have an array of people's names along with their knowledge of languages. What I want to do is pass a filter onto the language column and filter out any results that don't match.

This is the sample array

   var myArray = [["Steppen", "Spanish Polish"],
                  ["Wolf", "Spanish Polish Tagalog"],
                  ["Amanda", "Spanish"],
                  ["Ada", "Polish"],
                  ["Rhonda", "Spanish Tagalog"]];

As far as passing in filters, it could be either one language or many. Even if one language from a filter matches - the result should be returned. So for example, a filter of "Tagalog" should return - Wolf and Rhonda. A filter of "Spanish Polish" should return everyone - there's either a match in Spanish or Polish.

I wrote the filter function but for some reason it's getting stuck, when I pass the filter "Tagalog" it only iterates to the second cell in the array (Spanish Polish Tagalog) and repeats itself multiple times instead of going forward.

What am I doing wrong, should I be iterating differently?

 var userPassedFilter = new Array();
 userPassedFilter[0] = "Tagalog";

 newArray = consolidatedFilters(myArray, userPassedFilter);
 console.log(newArray);

 function consolidatedFilters(passedArray, passedFilter)
 {
 var filteredArray = passedArray.filter(    
    function(el)
    {
        for (var i = 0; i < passedArray.length; i++)
         {
            console.log("i is " + i);
             for (var j in passedFilter)
            {
                console.log("Passed Filter j " + passedFilter[j]);
                console.log("Passed Array  i " + passedArray[i][1]);        
                console.log("String Search " + passedArray[i][1].search(passedFilter[j]));

                if (passedArray[i][1].search(passedFilter[j]) != -1)
                {
                    return true;
                }
            }           
        }
         return false;
    }
 );     
 return filteredArray;
 }

解决方案

To me it seems like you're making it a little too complicated.

1. Iterating three times (filter, for loop, for in loop).
2. Using a for in loop for an array.
3. Using both new Array and [...].

I updated it a little and it looks like this is what you want: http://jsfiddle.net/pimvdb/RQ6an/.

var myArray = [["Steppen", "Spanish Polish"],
              ["Wolf", "Spanish Polish Tagalog"],
              ["Amanda", "Spanish"],
              ["Ada", "Polish"],
              ["Rhonda", "Spanish Tagalog"]];

var userPassedFilter = ["Tagalog"];

newArray = consolidatedFilters(myArray, userPassedFilter);
console.log(newArray);

function consolidatedFilters(passedArray, passedFilter) {
    var filteredArray = passedArray.filter(
    function(el) { // executed for each person
        for (var i = 0; i < passedFilter.length; i++) { // iterate over filter
            if (el[1].indexOf(passedFilter[i]) != -1) {
                return true; // if this person knows this language
            }
        }
        return false;
    }
    );     
    return filteredArray;
}

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