Question

我有一个数组，其中每一行都是一个包含2个值的数组，S / N和Description。

 array1 = [["xx9","Big Car"],["xx1","Small Car"],["xx9","Big Car"],["xx9"," Big Car"]];

正如您所看到的那样，有重复的内容。所以我编写的代码创建了一个count数组，该数组包含键值对并计算每个项目。计数长度为1。

count = [xx1: 1
         xx9: 3]

所以现在我想将两个数组合并为一个包含数量的新数组，在本例中最终输出如下所示。请注意，我还必须删除array1中的重复项。订单并不重要。

 final_array = [["1","xx1","Small Car"],["3",xx9","Big Car"]];

到目前为止，这是我的JS代码，但基本上 .forEach 循环不会对count数组起作用。任何人都可以帮助我，以便此代码适用于键值数组。

JS / Jquery的：

//Pull the Cart Data & Orangize
var savedList = JSON.parse(localStorage.getItem("partList"));
        console.log(savedList);
        determineQuantity(savedList); //Count quantity of each item

//Count the Quantity of each item in saved list & create key-value pairs
function determineQuantity(savedList){
    var newList = [];
    var count = [];
    for (var i=0, j = savedList.length; i<j; i++){
        count[savedList[i][0]] = (count[savedList[i][0]] || 0) + 1;
    };
        console.log(count);
    //Combine Quantity Array with SavedList Array
    count.forEach(function(item,index){ 
                    console.log("index = " + index + "  item = " + item);
          savedList.forEach(function(row){
                if ($.inArray(item,row == 0)){ //if found in this row at index 0
                     var new_part = [count[index],row[0],row[1]];
                            console.log("new_part = " + new_part);
                     newList.push(new_part);
                };  
        });
    });
    console.log(newList);
};

Answer 1

这是怎么回事。

var test = function(){
    var array = [["xx9","Big Car"],["xx1","Small Car"],["xx9","Big Car"],["xx9"," Big Car"]];
    var count = {xx1: 1, xx9: 3};
    var map = {};
    array.forEach(function(item){map[item[0]] = item[1]});
    var newArray = [];
    for(var key in count){
        if(!map[key])
            continue;
        newArray.push([count[key], key, map[key]]);
    }
    console.log(newArray);
}

首先，你算是一个对象，而不是一个数组。然后我们不需要每个

将2D数组与键值数组相结合

1 个答案: