JavaScript 在二维数组中存储数据答案

【问题标题】：JavaScript Storing Data in 2-D ArrayJavaScript 在二维数组中存储数据
【发布时间】：2018-02-12 23:24:44
【问题描述】：

我想检查数据数组中的元素，如果它在表格中包含该月份和数字，则显示为“O”。所以对于“Jan-1”，Array1[0][0] 应该显示为“O”，但下面的代码不起作用。有人能帮我吗？

var Data = ["Jan-1", "Feb-4", "Apr-5"];
var Month= ["Jan", "Feb", "Mar", "Apr", "May"];
var Number = ["1", "2", "3", "4", "5"];
var Array1 = [[]];

for (var k = 0; k < Data.length; k++) {
    var split = Data[k].split("-");
    for (var z = 0; z < Month.length; z++) {
        for (var s = 0; s < Number.length; s++) {
            if (Month[z] == split[0] && period[s] == split[1]) {
                Array1[z][s] = "O";
            } else {
                Array1[z][s] = "X";
            }             
        }
    }
}

number/month |  Jan  |  Feb  |  Mar  |  Apr  | May
------------------------------------------------------
     1       |   O   |   X   |   X   |   X   |   X  
     2       |   X   |   X   |   X   |   X   |   X  
     3       |   X   |   X   |   X   |   X   |   X  
     4       |   X   |   O   |   X   |   X   |   X  
     5       |   X   |   X   |   X   |   O   |   X

【问题讨论】：

这两个错了Array1[z][s] = "O";Array1[z][s] = "X";

标签： javascript arrays loops multidimensional-array tabular

【解决方案1】：

您应该循环遍历Number 和Month 数组，并且每次检查Month[j] + "-" + Number[i] 的组合是否在Data 数组中：

var Data = ["Jan-1", "Feb-4", "Apr-5"];

var Month = ["Jan", "Feb", "Mar", "Apr", "May"];
var Number = ["1", "2", "3", "4", "5"];

var result = [];


for (var i = 0; i < Number.length; i++) {                   // foreach number
  result[i] = [];                                           // create a row for this current number
  for (var j = 0; j <Month.length; j++) {                   // for each month
    if (Data.indexOf(Month[j] + "-" + Number[i]) !== -1) {  // check if the current combination (currentMonth-currentNumber) is in the Data array
      result[i][j] = "O";
    } else {
      result[i][j] = "X";
    }
  }
}

result.forEach(function(row) {
  console.log(row.join(" | "));
});

【讨论】：

【解决方案2】：

分析您的代码：

var Data = ["Jan-1", "Feb-4", "Apr-5"];
var Month= ["Jan", "Feb", "Mar", "Apr", "May"];
var Number = ["1", "2", "3", "4", "5"];
var Array1 = [  ["X","X","X","X","X"],  ["X","X","X","X","X"],  "X","X","X","X","X"],  ["X","X","X","X","X"],  ["X","X","X","X","X"]]; //Initialise all elements to "X" by default. We shall change only those indexes that match.

for (var k = 0; k < Data.length; k++) {
    var split = Data[k].split("-");
    for (var z = 0; z < Month.length; z++) {
        for (var s = 0; s < Number.length; s++) {
            if (Month[z] == split[0] && Number[s] == split[1]) {
                Array1[z][s] = "O";
            } else {
                //Array1[z][s] = "X"; Do not change here as the loop will go over the entire array once for each data. Hence previous matches would get lost!!
            }             
        }
    }
}

最后，console.table(Array1); 打印值。

评论：在初始化时，您将行声明为月，将列声明为天，但期望相反的输出。因此，要生成预期的输出，请在定义值时打印横向形式或更改：

if (Month[z] == split[0] && Number[s] == split[1]) {
            Array1[s][z] = "O";
        }

【讨论】：

【解决方案3】：

您可以获取具有给定数据的对象并迭代month和number以返回具有'O'或'X'信息的新数组。

var data = ["Jan-1", "Feb-4", "Apr-5"],
    month = ["Jan", "Feb", "Mar", "Apr", "May"],
    number = ["1", "2", "3", "4", "5"],
    result = [],
    hash = Object.create(null);

data.forEach(function(s) {
    var [m, d] = s.split('-');
    hash[m] = hash[m] || {};
    hash[m][d] = true;
});

result = month.map(function (m, z) {
    return number.map(function (s) {
        return (hash[m] || {})[s] ? 'O' : 'X';
    });
});

console.log(result);

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【讨论】：