fortran程序如何将两个数组中的元素，放入一个数组？
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参考技术A 这个简单！方法1：执行两个循环将两个给定数组逐个赋值到合并的数组中。
方法2：利用公用块，即将两个给定的数组和合并后的数组定义到同名的公用块中即可。
如何将数组元素组合成一个新数组？
【中文标题】如何将数组元素组合成一个新数组？【英文标题】：How to combine array elements into a new array? 【发布时间】：2019-12-05 08:40:05 【问题描述】：Javascript 新手（以及一般的编码，Excel 函数之外），希望有人能帮助我指出正确的方向。我在网上搜索过，但没有找到明确的答案。
我有两个数组 - 一个包含星期几，一个包含人们的名字和姓氏。我从带有人名的数组中选择一个随机索引号，并尝试将名字和姓氏分配给一天，然后将所有三个值放入一个新数组中。然后我将这个最终数组打印到 HTML 表中。
包含天数的数组是一维的，只是从星期日到星期六。我理解的包含人名的数组实际上是二维的，即他们的名字和姓氏。
我通过硬编码第三个数组来完成这项工作，但我想改为动态创建数组。我似乎无法让 array.push() 方法按我的意愿工作，我希望有人愿意提供帮助。
注意 - 我通过创建另一个数组来选择随机索引号，该数组在 0 和 people 数组的长度之间选择七个随机数。 push 方法在这种情况下可以正常工作。
我尝试过使用 array.push() 方法，但似乎无法按照我的预期工作。它要么返回一个空白值，要么返回一个看起来很乱的东西，尽管我认为它是来自 people 数组的随机字母。
//This is my working code. The script tag is after the body elements because I need to create some of the body elements before I run the script.
<body>
<table id="Week1Table" class="center">
<tr>
<th class="Week1">Week 1</th>
<th class="Week1">First Name</th>
<th class="Week1">Last Name</th>
</tr>
</table>
<table id="Week2Table" class="center">
<tr>
<th class="Week2">Week 2</th>
<th class="Week2">First Name</th>
<th class="Week2">Last Name</th>
</tr>
</table>
<script>
var people = [
["Andre", "the Giant"],
["Randy", "Savage"],
["Steve", "Austin"],
["The", "Rock"],
["Man", "Kind"],
["Steel", "Cage"],
["I", "Quit"]
];
var week1schedule = [
"Sunday",
"Monday",
"Tuesday",
"Wednesday",
"Thursday",
"Friday",
"Saturday"
];
var week2schedule = week1schedule;
var peoplePicker = []
while(peoplePicker.length < week1schedule.length*2)
var r = Math.floor(Math.random()*people.length);
if(peoplePicker.indexOf(r) === -1) peoplePicker.push(r);
;
//This is the final array, and the one that I would like to create dynamically rather than statically.
var peopleRandomizerTable = [];
peopleRandomizerTable[0] = week1schedule[0];
peopleRandomizerTable[1] = people[peoplePicker[0]][0];
peopleRandomizerTable[2] = people[peoplePicker[0]][1];
peopleRandomizerTable[3] = week1schedule[1];
peopleRandomizerTable[4] = people[peoplePicker[1]][0];
peopleRandomizerTable[5] = people[peoplePicker[1]][1];
peopleRandomizerTable[6] = week1schedule[2];
peopleRandomizerTable[7] = people[peoplePicker[2]][0];
peopleRandomizerTable[8] = people[peoplePicker[2]][1];
peopleRandomizerTable[9] = week1schedule[3];
peopleRandomizerTable[10] = people[peoplePicker[3]][0];
peopleRandomizerTable[11] = people[peoplePicker[3]][1];
peopleRandomizerTable[12] = week1schedule[4];
peopleRandomizerTable[13] = people[peoplePicker[4]][0];
peopleRandomizerTable[14] = people[peoplePicker[4]][1];
peopleRandomizerTable[15] = week1schedule[5];
peopleRandomizerTable[16] = people[peoplePicker[5]][0];
peopleRandomizerTable[17] = people[peoplePicker[5]][1];
peopleRandomizerTable[18] = week1schedule[6];
peopleRandomizerTable[19] = people[peoplePicker[6]][0];
peopleRandomizerTable[20] = people[peoplePicker[6]][1];
peopleRandomizerTable[21] = week2schedule[0];
peopleRandomizerTable[22] = people[peoplePicker[7]][0];
peopleRandomizerTable[23] = people[peoplePicker[7]][1];
peopleRandomizerTable[24] = week2schedule[1];
peopleRandomizerTable[25] = people[peoplePicker[8]][0];
peopleRandomizerTable[26] = people[peoplePicker[8]][1];
peopleRandomizerTable[27] = week2schedule[2];
peopleRandomizerTable[28] = people[peoplePicker[9]][0];
peopleRandomizerTable[29] = people[peoplePicker[9]][1];
peopleRandomizerTable[30] = week2schedule[3];
peopleRandomizerTable[31] = people[peoplePicker[10]][0];
peopleRandomizerTable[32] = people[peoplePicker[10]][1];
peopleRandomizerTable[33] = week2schedule[4];
peopleRandomizerTable[34] = people[peoplePicker[11]][0];
peopleRandomizerTable[35] = people[peoplePicker[11]][1];
peopleRandomizerTable[36] = week2schedule[5];
peopleRandomizerTable[37] = people[peoplePicker[12]][0];
peopleRandomizerTable[38] = people[peoplePicker[12]][1];
peopleRandomizerTable[39] = week2schedule[6];
peopleRandomizerTable[40] = people[peoplePicker[13]][0];
peopleRandomizerTable[41] = people[peoplePicker[13]][1];
var table1 = document.getElementById("Week1Table");
var table2 = document.getElementById("Week2Table");
var peopleRandomizerCount = 0;
//This populates the two tables. Two weeks; one table for each week.
for(var rowCount = 0; rowCount < 7; rowCount++)
// create a new row
var newRow = table1.insertRow(table1.length);
for(var columnCount = 0; columnCount < 3; columnCount++)
// create a new cell
var cell = newRow.insertCell(columnCount);
// add value to the cell
cell.innerHTML = peopleRandomizerTable[peopleRandomizerCount];
cell.className = "people";
peopleRandomizerCount++;
for(var rowCount = 0; rowCount < 7; rowCount++)
// create a new row
var newRow = table2.insertRow(table1.length);
for(var columnCount = 0; columnCount < 3; columnCount++)
// create a new cell
var cell = newRow.insertCell(columnCount);
// add value to the cell
cell.innerHTML = peopleRandomizerTable[peopleRandomizerCount];
cell.className = "people";
peopleRandomizerCount++;
</script>
</body>
【问题讨论】：
您可以使用 2 个 for 循环在代码中而不是手动创建该表，但我强烈建议您使用对象重写您的二维数组，而不是像 bhuvnesh 回答的那样。 另外，您不必创建另一个数组来存储随机不定值。在分配给表时，您可以通过创建一个函数并调用它（而不是一个额外的数组浮动）从随机索引中选择数据。 【参考方案1】：名字和姓氏不需要使用二维数组，
您可以简单地使用包含名字和姓氏的一维对象数组，如下所示：
let people = [
"firstName": "Andre",
"lastName": "the Giant"
,
"firstName": "Randy",
"lastName": "Savage"
]
现在我猜你可能知道如何组合 2 个一维数组。如果没有请告诉我
【讨论】：
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