Question

我正在尝试使用1d数组的内容创建2d数组。 2d数组具有正确的行数和列数，但是2d数组的每个索引都包含原始1d数组的全部。

如何旋转一维数组：

var OneD = [ "Bristol", "Cardiff", "Birmingham",
             "Luton", "Swansea", "Aberdeen",
             "Birmingham", "Manchester", "Southampton", 
             "Chester", "Swansea", "Brighton",
             "Portsmouth", "Bournemouth", "Glasgow", 
             "Newcastle", "Cardiff", "Bristol"];

进入2d数组：

twoD = [
        ["Bristol", "Cardiff", "Birmingham", "Luton", "Swansea", "Aberdeen"],
        ["Birmingham", "Manchester", "Southampton", "Chester", "Swansea", "Brighton"],
        ["Portsmouth", "Bournemouth", "Glasgow", "Newcastle", "Cardiff", "Bristol"]
 ];

我的代码

var twoD = [];
var rows = 3;
var cols = 6;

for (var i = 0; i < rows; i++) {
    twoD.push( [] );
}

for (var i = 0; i < rows; i++) {
    for (var j =  0; j < cols; j++) {
        twoD[i].push(oneD);
    }
}

console.log(twoD);

Answer 1

使用单个for循环。在每次迭代中，将cols编号添加到i，并从OneD数组中切出i和i + cols之间的项。将切片的项目推入twoD数组中。

var OneD = ["Bristol","Cardiff","Birmingham","Luton","Swansea","Aberdeen","Birmingham","Manchester","Southampton","Chester","Swansea","Brighton","Portsmouth","Bournemouth","Glasgow","Newcastle","Cardiff","Bristol"];

var twoD = [];
var cols = 6;

for(var i = 0; i < OneD.length; i += cols) {
  twoD.push(OneD.slice(i, i + cols));
}

console.log(twoD);

Answer 2

这样可以做到。但请务必处理遗漏的物品。这里的数组长度正好是6个项目的3个部分。

var OneD = ["Bristol", "Cardiff", "Birmingham",
  "Luton", "Swansea", "Aberdeen",
  "Birmingham", "Manchester", "Southampton",
  "Chester", "Swansea", "Brighton",
  "Portsmouth", "Bournemouth", "Glasgow",
  "Newcastle", "Cardiff", "Bristol"
];

var TwoD = [];
var cols = 6;
var rows = parseInt(OneD.length / cols, 10); // rows = 3
for (var i = 0; i < rows; i++) {
  TwoD.push(OneD.slice(i * cols, (i * cols + cols)));
}
console.log(TwoD);

（或）

var OneD = ["Bristol", "Cardiff", "Birmingham",
  "Luton", "Swansea", "Aberdeen",
  "Birmingham", "Manchester", "Southampton",
  "Chester", "Swansea", "Brighton",
  "Portsmouth", "Bournemouth", "Glasgow",
  "Newcastle", "Cardiff", "Bristol"
];

var TwoD = OneD.join(',').match(/([^,]*,[\w]*){5}[,]*/g).map(function(i) {return i.replace(/,$/, '').split(/,/);});

console.log(TwoD);

Answer 3

您的第二个循环每次都推入所有OneD。您需要索引数组。

var oneD = ["Bristol","Cardiff","Birmingham","Luton","Swansea","Aberdeen","Birmingham","Manchester","Southampton","Chester","Swansea","Brighton","Portsmouth","Bournemouth","Glasgow","Newcastle","Cardiff","Bristol"];

var twoD = [];
var cols = 6;
var rows = 3;

for (var i = 0; i < rows; i++) {
    twoD.push( [] );
}

for (var i = 0; i < rows; i++) {
    for (var j =  0; j < cols; j++) {
        twoD[i].push(oneD[i * cols + j]);
    }
}

console.log(twoD);

您还可以将两个循环组合为一个循环。并且通过使外循环按列数递增来摆脱索引计算。下面的版本也不要求1d数组的长度恰好为rows * cols。

var oneD = ["Bristol","Cardiff","Birmingham","Luton","Swansea","Aberdeen","Birmingham","Manchester","Southampton","Chester","Swansea","Brighton","Portsmouth","Bournemouth","Glasgow","Newcastle","Cardiff","Bristol"];

var twoD = [];
var cols = 6;
var rows = 3;

for (var i = 0; i < oneD.length; i += cols) {
    var row = [];
    var maxColIndex = Math.min(i+cols, oneD.length);
    for (var j =  i; j < maxColIndex; j++) {
        row.push(oneD[j]);
    }
    twoD.push(row);
}

console.log(twoD);

将数据从一维数组推送到二维数组

3 个答案: