在 JavaScript 中转置二维数组
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Transposing a 2D-array in JavaScript
提问by ckersch
I've got an array of arrays, something like:
我有一组数组，例如：
[
[1,2,3],
[1,2,3],
[1,2,3],
]
I would like to transpose it to get the following array:
我想转置它以获得以下数组：
[
[1,1,1],
[2,2,2],
[3,3,3],
]
It's not difficult to programmatically do so using loops:
使用循环以编程方式这样做并不困难：
function transposeArray(array, arrayLength){
var newArray = [];
for(var i = 0; i < array.length; i++){
newArray.push([]);
};
for(var i = 0; i < array.length; i++){
for(var j = 0; j < arrayLength; j++){
newArray[j].push(array[i][j]);
};
};
return newArray;
}
This, however, seems bulky, and I feel like there should be an easier way to do it. Is there?
然而，这看起来很笨重，我觉得应该有一种更简单的方法来做到这一点。在那儿？
回答by Fawad Ghafoor
array[0].map((_, colIndex) => array.map(row => row[colIndex]));
mapcalls a providedcallbackfunction once for each element in an array, in order, and constructs a new array from the results.callbackis invoked only for indexes of the array which have assigned values; it is not invoked for indexes which have been deleted or which have never been assigned values.
callbackis invoked with three arguments: the value of the element, the index of the element, and the Array object being traversed.[source]
mapcallback按顺序为数组中的每个元素调用一次提供的函数，并根据结果构造一个新数组。callback仅对已分配值的数组索引调用；不会为已删除或从未分配值的索引调用它。
callback使用三个参数调用：元素的值、元素的索引和被遍历的 Array 对象。[来源]
回答by Joe
回答by Mahdi Jadaliha
here is my implementation in modern browser (without dependency):
这是我在现代浏览器中的实现（无依赖）：
transpose = m => m[0].map((x,i) => m.map(x => x[i]))
回答by marcel
shortest way with lodash/underscoreand es6:
使用lodash/underscore和的最短方法es6：
_.zip(...matrix)
where matrixcould be:
matrix可能在哪里：
const matrix = [[1,2,3], [1,2,3], [1,2,3]];
回答by Yangshun Tay
Many good answers here! I consolidated them into one answer and updated some of the code for a more modern syntax:
这里有很多很好的答案！我将它们合并为一个答案并更新了一些代码以获得更现代的语法：
One-liners inspired by Fawad Ghafoorand óscar Gómez Alca?iz
One-liners 的灵感来自Fawad Ghafoor和óscar Gómez Alca?iz
function transpose(matrix) {
return matrix[0].map((col, i) => matrix.map(row => row[i]));
}
function transpose(matrix) {
return matrix[0].map((col, c) => matrix.map((row, r) => matrix[r][c]));
}
Functional approach style with reduce by Andrew Tatomyr
函数式方法风格 with reduce by Andrew Tatomyr
function transpose(matrix) {
return matrix.reduce((prev, next) => next.map((item, i) =>
(prev[i] || []).concat(next[i])
), []);
}
Lodash/Underscore by marcel
Lodash /下划线的烫发
function tranpose(matrix) {
return _.zip(...matrix);
}
// Without spread operator.
function transpose(matrix) {
return _.zip.apply(_, [[1,2,3], [1,2,3], [1,2,3]])
}
Vanilla approach
香草方法
function transpose(matrix) {
const rows = matrix.length, cols = matrix[0].length;
const grid = [];
for (let j = 0; j < cols; j++) {
grid[j] = Array(rows);
}
for (let i = 0; i < rows; i++) {
for (let j = 0; j < cols; j++) {
grid[j][i] = matrix[i][j];
}
}
return grid;
}
Vanilla in-place ES6 approach inspired by Emanuel Saringan
受Emanuel Saringan启发的 Vanilla 就地 ES6 方法
function transpose(matrix) {
for (var i = 0; i < matrix.length; i++) {
for (var j = 0; j < i; j++) {
const temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
}
// Using destructing
function transpose(matrix) {
for (var i = 0; i < matrix.length; i++) {
for (var j = 0; j < i; j++) {
[matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
}
}
}
回答by Andrew Tatomyr
Neat and pure:
干净利落：
[[0, 1], [2, 3], [4, 5]].reduce((prev, next) => next.map((item, i) =>
(prev[i] || []).concat(next[i])
), []); // [[0, 2, 4], [1, 3, 5]]
Previous solutions may lead to failure in case an empty array is provided.
如果提供空数组，以前的解决方案可能会导致失败。
Here it is as a function:
这是一个函数：
function transpose(array) {
return array.reduce((prev, next) => next.map((item, i) =>
(prev[i] || []).concat(next[i])
), []);
}
console.log(transpose([[0, 1], [2, 3], [4, 5]]));
Update.It can be written even better with spread operator:
更新。使用扩展运算符可以写得更好：
const transpose = matrix => matrix.reduce(
($, row) => row.map((_, i) => [...($[i] || []), row[i]]),
[]
)
回答by Emanuel Saringan
You can do it in in-place by doing only one pass:
您只需执行一次即可就地完成：
function transpose(arr,arrLen) {
for (var i = 0; i < arrLen; i++) {
for (var j = 0; j <i; j++) {
//swap element[i,j] and element[j,i]
var temp = arr[i][j];
arr[i][j] = arr[j][i];
arr[j][i] = temp;
}
}
}
回答by óscar Gómez Alca?iz
Just another variation using Array.map. Using indexes allows to transpose matrices where M != N:
只是使用Array.map. 使用索引允许转置矩阵，其中M != N：
// Get just the first row to iterate columns first
var t = matrix[0].map(function (col, c) {
// For each column, iterate all rows
return matrix.map(function (row, r) {
return matrix[r][c];
});
});
All there is to transposing is mapping the elements column-first, and then by row.
转置所要做的就是先按列映射元素，然后按行映射元素。
回答by Kevin Le - Khnle
If you have an option of using Ramda JS and ES6 syntax, then here's another way to do it:
如果您可以选择使用 Ramda JS 和 ES6 语法，那么这是另一种方法：
const transpose = a => R.map(c => R.map(r => r[c], a), R.keys(a[0]));
console.log(transpose([
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]
])); // => [[1,5,9],[2,6,10],[3,7,11],[4,8,12]]<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.22.1/ramda.min.js"></script>回答by Nina Scholz
Another approach by iterating the array from outside to inside and reduce the matrix by mapping inner values.
另一种方法是从外到内迭代数组，并通过映射内部值来减少矩阵。
const
transpose = array => array.reduce((r, a) => a.map((v, i) => [...(r[i] || []), v]), []),
matrix = [[1, 2, 3], [1, 2, 3], [1, 2, 3]];
console.log(transpose(matrix));