在 JavaScript 中旋转数组中的元素
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原文地址: http://stackoverflow.com/questions/1985260/
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Rotate the elements in an array in JavaScript
提问by Jean Vincent
I was wondering what was the most efficient way to rotate a JavaScript array.
我想知道旋转 JavaScript 数组的最有效方法是什么。
I came up with this solution, where a positive nrotates the array to the right, and a negative nto the left (-length < n < length) :
我想出了这个解决方案，其中正数n将数组向右旋转，负数n向左旋转( -length < n < length) ：
Array.prototype.rotateRight = function( n ) {
this.unshift( this.splice( n, this.length ) );
}
Which can then be used this way:
然后可以这样使用：
var months = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"];
months.rotate( new Date().getMonth() );
My original version above has a flaw, as pointed out by Christophin the comments bellow, a correct version is (the additional return allows chaining):
正如Christoph在下面的评论中指出的那样，我上面的原始版本有一个缺陷，一个正确的版本是（额外的返回允许链接）：
Array.prototype.rotateRight = function( n ) {
this.unshift.apply( this, this.splice( n, this.length ) );
return this;
}
Is there a more compact and/or faster solution, possibly in the context of a JavaScript framework? (none of the proposed versions bellow is either more compact or faster)
是否有更紧凑和/或更快的解决方案，可能在 JavaScript 框架的上下文中？（以下建议的版本都不是更紧凑或更快）
Is there any JavaScript framework out there with an array rotate built-in? (Still not answered by anyone)
有没有内置数组旋转的 JavaScript 框架？（仍然没有人回答）
采纳答案by Christoph
Type-safe, generic version which mutates the array:
类型安全的通用版本，它改变了数组：
Array.prototype.rotate = (function() {
// save references to array functions to make lookup faster
var push = Array.prototype.push,
splice = Array.prototype.splice;
return function(count) {
var len = this.length >>> 0, // convert to uint
count = count >> 0; // convert to int
// convert count to value in range [0, len)
count = ((count % len) + len) % len;
// use splice.call() instead of this.splice() to make function generic
push.apply(this, splice.call(this, 0, count));
return this;
};
})();
In the comments, Jean raised the issue that the code doesn't support overloading of push()and splice(). I don't think this is really useful (see comments), but a quick solution (somewhat of a hack, though) would be to replace the line
在评论中，Jean 提出了代码不支持push()和重载的问题splice()。我不认为这真的很有用（见评论），但一个快速的解决方案（虽然有点黑客）将替换该行
push.apply(this, splice.call(this, 0, count));
with this one:
有了这个：
(this.push || push).apply(this, (this.splice || splice).call(this, 0, count));
Using unshift()instead of push()is nearly twice as fast in Opera 10, whereas the differences in FF were negligible; the code:
在 Opera 10 中使用unshift()代替的push()速度几乎快两倍，而在 FF 中的差异可以忽略不计；编码：
Array.prototype.rotate = (function() {
var unshift = Array.prototype.unshift,
splice = Array.prototype.splice;
return function(count) {
var len = this.length >>> 0,
count = count >> 0;
unshift.apply(this, splice.call(this, count % len, len));
return this;
};
})();
回答by Yukulélé
You can use push(), pop(), shift()and unshift()methods:
您可以使用push()，pop()，shift()和unshift()方法：
function arrayRotate(arr, reverse) {
if (reverse) arr.unshift(arr.pop());
else arr.push(arr.shift());
return arr;
}
usage:
用法：
arrayRotate(['h','e','l','l','o']); // ['e','l','l','o','h'];
arrayRotate(['h','e','l','l','o'], true); // ['o','h','e','l','l'];
If you need countargument see my other answer: https://stackoverflow.com/a/33451102
如果您需要count争论，请参阅我的其他答案：https: //stackoverflow.com/a/33451102
回答by Gumbo
I would probably do something like this:
我可能会做这样的事情：
Array.prototype.rotate = function(n) {
return this.slice(n, this.length).concat(this.slice(0, n));
}
Edit????Here's a mutator version:
编辑????这是一个mutator版本：
Array.prototype.rotate = function(n) {
while (this.length && n < 0) n += this.length;
this.push.apply(this, this.splice(0, n));
return this;
}
回答by Yukulélé
This function works in both way and works with any number (even with number greater than array length):
此函数以两种方式工作并适用于任何数字（即使数字大于数组长度）：
function arrayRotate(arr, count) {
count -= arr.length * Math.floor(count / arr.length);
arr.push.apply(arr, arr.splice(0, count));
return arr;
}
usage:
用法：
for(let i = -6 ; i <= 6 ; i++) {
console.log(arrayRotate(["","","","",""], i), i);
}
result:
结果：
[ "", "", "", "", "" ] -6
[ "", "", "", "", "" ] -5
[ "", "", "", "", "" ] -4
[ "", "", "", "", "" ] -3
[ "", "", "", "", "" ] -2
[ "", "", "", "", "" ] -1
[ "", "", "", "", "" ] 0
[ "", "", "", "", "" ] 1
[ "", "", "", "", "" ] 2
[ "", "", "", "", "" ] 3
[ "", "", "", "", "" ] 4
[ "", "", "", "", "" ] 5
[ "", "", "", "", "" ] 6
回答by mickmackusa
So many of these answers seem over-complicated and difficult to read. I don't think I saw anyone using splice with concat...
这些答案中的许多似乎过于复杂且难以阅读。我想我没有看到有人使用带有 concat 的 splice ...
function rotateCalendar(){
var cal=["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"],
cal=cal.concat(cal.splice(0,new Date().getMonth()));
console.log(cal); // return cal;
}
console.log outputs (*generated in May):
console.log 输出（* 于 5 月生成）：
["May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec", "Jan", "Feb", "Mar", "Apr"]
As for compactness, I can offer a couple of generic one-liner functions (not counting the console.log | return portion). Just feed it the array and the target value in the arguments.
至于紧凑性，我可以提供几个通用的单行函数（不包括 console.log | return 部分）。只需将数组和参数中的目标值提供给它即可。
I combine these functions into one for a four-player card game program where the array is ['N','E','S','W']. I left them separate in case anyone wants to copy/paste for their needs. For my purposes, I use the functions when seeking whose turn is next to play/act during different phases of the game (Pinochle). I haven't bothered testing for speed, so if someone else wants to, feel free to let me know the results.
我将这些函数合并为一个四人纸牌游戏程序，其中数组为 ['N','E','S','W']。我将它们分开，以防有人想根据需要复制/粘贴。出于我的目的，我在游戏的不同阶段（Pinochle）寻找下一个轮到谁玩/行动时使用这些功能。我没有费心测试速度，所以如果其他人愿意，请随时让我知道结果。
*notice, the only difference between functions is the "+1".
*注意，功能之间的唯一区别是“+1”。
function rotateToFirst(arr,val){ // val is Trump Declarer's seat, first to play
arr=arr.concat(arr.splice(0,arr.indexOf(val)));
console.log(arr); // return arr;
}
function rotateToLast(arr,val){ // val is Dealer's seat, last to bid
arr=arr.concat(arr.splice(0,arr.indexOf(val)+1));
console.log(arr); // return arr;
}
combination function...
组合功能...
function rotateArray(arr,val,pos){
// set pos to 0 if moving val to first position, or 1 for last position
arr=arr.concat(arr.splice(0,arr.indexOf(val)+pos));
return arr;
}
var adjustedArray=rotateArray(['N','E','S','W'],'S',1);
adjustedArray=
调整数组=
W,N,E,S
回答by Aryeh Harris
Here is a very simple way to shift items in an array:
这是在数组中移动项目的一种非常简单的方法：
function rotate(array, stepsToShift) {
for (var i = 0; i < stepsToShift; i++) {
array.unshift(array.pop());
}
return array;
}
回答by Dudley Craig
Using ES6's spread for an immutable example ...
使用 ES6 的扩展作为不可变示例......
[...array.slice(1, array.length), array[0]]
and
和
[array[array.items.length -1], ...array.slice(0, array.length -1)]
It's probably not the most efficient though but it's concise.
虽然它可能不是最有效的，但它很简洁。
回答by molokoloco
@Christoph, you've done a clean code, but 60% slowestthan this one i found. Look at the result on jsPerf : http://jsperf.com/js-rotate-array/2[Edit] OK now there is more browsers an that not obvious witch methods the best
@Christoph，你已经完成了一个干净的代码，但比我发现的这个代码慢 60%。看看 jsPerf 上的结果：http://jsperf.com/js-rotate-array/2 [编辑] 好的，现在有更多的浏览器，一个不明显的女巫方法是最好的
var rotateArray = function(a, inc) {
for (var l = a.length, inc = (Math.abs(inc) >= l && (inc %= l), inc < 0 && (inc += l), inc), i, x; inc; inc = (Math.ceil(l / inc) - 1) * inc - l + (l = inc))
for (i = l; i > inc; x = a[--i], a[i] = a[i - inc], a[i - inc] = x);
return a;
};
var array = ['a','b','c','d','e','f','g','h','i'];
console.log(array);
console.log(rotateArray(array.slice(), -1)); // Clone array with slice() to keep original
回答by Vic99999
see http://jsperf.com/js-rotate-array/8
见http://jsperf.com/js-rotate-array/8
function reverse(a, from, to) {
--from;
while (++from < --to) {
var tmp = a[from];
a[from] = a[to];
a[to] = tmp;
}
}
function rotate(a, from, to, k) {
var n = to - from;
k = (k % n + n) % n;
if (k > 0) {
reverse(a, from, from + k);
reverse(a, from + k, to);
reverse(a, from, to);
}
}
回答by Dave Everitt
When I couldn't find a ready-made snippet to start a list of days with 'today', I did it like this (not quite generic, probably far less refined than the above examples, but did the job):
当我找不到一个现成的片段来开始一个带有“今天”的日期列表时，我是这样做的（不是很通用，可能远没有上面的例子那么精致，但完成了这项工作）：
//returns 7 day names with today first
function startday() {
const days = ['Sun','Mon','Tue','Wed','Thu','Fri','Sat'];
let today = new Date();
let start = today.getDay(); //gets day number
if (start == 0) { //if Sunday, days are in order
return days
}
else { //if not Sunday, start days with today
return days.slice(start).concat(days.slice(0,start))
}
}
Thanks to a little refactor by a better programmer than me it's a line or two shorter than my initial attempt, but any further comments on efficiency are welcome.
感谢一个比我更好的程序员的小重构，它比我最初的尝试短了一两行，但欢迎任何关于效率的进一步评论。
