javascript jQuery 从字符串数组中选择一个随机值
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jQuery pick a random value from a array of strings
提问by Alex
I know how do this in PHP, but in javascript arrays are weird.
我知道如何在 PHP 中执行此操作，但在 javascript 数组中很奇怪。
So I have a set of image transitions which have effects that use easing equations. For certain effects I want to pick up a random value from a array of multiple values:
所以我有一组图像过渡，它们具有使用缓动方程的效果。对于某些效果，我想从多个值的数组中选取一个随机值：
something like:
就像是：
easing: randomFrom(array('easeOutElastic', 'easeOutBounce', 'easeOutSince')),
easing: randomFrom(array('easeOutElastic', 'easeOutBounce', 'easeOutSince')),
回答by Dogbert
function randomFrom(array) {
return array[Math.floor(Math.random() * array.length)];
}
// in your code
easing: randomFrom(['easeOutElastic', 'easeOutBounce', 'easeOutSince']),
回答by Naftali aka Neal
Try this:
试试这个：
function randomFrom(arr){
var randomIndex = Math.floor(Math.random() * arr.length);
return arr[randomIndex];
}
回答by Aamir Afridi
For a demo I was trying to get a random easing string ;)
对于演示，我试图获得一个随机缓动字符串；)
function getRandomEasing() {
var easingArr = [];
for(easing in $.easing){
easingArr.push(easing);
}
var rnd = Math.floor(Math.random() * easingArr.length);
return easingArr[rnd];
}
var randomEasing = getRandomEasing();
Fun demo: http://jsfiddle.net/aamir/psWvy/
有趣的演示：http: //jsfiddle.net/aamir/psWvy/
回答by nonopolarity
I am a late comer to this question. If you don't mind adding a method to Array, you can:
我是一个迟到的人。如果您不介意向 中添加方法Array，您可以：
Array.prototype.pickARandomElement = function() {
return this[Math.floor(Math.random() * this.length)];
}
test run:
测试运行：
> [10, 123, 777].pickARandomElement()
10
> [10, 123, 777].pickARandomElement()
777
> [10, 123, 777].pickARandomElement()
777
> [10, 123, 777].pickARandomElement()
123
The following is just for fun:
以下内容仅供娱乐：
Array.prototype.pickARandomElement = function() {
return this.sort(function() { return Math.random() - 0.5; })[0];
}
it is like shuffling a deck of cards and returning the top card. But its time complexity is O(n log n)so I wouldn't really use it, and it is just for fun. The first solution here is a solution with O(1).
这就像洗一副牌并返回最上面的牌。但是它的时间复杂度太高了，O(n log n)所以我不会真正使用它，它只是为了好玩。这里的第一个解决方案是使用O(1).
