使用array.splice（）卡住洗牌组数组 [英] Stuck using array.splice() to shuffle a deck array

问题描述

我正试图将 Card 对象的套牌数组随机改组为 newDeck 的<$ c数组$ c> Card 对象，使用 array.splice（）。我想我的问题可能与可变范围有关，或者是对 array.splice（）的误解。

I'm trying to randomly shuffle a deck array of Card objects into a newDeck array of Card objects using array.splice(). I imagine my problem is either something to do with variable scope, or a misapprehension about array.splice().

var deck = [new Card(), new Card(), new Card(), new Card(), new Card(), new Card()];
var newDeck = [];
var shuffle = function(){
    var i = "";
    for (i = 0; i < deck.length; i++){      
        newDeck[i] = deck.splice(Math.floor(Math.random() * deck.length, 1));
}    
};

shuffle();

有没有更好的方法来洗牌了？

Is there a better way to shuffle a deck?

推荐答案

是的，关于 array.splice（）的理解有误。阅读文档：它将返回已删除元素的数组，如果您使用一张卡： [<对象卡>] 。此外，使用 deck.splice（Math.floor（Math.random（）* deck.length，1））时， 1 （我猜应该是要删除多少张卡）是 Math.floor 的参数，而不是 splice -您将删除索引之后的所有元素。因此，您似乎想要：

Yes, there is a misapprehension about array.splice(). Read the docs: It will return an array of the removed elements, in your case with one card: [<object Card>]. Also, with deck.splice(Math.floor(Math.random() * deck.length, 1)), the 1 (which i guess should be how many cards to remove) is an argument to Math.floor, not splice - you will remove all elements after the index. So, you seem to want:

function shuffle(deck) {
    var newDeck = [];
    for (var i = 0; i < deck.length; i++)    
        newDeck[i] = deck.splice(Math.floor(Math.random() * deck.length), 1)[0];
    return newDeck;
}    
shuffle([new Card(), new Card(), new Card(), new Card(), new Card(), new Card()]);

您还要求其他方法进行洗牌：可以使用非常常见的

You have asked for other ways to shuffle: You could use the very common

deck.sort(function(a, b){ return 0.5-Math.random(); })

或 underscore.js中的算法：

function shuffle(deck) {
    var newDeck = [];
    for (var i = 0; i < deck.length; i++) {    
        var rand = Math.floor(Math.random() * (i + 1));  
        newDeck[i] = newDeck[rand];
        newDeck[rand] = deck[i];
    }
    return newDeck;
}  

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