Question

我写了这个Matlab代码来生成随机[1 0]和[2 0]：

的向量

nTrials = 8; 
Seq1 = [1 0]; % sequence 1
Seq2 = [2 0]; % sequence 2

a=repmat(Seq1,(nTrials/4),1); 
b=repmat(Seq2,(nTrials/4),1); 
abcd = vertcat(a,b); % concatenate all three couples
CouplesOrderStruct = abcd(randperm(size(abcd,1)),:); % couples in columns
vector = (reshape(CouplesOrderStruct.',[],1))';

结果是像[1 0 2 0 2 0 1 0]

这样的矢量

代码解释：

我有两个数字序列，1-0和2-0，我想在我的矢量中随机化。

首先，在a = repmat(Seq1,(nTrials/4),1); b=repmat(Seq2,(nTrials/4),1);我创建了固定数量的序列
其次，我把a和b放在一起：abcd = vertcat(a,b); % concatenate all three couples
第三，我在CouplesOrderStruct = abcd(randperm(size(abcd,1)),:);

结果是具有相同量的1-0和2-0的矢量，但是以随机顺序

有没有办法用JavaScript获得相同的结果？

Answer 1

Sooo我刚刚为你构建了一个很好的小文档：

function randomRepeatSequence(sequences, times) {
    // times has to be a multiple of sequences.length
    if (times % sequences.length !== 0)
        return console.log("times has to be a multiple of sequences.length");

    // Remap our sequence-array so we can store the count it has been used
    var seqmap = [];
    for (var seqid = 0; seqid < sequences.length; seqid++)
        // Push the current sequence n times; n = times/sequences.length
        for (var idx = 0; idx < times/sequences.length; idx++)
            seqmap.push(sequences[seqid]);

    var resultmap = [];
    // Now just select and remove a random sequence from our seqmap, until it is empty
    while (!seqmap.length == 0) {
        // Select a random element
        var randomidx = Math.floor(Math.random()*seqmap.length);
        var currentElement = seqmap[randomidx];
        // remove the random element from seqmap...
        seqmap.splice(randomidx, 1);
        // .. and push it to the resultmap
        resultmap.push(currentElement);
    }

    // now our resultmap looks like [[1],[2],[3]]... just flatten it!
    var result = resultmap.reduce( function(a, b) {
        return a.concat(b);
    });

    return result;      
}

您可以像

一样使用它

console.log(randomRepeatSequence([[1,0], [2,0]], 4));

或者，更好地理解：

var seqlist = [
    [1, 0],
    [2, 0]
]
randomRepeatSequence(seqlist, 4)

请注意，times参数只需要使用必须使用的序列数量，而不是结果的长度。但你只需要像

那样轻松地计算出来

randomRepeatSequence(seqlist, 8/seqlist[0].length)

（给出4，因为seqlist [0] .length = 2和8/2是4）

原始答案

您的结果是例如

vector = 2 0 1 0 2 0 1 0

我猜seq1和seq2应该包含相同的次数。我决定采用一种易于理解的方法，即使我可以做得更短：

var trials = 8; // has to be even

var seq1 = [1, 0];
var seq2 = [2, 0];

// "Build" a sequence list
var seqlist = [
    seq1, seq1,
    seq2, seq2
]

var list = []

for (var n = 0; n < trials/2; n++) {
    // search a random entry
    var index = Math.floor(Math.random()*seqlist.length);
    var toUseSeq = seqlist[index];

    // delete the entry
    seqlist.splice(index, 1);

    list.push(toUseSeq);
}
// flatten result array

var result = list.reduce( function(a, b) {
    return a.concat(b);
});

console.log(result);

执行此操作会给我一个控制台输出：

[ 2, 0, 1, 0, 2, 0, 1, 0 ] 
[ 2, 0, 1, 0, 2, 0, 1, 0 ]
[ 1, 0, 1, 0, 2, 0, 2, 0 ]
[ 1, 0, 2, 0, 1, 0, 2, 0 ]

Javascript中的随机重复序列

1 个答案: