假设我在 javascript 中有一个数组,例如
[[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]
我可以随机打乱子数组的元素以获得这样的东西吗
[[16,4,10],[8,3,9],[1,14,18],[2,5,7],[6,17,11],[12,13,15]]
编辑:所有子数组的长度相同。并且新数组将具有与旧数组相同长度的子数组。
假设我在 javascript 中有一个数组,例如
[[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]
我可以随机打乱子数组的元素以获得这样的东西吗
[[16,4,10],[8,3,9],[1,14,18],[2,5,7],[6,17,11],[12,13,15]]
编辑:所有子数组的长度相同。并且新数组将具有与旧数组相同长度的子数组。
我会为此使用下划线:http: //underscorejs.org/
v = [[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]
v2 = _.shuffle(_.flatten(v))
v3 = _.groupBy(v2, function(item, i) {
return i % v.length;
})
您可以展平您的阵列,将其洗牌,然后再次拆分:
var flat = [].concat.apply([], myArray);
arrayShuffle(flat);
var newArray = [],
sublen = myArray[0].length;
for (var i=0; i<flat.length; i+= sublen)
newArray.push(flat.slice(i, i+sublen));
或者您修改现有的洗牌算法之一以使用子数组中的项目。例如,Fisher-Yates-shuffle:
function shuffle2dArray(array) {
if (!array.length) return array;
var sublen = array[0].length,
len = array.length * sublen;
for (var i = len - 1; i > 0; i--) {
var j = Math.floor(Math.random() * (i + 1));
var ix = ~~(i/sublen),
iy = i % sublen,
jx = ~~(j/sublen),
jy = j % sublen;
var temp = array[ix][iy];
array[ix][iy] = array[jx][jy];
array[jx][jy] = temp;
}
return array;
}
这会随机播放,它不会返回新数组:
function shuffleMatrix (m) {
if (m.length == 0 || m[0].length == 0) {
// no rows or columns, just return it
return m;
}
function swap(i1, j1, i2, j2) {
if (i1 != i2 || j1 != j2) {
var temp = m[i1][j1];
m[i1][j1] = m[i2][j2];
m[i2][j2] = temp;
}
var rows = m.length;
var cols = m[0].length;
for (i = 0; i < rows; i++) {
for (j = 0; j < cols; j++) {
var new_i = Math.random()*rows;
var new_j = Math.random()*cols;
swap(i, j, new_i, new_j);
}
}
}
function arrayShuffle(){
var tmp, rand;
for(var i =0; i < this.length; i++){
rand = Math.floor(Math.random() * this.length);
tmp = this[i];
this[i] = this[rand];
this[rand] =tmp;
}
}
Array.prototype.shuffle =arrayShuffle;
And then
for(var i in arrays){
arrays[i].shuffle()
}
same thing to position of arrays inside the toplevel array.
//EDIT: Missreading :)
function shuffleAnyNumber (arrays) {
var numbers = new Array(); //collection of all numbers
for(var i in this){
for(var j in arrays[i]){
numbers.push( arrays[i][j]); //collect numbers out of the given arrays
}
}
numbers.shuffle(); //Method shown above
var output = new Array();
var tempArray= new Array();
//putting it together
for(var i in numbers){
if(tempArray.length == 3){
output.push(tempArray);
tempArray = new Array();
} else {
tempArray.push(numbers[i]);
}
}
return output;
}
i ' d say it will work like that.
我 100% 支持 Ilan Berci 的回答。但是,您可以使用下划线以更实用的样式编写它:
var v = [[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]];
return _(v).chain().flatten().shuffle().groupBy(function(item, i) {
return i % v.length;
}).values().value();
去下划线!!