JavaScript中多个数组的笛卡尔积 [英] Cartesian product of multiple arrays in JavaScript

问题描述

举例来说，

  cartesian（[1,2]，[10,20]，[100,200,300]）//应该是
 // [[1,10,100]，[1,10,200]， [1,10,300]，[2,10,100]，[2,10,200] ...] 
  

解决方案

使用 reduce 和<$>解决问题的功能性解决方案（不含可变变量！） c $ c> flatten ，由 underscore.js 提供：

  function cartesianProductOf（）{
 return _.reduce（arguments，function（a，b）{
 return _.flatten（_。map（a，function（x）{
 return _.map（b，function（y）{
 return x.concat（[y]）; 
}）; 
}），true）; 
 }，[[]]）; 
}; 
 
 cartesianProductOf（[1，2]，[3，4]，['a'，'b']）; // [[1,3，a]，[1,3，b]，[1,4，a]，[1,4，b]，[2,3，a ]，[2,3，b]，[2,4，a]，[2,4，b]] 
  

备注：此解决方案的灵感来源于 http://cwestblog.com/2011/05/02/cartesian-product-of-multiple-arrays/

How would you implement the Cartesian product of multiple arrays in JavaScript?

As an example,

cartesian([1,2],[10,20],[100,200,300]) //should be
// [[1,10,100],[1,10,200],[1,10,300],[2,10,100],[2,10,200]...]

解决方案

Here is a functional solution to the problem (without any mutable variable!) using reduce and flatten, provided by underscore.js:

function cartesianProductOf() {
    return _.reduce(arguments, function(a, b) {
        return _.flatten(_.map(a, function(x) {
            return _.map(b, function(y) {
                return x.concat([y]);
            });
        }), true);
    }, [ [] ]);
};

cartesianProductOf([1, 2], [3, 4], ['a', 'b']); // [[1,3,"a"],[1,3,"b"],[1,4,"a"],[1,4,"b"],[2,3,"a"],[2,3,"b"],[2,4,"a"],[2,4,"b"]] 

Remark: This solution was inspired by http://cwestblog.com/2011/05/02/cartesian-product-of-multiple-arrays/

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