javascript - How many one-dimensional arrays can be formed from the elements of 4 one-dimensional arrays?

Question

For example,

[1,2,3,4], [5], [7,8], [9]

Generate a one-dimensional array, which should be 8 types.

In fact, it looks like a branch ~

[1,5,7,9],
[2,5,7,9],
[1,5,7,9],

/***
** Omitted ...
***/

[4,5,8,9]

What if there is one more element?

[1,2,3,4], [5,6], [7,8], [9]

How to list all possible combinations?

Answer 1

#这就是求笛卡尔积

from itertools import product
print list(product([1,2,3,4], [5, 6], [7,8], [9]))

Answer 2

4*2*2, 16 kinds.
Permutations and combinations, since each array must have one element, then for
array 1: c(4,1)=4,
array 2: c(2,1) =2,
Array 3: c(2,1)=2,
Array 4: c(1,1)=1,
Multiply each one again and pay attention to repeated elements.

Answer 3

    $a1 = array(1,2,3,4);
    $a2 = array(5);
    $a3 = array(7,8);
    $a4 = array(9);
    foreach ($a1 as $k1=>$v1){
        foreach ($a2 as $k2=>$v2){
            foreach ($a3 as $k3=>$v3){
                foreach ($a4 as $k4=>$v4){
                    $a = array($v1,$v2,$v3,$v4);
                    print_r($a);
                    echo "<br>";
                } 
            } 
        } 
    }

Answer 4

Generally speaking, if there are several groups, several loops are needed. Just combine the elements inside one by one. This is the conventional solution of js. I don’t know if there is a better one.

var a = [1,2,3,4],
    b = [5,6],
    c = [7,8],
    d = [9];

for(var i = 0; i < a.length; i++) {
    for(var j = 0; j < b.length; j++) {
        for(var k = 0; k < c.length; k++) {
            for(var l = 0; l < d.length; l++) {
                console.log(a[i] + '-' + b[j] + '-' + c[k] + '-' + d[l]);
            }
        }
    }
}

Answer 5

function list(arr){
    function newArr(arr){
      var num=1;
      arr.forEach((x)=>{
        num*=x.length
      })
      var tem=[]
      for(var i=0;i<num;i++){
        tem[i]=[];
      }
      return tem
    }
    function numFn(num){
        var pp=[];
        var nums=list.length;
        arr.forEach(function(x){
          nums=nums/x.length
          pp.push(x[nums==1?num%x.length:parseInt(num/nums)%x.length]);
       
        })
        return pp
    }
    var list=newArr(arr);
    for(var j=0;j<list.length;j++){
       list[j]=numFn(j)
    }
    return list
  }