关于循环规格组合可能性思路的问题

Question

尺寸有1个参数
孔数和孔距 有3个参数
J值有 2个参数
ET值 有1个参数
中心孔有 2个参数

一共会有12种组合的可能 求思路 谢谢啦

Answer 1

尝试以下代码function cross(inputs){    let result = inputs[0].map(item => [item]);    for(let cross_item of inputs.slice(1)){        result = cross_item.reduce((acc,target)=>(result.forEach(result_item=>acc.push([...result_item,target])),acc),[]);    }    return result;}console.log(cross([['n','o','t'], ['i', 'n'], ['a'],['(',')']]));----------------其实可以再短一点------------------function cross(inputs){    return inputs.reduce((result,cross_item)=> cross_item.reduce((acc, target) => (result.forEach(result_item => acc.push([...result_item, target])), acc), []))}

Answer 2

放进二维数组里 递归组合let arr=[&nbsp; &nbsp; [11,12],&nbsp; &nbsp; [21,22,23],&nbsp; &nbsp; [31],&nbsp; &nbsp; [41,42]]f(arr,0,[])function f(arr,k,temp){&nbsp; &nbsp; if(k==arr.length){&nbsp; &nbsp; &nbsp; &nbsp; console.log(JSON.stringify(temp))&nbsp; &nbsp; &nbsp; &nbsp; return&nbsp; &nbsp; }&nbsp; &nbsp; for(let i=0;i<arr[k].length;i++){&nbsp; &nbsp; &nbsp; &nbsp; temp[k]=arr[k][i]&nbsp; &nbsp; &nbsp; &nbsp; f(arr,k+1,temp)&nbsp; &nbsp; }}

Answer 3

笛卡尔积&nbsp;了解一下生成给予属性的所有可能组合方式