使用减少,地图和传播的笛卡尔乘积将需要平坦化的结果分组 [英] Cartesian product using reduce, map and spread is grouping the result which need to be flattened
问题描述
构造函数(arr){
super(arr);
}
union(set){
return new MathSet([... this,... set])
}
intersection(set){
返回新的MathSet([... this] .filter(x => set.has(x)));
}
差(集){
返回新的MathSet([... this] .filter(x =>!set.has(x)));
}
cartesian(set){
return new MathSet([... this] .reduce((acc,i)=> [... acc,[... set ] .map(j => [i,j])],[]))
}
}
let x = new MathSet([1,2,3 ]);
let y = new MathSet([1,2,3,4,5]);
console.log(JSON.stringify([... x.cartesian(y)]));
// [
// [[1,1],[1,2],[1,3],[1,4],[1,5]],
/ / [[2,1],[2,2],[2,3],[2,4],[2,5]],
// [[3,1],[3,2 ],[3,3],[3,4],[3,5]]
//]
使用 cartesian
函数预期结果是上述数组的平面版本( [[1,1],[1,2 ],[1,3],[1,4],[1,5],[2,1],[2,2],[2,3],[2,4],[2,5], [3,1],[3,2],[3,3],[3,4],[3,5]]
),但是你可以看到它被分为三个阵列。减少是保持连接早期结果与新结果的扩展版本。任何猜测我做错了什么?
要平铺数组,您只需要一个扩展:
返回新的MathSet([... this] .reduce((acc,i)=> [... acc,... [ ... set] .map(j => [i,j])],[]))
// ^^^
(或 acc.concat(Array.from(set,j => [i,j]))
)
class MathSet extends Set{
constructor(arr){
super(arr);
}
union(set){
return new MathSet([...this, ...set])
}
intersection(set){
return new MathSet([...this].filter(x => set.has(x)));
}
difference(set){
return new MathSet([...this].filter(x => !set.has(x)));
}
cartesian(set){
return new MathSet( [...this].reduce((acc, i)=> [...acc, [...set].map(j=>[i,j])], []) )
}
}
let x = new MathSet([1,2,3]);
let y = new MathSet([1,2,3,4,5]);
console.log(JSON.stringify([...x.cartesian(y)]));
//[
// [[1,1],[1,2],[1,3],[1,4],[1,5]],
// [[2,1],[2,2],[2,3],[2,4],[2,5]],
// [[3,1],[3,2],[3,3],[3,4],[3,5]]
// ]
With cartesian
function expected result is a flattened version of the above array ([[1,1],[1,2],[1,3],[1,4],[1,5],[2,1],[2,2],[2,3],[2,4],[2,5],[3,1],[3,2],[3,3],[3,4],[3,5]]
), but as you can see somehow its getting grouped into three arrays. The reduce is keeping on concatenating earlier result with spreaded version of new results. Any guess on what I am doing wrong?
To flatten the arrays you need just one more spread:
return new MathSet( [...this].reduce((acc, i)=> [...acc, ...[...set].map(j=>[i,j])], []) )
// ^^^
(or acc.concat(Array.from(set, j=>[i,j]))
)
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